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Unformatted text preview: MINITRY OF EDUCATION AND TRANNING PHYSICS
12 2000 TABLE OF CONTENTS
Part I. OSCILLATIONS AND WAVES ...................................................................................................8
Chapter I – Mechanical Oscillations ............................................................................................................8
§1. Periodic and simple harmonic motions. Oscillation of a massspring system. ..................................8
1. Oscillations ......................................................................................................................................8
2. Periodic motion ...............................................................................................................................8
3. Massspring system. Simple harmonic motion................................................................................8
§2. Exploring a simple harmonic motion...............................................................................................10
Uniform circular motion and simple harmonic motion .....................................................................10
2. Angular phase and angular frequency of a simple harmonic motion ............................................11
3. Free motion....................................................................................................................................11
4. Velocity and acceleration in a simple harmonic motion................................................................11
5. Oscillation of a simple pendulum ..................................................................................................12
§3. Energy in a simple harmonic motion ...............................................................................................13
1. Energy changes during oscillation.................................................................................................13
2. Conservation of mechanical energy during oscillation .................................................................14
§4.  §5. The combination of oscillations...............................................................................................15
1. Examples of the combination of oscillations.................................................................................15
2. Phasedifferences between oscillations .........................................................................................15
3. Vectordiagram method.................................................................................................................16
4. The combination of two oscillations of same directions and frequencies .....................................16
5. Amplitude and initial phase of the combinatorial oscillation ........................................................17
§6.  §7. Underdamped and forced oscillations .....................................................................................18
1. Underdamped oscillation...............................................................................................................18
2. Forced oscillation ..........................................................................................................................18
3. Resonance......................................................................................................................................19
4. Applying and surmounting resonant phenomenon ........................................................................19
5. Selfoscillation...............................................................................................................................20
Summary of Chapter I............................................................................................................................20
Chapter II – Mechanical wave. Acoustics..................................................................................................22
§8. Wave in mechanics ..........................................................................................................................22
1. Natural mechanical waves .............................................................................................................22
2. Oscillation phase transmission. Wavelength. ................................................................................22
3. Period, frequency and velocity of waves.......................................................................................23
4. Amplitude and energy of waves ....................................................................................................23
§9.  §10. Sound wave............................................................................................................................24
Sound wave and the sensation of sound ............................................................................................24
2. Sound transmission. Speed of sound .............................................................................................25
3. Sound altitude................................................................................................................................25
Timbre ...............................................................................................................................................25
5. Sound energy .................................................................................................................................26
6. Sound loudness..............................................................................................................................26
7. Sound source – Resonant box........................................................................................................27
§11. Wave interference ..........................................................................................................................28
1. Interferential phenomenon.............................................................................................................28
Translated by VNNTU – Dec. 2001 Page 2 2. Theory of interference ...................................................................................................................28
3. Standing wave................................................................................................................................29
Summary of Chapter II ..........................................................................................................................31
Chapter III – Electric oscillation, Alternating current................................................................................32
§12. Harmonic oscillation voltage. Alternating current.........................................................................32
1. Harmonic oscillation voltage.........................................................................................................32
2. Alternating current.........................................................................................................................32
3. Root mean square (rms) value of intensity and voltage.................................................................33
§13.  §14. Alternating current in a circuit containing only resistance, inductance or capacitance.......34
1. Relation between current and voltage............................................................................................34
2. Ohm’s law for an AC circuit containing only resistance...............................................................34
1. Effect of capacitors to the alternating current................................................................................34
2. Relation between current and voltage............................................................................................35
3. Ohm’s law for an AC circuit containing only capacitance............................................................35
1. Effects of an inductor to the alternating current ............................................................................36
2. Relation between current and voltage............................................................................................36
3. Ohm’s Law for an AC circuit with inductors ................................................................................36
§15. Alternating current in an RLC circuit ............................................................................................37
Electric current and voltage in an RLC circuit ..................................................................................37
Relation between current and voltage in an RLC circuit...................................................................38
3. Ohm’s Law for an RLC circuit......................................................................................................38
4. Resonance in an RLC circuit .........................................................................................................39
§16. Power of the alternating current.....................................................................................................39
1. Power of the alternating current ....................................................................................................39
2. Significance of the power coefficient ............................................................................................40
§17. Problems on AC circuits ................................................................................................................41
Problem 1...........................................................................................................................................41
Problem 2...........................................................................................................................................41
§18. Singlephase AC generator ............................................................................................................42
1. Operational principle of singlephase AC generators....................................................................42
2. Structure of an AC generator.........................................................................................................42
§19. Threephase alternating current .....................................................................................................43
1. Operational principle of threephase AC power generators ..........................................................43
2. WYE connection ...........................................................................................................................44
3. Delta connection ............................................................................................................................45
§20. Asynchronous threephase motors .................................................................................................45
Operational principle of asynchronous threephase motors ..............................................................45
Rotating magnetic field of threephase current .................................................................................46
3. Structure of an asynchronous threephase motor...........................................................................46
§21. Transformers. Electricity transmission ..........................................................................................47
1. Operational principle and structure of transformers ......................................................................47
2. Transformation of current and voltage via transformer.................................................................47
3. Transmission of power ..................................................................................................................48
§22. Generation of direct current ...........................................................................................................49
1. Benefits of direct current ...............................................................................................................49
Halfcycle rectifying method.............................................................................................................49
Translated by VNNTU – Dec. 2001 Page 3 3. Two halfcycle rectifying method .................................................................................................49
4. Operational principle of DC power generators..............................................................................50
Summary of Chapter III .........................................................................................................................50
Chapter IV – Electromagnetic oscillation. Electromagnetic wave.............................................................52
§23. Oscillation circuits. Electromagnetic oscillation ...........................................................................52
1. Fluctuation of charges in an oscillation circuit..............................................................................52
2. Electromagnetic oscillation in an oscillation circuit......................................................................53
§24. Alternating current, highfrequency electromagnetic oscillation, and mechanical oscillation ......54
1. Electric oscillation in an alternating current..................................................................................54
2. Highfrequency electromagnetic oscillation..................................................................................54
3. Electromagnetic oscillation and mechanical oscillation................................................................54
§25. Electromagnetic field .....................................................................................................................57
1. Fluctuated electric field and fluctuated magnetic field..................................................................57
2. Electromagnetic field.....................................................................................................................57
3. Transmission of electromagnetic interaction.................................................................................57
§26. Electromagnetic waves ..................................................................................................................58
1. Electromagnetic waves ..................................................................................................................58
2. Properties of electromagnetic waves .............................................................................................58
3. Electromagnetic waves and wireless communication ...................................................................58
§27. Transmitting and receiving electromagnetic waves.......................................................................59
Periodicoscillation transmitters using transistors .............................................................................59
2. Open oscillation circuit. Antenna ..................................................................................................60
Principles of transmitting and receiving electromagnetic waves.......................................................60
§28.  §29. A glance at radio transmitters and receivers........................................................................61
1. Principle of oscillation amplification.............................................................................................61
2. Principle of amplitude modulation ................................................................................................62
3. Operational principle of radio transmitters....................................................................................62
4. Operational principle of radio receivers ........................................................................................63
Summary of Chapter IV.........................................................................................................................64
Part II.
OPTICS.....................................................................................................................................66
Chapter V – Light reflection and refraction ...............................................................................................66
§30. Light transmission. Light reflection. Plane mirror.........................................................................66
1. Light propagation ..........................................................................................................................66
2. Light reflection ..............................................................................................................................67
3. Plane mirror ...................................................................................................................................67
§31. Concave spherical mirrors .............................................................................................................68
Definitions .........................................................................................................................................68
2. Reflection of a light in a concave spherical mirror........................................................................69
3. Formation of images by concave spherical mirror ........................................................................69
4. Main focal point. Focal length.......................................................................................................70
5. Method to draw an object’s image obtaining from a concave spherical mirror ............................70
§32. Convex spherical mirrors. Convex spherical mirror equations. Applications of convex spherical
mirrors....................................................................................................................................................72
1. Convex spherical mirror ................................................................................................................72
2. Convex spherical mirror equations................................................................................................72
3. Applications of convex spherical mirrors......................................................................................74
Translated by VNNTU – Dec. 2001 Page 4 §33. Light refraction ..............................................................................................................................75
Light refraction phenomenon ............................................................................................................75
2. The law of light refraction .............................................................................................................75
3. Index of refraction (refractive index) ............................................................................................76
§34. Total internal reflection..................................................................................................................77
Total internal reflection .....................................................................................................................77
2. Conditions to achieve total internal reflection...............................................................................78
3. Critical angle .................................................................................................................................78
4. Applications of total internal reflection.........................................................................................79
§35. Prism ..............................................................................................................................................80
1. Definition.......................................................................................................................................80
2. Path of a monochromatic ray through a prism. Angle of deviation...............................................80
3. Prism equations .............................................................................................................................80
4. Minimum deviation angle..............................................................................................................80
§36. Thin lenses .....................................................................................................................................82
1. Definition.......................................................................................................................................82
2. Main focal point. Optical center. Focal length ..............................................................................82
3. Supplemental focal points. Focal plane .........................................................................................83
4. Lens power ....................................................................................................................................84
§37. Image of an object through lenses. Lenses equations ....................................................................85
1. Observing an object’s image through a lens..................................................................................85
2. Method to draw an object’s image through a lens.........................................................................85
3. Lens equation.................................................................................................................................86
4. Lateral magnification.....................................................................................................................87
The human eye and optical instruments .....................................................................................................91
§38. Camera and the human eye ............................................................................................................91
1. Camera...........................................................................................................................................91
2. The human eye...............................................................................................................................91
§39. Eye’s defects and correcting methods............................................................................................94
1. Nearsightedness (myopia) ............................................................................................................94
Farsightedness (hyperopia)................................................................................................................95
1.State the characteristics of nearsighted eye and the correcting method. .......................................95
§40. Magnifying glass............................................................................................................................95
1. Definition.......................................................................................................................................95
2. Near point and infinite point..........................................................................................................96
3. Angular magnification...................................................................................................................96
§41. Microscope and telescope ..............................................................................................................98
1. Microscope ....................................................................................................................................98
2. Telescope.......................................................................................................................................99
Chapter VII – The wavenature of light ...................................................................................................103
§42. Light dispersion phenomenon......................................................................................................103
1. Experiment on light dispersion phenomenon ..............................................................................103
2. Experiment on monochromatic light ...........................................................................................103
3. Synthesizing white light ..............................................................................................................104
4. Dependence of the index of refraction of a transparent medium on the color of the light ..........104
§43. Light interference phenomenon ...................................................................................................105
Translated by VNNTU – Dec. 2001 Page 5 1. Young's experiment on light interference phenomenon ..............................................................105
2. Explanation of the phenomenon ..................................................................................................105
3. Conclusion...................................................................................................................................106
1. Describe the experiment on the interference of light?.................................................................106
§44. Measuring the wavelength of the light. The wavelength and the color of the light.....................106
1. Interference fringe distance .........................................................................................................106
2. The wavelength and the color of the light ...................................................................................108
§45. Spectrometer. Continuous spectrum ............................................................................................108
1. Relation between the index of refraction of a medium and the wavelength of the light .............108
2. Spectrometer................................................................................................................................109
3. Continuous spectrums: ................................................................................................................109
§46. Line spectrum...............................................................................................................................110
1. Emission line spectrum................................................................................................................110
2. Absorption line spectrum.............................................................................................................111
3. The spectroscopic analysis approach and its advantages.............................................................112
§47. Infrared and ultraviolet rays.........................................................................................................112
1. Experiments to discover infrared and ultraviolet rays .................................................................112
2. The infrared ray ...........................................................................................................................113
3. The ultraviolet ray .......................................................................................................................113
§48. Xrays...........................................................................................................................................114
1. Xray tube....................................................................................................................................114
2. The nature of Xrays....................................................................................................................114
3. Properties and uses of Xrays ......................................................................................................114
4. Electromagnetic waves scale .......................................................................................................115
Chapter VIII – Light quantum..................................................................................................................118
§49. The photoelectric effect ...............................................................................................................118
1. Hertz’s experiment ......................................................................................................................118
The experiment with a photocell .....................................................................................................118
§50. The quantum hypothesis and photoelectric laws .........................................................................120
1. Photoelectric laws........................................................................................................................120
2. The quantum hypothesis..............................................................................................................120
3. Explaining photoelectric laws by using the quantum hypothesis ................................................121
4. Waveparticle duality of the light ................................................................................................122
§51. Light dependant resistor and photoelectric battery ......................................................................123
1. The photoconduction phenomenon..............................................................................................123
2. Light dependant resistor (LDR)...................................................................................................123
3. The photoelectric battery .............................................................................................................124
§52. Optical phenomena relating to the quantum property of the light ...............................................125
1. The luminescence ........................................................................................................................125
2. Photochemical reactions ..............................................................................................................125
§53. Application of the quantum hypothesis to hydrogen atom ..........................................................126
1. The Bohr model of the atom........................................................................................................126
2. Using the Bohr model to explain hydrogenous line spectrum.....................................................127
Summary of Chapter VIII ....................................................................................................................128
Part III. NUCLEAR PHYSICS ............................................................................................................130
Chapter IX – Basic knowledge on the atomic nucleus.............................................................................130
Translated by VNNTU – Dec. 2001 Page 6 §54. Structure of the nucleus. The unit for atomic...............................................................................130
1. Structure of the nucleus ...............................................................................................................130
2. Nuclear forces..............................................................................................................................130
3. Isotopes........................................................................................................................................130
4. The unified atomic mass unit.......................................................................................................131
§55. Radioactivity ................................................................................................................................132
Radioactivity....................................................................................................................................132
2. The radioactive decay law ...........................................................................................................133
§56. Nuclear reactions .........................................................................................................................134
1. Nuclear reactions .........................................................................................................................134
2. Conservation laws in nuclear reactions .......................................................................................134
3. Application of conservative laws to radioactivity. Transmutation rules .....................................135
§57. Artificial nuclear reactions. Applications of isotopes ..................................................................136
1. Artificial nuclear reactions ..........................................................................................................136
2. Particle accelerators.....................................................................................................................136
§58. Einstein’s relation between mass and energy..............................................................................138
1. Einstein’s axioms.........................................................................................................................138
§59. THE LOSS OF MASS. NUCLEAR ENERGY ...........................................................................140
1. The loss of mass and binding energy...........................................................................................140
Exothermic and endothermic nuclear reactions...............................................................................140
3. Two exothermic nuclear reactions...............................................................................................141
§60. Nuclear fission. Nuclear reaction plants ......................................................................................142
1. Chain nuclear reactions ...............................................................................................................142
Nuclear reaction plants ....................................................................................................................143
§61. THERMONUCLEAR REACTION.............................................................................................144
Supplemental reading: Primary Particles ..........................................................................................145
1. Properties of the primary particles: .............................................................................................145
2. Antiparticles. Antimatter. ............................................................................................................146
3. Fundamental interactions. Classification of primary particles. ...................................................146
4. Quarks..........................................................................................................................................147
SUMMARY of chapter IX...................................................................................................................147
Part IV. PRACTICAL EXPERIMENTAL EXERCISES ....................................................................150
Experimental exercise 1 – Clarification of the law on the simple pendulum’s oscillation.
Determination of the gravity acceleration............................................................................................150
Experimental exercise 2 – Determination of Sound wavelengths and frequencies .............................152
Experimental exercise 3 – The alternating current circuit with R, L, C ..............................................154
Experimental exercise 4 – The refraction index of glass .....................................................................156
Experimental exercise 5 – Observation of light dispersion and interference phenomena ...................158
COMBINED EXPERIMENTAL EXERCISES.......................................................................................160
Experimental exercise A – Determination of Capacitance and inductance (2 sections)......................160
Experimental exercise B – Characteristics and applications of transistors (2 sections) ......................163
Experimental exercise C – Determination of focal length of lenses (2 sections) ................................166 Translated by VNNTU – Dec. 2001 Page 7 Part I. OSCILLATIONS AND WAVES Chapter I – MECHANICAL OSCILLA TIONS §1. PERIODIC AND SIMPLE HARMONIC MOTIONS. OSCILLATION OF A MASSSPRING SYSTEM.
1. Oscillations
Flower stirs in the branch as the wind breezes. Pendulum of the clock swings to the left and right. On the
rippled lake, a small piece of wood bobs and rolls. The string of the guitar vibrates when it is played.
In the examples above, things move in a small space, not too far away from a certain equilibrium
position. The movement likes that is called the oscillation. An oscillation, or vibration, is a limited
motion on a space, repeating back and forth many times around an equilibrium position.
That position often is where thing is at rest (does not move): when there is no wind, a clock does not
work, a smooth lake, non vibrating guitar’s strings. 2. Periodic motion
Observing the oscillation of a pendulum of the clock, after a certain period of time of 0.5s, it passes
through a lowest position from the left to the right. The oscillation like that is called the periodic
oscillation. Periodic oscillation is the oscillation whose state is repeated as it was after a constant period
of time. The smallest period of time of T after that states of oscillation are repeated as they were is called
the period of periodic oscillation.
The quantity f = 1
showing the number of oscillations (i.e. how many times a state of oscillation is
T repeated as it were) per unit of time is called the frequency. Frequency is usually specified in hertz (Hz).
In the example above, the period of the pendulum is T = 0.5s so its frequency is f = 1
= 2Hz, it means
0.5 that the pendulum carries out 2 oscillations in a second.
The vibration of the guitar’s strings do not permanently maintained. It is damped then ended. But if it is
observed in a very small period of time, it is approximately a harmonic oscillation. 3. Massspring system. Simple harmonic motion
Considering a massspring system consisted of a small ball of m kg attached rigidly to a spring of
negligible mass, put in the horizontal plane as shown (figure 1.1a). There is a small hole through the ball
so it can be translated along a fixed rod in the same plane. We choose a datum axis that coincides with the rod, is directed from the left to the right, and the origin O
is the equilibrium position of the ball (position where the ball is at rest). A ball is deflected to the right by
a force F then released (figure 1.1b; the spring is not shown). It is observed that the ball moves toward
Translated by VNNTU – Dec. 2001 Page 8 the point O, passes through O. This translation is repeated many times, i.e. the ball oscillates around the
equilibrium position O.
This phenomenon is analyzed as following: when the ball is pulled to an ordinate x, forces exert into it
consist of the pulling force F’, the elastic force F of the spring, the gravity force and the reacting force of
the rod to the ball (these two forces are not shown in the figure). The gravity and reacting force are in the
vertical plane, equal to each other and opposite in the direction so they have no affect on the horizontal
translation of the ball. At the time the ball is released, there is only an elastic force exert on it.
Within the limitation of elasticity of the spring, the force F is always proportional with the displacement
x of the ball from the equilibrium position (is also the deflection of the spring), and directs toward the
point O. Since F is along the coordinate axis, it can be written as:
F =  kx (11) Here k is the spring constant (stiffness) of the spring, and the minus sign indicates that the force F is
acting in opposite direction compared with the deflection x of the ball.
According to Newton’s second law, it can be written as F = ma, or ma =  kx. Thus a = −
It is known that a velocity and acceleration are defined by v =
investigated in a very small period of time ∆t, then k
x.
m ∆x
∆v
and a =
. If the motion is
∆t
∆t ∆x
becomes a derivative of x with respect to time
∆t ∆v
!
becomes a derivative of x respecting to time t, a = v , i.e. a second order
∆t
derivative of x with respect to time variable: a = !! .
x
!
variable, v = x ; similarly, Therefore we have !! = −
x
Let ω = k
x
m (12) k
x
then !! + ω2x = 0
m (12a) It can be proved having a solution of x = Asin(ωt+ϕ)
where A and ϕ are constants and ω = (13) k
.
m Really , taking derivative of the displacement x (13) with respect to the time variable we have the velocity of the
!
ball : v = x = ωAcos( ωt + ϕ)
(14)
Taking derivative of the velocity v (14) with respect to the time, we get the acceleration of the ball:
a= !! = ω2Asin( ωt + ϕ)
x Replacing the value of x into (15) we get: (15) !! =  ω2x
x (16) (16) has the same format as (12a), it shows that (13) is the solution of (12a), in another way, the equation of the
oscillated ball is x = Asin( ωt +ϕ). Since sine function is a periodic function, it is said that the oscillation of the ball (i.e. the oscillation of
the massspring system) is a simple harmonic motion (SHM). Note that a cosine expression can be
transformed to a sine expression such a way that: Acos(ωt+ϕ) = Asin(ωt+ϕ+π/2)
Therefore, it can be defined that a SHM is an oscillation that can be described by a sinusoidal (or
cosinusoidal) function, where A, ω, ϕ are constants.
In the equation (13), x is the displacement of the oscillation, showings precisely the deflection of the
ball from the equilibrium position. A is the amplitude of the oscillation. It is the maximum value of Translated by VNNTU – Dec. 2001 Page 9 displacement, occurred when sin(ωt+ϕ) has the maximum value of 1. The meanings of ω, ϕ and ωt+ϕ
will be clarified at §2.
It is known that sine function is a periodic function with the period of 2π. Thus, it can be written as
x = Asin(ωt +ϕ) = Asin(ωt +ϕ +2π), or x = Asin[ω(t + 2π
) + ϕ]
ω 2π
) have the same value at time t. The period of
ω
2π
1ω
is called the cycle of the SHM. The reciprocal of T, f = =
is called the frequency of
time T =
T 2π
ω
It means that the displacement of the ball at time (t + the SHM. Particularly, for the massspring system, we have
T= m
2π
= 2π
k
ω (14) Now the system is taken out from the rod and hung up vertically (figure 1.1c). If the ball is pulled down
then released, it will oscillate in the vertical direction. That is also a massspring system. Everything have
been said about a horizontally oscillated spring system can be applied to a vertically oscillated spring
system as well. In this case, the equilibrium position is no longer the point O that corresponds with the
time the spring was not deflected, but is a point O’ that corresponds to the time the spring was deflected
due to the gravity of the ball.
Questions
1. Make statement about the definitions of oscillation, periodic oscillation and harmonic oscillation?
2. Differentiate between periodic and general oscillation, between periodic and harmonic oscillation?
3. Make statement about the definitions of time constant, frequency, displacement, amplitude of
harmonic oscillation?
4. Give more example about oscillation and harmonic oscillation? §2. EXPLORING A SIMPLE HARMONIC MOTION
1. Uniform circular motion and simple harmonic motion
Let consider a point M moves in a circle of central point O and radius
A (figure 1.2). The angular velocity of point M is ω (measured in
rad/s). A point C in the circle is chosen to be an origin. At the initial
time t = 0, the position of the moving point is M0, is specified by an
angle of ϕ. At an arbitrary time, the position of the moving point is Mt
specified by an angle of (ωt + ϕ).
We project the path of point M onto an axis x’x pass through point O
and perpendicular to OC. At time t, the projection of point M onto x’x
axis is point P which has the ordinate of x = OP. Since OP is the
projection of OMt onto the x’x axis so we have:
x = OMtsin(ωt +ϕ)
x = Asin(ωt +ϕ) (1.8) (1.8) has the same format as (1.3) so we can conclude that the motion of point P on the x’x axis is a
SHM. In the other way, a simple harmonic oscillation can be considered as the projection of an uniform
circular motion onto any straight line in the same plane. Translated by VNNTU – Dec. 2001 Page 10 2. Angular phase and angular frequency of a simple harmonic motion
From figure 1.2, the angular (ωt +ϕ) specifies the position of point P at the time t, it is called the phase
(or angular phase) of the oscillation at the time t. The angle of ϕ specifies the position of P at the initial
time t = 0, and is called the initial phase (or initial angular phase) of the oscillation. The angular velocity
ω allows us to determine f = ω
, which is the number of circle of M in a unit of time, and is also the
2π number of oscillation of P in a unit of time. We know that f is the frequency of the oscillation, therefore
ω is called the angular frequency (circular frequency) of the oscillation. Here ϕ, ω and (ωt +ϕ) are
specified angles and can be measured directly.
In equation (13) for the massspring system, the quantities ϕ, ω and (ωt+ϕ) have the same names but
they are not the real angles which can be experimentally measured. They are intermediary quantities
which allows determining the frequency and states of the oscillation. 3. Free motion
Let’s analyze more detail the motion of the massspring system described in §1 (figure 1.1).
The maximum displacement the ball can reach is the amplitude A. The time when the ball is released and
start to move is chosen to be the initial time t = 0. At that time x = A. In order to have the equation
x = Asin(ωt +ϕ) satisfied, we must have sin(ωt +ϕ) = 1, and since ωt =0 so ϕ = π/2.
Therefore, the oscillation equation of the ball is x = Asin(ωt +π/2) (19) So we have determined the amplitude, initial phase and the cycle of the oscillation. The amplitude and
initial phase depends on the initial conditions, i.e. the way to excite the oscillation and the way to choose
the space and temporal coordinate. The period depends only on the mass of the ball and the spring
constant, not on other factors. If the initial conditions are changed then the amplitude A and initial phase
ϕ will be changed as well but ω, T are constant.
An oscillation the period of which depends only on the system’s characteristics (here is a massspring
system), and not on other stimulating factors, is called a free oscillation. A system that can implement a
free oscillation by itself is called a selfoscillation system. After being stimulated, a selfoscillation
system will proceed with its own frequency. The oscillation of a massspring system is a free oscillation. 4. Velocity and acceleration in a simple harmonic motion
As we know from §1, !
v = x = ωAcos(ωt +π/2) = ωAsin(ωt +π) (110) !x
a = v = !! =  ω2Asin(ωt +π/2) = ω2Asin(ωt π/2) (111) In figure 1.3, there are curves presenting functions (19), (110) and (111). It is observed that after each
cycle T = 2π
, the values of displacement, velocity and acceleration are in same values as before, and the
ω behaviors of curves are also unchanged. The phase of the oscillation (ωt +ϕ) not only determines the
position of the oscillating thing but also allow to specify the value and behavior of the velocity and
acceleration. The phase of the oscillation determine the state of the oscillation. Similarly, the initial
phase ϕ specifies the initial state of the oscillation. When the ball is in the harmonic oscillation, its velocity and acceleration fluctuate following a sinusoidal
or cosinusoidal function, i.e. they fluctuate harmonically with the same frequency of the ball. Translated by VNNTU – Dec. 2001 Page 11 5. Oscillation of a simple pendulum
A simple pendulum consists of a ball attached at one end of a string. The ball has a mass of m and its
size is very small in comparison with the length of the string. The string is inelastic (constant length) and
has a negligible mass. The ball can be seen as a point of mass m attached to a nomass string. When it is
hung at point Q, its equilibrium position is QO (figure 1.4). The ball is pushed follow an arc from O to P
corresponding to a deflection angle of α. We only investigate the case in which α is small enough to have
"
the arc OP coincide with the chord OP and sinα can be approximated as α (in radian). If α ≤10o then the
error is not greater than 6/1000.
Thus we have: sinα ≈ α = s
l (112) Now, the ball is released and it swings itself. The force acting on the ball include of the gravity Ft = mg,
the tend force T of the string. The force Ft is resolved into 2 components: F’ in the direction of the string
and F is perpendicular to the string. The component F’ balances the tend force T, hence the ball does not
∩ move in the string direction. The direction of component F is tangential to the arc OP , but since α is
very small it can be considered lying along the chord OP and direct to point O.
From Newton’s second law, it can be written as :
ma = F (113) Point O is chosen to be the origin while the chord OP is taken as coordinate axis. Since a and F are in OP
axis and the direction of F is opposite with the ordinate s = OP so s
l ma = F = Ft sinα =  mgα = mg , or
a= g
g
s; and !! =  s
s
l
l Equation (1.14) has the same format as (1.2). Here (114) g
k
plays the role of
, and s plays the role of x.
l
m Therefore (1.14) has the same meaning with (1.2). We can applied the analysis process in §1 and §2 as
well then it can be concluded that the motion of a single pendulum is a harmonic oscillation with angular
frequency ω = g
.
l Translated by VNNTU – Dec. 2001 Page 12 The time constant of the pendulum is:
T= 2π
g
= 2π
l
ω (115) For small oscillation, i.e. with α ≤10o, the cycle of a simple pendulum is not dependent on the oscillation
amplitude. All the discussion have been made for a massspring system in §2 can be applied to the single
pendulum as well.
The period of the single pendulum depends on the gravity constant g. At a specified location to the earth
(g is constant), the oscillation of the single pendulum can be regarded as free oscillation.
Calculations in details have proved that when the ball is moving, the tend force T has a magnitude T > F’. The
result is that the ball is exerted by a force of (T  F’) directed to Q. This force cause a centripetal acceleration so
that the ball travels in a circle path while the acceleration in the OP direction maintains a = gs/l.
In the calculation above, the change of force T is not taken in to account, but the result is still valid. Questions
1. Make statement about definition of phase and initial phase of periodic oscillation?
2. What is angular frequency? What is the relationship between angular frequency ω and the frequency f?
3. What kind of oscillation that can be called free oscillation?
4. Why is the formula (1.15) valid only for small oscillations?
5. The displacement of an object (measured in cm) fluctuates is described by x = 4cos4πt. Calculate the
frequency of this oscillation. Determine the displacement and velocity after it starts to oscillate in 5
seconds?
6. A single pendulum has a period of 1.5s when it oscillates at a place where the gravity constant is
9.8m/s2. Determine the length if the string?
7. Determine the time constant of the pendulum in exercise 6 when it is brought to the Moon, knowing
that the gravity constant in the Moon is 5.9 times smaller than the Earth.
Hints: 5) 2Hz; 4cm; 0cm/s; 6) 0.56m; 7) 3.6s. §3. ENERGY IN A SIMPLE HARMONIC MOTION
1. Energy changes during oscillation
When the ball of the massspring system is pulled from point O to point P (figure1.5, the spring is not
shown), the force has done a work to elongate the spring, this work is passed to the ball as potential
energy. At that time, the elastic force of the spring has a maximum value so the potential energy also has
a maximum magnitude.
When the force is not exerted, the spring compressed, the
elastic force directs the ball toward point O. Its velocity is
increasing, the kinetic energy is increasing while the
potential energy is decreasing.
When the ball is at the equilibrium position O, the elastic force and the potential energy are zero, its
velocity and kinetic energy reach maximum value. The ball continues to move due to the inertial motion,
the spring is contracted, the elastic force appears in opposite direction and grows up, and the velocity is
decreasing.
When the ball reaches to point P’, the spring is contracted to the shortest length, the elastic force reaches
the maximum value and the ball is stopped. Its kinetic energy is zero, the potential energy has maximum
magnitude and stop increasing.
After that, the spring is stretching, the elastic force is decreasing, the ball is pushed to point O. Its kinetic
energy is increasing while the potential energy is decreasing. Translated by VNNTU – Dec. 2001 Page 13 During the oscillation process of the massspring system, there is always a transformation between
kinetic and potential energy: when the kinetic energy is increasing then the other is decreasing and vice
versa. 2. Conservation of mechanical energy during oscillation
We will analyze quantitatively a process of energetic transformation of a massspring system.
1
The kinetic energy of the ball is Ed = mv2.
2
Replacing v by its expression from (110): v = ωAcos(ωt +π/2), we have :
1
π
Ed = mω2A2cos2(ωt + )
2
2 (116) It was proved that the potential energy of a ball is equal to the work done of the elastic force in order to
1
bring it from the position x to the equilibrium: Et = kx2
2
Replacing x by its expression from (19): x = Asin(ωt +π/2) and replace k by mω2, we get
1
π
Et = mω2A2sin2(ωt + )
2
2 (117) (116) and (117) are the expressions of kinetic and potential energy of the ball at an arbitrary time t, and
the total energy of the ball at this time is
E = Et + Ed
E= 1
π
π
mω2A2[sin2(ωt + ) + cos2(ωt + )]
2
2
2 E= 1
mω2A2 = constant
2 (118) The total energy of oscillation is conserved. During the oscillation process, the total energy is unchanged
and proportional with the square of amplitude.
There is a transformation between potential and kinetic energy. Based on (118), we rewrite (116) and
(117) in new forms:
Ed = Ecos2(ωt +
Et = Esin2(ωt + π
)
2 π
)
2 (119a)
(119b) Note that the kinetic energy of a single pendulum is dependent on the initial excitation. If it is excited by
a powerful force to make a bigger displacement then the amplitude is bigger hence the total energy is also
bigger. Certainly, we just can increase the amplitude to a limited value within elastic limitation of the
spring.
Questions
1. Describe quantitatively the process of transformation of total energy of a single pendulum?
2. How to increase the total energy of single pendulum and at what value it can be increase?
3. How many time is the total energy of a pendulum changed if its frequency is increased 3 times while
the amplitude is reduced 2 times? Translated by VNNTU – Dec. 2001 Page 14 §4.  §5. THE COMBINATION OF OSCILLATIONS
1. Examples of the combination of oscillations
In the real life as well as in science and technology, there are cases in which the oscillation of an object is
a combination of many different oscillations. When the hammock is hung on a ship, it will swing with its
own frequency. However, the ship is also oscillated due to wave. Finally, the oscillation of the hammock
is a combination of two components: its own oscillation and the oscillation of the ship.
Generally, partial oscillations can have different directions, amplitudes, frequencies and phases.
Therefore, it is very complicated and difficult to determine the combined oscillation. We will only deal
with simple situations that are usually encountered in science and technology. 2. Phasedifferences between oscillations
Two oscillations which have the same frequency, generally can have different phases. For example, two
identical massspring systems are hung next to each other, they have the same angular frequency of ω.
The balls are pulled to displacements x1 = A1 and x2 = A2 respectively. At the time t = 0, ball 1 is released
to start moving. At the time when ball 1 passes through its equilibrium position, ball 2 is released and
start traveling.
It takes a quarter of period for ball 1 to travel from position x1 = A1 to the equilibrium position. So that
the oscillation of ball 2 is retarded a mount of T
compared with ball 1.
4 We find the oscillating equations of two ball in the form of
x1 = A1sin(ωt + ϕ1)
x2 = A2sin(ωt + ϕ2)
As we have known (from §2), the oscillation equation of ball 1 is:
x1 = A1sin(ωt + π/2)
For ball 2, at the time t = A2 = A2 sin( (120) T
then x2 = A2, thus:
4 ωT
ωT
ωT
π
π ω T π 2π T
. =0
+ ϕ2 ) = 1 ;
+ ϕ2 = ; ϕ2 = −
+ ϕ2 ) ; sin(
=−
4
4
4
2
2
4
2 T4 The oscillation equation of ball 2 is
x2 = A2sinωt (121) In general, the phase difference between two oscillations that have the same frequency is:
(ωt + ϕ1)  (ωt + ϕ2) = ϕ1  ϕ2
The phase difference is a constant quantity and equal to the difference between the initial phases. It is
called the phase difference ∆ϕ and these two oscillation are called phasedifferent oscillations. When
∆ϕ = ϕ1  ϕ2 > 0, it is said that oscillation 1 is a lead in phase to oscillation 2 or oscillation 2 is a lag in
phase to oscillation1. When ∆ϕ = ϕ1  ϕ2 < 0, it is said in a contrast way.
In this example, it is said that ball 1 leads ball2 by an angle of π
or ball 2 lags ball 1 by an angle of
2 π
(note that these angles only appear in calculations but are not real angles that can measured by angular
2
ruler).
The phase difference is a characteristic quantity for the discrepancy between two oscillations that have
the same frequency. If the phase difference is zero, or generally is 2nπ then they are in phase. If it is π or
(2n+1)π then they are out of phase (n is an arbitrary integer, n = 0; ±1; ±2; ±3; ...)
Translated by VNNTU – Dec. 2001 Page 15 3. Vectordiagram method
In order to combine two harmonic oscillations with the same directions, frequencies but different
amplitudes and phases, it is usually used a very convenient method called Fresnel’s vector diagram. This
method is based on a property having been discussed in §2: a simple harmonic oscillation can be treated
as the projection of a uniform circular motion on to a straight line in the same plane.
According to this method, each oscillation can be represented by a
vector. Suppose that an oscillation x = A sin(ωt+ϕ) need to be
represented. A horizontal axis (∆) and a vertical axis x’x that
intersects (∆) at point O are built (figure 1.6). A vector A whose
origin is at point O, magnitude is proportional with amplitude A and
makes with axis (∆) an angle of initial phase ϕ. At the time t = 0,
vector A (its head is M0) is rotated in positive direction
(conventionally is counterclockwise) with angular velocity of ω.
When the head M of vector A is projected on to x’x axis then the
motion of the projection P on x’x is a harmonic oscillation. At any
time t, the head of A is M, its projection on x’x is P and we have:
x = OP = Asin(ωt + ϕ)
That is the simple harmonic oscillation necessary to express. It is said that a simple harmonic oscillation
x = Asin(ωt + ϕ) is represented by a vector A. 4. The combination of two oscillations of same directions and frequencies
Suppose that one object (e.g. a massspring system hung in a moving train) simultaneously takes part in
two oscillations of the same directions and the same frequency ω, but they have different amplitudes A1,
A2 as well as initial phases ϕ1 , ϕ2
x1 = A1sin(ωt + ϕ1) (122) x2 = A2sin(ωt + ϕ2) (123) The resultant motion is the combination of two components (122) and (123). The Fresnel’s vector
diagram method will be applied to find the equation of resultant motion.
Two axes (∆) and x’x are drawn as in figure 1.7. Draw a vector,
namely A1, whose magnitude is proportional with amplitude A1
makes an angle of ϕ1 with (∆) (figure 1.7). Similarly, draw
vector A2 whose magnitude is proportional with amplitude A2
makes an angle of ϕ2 with (∆). Draw vector A which is the
resultant vector of A1 and A2, this vector makes an angle of ϕ
with (∆).
In figure 1.7, the angle between A1 and A2 is (ϕ2  ϕ1) (the phase
difference of two components x1 and x2). Since ϕ1 and ϕ2 are
constant then (ϕ2  ϕ1) is also constant.
Now rotate A1 and A2 around point O in positive direction with
the same frequency of ω. Then a trapezoid OM1MM2 is not
"
deformed since both sides OM1, OM2 and the angle M OM are unchanged. Therefore A has a constant
2 1 magnitude and rotates around O in positive direction with angular velocity ω of A1 and A2.
Since the resultant of projections of components onto an axis is the projection of the resultant vector
projected on that axis. So motion of P (projection of M) on x’x is the combination of P1 (projection of
M1) and P2 (projection of M2) on x’x axis, it is also a harmonic oscillation. A is the resultant vector of A1
and A2, and it also represents the combined oscillation and its initial phase is ϕ (figure 1.7).
Similarly, if it is necessary to combine various oscillations x1, x2, x3… it is recommended to draw the
resultant vector A of A1, A2, A3 ...
Translated by VNNTU – Dec. 2001 Page 16 Figure 1.7 is called a vector diagram. 5. Amplitude and initial phase of the combinatorial oscillation
The equation of resultant motion is x = x1 + x2 = Asin(ωt +ϕ) (124) where A is proportional with the magnitude of amplitude vector A.
It is necessary to evaluate A and ϕ in (124). For triangle OMM2 in figure 1.7, we have:
"
OM 2 = OM 2 + M 2 M 2 − 2OM 2 .M 2 M cos OM 2 M , or
2
"
A2 = A12 + A12 – 2A1A2cos OM 2 M
"
"
Since OM 2 M and M 2 OM are supplementary angles thus:
"
"
cos OM 2 M =  cos M 2 OM =  cos(ϕ2  ϕ1) (125) From figure1.7 we can deduce that
tgϕ = MP ' OP A1 sin ϕ1 + A 2 sin ϕ2
=
=
OP ' OP ' A1 cos ϕ1 + A 2 cos ϕ2 (126) Finally, the total oscillation is a harmonic oscillation described by (124), has the same frequency with
constituent oscillation, has amplitude specified by (125) and initial phase determined by (126)
From (125), the amplitude of total oscillation is dependent on the phase difference (ϕ2  ϕ1) of
constituent oscillations.
If the constituencies are in phase (ϕ2  ϕ1 = 2nπ) then cos(ϕ2  ϕ1) = 1, the resultant amplitude has
maximum value of A = A1 + A2.
If the constituencies are out of phase (ϕ2  ϕ1 = (2n+1)π) then cos(ϕ2  ϕ1) = 1, the resultant amplitude
has minimum value of A = A1  A2.
If the phase difference is arbitrary then the resultant amplitude satisfies A1 A2 < A < A1+A2.
Questions
1.What is the phase difference?
2.How are in phase oscillations, out of phase oscillation, leading phase oscillation, lagging phase
oscillation?
3.From figure 1.3 in §2, compared with the oscillation of a massspring system, are the velocity and
acceleration of ball lagging or leading and how much are they?
4.Briefly state the Fresnel’s vectordiagram method?
5.Two harmonic oscillations have the same direction and the same frequency f = 50Hz, and have the
amplitudes A1 = 2a, A2 = a and the initial phases ϕ1 = π
, ϕ2 = π.
3 a) Write the equations of these two oscillations.
b) Draw in the same diagram vectors A1, A2 , A, of these two components and of the resultant oscillation.
c) Calculate the initial phase of the resultant.
Hints: 5) ϕ = π/2. Translated by VNNTU – Dec. 2001 Page 17 §6.  §7. UNDERDAMPED AND FORCED OSCILLATIONS
1. Underdamped oscillation
In research of harmonic oscillation of massspring system, simple pendulum and other things, it is
observed that their frequencies and amplitudes are timeindependent quantities. It means that they will
repeats back and forth forever, never ended. But in fact, amplitudes of free oscillations will be damped
then ended because generally they move in a certain medium and are effected by frictions. Depending on
how much the friction is, the oscillation will be damped fast or slowly. Such oscillations are called
underdamped oscillations. An underdamped oscillation does not have harmonic properties, therefore
when talking about the amplitude, frequency, or cycle of an underdamped oscillation, it implies
approximation.
When a massspring system oscillates in the air, the air friction make it be damped.
But since it is small so it takes quite a long time to ended. Therefore, if this
oscillation is examined in a short time, the damping is negligible and it can be seen
as a harmonic oscillation.
Let a pendulum fluctuates in a container of water (figure 1.8). The friction of water
is strong enough so it will be damped fairly quickly and it will stop at the
equilibrium position (figure 1.9a).
Replacing with a container of lubricating oil, if its friction is large enough there is
no fluctuation. The ball passes through an equilibrium position (one time only) then
returns and stops there. (figure 1.9b)
If the friction of lubricating oil is much stronger, the
ball even can not passes through the equilibrium
position and stops immediately (figure 1.9c)
In the real life and technology, in some cases the
underdamping is harmful and it is required to
overcome this phenomenon (i.e. for clock pendulums).
In contrast, in some cases it is useful and needed so
people make use of that. For example, we all know
that the surface of the road is not fairly flat, and the
faster the vehicle travels, the more vibrative it is,
hence vehicles and motorcycles must have buffer
springs. When there is a gap, the spring is compressed
or stretched. After that, the frame continues vibrating
like a spring, makes travelers tired and uncomfortable.
In order to make it damped faster, vehicles are
equipped with a special equipment. It consist of a
piston that can travel in a vertical cylinder contain
lubricating oil, piston is assembled to the frame and
the cylinder is mounted to a shaft of wheel. When the
frame vibrates on buffer springs, the piston is also
fluctuated inside the cylinder. The lubricating oil
make the vibration damped faster and so the vibration
of the frames does. 2. Forced oscillation
In order to make an oscillation not be damped, the simplest way is exerting on it an externally periodic
force. This force supplies energy to the whole system to compensate the frictional losses.
It is known that a massspring system and simple pendulum are free oscillations. If there is no friction,
they will oscillate ever and ever with their own frequencies. However, it is ideal. In fact, external
environment exert on the ball a strong or light frictional force Fms , make the vibration damped (see the
diagram in figure 1.9).
An externally periodic force is applied to the ball called the forced force:
Translated by VNNTU – Dec. 2001 Page 18 Fn = Hsin(ωt +ϕ)
where H is the amplitude and ω is the frequency of the forced force. Generally, the frequency
of the external excitation f = ω
is different from that of the free oscillation f0 of the ball.
2π Theoretical calculations resulted in: during a period of initial time ∆t, total vibration of the system is a
complicated, a combination of many free vibrations as well as external excitation. After that, free
vibrations are ended, the ball oscillates due to the external excitation. Its frequency is the frequency of the
external force and the amplitude is dependent on a relationship between the externally excited frequency
f and free frequency f0. That is why a vibration after along time is a forced oscillation. If the excitation is
maintained for a long time then the forced vibration is also maintained during that time.
The complicated oscillation time ∆t is always very small compared with the forced oscillating time
afterward. It can be said that after ∆t, the system ‘forgot’ its free vibration. Therefore, in fact, it is usually
studied the forced oscillation after ∆t and it is unnecessary to care about a complicated vibrations during
∆t. 3. Resonance
This phenomenon can be examined by an experiment (figure 1.10).
A is a pendulum consisted of a mass of m fixed on to the metal rod. N is a
light and thin slab by assemble composite. The frequency f0 of the pendulum
when it does not assembled slab N is directly determined by a chronometer.
B is another pendulum consisted of a mass of M >> m that can easily slide on
to a thin calibrated metal rod. The frequency f is determined corresponding
to each position of pendulum B on the rod by a chronometer.
These two pendulums, A (that is not assembled to slab N) and B, are hung
next to each other, two rod are joined by a light spring L. Pendulum B is
allowed to swing in the plane that is perpendicular to the plane of paper. The
frequency f of pendulum B is transferred to pendulum A as a forced
excitation by the spring. This force makes pendulum A vibrating, and after a
time, pendulum A has forced oscillation with the frequency of f. Changing in
position of pendulum B results in the change in frequency f as well, and it is
observed that the vibration of pendulum A reaches the maximum value when
f ≈ f0, but when f is smaller or greater than f0 then the amplitude of pendulum
A is decreasing dramatically.
The phenomenon of the amplitude of forced oscillation is increased dramatically to a maximum value
when the frequency of forced excitation is equal to the free frequency of system is called the resonance.
Now, slab N is assembled to pendulum A to increase the atmospheric friction. Repeating the whole
process above, it is shown that pendulum A has a resonant oscillation at f = f0, but its amplitude is
smaller than that in the case of no assembly with slab N. In this case, because energy provided by forced
excitation is mainly used to compensate frictional losses, thus it does not make the amplitude increase
significantly. The resonance exhibits clearest when the friction is insignificant. 4. Applying and surmounting resonant phenomenon
Resonance is the most encountered phenomenon in life and technology, it can be harmful and useful for
people.
A child can use a small force to swing an adult’s hammock when the hammock reaches the highest level.
Continuing swing like that after certain time, i.e. exerting on the hammock a periodic force whose
frequency is the same as hammock’s frequency, a child can swing it to higher level, i.e. hammock’s
amplitude is bigger. It is impossible for the child to push a hammock to that level.
In many cases, resonance is harmful and need to be overcome. Every elastic thing are oscillations and
they have their free frequencies. They can be a bridge, frame, shaft…Due to some reasons, they vibrate
Translated by VNNTU – Dec. 2001 Page 19 resonantly with the other (e.g. a big electric generator), and they will vibrate dramatically and can be
broken, collapsed that is the concern of engineers.
At the middle of XIX century, there was a troop paraded on to a bridge, it was vibrated dramatically and
broken, make a lot of human losses. That is because the parading frequency of the troop coincided
accidental with the free oscillating frequency of the bridge and it made resonance. After this accident,
army regulation of some countries do not allow parade on the bridge. 5. Selfoscillation
There is another way to maintain oscillation, keep it not be damped and there is no external excitation. A
simple example is the pendulum clock.
The pendulum of the clock swings freely with its specified cycle T. Due to the friction with the air and at
the shaft, its fluctuation will be damped if it is not compensated the losses.
When the clock is wound up, it is accumulated a certain amount of potential energy. The spring is related
with the pendulum by a system of cogwheel and proper mechanism. When the pendulum reaches to the
maximum displacement, after half of cycle, the spring is stretched a little bit and a part of this potential
energy is transferred to the pendulum through agent mechanism. This amount of energy is sufficient to
compensate the frictional losses. Therefore, the pendulum can swing for a long time with the same
frequency and amplitude. In the watch and table clock, spiral pendulum plays the role of the pendulum
clock.
The vibration that can be maintained without external excitation is called selfoscillation. A system, such
as a pendulum clock, consists of oscillating mass, energy source and energy transfer mechanism is called
self oscillation system.
Note that in the forced oscillation, oscillating frequency is the externally excited force and the amplitude
depends on the external amplitude. But in the self oscillation, the frequency and amplitude is unchanged
from their original value as well as free oscillation.
Questions
1. In what conditions the oscillation is underdamped oscillation? How does the amplitude of
underdamped oscillation change?
2. How to make a forced oscillation? Why it has the name like that?
3. What is the resonant phenomenon? When does it happen?
4. A motorcycle is traveling on the road, consequently there is a small gap after a distance of 9m. The
free frequency of the frame of the motorcycle is on the buffer springs is 1.5s. At what speed, the
motorcycle is vibrated most dramatically?
SUMMARY OF CHAPTER I
1.Oscillation is a space limited motion, repeat back and forth around an equilibrium position.
In all kind of oscillations, periodic oscillation is a kind in which the state of motion is repeated as it was
after a certain period of time.
Cycle T is the smallest time after that the state of motion is repeated as it was. Frequency f = 1
is the
T number of oscillation in a unit of time. Its unit is Hertz (Hz).
In all kind of periodic oscillations, harmonic oscillation is described by a sinusoidal or cosinusoidal law:
x = Asin(ωt+ϕ) or x = Acos(ωt+ϕ). The displacement x is the deflection from an equilibrium position.
The amplitude is the maximum displacement. The angular frequency ω is a quantity to specify the
frequency f = ω
2π
and the cycle T =
of the harmonic oscillation. The phase (ωt+ϕ) specifies the
ω
2π state of oscillation at any time of time t. The initial phase ϕ specifies the initial state, i.e. at the time t = 0. Translated by VNNTU – Dec. 2001 Page 20 A harmonic oscillation can be regarded as the projection of a uniform circle motion in the projectile
plane. The angular velocity of the circle motion is the angular frequency of harmonic oscillation. In the
vector diagram method of Fresnel, each oscillation is represented by a vector rotates in the datum plane
in the positive direction, and the total oscillation is the projection of the motion of the head of resultant
vector on a straight line in the same plane.
2. The cycle of a massspring system is T = 2π m
depending only on the characteristics of the
k pendulum. It is called free oscillation and its cycle is called free cycle. The simple pendulum has a cycle
of T = 2π g
depending on the gravity acceleration. When the simple pendulum is placed in a specific
l location (g is constant), it can be seen its fluctuation as free oscillation.
When the pendulum oscillates harmonically, the velocity and acceleration of the ball behave follow a
sine or cosine law, i.e. they behave with the same frequency of the ball.
During the oscillating process, there is a transformation from kinetic energy to potential energy and vice
versa, but the mechanical energy is constant and proportional to the square of amplitude of the ball.
3. The phase difference of two oscillation is the difference of initial phase and is call the phase difference
∆ϕ= ϕ1  ϕ2. Two oscillations are in phase if ∆ϕ = 2nπ, are out of phase if ∆ϕ = (2n+1)π.
The combination of two harmonic oscillations with the same directions and frequencies but different
amplitudes is a harmonic oscillation with the same frequency. However, the amplitude and initial phase
of total oscillation is dependent on the phase difference of two components. If two components are in
phase then the total amplitude is maximum of A=A1 + A2 . if they are out of phase then it is minimum of
A= A1 – A2 .
4.In fact, every vibration is underdamped vibration, because the environmental frictions dissipate the
oscillating energy. In order for an oscillating system which has free frequency f0 is not damped, it is
applied a externally harmonic force of frequency f, called forced excitation. Forced oscillation has the
same frequency f with external excitation. The resonance happens when f = f0, the amplitude of forced
oscillation is increased dramatically to the maximum value. Bigger maximum amplitude is, smaller
environmental friction is.
In life and science, technology, the resonance can be either harmful or useful. Translated by VNNTU – Dec. 2001 Page 21 Chapter II – MECHANICAL WAVE. A COUSTICS §8. WAVE IN MECHANICS
1. Natural mechanical waves
When we throw a rock into a still water surface, we can observe a number of circular water waves
spreading out to every direction from the place where the stone is thrown. If we drop a cork or a leaf
down to the water surface, it will also rise up and down in response to the stimulated water waves. But it
only fluctuate in one vertical direction, instead of moving horizontally with the circular water wave.
We can explain the observation as follows. Among the water molecules, there is a coalescent force that
make them united together. When a water molecule, say A, rises up, the coalescent force makes the
nearby molecules to go up also, but a few time later. It is also these forces that helps to draw the water
molecule A back to its previously resting place. These forces acting very much the same role as the
elastic force does in an elastic pendulum. In conclusion, each molecule oscillating in a vertical direction
will tend to make the nearby molecules to oscillate in the vertical mode likewise and this mechanic
makes the oscillation to spread faring away.
Mechanical waves are mechanical oscillations that spread out with time in a material medium.
Note that in mechanical waves only the oscillation states, i.e. the phases of the oscillation, is spreading
away, while the medium’s small mediums are only fluctuating around its original resting balance place.
The water wave is one type of waves that can be observed by normal eye. In reality, using appropriate
equipment, scientists can observe waves in all other types of material – say it in solid, liquid or in gas
form. For example, if dropping some grains of sand into the surface of a wide big iron board, then using a
hammer to smash hard in one far end of the iron surface, we can still see the grains of sand bumping up.
This is because of the waves spreading through the iron board. Unfortunately, we cannot see this type of
waves with baredeyes.
In the example of the water waves, the direction of the oscillations of the mediums’ elements is
perpendicular to the direction in which the waves travel. Such a wave is called a transverse waves. There
exists another type of wave, known as a longitudinal wave, in which the oscillation of particles of the
medium is along the same direction as the motion of the wave. Longitudinal waves will be discussed in
details in this chapter. 2. Oscillation phase transmission. Wavelength.
A stone thrown into a water surface can create only a few
small waves, the oscillation will soon die out. To make
better research in mechanical waves, a small equipment is
created to help making the waves last longer. Using a thin
pieces of elastic metal, at one end sticking in a small ball or
needle. Place the metal piece so that the marble slightly
touches the surface of a large water tray (figure 2.1). Then
we just need to flip on the right end of the metal piece to
make the ball harmonically vibrate with period T. Then all
water molecules contacting with the ball will also vibrate
with period T in a relatively long time. On the water
surface, a number of circular waves will start to spread in
all directions.
We can simplified by imagining viewing the water tray through a vertical projection through P. The cut
trace will have the form displayed in figure 2.2. It is the truly form of the water waves in different instant
of time.
Suppose at t = 0 the waves have the form displayed in figure 2.2a. We can see that points A, E and I
vibrate in phase: they all go through their equilibrium positions and move downwards. Points C and G
are out of phase to points A, E and I: they also go through their equilibrium positions but moving
Translated by VNNTU – Dec. 2001 Page 22 upwards. At the time of t = T/4 (figure 2.2b) the oscillating phase of point A transmitted to point B, and
at different times of t = T/2; (figure 2.2c), t = 3T/4 (figure 2.2d) and t = T (figure 2.2e) the phase
transmitted to points C, D and E respectively. It should be noted that the oscillating phase is transmitted
in a horizontal direction, while water elements only fluctuating vertically. In figure 2.2 we can see that points A, E and I are always in phase with each other. The distance between
two successive inphase points along the direction of wave transmission is called the wavelength,
denoted by λ (the Greek letter lambda). In general, those points the distance between which is a multiple
of the wavelength will oscillate in phase.
The distance between points I and G is a half of the wavelength and they are in opposite phase of each
other. In general, those points the distance between which is an odd multiple of a half of the wavelength
will oscillating out of phase. 3. Period, frequency and velocity of waves
At every points through which the mechanical waves go, the medium elements oscillate with the same
period, which equals to the period T of the wave source. This period is called the period of wave. The
reciprocal of the period, f = 1/T, is called the frequency of wave.
In the above example, we can see that after each period T the oscillation phase travels through a distance
equal to the wavelength. Thus, we can also say that: the wavelength is the distance the wave travels in a
period T.
The speed of wave transmission is called the wave velocity. Since in a period T the wave travels through
a distance equal to the wavelength λ, we have the following relation:
λ=vT
or λ= v
f (21)
(22) 4. Amplitude and energy of waves
Once the wave reached a point, it makes the medium elements at that point oscillate with a particular
amplitude. This amplitude is called the wave amplitude at the specific point in question.
We have known that the energy of a harmonic oscillation is proportional to the square of its amplitude.
The wave makes the elements oscillating, thus provides them an energy. We say: the process of wave
transmission is also the process of transferring energy.
Translated by VNNTU – Dec. 2001 Page 23 For waves that originating from one point and spreading out in a surface, the wave energy is also
stretched in to a circle that keep expanding. Since the length of the circle is proportional to its diameter,
when the wave spreads far away its energy also decreases proportional to the traveling distance. For
waves that originating from one point and spreading out in a space, the wave energy reduces proportional
to the square of the traveling distance.
In an ideal case when the wave is transmitted in one straight line, the wave energy will not be reduced
along the direction of wave propagation and the wave amplitude is the same at every point the wave goes
through.
Questions
1. What is a wave?
2. For a wave, what is transmitted, what is not?
3. Define the transverse wave and the longitudinal wave?
4. State two definitions of wavelength. If the wave velocity is constant, then what is the relationship
between the wavelength and the wave frequency?
5. In figure 2.2, which points oscillate in phase and out of phase to point H? §9.  §10. SOUND WAVE
1. Sound wave and the sensation of sound
Take a thin and long iron bar, then firmly clamp the below end of it (figure 2.3a).
Flip on the other end of the bar, we can see the iron bar to swinging back and
forth. Lowering the clipped end of the bar, so that the swinging part is shortened.
Again we flip on the free end of the bar. We can observe that the bar also
oscillating but with a faster frequency than before. When the swinging part of the
iron bar is shortened to some extend (that means the iron bar oscillating frequency
has been raised to some specific degree) we start hearing the small little noise
“uh”, i.e. the vibration of the iron bar starts creating sounds. Thus, we conclude,
the oscillating of the iron bar do sometime create sounds, and sometimes do not.
This phenomenon can be explained as follows: When the iron bar swing into one side, it makes the air
layer right before it was compressed, and it also makes the air layer right after it being relaxed. This
compressing and relaxing of the air happens periodically, and has created in the air a mechanical wave
which have a frequency equals to that of the oscillation of the iron bar. This wave spread towards our
ears, and forced our eardrums starting to be compressed and relaxed with the same frequency. It is the
fluctuating of our eardrums at a certain range of frequency that helped us to recognize sounds.
The human ears can detect only oscillations having frequencies in the range from 16Hz to 20,000Hz. Any
oscillation with the frequency in the range of 16Hz – 20,000Hz is called the sound wave, or shortly the
sound. The science that study sounds is called acoustics.
Sounds can spread in any medium, say it solid, liquid or gas/air. For example: if we have some
experiences, we can here the sounds of a horse herd trotting or the sounds of a train running far away,
which we cannot hear from the sound spreading through the air. The reason is that, sound waves
spreading in the air was barred by a lot of barricade and was faded away very rapidly.
Mechanical waves with frequency greater than 20,000Hz is called ultrasonic waves. Some natural
species can emit and detect ultrasonic. Those waves which has its frequency smaller than 16Hz is called
infrasonic waves. Human by using appropriate equipment can also create and detect these sounds, and
these has a lot of applications in science and technology.
In terms of physic nature, sound waves, ultrasonic and infrasonic sounds are not different to each other
and to other mechanical waves. The classification is made based on the sensation ability of the human ear
in detecting different waves. This is determined by the physiological characteristics of the human ear.
Thus, in acoustics, scientists do distinguish the physical and physiological characteristics of sounds. Translated by VNNTU – Dec. 2001 Page 24 2. Sound transmission. Speed of sound
Sound waves can spread in any medium, but its soundtransmission speed (the speed of sound) depends
on the elasticity and density of the medium.
In general, the speed of sound in a solid medium is greater than that in the liquid, and spreading speeds in
a liquid medium is greater than that of a gas/air medium. The speeds also change in response to the
change in the temperature of the medium.
Below are speeds of sound in some substances:
Solid and liquid (t = 20oC) Gas (in atmospheric pressure) Carbonized steel 6,100m/s Air (t = 0oC) 332 m/s Iron 5,850 m/s Steam (t = 135oC) 494 m/s Rubber 1,479 m/s Water 1,500 m/s Those mediums like cotton or sponge is not good in transmitting sound waves as their elastic
characteristic is bad. They are used as soundproof material.
Sound waves cannot be transmitted in vacuum. We can prove that by putting an electronic bell into a
glass container of a vacuum pump. When we start to extract the air out of the container, we can here that
the bell’s sounds also start to fading, and when there is no air left in the container, the sounds of the bell
has disappeared. 3. Sound altitude
Among the sounds that we can detect, there are sounds that its frequency is specified, e.g. the sound of a
singer singing a song, or the sounds of a musical instrument. This is called musical sounds. There are
also sounds that do not have specific frequency, like the sound of a diesel engine, or the sound of a herd
of horse running. These sounds are called interference. In its nature, these sounds are the combination of
a number of oscillations that having very different frequency and amplitude. We’ll only study the musical
sounds.
With only one tune of a song, but if it was song by either soprano or tenor, can give us very different
experiences. Sounds with different frequencies present us different sound senses. Those sounds that have
a high frequency is called highpitch sounds or treble. Those sounds that have lower frequency is called
lowpitch sounds or bass. The pitch of a sound is a physiological characteristic of a sound, it is based
upon a physical characteristic: the sound frequency. 4. Timbre
Even when two singers sings the same tune in the same
pitch, we can distinguish the sound of each singer. Or when
the guitar, the flute, the clarinet is playing the same musical
tune, we can still make distinctions between those different
musical instruments. Each person, each musical equipment
create sounds that have different characteristics that we, by
our hearing senses, can distinguish. These characteristics of
sounds is called timbre.
Timbre is a physiological characteristics of sounds. This
characteristic is based on physical characteristics of sound:
the frequency and the amplitude. Experiments have proved
that when a man, or a musical instrument, produces a sound
wave with frequency f1, he or it also produces other sound
waves of frequencies f2 = 2f1, f3 = 3f1, f4 = 4f1, etc.
The sound waves with frequency f1 is called the
fundamental sound, or the first harmonic. Other sounds
Translated by VNNTU – Dec. 2001 Page 25 having frequency of f2, f3, f4, … are called the second harmonic, the third harmonic, the forth harmonic…
Depending on the structure of the music instrument, and the structure of the human mouth and throat,
then the relative amplitude of the different harmonics is different. Because of this, the sound emitted is a
combination of the different harmonics. It still has frequency f1 but is no longer a sinusoidal curve,
instead it becomes a complicated periodic curve. Each form of the curve represents a different timbre. In
figure 2.4, the curves represent timbres of a piano and a clarinet, corresponding to the same fundamental
sound. They have the same period, but the shape of curves is different.
Depending on the sensitivity of the ears, we can distinguish different singing voices, or different sounds
of different music instruments. 5. Sound energy
Like any other mechanical waves, sound wave retains in it an energy that proportional to the square of
the amplitude of the wave. This energy is transmitted from the source of the sound to our ears.
The intensity of sound is the energy that a sound transports per unit time across unit area. The unit of the
sound intensity is watt per square meter (W/m2).
To the human ear, the absolute value of a sound intensity I is not as important as its relative value in
comparison with a certain value I0 which is selected as the standard sound intensity. The intensity level,
L, of any sound is defined as the (decimal) logarithm of the ratio I/I0 L(B) = lg I
I0 (24) The unit of the intensity level is bel (B). Thus if L = 1, 2, 3, 4B… then the sound intensity I is 10, 102,
103, 104… times the standard sound intensity I0.
Actually, the unit of decibel (dB) is usually used. A decibel is equal to 1/10 of bel, i.e. a measurement in
decibel would be ten times the measurement in bel. Thus, equation (23) would become L(dB) = 10 lg I
I0 (24) When L = 1dB, then I = 1.26 I0. It is the smallest sound intensity that the human ear can detect. 6. Sound loudness
To create the feeling of a sound, the sound intensity must greater than a certain minimum value which is
called the threshold of hearing. Due to the physiological characteristics of human ears, the threshold of
hearing changes according to sound frequencies. With the sound frequencies in the range of 1000 –
5000Hz, the threshold of hearing is about 1012 W/m2. With the frequency of 50Hz, the threshold of
hearing is 105 times bigger.
Thus, a sound wave of frequency 1000Hz and of intensity 107 W/m2 is considered ‘very loud’ to the
human ear, while another sound of frequency 50Hz is considered as a ‘whispering’ sound. Hence, the
loudness of a sound, or volume, is not coincide with the intensity of sound.
Human ears is most sensitive with sounds in the range of 1000 – 5000Hz, and normally more sensitive
with highpitch than lowpitch. Therefore, broadcasting announcers are often female than male. And
that’s the reason why when we lower the volume of a radio we cannot hear the lowpitch sounds
anymore.
If the sound intensity increase to as high as 10W/m2, the human ear will be suffering from stinging no
matter which frequency is, and the sound is no longer considered normal. That highest level of the sound
intensity is called the threshold of hurting. The range between the threshold of hearing and the
threshold of hurting is called the audible range of the sound.
While calculating the intensity level by using equation (24), the standard intensity I0 is selected as the
threshold of hearing of a sound wave of frequency 1000Hz.
Below are some sound’s intensity that should be noted:
Translated by VNNTU – Dec. 2001 Page 26 The threshold of hearing 0dB Noises in a room 30dB Noises in a busy supermarket 60dB Noises on the street 90dB The sound of a big lightning 120dB The threshold of hurting 130dB Sounds with high intensity make stresses and tiredness to human. Living or working long time in a place
of high soundintensity reduces the sharpness of the ears, and badly affects the human health and mind.
The performances of rock music with the speaker’s volume of around 100dB also bring bad
consequences to the audiences. 7. Sound source – Resonant box
In our surroundings, there are many sources of sound. The sound source can be an interference/noise one,
e.g. the car engine on the nearby streets, the wind squeezing through the leaves, etc.; or it can be a
musical sound source. They distinguish two main types of musical sound sources: producing sounds
using strings (stringed instruments), and producing sounds using air columns (wind instruments).
When a stretched string is stimulated by flipping on it, it vibrates with a specified frequency, which is
determined by the length, the section area, the strain and the material of string. Musical strings have very
small section areas so it can only make whirling oscillations in the nearby air space, and can hardly
produce a sound. A musical string is normally stretched over a wooden or leather surface, when it
oscillating it also makes the surface vibrate with the same frequency. Since the surface has a large area, it
can create considerable compressive and expanding air regions and therefore produce sound waves.
We also know that when a string vibrates and produces a fundamental sound, it also produces other
harmonics. Each type of musical instruments has a gourd with a certain shape, acting as a resonant box,
i.e. an empty object that has the capability of resonance to a number of different frequencies, and thus it
can intensify sound waves of those frequency. Depending on the shape and the material of the gourd,
each type of musical instruments can intensify certain harmonics, and therefore can produce its own
characteristic timbre.
The human vocalization organ operates in the same way as a string instrument. The vocal cords act like
musical strings, while the larynx, the mouth space and the nose space play the role of a resonant box.
Especially, by adjusting the positions of the lower jaw, lips and the tongue, it is possible to change the
shape of the mouth space and therefore change the timbre accordingly. Hence, the voice of a person is
very rich in timbre, and one can imitate voices of others or voices of musical instruments quite
successfully.
For flutes and clarinets, when we are blowing the air into these instruments, the air columns inside these
instruments oscillate with fundamental and harmonic frequencies. The bodies of these instruments have
different shapes and materials, they act as a resonant box and produce timbres characterizing each type of
instruments.
Questions
1. What is sound waves, and how it created sound senses? Sound waves can travel in which medium?
2. What is the timbre? And why do we have timbres?
3. What is the audible range and what are its thresholds?
4. Distinguish the sound intensity and the sound loudness.
5. How the stringed instrument created sounds?
6. One person does hammering heavily onto the train rails. At a distance of 1,090m away, another person
hears the hammering sound through the rails 3s before hearing it from the air. Calculate the sound speed
in the rails if we know that the sound speed in air is 340m/s.
Translated by VNNTU – Dec. 2001 Page 27 7. One metal surface vibrates with a frequency of 200Hz. It produces a sound wave of wavelength 7.17m
in water. Calculate the sound speed in water.
Hints: 6) ≈5300m/s; 7) 1434m/s §11. WAVE INTERFERENCE
1. Interferential phenomenon
In real life, sometimes there are two or more waves
originating from different sources and interfere each other
at a specific point. Such a situation creates a characteristic
phenomenon of wave, namely wave interference, which
will be studied in this section.
Using experimental equipment similar to that is described
in §8 (figure 2.1) but here a light bar is used instead of the
ball. In each end of the bar a small ball is attached
contacting the water surface (figure 2.5). When the bar P is
stimulated to oscillate, two balls at points A and B will
produce on the water surface two systems of circular waves
originated from two corresponding points. Two waves
spread out, and mix up with each other.
After a while, when the features of the waves are stabilized, we can observed on the water surface that
there is a group of curves of which the oscillation amplitude is maximum, coming between them is
another group of curves in which the water surface does not vibrate. These curves keep standing, unlike
waves under observation previously which spread out. 2. Theory of interference
Suppose that A and B are two sources of oscillation with the same
frequency and phase, and their waves both travel to a point M on
the same plane with A and B following two corresponding paths
d1 and d2 (figure 2.6). Two sources of wave having the same
frequency and the same phase, or having the same frequency and
a constant difference in phase are called constructive sources,
and their corresponding waves are called constructive waves.
In the example described above, the two sources do not vibrate
independently. They always oscillate with the same frequency and
phase as that of the bar P, and thus they are exactly two
constructive sources.
Suppose that the equation of oscillations at both A and B is u = asinωt. If the distance l between A and B
is negligible in comparison with lengths of paths d1 and d2, then the amplitudes of waves traveling to
point M can be safely considered to be equals.
Let’s denote v as the traveling speed of the wave, then (d1/v) is the time needed for the oscillation to
travel from A to M. As such, the oscillation at M at the instant t is in phase with the oscillation at the
source A at the instant (t  d1/v). Thus the equation describing the oscillation at point M originating from
source A is: u A = a M sin ω( t − d1
ω
) = a M sin(ωt − d 1 )
v
v (25) Similarly, the equation describing the oscillation at point M originating from source B is: u B = a M sin ω( t − d2
ω
) = a M sin(ωt − d 2 )
v
v (26) The oscillation at point M is a combination of two oscillations described by (25) and (26), having the
Translated by VNNTU – Dec. 2001
Page 28 same frequency but different phase. The phase deviation equals:
∆ϕ = ω
ω
d1 − d 2 = d
v
v where d = d 1 − d 2 is the difference in path lengths. The absolute sign means that no matter which
among two paths d1 and d2 is longer.
Since ω = 2π
d
λ
and v = , we have ∆ϕ = 2π
T
T
λ (27) thus we have the conclusion: at the points that the difference in the wave’s paths equals an integer
multiple of the wavelength, d = nλ (n = 0, 1, 2, ...), the difference in phase will equals 2nπ – which means
the two oscillating components are in phase with each other – will making the amplitude of the combined
oscillation is twice as that of each component. The oscillation at these points are maximum. Using
mathematical tools, it can be proved that the locus of these points is a family of hyperbolic curves that
receive points A and B as two focuses, and also included the equidistant line of AB (figure 2.7,
continuous lines).
Similarly, at those points that the difference in the waves’ paths d
equals an odd number time of half of the wavelength,
d=(2n+1)λ/2, then the phase difference of the two wave
components is (2n+1)π – which means the two oscillating
components are in out of phase to each other  thus making the
oscillation amplitude at these points equals zero. In other words,
the water elements at this points do not fluctuate at all. The locus
of these points is also a family of hyperbolic curves having two
points A, B as its own focuses (figure 2.7, slash lines).
At all other points, the amplitude of the oscillation stays
somewhere between those two above mentioned numbers.
The above observation is called the wave interference. The wave
interference is the combination of two or more coordinate waves
in space, in which there are fixed places where the oscillation
amplitudes are either strengthened or dismissed.
The combination of three or more waves will create a complex
picture of interference that goes beyond the scope of this book. 3. Standing wave
Prepare a string of length 2m, one end (M) was tied to
the wall, the other end (P) is kept in the our hand (figure
2.8). Stretch the string and swing our hands back and
forth, making point P oscillate. Adjust slightly the
frequency of oscillation of point P until we can see a
stabilized oscillation of the string, in which some points
in the string are fluctuating very heavily while there are
other points where the string does not seems to have in
fluctuation.
The observation can be explained as follows: the oscillation of point P travels along the string from P to
M. Since M is tied, there will be a reflected waves traveling from M to P. These incident and reflected
waves satisfy the condition of coordinate waves. Here, point M does not oscillate, i.e. at this point two
waves are out of phase to each other. The result is that, in the string there is a combination of two
coordinate waves with the same frequency but opposite phases at point M (two points P and M can be
considered as two sources of coordinate waves).
To better investigate the experiment, let’s consider a string with two tied end points A and B, over which
there will be two coordinate waves traveling in opposite directions (from A to B, and from B to A). The
Translated by VNNTU – Dec. 2001 Page 29 case here is very similar to the experiment we have just described above, but both two end points are tied
down and not oscillate at all.
Take t = 0 at the instant when two waves are in opposite phase to each other at a certain point M in the
string. The string AB will have a form as described in figure 2.9a. The first wave going from left to right
while the second wave going from right to left. The combined oscillation amplitude at every points at this
instant is 0. At t = T/4, each wave has traveled a distance of λ/4, and the combined oscillation of two
waves on the string AB has the form as described in figure 2.9b. Similarly, at t = T/2 and t = 3T/4 the
combined wave has the form as depicted in figures 2.9c and 2.9d respectively. Observing the string AB over time, we can see that point M and other points that distancing it an integer
multiple of half of the wavelength always remain stationary. They are called nodes. Point N and other
points distancing it an integer multiple of half of the wavelength are the points that fluctuate the most.
They are called antinodes (or bells) of the wave. The locations of nodes and antinodes do not change
over time. The distance between two successive nodes or antinodes equals half of the wavelength (λ/2).
The wave of which all nodes and antinodes are fixed in the space is called the standing wave. A standing
wave does not travel in the space. It should be noted that even though two component waves still travel
following two different directions, the combined wave still “stand”.
In the case of longitudinal waves, although the picture of the stand waves is a little bit different, it still
have nodes and antinodes, and their distance is still λ/2. In the string of the string musical instrument, the
standing wave is a transverse one, while in the flute or clarinet, the standing wave is a longitudinal one.
The phenomenon of standing waves allows us to observe a wavelength by normal eyes, and accurately
measure a wavelength as well. For sound waves and other types of waves, measuring the wave frequency
is rather simple. Since the frequency f, the wave speed v and the wavelength λ are related through the
correlation v = λf, based on the standing wave phenomenon it is possible to calculate the wave speed by
measuring λ and f.
Questions
1. What is the coordinate source?
2. What is the wave interference? Show how to create the interference of water waves.
3. Show how to create a standing wave in a stretched elastic string. Which points are nodes, and which
points are antinodes?
4. How to determine the wave speed on a string with a standing wave?
5. One musical string of length 60cm produces a sound note of frequency 100Hz. Observing the string, it
is recorded that there are 4 nodes (including two ends) and 3 antinodes. Calculate the wave speed on this
string.
Hints: 5) 40m/s
Translated by VNNTU – Dec. 2001 Page 30 SUMMARY OF CHAPTER II
1. Mechanical waves are mechanical oscillations that spread out with time in a material medium. The
mechanical wave can travel in either solid, liquid, or gas mediums. Wave transmission means the
transmission of oscillation phase, in which the medium particles do not travel but vibrate around
equilibrium positions. The longitudinal wave has its direction of oscillation coincided to the traveling
direction of wave, while the transverse wave has its direction of oscillation perpendicular to the traveling
direction of wave.
The period of a wave is a common period of oscillation for all substance elements through which the
wave travels, and is equal to the oscillation period of the wave source. The reciprocal of wave period is
the frequency of wave. The wave speed is the speed of phase transmission. The wavelength is the
distance between two successive inphase points along the direction of wave transmission. It is also the
distance the wave travels in a period.
Between the wave speed v, the wavelength λ and the frequency of wave f (or the period of wave T) exist
v
the correlation λ = vT or λ = .
f
The amplitude of oscillation of medium elements at the point the wave travels though is called the wave
amplitude at this point. The transmission of wave is also the transmission of energy. The wave energy at
a point is proportional to the square of the wave amplitude at this point. Generally, the farther the wave
travels from the wave source, the smaller the wave amplitude and the wave energy are.
2. Sound waves are those longitudinal waves traveling in a material medium, having frequencies in a
range of 16Hz – 20,000Hz, and causing acoustic senses inside human ears. Sound waves have both
physical and biophysical characteristics. From the physical point of view, both sound waves, ultrasonic
waves and infrasonic waves are similar to other types of mechanical wave. The biophysical
characteristics of sound waves depend on the structure of human ear.
Musical sounds are those which have specific frequencies. Noises are those which have no specific
frequencies. The sound altitude is a biophysical characteristic of sound, characterized by the frequency of
sound. Sounds created by human or musical instruments are combinations of the fundamental sound and
other harmonics, creating the timbre which is a biophysical characteristic of sound.
The loudness of sound is a biophysical characteristic of sound, depending on the intensity of sound. Each
sound frequency corresponds to a threshold of hearing, therefore two sounds with the same intensity but
different frequencies will have different loudness. The intensity level is measured in decibel (dB). An
audible sounds has an intensity level in a range of 0dB to 130dB.
There are two main sources of music: vibrating strings in stringed instruments and air columns in wind
instruments.
3. The wave interference is the combination of two or more coordinate waves in space, in which there are
fixed places where the oscillation amplitudes are either strengthened or dismissed. Coordinate waves are
those produced by coordinate sources of oscillation, i.e. sources having the same frequency and the same
phase, or having the same frequency and a constant phase difference. In a plane, two coordinate sources
with the same frequency and the same phase produce an interferential image in which the points of
maximum oscillation and the points of nonoscillation lie on two alternate family of hyperbolic curves.
The points of maximum oscillation are those the difference in traveling distances from two sources to
which equals an integer multiple of the wavelength. The points of non oscillation are those the difference
in traveling distances from two sources to which equals an odd integer multiple of the wavelength.
When an incident wave and its reflected wave travel in the same path, they will interfere with each other
and produce a standing wave with nodes and antinodes. The distance between two successive nodes or
antinodes equals a half of the wavelength. Based on the phenomenon of standing wave, it is possible to
easily measure the wavelength λ and to determine the speed of wave v when the wavelength λ and the
frequency f are all known. Translated by VNNTU – Dec. 2001 Page 31 Chapter III – ELECTRIC OSCILLATIO N, ALTERNATING CURRENT §12. HARMONIC OSCILLATION VOLTAGE. ALTERNATING CURRENT
1. Harmonic oscillation voltage
Let a metal loop of area S and turn N uniformly rotate around its
symmetrical axis xx’ in a uniform magnetic field B whose direction
is perpendicular to xx’ (figure 3.1). The angular velocity of the loop
is ω.
Suppose that at time t = 0 the normal On of the frame is parallel to
the direction of the magnetic field. The magnetic flux according to
one turn is BS.
At time t > 0, the normal On makes an angle ωt with vector B, the
magnetic flux according to one turn is:
Φ = BScosωt
The magnetic flux through each turn varies with time, inducing an
induced electromotive force (emf) with magnitude:
│e1│ = ∆Φ
∆t The polarity of the induced emf is such that it tends to produce a current that creates a magnetic flux to
oppose the change in magnetic flux through the area enclosed by the current loop. It determines the
direction of the induced emf and current.
e1 =  ∆Φ
∆t (31) To calculate the emf of the loop at a time t, we consider the variation of Φ within an infinitely small time
interval ∆t. It can be written that:
e1 =  Φ’ = ωBSsinωt
The total induced emf in the loop is given by:
e = ωNBSsinωt (32) where ω, N, B, S are constant with respect to time. It can be concluded that the emf exists in the loop is a
variable harmonic one. If two points A, B are connected to a external circuit, an electric current will
appear in a closed loop. The loop acts as an electric generator, an the induced emf is the emf of the
electric source.
As the emf varies harmonically with an angular frequency of ω, the voltage of the external circuit also
varies harmonically with the same angular frequency ω. With suitable initial conditions, the equation of
voltage has a simple form
u = U0sinωt (33) where u is U0 is the maximum, ω is the angular frequency which equals to the angular speed of the loop. 2. Alternating current
In practice, electric generators are used based on the above principles, with the help of many loops in
series which can produce an emf big enough. A wire system connects the generators to other places to
consume electricity. There, a potential difference of simple harmonic oscillation is maintained at the
electric knifeswitch or electric plug:
u = U0sinωt
Translated by VNNTU – Dec. 2001 (34)
Page 32 The electric circuits are usually composed of bulb, electric engine, electric cooker, etc. there are also
resistors, capacitors and inductors. They are damped oscillation circuits, and when being connected to the
plugs, the potential difference produces a forced oscillation electric current with angular frequency ω:
i = Iosin(ωt + φ) (35) The phase difference φ between i and u depends on the properties of the circuit. As the velocity of
electric field flowing in wires is so high, nearly equal to velocity of light, at a certain time t, the electric
fields at all points of the series circuit are the same. Hence, the currents at any point are the same.
The current described in (35) is a harmonic fluctuating one. It is called the alternating current. In fact,
there are currents with direction changed but they are not harmonic. When saying about alternating
current, it should be understood that we are talking about harmonic oscillation current. 3. Root mean square (rms) value of intensity and voltage
The alternating current we used is of frequency 50Hz (60Hz in some countries), the intensity varies
rapidly with respect to time. When using the alternating current, we are not interested in the effect of the
electric current at a certain time, but in the effect in a long time.
Let an alternating current i = I0sinωt go through a resistor R for quite a long time t, and then measure the
thermal energy the resistor R emitted. There is a relation between Q, I0, R and t as follows
I2
Q=R 0 t
2
Now, if there is a direct current I going through this resistor R in the same time t, emitting the same
thermal energy Q, then the relation between I, R, t, Q is:
Q = RI2t
Comparing these two equations we got I = I2
I
0
; or I = 0
2
2 (36) It means that the alternating current i = I0sinωt is equivalent to a direct current I = I0 on the heat
2
emission effects. Theoretical calculations also give the same result. The intensity I determined by (36) is
called the rms (root mean square) value of the alternating current (sometimes called the effective value).
The rms value of an alternating current is equal to the intensity of a direct current which would produce
the same thermal energy when they go through the same resistance R in the same time interval. The rms
current is equal to the peak current divided by 2 . Similarly, the rms value of an emf of an alternating electric source is E =
and the rms value of the voltage is U = U0
2 E0
2 (37)
(38) To measure AC currents and voltages, it is impossible to use DC ammeters and DC voltmeters which are
based on rotating frame principle. Other types of measuring instruments, namely AC ammeters and AC
voltmeters, should be used instead. These instruments will measure the rms values of alternating current
and voltage. The principles of operation of these instruments are based on the effects whose magnitude is
proportional to the square of current intensity.
Questions:
1. What is the alternating current?
2. Give the definitions of rms intensity of an alternating current.
3. What are the rms value of current, emf and voltage? Translated by VNNTU – Dec. 2001 Page 33 4. Write the oscillation equations of alternating voltage in the case that the rms voltage and the frequency
are: a) 220 V, 50 Hz; b) 127 V, 60 Hz.
Hints: 4 a) u = 311sin100πt (V); b) u = 180sin120πt (V). §13.  §14. ALTERNATING CURRENT IN A CIRCUIT CONTAINING ONLY RESISTANCE, INDUCTANCE OR
CAPACITANCE Normally, an AC circuit in a household equipment contains both resistance, inductance and capacitance.
However, here we will study those circuits containing only resistance, inductance or capacitance, before
continue with a common case.
B  ALTERNATING CURRENT IN A CIRCUIT CONTAINING ONLY RESISTANCE 1. Relation between current and voltage
Consider a circuit containing a resistor R (figure 3.2), and applied to two ends
of this circuit is an alternating voltage
u = U0sinωt (39) For a very small period of time ∆t, the current is considered unchanged.
According to Ohm’s law:
i= u U0
=
sinωt
R
R Because U0 and R are not changed in time, let I0 = (310)
U0
and rewrite (310) to
R i = I0sinωt (311) Comparing (39) and (311), we can see that the voltage applied
to two end of a circuit containing only resistance fluctuate
harmonically and in phase with the current. Figure 3.3 shows the
vector diagram demonstrating the relationship between voltage u
and current i. Axis Ox is called the axis of current, since the
direction of vector I0 is coincident with that of axis Ox. In this
case vector U0 lies on the axis of current. 2. Ohm’s law for an AC circuit containing only resistance
In the equation I0 = U0
, if the two sides are divided by
R 2: I= U
R (312) where I and U are rms values of current and voltage respectively. Formula (312) shows Ohm’s law for
an AC circuit containing only resistance, in the same form as for a DC circuit. Note that (310) shows the
relationship between i and u which is valueless in the real life, while (312) shows the relationship
between the rms values I and U that we interest in when using the alternating current.
C  ALTERNATING CURRENT IN A CIRCUIT CONTAINING ONLY CAPACITANCE 1. Effect of capacitors to the alternating current
Consider a circuit as shown in figure 3.4. There is an alternating
voltage between A and B. When K is switched to M, the bulb D is
lightened. Then switch K to N, the bulb is also lightened, but not as
bright as when K is switched to M. If we replace the AC voltage by a
DC voltage when K is switched to N, the bulb D can not be lightened.
It shows that the capacitor prevents the direct current from going
through it, but allows the alternating current to go through. The
Translated by VNNTU – Dec. 2001 Page 34 capacitor also has a preventive effect to the alternating current, i.e. it has a resistance, which is called the
capacitive impedance to distinguish with the normal resistance. The issue of an alternating current going
through a capacitor will be discussed farther in §25. 2. Relation between current and voltage
Connect two terminals A, B of a capacitor C (figure 3.5) with an alternating voltage u = U0sinωt (313)
The amount of charge q of the capacitor at time t is q = Cu = CU0sinωt.
The amount of charge of capacitor varies harmonically with an angular
frequency ω, i.e. there are always electrons flowing from one terminal to a
conducting plate of the capacitor, or viceversa. In other words, there is an
alternating current in AB. Considering a very small time interval ∆t, the
current i will become the derivative of q with respect to time t:
!
i = q = ωCU0cosωt = ωCU0sin(ωt + π
)
2 Let ωCU0 = I0, the above expression can be written as: i = I0sin(ωt + π
)
2 (314) Comparing (313) and (314), it can be seen that the current also varies
π
in
harmonically with the angular frequency ω, but leads the voltage
2
phase. In other words, the voltage between two ends of a circuit
containing only conductance varies harmonically with a lag of π/2 in
phase compared with the current.
By changing the time origin we can rewrite (313) and (314) to
i = I0sinωt
u = U0sin(ωt – (314a)
π
)
2 (313a) Figure 3.6 shows the relationship between the voltage and the current. In this case vector U0 is
perpendicular to the axis of current, and has a direction downward. 3. Ohm’s law for an AC circuit containing only capacitance
When two sides of the expression Io = ωCU0 are divided by
I = ωCU 2 , it leads to a new expression (315) It is the expression of Ohm’s law for circuits containing only capacitance, where I and U are the rms
values of current and voltage respectively.
1
1
is called the capacitive impedance of the circuit: ZC =
. The capacitive
ωC
ωC
impedance depends on the angular frequency ω of the current, and plays the role of resistance in Ohm’s
law for direct currents. The expression (315) can be rewritten as:
The quantity I= U
ZC (315a) According to (315), the higher the angular frequency ω is, the larger the rms value I is. Hence, if the
frequency is high, it will be easier for the current to go through the capacitor, and viceversa. If ω = 0 (i.e.
f = 0) then I = 0. The direct current cannot go through the capacitor. Translated by VNNTU – Dec. 2001 Page 35 D  ALTERNATING CURRENT IN A CIRCUIT WITH INDUCTORS 1. Effects of an inductor to the alternating current
Consider a circuit as shown in figure 3.7. There is an alternating
voltage between two terminals A, B. The resistance of the inductor is
negligible.
When K is switched to M, the bulb D is lightened. Then k is switched
to N, the bulb is also lightened, but not as bright as before.
It means that the inductor has a preventive effect to the alternating
current, i.e. it has a resistance, which is called the inductive
impedance to distinguish with the normal resistance. 2. Relation between current and voltage
Apply an alternating voltage to two terminals of an inductor of inductance L and negligible resistance
(figure 3.8). The alternating voltage produces an alternating current:
i = I0sinωt (316) Suppose that at time t, the current through L is increasing. L then
plays the role of a consumer with: ∆i
∆t e=L •
∆i
becomes the derivative i of i with respect to t. As i is
∆t
•
•
π
increasing, i > 0 and hence: e = L i = ωLI0cosωt = ωLI0sin(ωt + ).
2 Consider a very small time interval ∆t, At time t, the Ohm’s law for circuit AB is as follows
u = (R + r’)i + e
where R + r’ = 0. Thus,
u = e = ωLI0sin(ωt + π
)
2 Let ωLI0 = U0, we obtain: u = U0sin(ωt + π
)
2 (317) Complete calculations show that (317) is satisfied at any arbitrary time.
Comparing (316) and (317), it can be seen that the voltage between two
ends of a circuit containing only inductance varies harmonically with a
lead of π/2 in phase compared with the current.
Figure 3.9 shows the relationship between the voltage and the current. In
this case vector U0 is perpendicular to the axis of current, and has a
direction upward. 3. Ohm’s Law for an AC circuit with inductors
From the equation ωLI0 = U0, we deduce: I0 =
Dividing two sides by 2 , we get: I = Translated by VNNTU – Dec. 2001 U
ωL U0
ωL
(318) Page 36 It is the expression of Ohm’s law for circuits containing only inductance. I and U are rms values of
current and voltage respectively, ωL is the inductive impedance, ZL = ωL. The inductive impedance
depends on the angular frequency ω, and plays the role of resistance in Ohm’s law for the direct current.
The expression of Ohm’s law, (318), can be rewritten as:
I= U
ZL (318a) An inductor does not prevent the direct current but the alternating current. The higher the frequency is,
the more the current is prevented.
The inductor without resistance is just an ideal definition. In practice, every inductor has a resistance.
A circuit containing an inductor of inductance L and resistance R should be seen as a circuit with an
inductor L without resistance and a resistance R without inductance connected in series, as there is an
only current flowing from one end of the inductor to the other. Ohm’s law for this case is discussed in
§15.
Questions
1. In a circuit with only resistance, how does the voltage vary with current. Write the equation of Ohm’s
law for it.
2. How does the voltage vary with current in a circuit containing: a) only inductance; b) only capacitance
3. Write the expression of Ohm’s law for a circuit containing: a) only inductance; b) only capacitance
4. Write the equations demonstrating the magnitudes of capacitive impedance and inductive impedance.
What are their effects on alternating currents with different frequencies?
5. A circuit contains an inductor with inductance L = 0.8 H. The resistance of the circuit is negligible.
There is an alternating voltage 220V, 50Hz between the two terminals. Calculate the capacitive
impedance and the current intensity.
6. A circuit has a capacitor, capacitance C = 20μF. The resistance is negligible. There is an alternating
voltage 127V, 60 Hz between two terminals of the circuit. Calculate the capacitive impedance and the
current intensity. §15. ALTERNATING CURRENT IN AN RLC CIRCUIT
An electric cooker with both the resistance and the inductive impedance is equivalent to a cooker with a
resistor and an inductor connected in series. A normal electric fan consists of resistors, inductors and
capacitors connected complicatedly. In this section, we concentrate on the case of a simple nonbranched
circuit that contains a resistor R, an inductor L and a capacitor C connected in series, namely the RLC
circuit. 1. Electric current and voltage in an RLC circuit
Figure 3.10 shows a RLC series circuit with a voltage between
two terminals and an alternating current. From §12, it’s known
that at any time t, the intensities at any point of the series circuit
are the same. Suppose that the intensity of the alternating current
at time t is:
i = I0sinωt
We have known the method to determine voltages uR (between A and M), uL (between M and N), and uC
(between N and B). These voltages vary sinusoidally with the same angular frequency ω; uR and current i
π
π
are in phase; uL lead of phase to current i; and uC lags of phase to current i. Their descriptive
2
2
equations are uR = URosinωt; uL = ULosin(ωt + Translated by VNNTU – Dec. 2001 π
π
); and uC = UCosin(ωt – ) in turn.
2
2 Page 37 As this is a series circuit, the voltage between two terminals A and B at time t is:
u = uR + uL + uC
u = URosinωt + ULosin(ωt + π
π
) + UCosin(ωt – )
2
2 (319) 2. Relation between current and voltage in an RLC circuit
The oscillation of voltage u is the combination of three
oscillations of uR, uL and uC. By using Fresnel diagram, the
equation of oscillation could be found.
In the same vector diagram, draw three vectors URo, ULo, UCo
(figure 3.11). As ULo and UCo are in opposite direction, the
summation ULo + UCo has a magnitude of │ULo – UCo│ The
direction is upward if ULo > UCo , downward if ULo < UCo and
equals to zero if ULo = UCo. Vector U0 is the summation of
URo + ULo + UCo, making an angle φ with the current axis.
The equation of the voltage u is as follow:
u = U0sin (ωt + φ)
In the rightangled triangle OSP:
OP = │URo│ = URo, SP = OQ = │ULo + UCo│ = │ULo  UCo│
U0 = OS = OP 2 + PS2 = It’s known that URo = I0R, ULo = I0ωL and UCo =
U0 = Io R 2 + (ωL − 12
)
ωC 2 U Ro + (U Lo − U Co ) 2 (320) I0
. Substituting into (320):
ωC
(321) Moreover, PS
tgφ =
, tgφ =
OP ωL − 1
ωC R (322) Hence, the voltage between two terminals of the series circuit RLC varies sinusoidally with a phase
difference φ from the current. The maximum voltage is determined from (321) and the phase difference 1
, voltage and current are out of phase with voltage leading
ωC
1
1
current. If ωL <
, voltage and current are out of phase with voltage lagging current. If ωL =
, the
ωC
ωC φ is determined from (322). If ωL > voltage and current are in phase. 3. Ohm’s Law for an RLC circuit
Let both sides of (321) divide by
2
Z = R + ( ωL − 2 and let 12
)
ωC (323) We deduce that U = IZ, or
I= U
Z Translated by VNNTU – Dec. 2001 (324)
Page 38 (324) is the expression of Ohm’s law for a RLC circuit. Here, I and U are rms values of intensity and
voltage respectively, Z is the impedance playing the role of a resistance as in Ohm’s law for direct
current. 4. Resonance in an RLC circuit
Let U be the rms value and ω be the frequency of the alternating voltage between two terminals of a RLC
series circuit. According to (324), the rms value I will be minimum when the impedance Z is maximum.
According to (323), Z will be minimum when (ωL ω2 = 1
1
) = 0, i.e. ωL =
, or
ωC
ωC 1 (325) LC Therefore, if the value of L and C satisfy condition in (325), the intensity of the current in RLC series
circuit will have a maximum value. The voltage then will be in phase with current. It’s the resonance in
RLC series circuit.
Questions
1. Draw a vector diagram for a RLC circuit when ULo = 1
UCo. In this case, the voltage u leads or lags in
2 phase to the current i?
2. Write the formulae to calculate the impedance of a RLC series circuit. In which case the impedance
reaches maximum? What will happen then?
3. In what conditions u and i are out of phase and in phase. What happen in the RLC circuit when u and i
are in phase?
4. An RLC series circuit has R = 140Ω, L = 1H and C = 25μF. The alternating current in the circuit has a
frequency f = 50 Hz and I = 0.5 A (rms value). Calculate the impedance of the circuit and the voltage. §16. POWER OF THE ALTERNATING CURRENT
1. Power of the alternating current
Applied to two terminals of a circuit a certain voltage. Using a voltmeter and an ammeter to measure the
rms value of voltage and current, U and I respectively. Using a wattmeter to measure the power P
absorbed by the circuit.
If the circuit contains resistors only, the relationship between U, I and P is expressed as follow;
P = UI
If a capacitor or an inductor is connected to the circuit, the power decreases:
P < UI
It can be written that:
p = kUI
where k < 1 is a coefficient showing the decreasing of the power. Experiment leads to the relationship
between k and phase difference φ:
k = cosφ
When the circuit has only resistors, I and U are in phase and cosφ = 1, in both cases we can write:
p = UIcosφ (326) The quantity cosφ is called the power factor of the current. As shown in Figure 3.11: Translated by VNNTU – Dec. 2001 Page 39 cosφ = OP U Ro
=
OS U o But URo = IoR and Uo = IoZ. Hence:
cosφ = R
Z (327) 2. Significance of the power coefficient
According to (326), when U and I are constant the power P will increase if cosφ increases. Let us
consider specific cases:
a) when cosφ = 1, i.e. φ = 0: In this case, the circuit has only resistors, or there is a resonance in the
circuit. The power consumed is maximum and equals to UI.
b) when cosφ = 0, i.e. φ = ± π
: In this case, the circuit does not have resistors, but only has capacitors or
2 inductors or both of them. The power consumed is minimum and equals to zero. A large power can be
provided, making U and I become large, but no power can be absorbed.
c) when 0 < cosφ < 1, i.e. – π
π
< φ < 0 or 0 < φ < :
2
2 The power absorbed UIcosφ is less than the power provided UI. To enhance the effectiveness of
consuming, cosφ should be enlarged. Hence, the circuit can use most of the power provided.
In fact, systems with cosφ < 0.85 are normally not be used. Usually, electric engines have inductive
impedance larger than resistance, cosφ therefore is small. A capacitor is connected in parallel to increase
cosφ. For a RLC circuit, when it has only R and L, φ can be quite large (figure 3.12a). The connection of
C to the circuit could make φ decrease, i.e. make cosφ increase (figure 3.12b). Questions:
1. In what cases does the power factor cosφ = 1? Draw the correspond diagram.
2. In what cases does the power factor cosφ = 0? Draw the correspond diagram.
3. A coil with inductance L = 0.2H and resistance R = 10Ω is effected by alternating voltage 220 V. The
frequency is 50 Hz. Calculate the intensity of the current and the thermal energy emitted in 5 seconds. Translated by VNNTU – Dec. 2001 Page 40 §17. PROBLEMS ON AC CIRCUITS
1. Problem 1
A bulb D is connected to an AC circuit 127 V, 50 Hz through a coil of inductance
L = 0.05H and resistance RL = 1Ω (figure 3.13). The intensity of the current
through the coil is 2A.
a) Calculate the impedance.
b) Calculate the voltage between the two terminals of the coil and between the
bulb.
c) Calculate the power absorbed by circuit AB.
Solution:
Remark: The coil with both inductance and resistance is
equivalent to an inductor L (without resistance) and a resistor RL
connected in series.
Let R be the resistance of the bulb D. Figure 3.13 can be
replaced by Figure 3.14.
a) The impedance of the circuit:
Z= U 127
=
= 63.5Ω
2
I b) As Z2 = (RL + R)2 + ω2L2,
RL + R =
R= Z2 − ω2 L2 Z2 − ω2 L2 – RL = 63.52 − (100π )2 0.052 – 1 = 60.5Ω Applying Ohm’s law for circuit BM:
UR = RI = 60.5 x 2 = 121 V
Applying Ohm’s law for circuit AM:
2 2
2
2
22
ULR = ZLR.I = I R L + ω L = 2 1 + (100π ) 0.05 = 31.5 V c) The power consumed by the circuit:
P = UIcosφ
where cosφ = R L + R 1 + 60.5
=
= 0.969
Z
63.5 Thus P = 127 × 2 × 0.969 = 246W 2. Problem 2
A circuit AB contains a resistor R = 100Ω, a coil with
inductance L = 0.5H and a capacitor with variable
capacitance. The voltage between two terminals A, B has a
rms value U = 220 V and a frequency f = 50Hz. The
capacitance C is adjusted to be 10μF.
a) Calculate the circuit impedance.
b) Calculate the rms value of current.
c) Calculate the capacitance so as to get a maximum value of the rms intensity.
d) Calculate the power factor corresponding to the two capacitances above.
Translated by VNNTU – Dec. 2001 Page 41 Solution:
a. ω = 2 π f = 2 π 50 = 100 π : 10μF = 105F
The circuit impedance: Z = R 2 + ( ωL − 12
1
) = 1002 + (100π 0.5 −
) 2 = 189.7Ω
−5
100π 10
ωC b. The rms value of the current intensity:
I= U
220
=
= 1.16A
Z 189.7 c. I is maximum when there is a resonance in the circuit:
ω2 =
C= 1
LC
1
1
=
= 2x105F = 20μF
22
2
ω L 100 π 0.5 d. When C = 10 μF:
cosφ = R
100
=
= 0.527
Z 189.7 When C = 20μF (resonance):
cosφ = R
=1
R §18. SINGLEPHASE AC GENERATOR
Among all kinds of energy, it is most convenient to use electric energy. Electric energy is easy to be
transmitted and delivered. It’s easy to change into other kinds of energy by using simple systems.
The main source of electric energy which is widely used nowadays is the induction generator, it will be
discussed in detail in this section. This device converts mechanical energy to electricity, and creates a
stable, highintensity current which is convenient to use. 1. Operational principle of singlephase AC generators
The AC induction generator operates on the principle of electromagnetic induction, which is introduced
in §12. When a coil of wire is rotated in a magnetic field so that the flux through the coil fluctuates
harmonically, there will be an induced emf generated in the coil; and this emf will create a harmonic
alternating current in the external circuit.
The induced emf is very small. To create a big enough induced emf for industrial and living purposes,
many coils are arranged in the generator; each coil consists of many turns and electromagnets.
The coils in generator are connected in series, and the two ends are connected to the external circuit by a
separated system. 2. Structure of an AC generator
As two ends A and B of the coil (figure 3.16) also revolve around rotating axis xx’ of the coil, they can
not be connected directly to the external circuit. Instead, the coil is connected to two slip rings which are
put coaxially with the coil: A is connected to ring 1 and B is connected to ring 2. Two fixed brushes, a
and b, press against two slip rings and are connected to the external circuit. When the coil rotates, the
brushes slip on the rings, the current flows through brushes and rings to the external circuit.
In the generator, the component which produces magnetic field is called the field winding, and the
component which produces current is called the armature. For small generators, like dynamo in bicycle,
Translated by VNNTU – Dec. 2001 Page 42 permanent magnets can be used as the field winding. But normally, electromagnet is used to produce
strong magnetic fields.
The coils of the armature and field winding are wound on
common iron cores so that the flux can be increased. The
structures of armature and field winding are illustrated in
figure 3.17. To minimize Eddy current, actual cores are
usually laminated, with many thin sheets arranged parallel to
the flux.
The armature as well as the field winding can either be in
motion or not. The stationary outer portion is called the stator,
and the one in motion is called rotor.
The common alternating current has a frequency of 50Hz. If
the generator has a coil and a magnet, rotor has to rotate with
the angular velocity of 50 rounds per second, i.e. 3000 rounds
per minute. To decrease the number of rounds 2, 3,... n times,
the number of coils and pairs of pole increase 2, 3,... n times
(the number of coils is always equal to the pairs of pole. If
the generator has p pairs of pole rotating at a velocity of n
rounds per minute, the frequency of the current is:
f= np
60 Generators with above construction are called singlephase
alternating current generator. Their current is called singlephase alternating current.
Questions:
1. Describe the principle of operation of a singlephase
alternating current generator.
2. Define the armature and field winding of an alternating
current generator. Which one is stator and which one is
rotor.
3. Describe the operation of slip rings and brushes. §19. THREEPHASE ALTERNATING CURRENT
Normally, threephase alternating current is used in industry. It is a system of three alternating currents
having the same amplitude and frequency but are 120 degrees out of phase, i.e. have onethird period
difference of time. Threephase alternating currents are made by threephase generators. 1. Operational principle of threephase AC power generators
Similarly, the principle of operation of a threephase generator is like that of a singlephase generator.
The difference is how the coils of armature are arranged. Figure 3.18 shows structure diagram of a threephase generator. Three coils of armature are arranged on stator with a difference of onethird of the
circle. As shown in this Figure, the flux through coil 1 has a maximum value. When rotor rotates in the
arrow direction with period T, after a time of T/3, the flux through coil 2 get maximum value, and after
another T/3, the flux through coil 3 get maximum value. Hence, the flux through the 3 coils is onethird
period different with respect to time, i.e. 120 degrees out of phase. Similarly, the induced emf between
two terminals of coils is also 120 degrees out of phase. If the coil terminals are connected to 3 identically
external circuits, the three currents are also 120 degrees out of phase. The equations can be as follow:
i1 = I0sinωt Translated by VNNTU – Dec. 2001 (328) Page 43 i2 = I0sin(ωt i3 = I0sin(ωt + 2π
)
3
2π
)
3 (329)
(330) Figure 3.19 show the waveforms of (328), (329) and (330).
If each phase is connected to a separately external circuit, an alternating current like that of a singlephase generator will appear. The function of a threephase generator is then same as that of a singlephase generator. Only when 3 phases are used, can the threephase generator has more advantages than
the singlephase generator. To obtain this, there are two ways of connecting a threephase circuit: delta
(triangular) arrangement and WYE (star) arrangement. 2. WYE connection
Three points A1, A2, A3 are connected to an external circuit by three different wires, namely phase lines
(figure 3.20). Three ends B1, B2, B3 are first connected to each other, and then connected to the external
circuit by a common wire called the neutral line.
The loads are also in WYE arrangement, three ends A’1, A’2, A’3 are connected to phase lines; common
ends B’1, B’2, B’3 are connected to the neutral line. Recall that when the loads are perfectly balance, the neutral current is zero :
i = i1 + i2 + i3 = 0
where i1, i2 and i3 are the current intensities of phase lines (determined by (328), (329), (320)); i is the
current intensity of neutral line.
However, in fact the loads are not equal (unbalanced), a neutral line is essential to provide a return
current path, because the neutral current has a nonzero value. A neutral line usually has a small cross area
and connected to the ground.
Translated by VNNTU – Dec. 2001 Page 44 The circuits of home devices use one phase of the threephase network, therefore they contain a phase
line and a neutral line. 3. Delta connection
As shown in figure 3.21, the end point of coil 1 (B1) is connected to the beginning point of coil 2 (A2),
the end point of coil 2 (B2) is connected to the beginning point of coil 3 (A3) and the end point of coil 3
(B3) is connected to the beginning point of coil 1 (A1). There is no neutral wire in the delta arrangement. As a result, loads should be more symmetrical.
In some cases, a deltaconnected load can be connected to a WYEconnected generator, and viceversa.
In a Yconnected generator, the voltage across the winding are called the phase voltage Up and the
voltages across the lines are called the line voltage Ud:
Ud = 3 Up The phase voltages are 120 degrees apart. We use vector diagram to
calculate the voltage of A1B1 and A2B2 (figure 3.20). As shown in the
vector diagram (figure 3.22), the angle between two vectors Uo2 and –Uo1
is 60 degrees.
U0 = 3 Uo2 Ud = 3 Up Questions:
1. What are the differences in construction between singlephase generators and threephase generators.
2. Draw diagrams for WYEconnected and deltaconnected circuits.
3. Can a threephase electric motor 127V or 220V be connected to a 110V WYEconnected circuit?
Explain why. §20. ASYNCHRONOUS THREEPHASE MOTORS
1. Operational principle of asynchronous threephase motors
Electric energy of alternating currents is
converted to mechanical energy by using AC
electric engines. The most commonly used is
based on the electromagnetic induction and the
rotating magnetic field.
A permanent magnet can rotate around its axis
xx’ (figure 3.23). Between two poles there is a
closed wireloop revolving around its
symmetrical axis, which is in line with xx’. Translated by VNNTU – Dec. 2001 Page 45 The magnet rotates in the direction of the arrow with a constant angular velocity ω, making the magnetic
field between its two poles also rotate with angular velocity ω. The wireloop begins to rotate more and
more rapidly with the same direction as that of the magnet until it reaches a velocity of ω0 and continues
two rotate with this velocity.
When the magnet begins to rotate, the flux through the magnet changes, producing an induced current,
which opposes the change in flux. An electromagnetic force acts on the wireloop, making it rotate in the
same direction as the magnet so that it opposes the relative position in accordance with magnet. If the
wireloop reaches a velocity of ω, the flux does not change any more, the induced current and the
electromagnetic force becomes zero. In fact, it can only reach a stable velocity ω0 smaller than ω.
Engines working with the above principle are called asynchronous engines. The angular velocity can
vary in a wide range when the rotting velocity ω of the magnetic field is constant. As a matter of fact,
when the external loads change, it still works normally. That is the advantage of the asynchronous motor. 2. Rotating magnetic field of threephase current
In the above example, to create a rotating magnetic field, we have to
use a permanent magnet. An advantage of threephase current is that
it can produce rotating magnetic field without revolving
electromagnets.
Rotating magnetic field is produced by letting threephase current
flow into 3 electromagnets which are 120 degrees apart (figure
3.24). The arrangement of coils is similar to that in the threephase
generators (figure 3.18). But in electric engines (figure 3.24),
external currents are used to flow into coils 1, 2, 3.
The flux in the coils also varies sinusoidally. Figure 3.25 show the
waveforms of magnetic field B1, B2, B3 of the 3 coils.
Suppose that at a certain time t (for example, when t = T/4), the flux
through coil 1 has a positively maximum value and has a direction
outside the coil. As shown in figure 3.25, the flux of coils 2 and 3 is negative: B2 = B3 = ½ B1, i.e. its
direction is toward the coils. the resultant magnetic field has the same direction as B1 (figure 3.24). Similarly, after 1/3 period, the flux through coil 2 has a maximum value and its direction is outward. And
after another 1/3 period, the resultant magnetic field is from coil 3 outward. In short, the resultant
magnetic field rotates around center O with the same frequency as that of the current. 3. Structure of an asynchronous threephase motor
A threephase asynchronous engine is constructed based on the above principle. It has two main parts: the
stator and the rotor.
A stator consists of coils wound on iron cores which are arranged on a circle to produce rotating
magnetic field. A rotor, which is in a cylinder shape, acts as a coil wound on steel core.
When connected to a threephase circuit, rotating magnetic field produced by the stator makes the rotor
revolve around the axis. The motion of rotor is used to run other systems.
Translated by VNNTU – Dec. 2001 Page 46 Based on the principle of operation of threephase asynchronous engines, singlephase asynchronous
engines can be made. The stator of these devices consists of two coils situated 90 degrees apart, one is
connected directly to the circuit, the other is connected to the circuit through a capacitor. It makes the two
currents 90 degrees out of phase and thus a rotating magnetic field is produced. However, singlephase
asynchronous engines can only reach the power of less than hundreds watts. They are usually used for
home devices such as electric fans, vacuum cleaners, or electric pumps.
Questions:
1. Principle of operation of asynchronous motor?
2. The formation of rotating magnetic field by using threephase current.
3. Draw the stator construction diagram of a singlephase asynchronous and describe the formation of
rotating magnetic field by using direct current. §21. TRANSFORMERS. ELECTRICITY TRANSMISSION
1. Operational principle and structure of transformers
A great advantage of AC is the possibility of increasing or reducing the voltage easily almost without
loss of energy. The instrument allowing to change the voltage of AC is called the transformer.
A transformer consists of two coils winded on a common
core composed by electrical technical steel. The turn is
usually made of copper with small resistance. The core
consists of thin laminated iron blades in rectangle or
round shape (figure 3.26a). The transformer is
symbolized as in figure 3.26b.
The principle of operation of a transformer is based on
electromagnetic induction. One coil is connected to the
AC power source and is called the primary coil. The
other is connected to the loads and is called the
secondary coil. The current flowing in the primary coil
produces a fluctuated magnetic field. The varying flux
passes through the secondary coil and produces an
induced current flowing in the secondary coil and the loads. 2. Transformation of current and voltage via transformer
Consider a transformer having a primary coil with N loops and
secondary coil with N’ loops (figure 3.27).
When the primary coil is connected to an ac circuit with voltage U, the
alternating current I flowing in the primary coil produces a periodical
oscillated magnetic field. At nay time t, the flux through the cross area
has the same instantaneous value.
In a small time t, the induced emf is e0 = −
e = Ne0 ∆Φ
. Therefore, the instantaneous emf of the primary coil is:
∆t
(331) The emf of the secondary coil:
e’ = N’e0
From (331) and (332), we have (332) e' N'
=
. As e’ and e does not change over time, it can be replaced by
e
N rms values: Translated by VNNTU – Dec. 2001 Page 47 E' N'
=
E
N (333) Since the resistance of primary coil is very small, the voltage between two ends of the primary coil is
approximately equal to E. When the secondary circuit is open, the voltage U’ between two ends is equal
to E’. Then (333) becomes: U' N'
=
U
N (334) The ratio between the voltages applied to two ends of the primary coil and applied to two ends of the
secondary coil is equal to the ratio between numbers of turns of the primary and secondary coils. If the
secondary coil has more loops than the primary coil does (N’ > N), then U’ > U, the transformer is called
stepup transformer. If N’ < N, then U’ < U, it is a stepdown transformer.
When the secondary circuit is connected to the load and become a closed circuit, the voltage u’ is smaller
than e’. However, we can still use (334) as an approximate formula. If the loss made by eddy current, the
emitter energy is negligible, electric energy is conservative; the powers of the two coils are the same:
UI = U’I’
and: I U'
=
I' U (335) By using the transformer, the more the voltage increases the more the current decreases, and vice versa. 3. Transmission of power
Electric energy is used everywhere, nut it can only be produced at some places. After being produced,
electric energy should be transmitted. Transformers play an important part in transmitting electric energy.
Suppose that we have to transmit a power P in a long way. The equation showing the relationship
between P, U and I:
P = UI
Let ΔP be the loss of energy. We have:
∆P = I2R = P2 R
U2 (336) The problem is that ΔP should be as small as possible. If ΔP decreases by 100 times, R can be decreased
by 100 times or U can be increased by 100 times. In order to decrease R by 100 times, we have to
increase the cross area of wire by 100 times, that means the weight also increases by 100 times, that’s
very costly. We can increase U 10 times simply by using transformers; with power P nearly remains
unchanged. The generators in power plants can produce the alternating current of voltage 24kV. After going through
the transformers of plant, the voltage can be stepped up to 500 kV (figure 3.28). Such a current is called
Translated by VNNTU – Dec. 2001 Page 48 the high voltage current. The longer the transmitting way is, the larger the voltage will be. Then the
voltage is lowered to 635kV by stepdown transformers located near consumers. Finally, the voltage is
stepped down to 110V, 127V or 220V for home use. Also, in some home appliances (radios, televisions,
electric meters, etc.) there are small transformers which step down the voltage to 36V, 12V, 6V.
Questions
1. Present the change of voltage through the transformer.
2. Present the change of current through the transformer.
3. What are advantages of transformer in transmission and use of electric power?
4. The primary coil of a transformer has 900 turns of wire, and it is connected to a network of 127V. The
secondary coil has a voltage of 6.3V and is connected to a system of lamps with a current 3A. Calculate
the number of turns of the secondary coil and the current in the primary coil. §22. GENERATION OF DIRECT CURRENT
1. Benefits of direct current
Alternating current is widely used in daily life and industry. However, in some specific case, DC is
irreplaceable. DC is used in industry to electroplate, to cast, to charge battery… DC motors are used to
run electrical train… because of their advantage of great starting time and easily changing of speed.
DC supplied by battery do not have high power and voltage. DC is also
supplied by DC generator. But for the same power, a DC generator is more
expensive to construct then AC generator and DC cannot be transmitted on a
long distance as AC.
The most economical and popular method to get DC is rectifying AC, convert
AC to DC. 2. Halfcycle rectifying method
The instrument that only let the current to get through in one
direction is called rectifier. The simplest rectifier is a diode
connected in series with the load on an AC circuit (figure
3.29)
In a half period, the current has the direction as shown in the
figure, diode D let the current go through it and through the
load R. In the next half period, when B is positive pole and A
is negative, diode D does not allow the current get through it
and the load R also does not have the current get through it.
In figure 3.30a is the representative curve of current through
load R.
The halfcycle rectified current is a blinked current and not
convenient for using but can be used to charge the battery. 3. Two halfcycle rectifying method
We can use both two halfperiod of the AC and have a less blinked DC by
using 4 diodes D1, D2, D3, D4 connect as shown in diagram 3.31.
In the first halfperiod, A is positive, the current goes from A to M, then
to N (cannot goes through MP because D4 open the circuit), from N it
goes through the load R to P (cannot goes through NQ because of D2),
from P to Q then back to B.
In the second halfperiod, B is positive, similarity the current goes
through BQNRPMA. In both halfperiod, there is a current going through
the load R.
Translated by VNNTU – Dec. 2001 Page 49 The curve representing the current going through the load R (figure 3.30b) shows that the blink was
reduced but the current in each halfperiod still varies in widely range. Using an instrument called filter,
we can make the blink reduced considerably (figure 3.30c). 4. Operational principle of DC power generators
The operation principle of DC generator is similar to the principle of the onephase AC generator (§18).
The difference is in the way to lead to current to the outer circuit.
In both generators, the current in the wire loops oscillate
harmonically and change the direction after each halfperiod. In AC generator, the current is lead out by two slip
rings (figure 3.16), and the current in the external circuit
also changes direction after each halfcycle. In DC
generator, they use two commutators to replace the slips
rings, between which there is an insulated slit (figure 3.32).
When the wire loop has a position as shown in figure
3.32a, B is positive pole, A is negative. The current from B
goes through slip ring 2 and out of brush a, returns to brush
b. After a halfcycle, the current changes its direction, A
becomes positive and B becomes negative (figure 3.32b). They arrange the slip rings so that at this right
time, slip ring 2 leaves brush a and slip ring 1 touches it. Now the current from A goes through slip ring
1, goes out on brush a and backs to brush b. As a result, brush a is always a positive pole and brush b is
always a negative pole of the generator, and the current of the external circuit is a direct current.
If the generator has only one wire loop, the current is flickering like the current made by two halfcycle
rectifying method. In industry, they construct the generators that have a great number of loops arranged
in series, creating an almost nonflickering direct current.
Questions:
1. Describe the propagation of the current through two halfcycle rectifier.
2. Describe the operation of slip rings and brushes.
SUMMARY OF CHAPTER III
1. Alternating current is the current of which intensity varies harmonically as in the equation
I = Iosin(ωt + φ), in which i is the instantaneous intensity and Io is the maximum intensity. The rms value
of alternating current equals to the intensity of a direction current so that they would create the same
amount of heat when they pass through the same resistor. The intensity is I = I0
.
2 The potential difference between two terminals of an AC circuit also varies harmonically with the same
frequency of the current, but in general, it is not in phase with the current. The rms voltage has a
magnitude U = U0
.
2 2. Voltage between two terminals of a purely resistive circuit varies harmonically and is in phase with the
current. Ohm’s law for that kind of circuit has the form: I = U/R I = U
.
ZL The voltage between two terminals of a purely inductive circuit varies harmonically and leads the current
by 90 degrees. Ohm’s law for this circuit has the form: I = U
.
ZL ZL = ωL is the inductance of the circuit. Translated by VNNTU – Dec. 2001 Page 50 The voltage between two terminals of the purely capacitive circuit varies harmonically, lags the current
by 90 degrees. Ohm’s law for this circuit has the form: I = ZC = U
.
ZC 1
is the capacitance of the circuit.
ωC Voltage between two terminals of a RLC circuit varies harmonically with phase difference φ with respect
to the current. Ohm’s law for this circuit has the form: I = Z = R 2 + (ω L − U
Z 12
) is the total impedance of the circuit.
ωC 2
Resonance occurs in a RLC circuit when ω = 1
U
. The current then has a maximum value of
.
LC
R 3. The current in the circuit has a power of P = UIcosφ. In RLC circuit, cosφ = R
is the power factor of
Z the current. In reality, generators are designed such that cosφ >= 0.85
4. The alternating current is produced by singlephase generators and threepkhase generators. They
operate by electromagnetic induction.
A singlephase generator consists of the field winding, which normally is an electromagnet or a
permanent magnet, the armature which is composed of turns in which the induced current flows.
The current goes to the external circuit through slip rings and brush. In high capacity generators, stator is
the armature and rotor is the induced part. The current goes directly without slip rings and brush to the
external circuit.
The principle of operation of threephase generators is similar to that of singlephase generators. But the
turns of armature are arranged in three groups in which three currents are 120 degrees out of phase. The
current from the generator is brought to the external circuit by using star connection or delta connection.
The loads are also arranged in the same way as that of the generators.
5. The direct current is produced by alternating current rectifying, or by a dc generator.
In the normal rectifying method, semiconductordiodes are used. To improve the stability of the rectified
current, a reasonable system of four diodes is used.
Basically, dc generators have the same principle of operation as that of ac generators, but slip rings are
replaced by commutators. To create a stable current, the armature consists of a great number of turns.
The current produced is the sum of many out of phase currents.
6. Alternating current motors operate by using a rotating magnetic field produced in stator, which affects
the induced current in rotor and makes the rotor rotate. That threephase current itself can produce a
rotating magnetic field when a threephase current flow in three magnets which are arranged 120 degrees
out of phase. For singlephase current, the rotating magnetic field in stator is produced by two
electromagnets arranged 90 degrees out of phase on a circle, one is directly connected to the load, the
other is connected to the load through a capacitor. Threephase and singlephase motors are asynchronous
motors, in which the rotating velocity of rotor is smaller than that of the magnetic field.
7. The great advantage of an alternating current is that we can use the transformer to reduce or increase
the voltage without electric dissipation. Between the current I and I’, the voltage U and U’, the number of
turns N and N’ in primary and secondary loops have a relation: U I' N
==
U' I N'
The transformer is applied in remote electricity transmission.
Translated by VNNTU – Dec. 2001 Page 51 Chapter IV – ELECTROMAGNETIC O SCILLATION. ELECTROMAGNETIC WAVE §23. OSCILLATION CIRCUITS. ELECTROMAGNETIC OSCILLATION
1. Fluctuation of charges in an oscillation circuit
Consider a circuit as shown in figure 4.1a,
knowing the capacitance C of the capacitor, the
inductance L of the coil, and knowing that the
resistance is negligible. When switch K to A,
the voltage source P charges the capacitor. The
charge q of the capacitor increases from 0 to a
maximum value Q0, and after that the capacitor
stop being charged.
Now switch K from A to B, we have a closed
circuit containing L and C, called the
oscillating circuit. The capacitor discharges
and there is a current through the inductor in
the direction of the arrow (figure 4.1b)
Suppose at the time t, the left plate of the capacitor have a charge of +q and the capacitor is discharging.
The capacitor has the role of a voltage source and the inductor has the role of a receiver.
!
In a small duration ∆t after t, the charge of the left panel reduces an amount of ∆q and q =
!
derivative of q. Since ∆q < 0 and q < 0, the current through L is:
I= ∆q
!
= q
∆t ∆q
is the
∆t (41) The increasing current i produce a induced emf in the coil with the role of a back emf
e=L│ •
∆i
│ = L│ i │
∆t • • As i > 0, we can simply write as follows: e = L i
• Combine with (41), we have: e = L i = L !!
q (42) At the time t, we can assume that the instant current i remains unchanged. Apply Ohm’s law for the part
of circuit DB containing the inductor. Note that resistance of this circuit is negligible, we have
u = (R + r)i + e = e
The voltage between two ends of the inductor is the voltage between two ends of the capacitor with the
q
1
, and combining it with (42) we have L !! = , or
q
role of voltage source. Therefore, e = u =
C
LC
!! +
q 1
q=0
LC (43) The full calculations show that (43) is true at any time of t.
As shown in §1 for the massspring system, we claim that the solution of equation (43) has the form
1
q = Q0sin(ωt + φ) with ω =
. That means the charge of the capacitor in the oscillating circuit
LC
1
changes harmonically with an angle frequency ω =
.
LC
Translated by VNNTU – Dec. 2001 Page 52 2. Electromagnetic oscillation in an oscillation circuit
With appropriate initial condition, the solution of (43) has the simple form:
q = Q0sinωt (44) In a short time interval ∆t, assuming that the instantaneous charge and instantaneous voltage of the
capacitor is constant, we can write:
u= Q
q
= 0 sinωt
C
C (45) Instantaneous energy of the capacitor:
Q2
1
qu = 0 sin2ωt
2
2C wd = wd = W0d sin2ωt
in which (46) Q2
0
2C (47) W0d = Instantaneous current through the inductor
!
i =  q =  ωQ0cosωt =  I0cosωt (48) where I0 = ωQ0.
Instantaneous energy of the inductor
wt = 121
1
Li = LI02cos2ωt = Lω2Q02cos2ωt
2
2
2 Q2
1
, we have wt = 0 cos2ωt, or
Since ω =
LC
2C
2 wt = W0tcos2ωt
in which W0t = 2
LI0 Q2
=0
2
2C (49)
(410) Compare (47) and (410), we can see that the maximum electrical energy has the same value with the
maximum magnetic energy.
Therefore: W0d = W0t = W0
We can rewrite (46) and (49) as
wd = W0 sin2ωt
2 wt = W0 cos ωt
And (46a)
(49a) wd + wt = W0sin2ωt + W0cos2ωt = W0 = const From the above result, we can have the conclusions:
a) Energy of oscillating circuit consists of electrical energy of the capacitor and magnetic energy of the
inductor.
b) Electrical energy and magnetic energy both change repeatedly with the same frequency
c) At any time, the total energy is constant, or in other words, the energy of oscillation circuit is
conservative.
The oscillation of the oscillation circuit has the above characteristics so it is called the electromagnetic
1
depends only on the properties of the circuit, therefore
oscillation. The oscillation frequency ω =
LC
the oscillation of the circuit is a free oscillation and ω is the private oscillation frequency of the circuit.
Translated by VNNTU – Dec. 2001 Page 53 Question
1. Repeat the way to find out the formula of the instantaneous energy of the capacitor and the inductor.
2. Repeat the way to prove the conservation of energy in oscillating circuit.
3. Why the oscillation of the circuit is called electromagnetic oscillation. §24. ALTERNATING CURRENT, HIGHFREQUENCY ELECTROMAGNETIC OSCILLATION, AND
MECHANICAL OSCILLATION Oscillating circuit without resistance is only an ideal concept. In fact, the circuit always has the resistance
of the inductor and the wire (although very small) and other resistance. Theory and experiment shows
that the pure resistance of the circuit makes the oscillation becomes damped, but not affects to the
1
(similar to the resistance of the environment to the oscillation of the
oscillation frequency ω =
LC
pendulum). 1. Electric oscillation in an alternating current
If we take a RLC circuit that we’ve investigated when studying
AC (figure 4.2), charge the capacitor C, then connect the two
ends AB to have a closed circuit, we also have an oscillating
circuit. AC circuit in nature is also an oscillating circuit with
1
the private oscillation frequency ω0 =
.
LC
Q2 1
1
1
, we have =L ω02 and W0 = 0 = Q02 Lω02, the energy of the circuit is proportional
LC
2C 2
C
to ω02. For RLC circuits, ω0 is very small, so the energy is also small, the oscillation of the circuit
becomes damped very fast. The alternating current in the RLC circuit is the forced oscillating circuit, the
frequency ω is forced frequency.
Since ω2 = 2. Highfrequency electromagnetic oscillation
The alternating current has a frequency of 50 Hz or 60 Hz. In science and technology, especially in radar
technology, they also used oscillating circuits with frequencies of thousands of hertz or higher. The
oscillation of such circuits is called highfrequency electromagnetic oscillation.
The circuits with highfrequency electromagnetic oscillation are used in radar generators and receivers
(will be discussed in §27). The internal oscillations are used to generate and receive radar signals. 3. Electromagnetic oscillation and mechanical oscillation
From equations (4.3) and (1.2), we can see that the forms of the electromagnetic and mechanical
oscillation are similar. Compare the two oscillations in a half of a period, we can obtain:
a) At t = 0:
The charge of the capacitor reaches the maximum
value; the current flows in the circuit with the
initial value of 0. The electric energy has its
maximum value, while the magnetic energy is 0. The distance of the ball gets the maximum value;
the ball starts moving with the initial velocity of
0. The potential energy has its maximum value,
while the kinetic energy equals 0. b) In the interval 0 < t < T/4
The charge decreases, with the current increases,
the electric energy decreases, magnetic energy
increases.
Translated by VNNTU – Dec. 2001 The distance of the ball decreases, with the
velocity increases, the potential energy decreases,
the kinetic energy increases.
Page 54 c) At t = T/4
The charge equals 0, the current is maximum, the
electric energy is 0, the magnetic energy is
maximum The distance of the ball is 0, the velocity gets
maximum value. The potential energy is 0, the
kinetic energy is maximum. d) In the interval T/4 < t < T/2
The charge increases but changes sign in the
plates of the capacitor, the current decreases, the
electric energy increases, the magnetic energy
decreases. The distance of the ball increases, but the ball
goes to the other side of the equilibrium, the
velocity decreases, the potential energy increases,
the kinetic energy decreases. e) At t = T/2
The charge is maximum but changes sign in the
plates of the capacitor, the current equals 0. The
electric energy is maximum, the magnetic energy
is 0. The distance of the ball is maximum, but the ball
is in the other side of the equilibrium, the velocity
is 0. The potential energy is maximum, the kinetic
energy is 0. Questions
Making a table describing the variations of the charge q and the current i in an electromagnetic oscillation
T
circuit from time t = to time t = T. From this table and the table of comparison (above), describe the
2
oscillation of the alternating current i in a cycle. Translated by VNNTU – Dec. 2001 Page 55 Table in page 91: Comparison between electromagnetic oscillation and mechanical oscillation Electromagnetic oscillation
Time Oscillation
state t=0 0<t< t= 2
mvmax
2 T
4 T
T
<t<
4
2
t= T
2 Translated by VNNTU – Dec. 2001 Mechanical oscillation Charge q Current i Electric
energy Magnetic
energy Q0 0 Q02
2C q < Q0 i = q’ 0 Oscillation
state Distance x Velocity Potential
energy Kinetic
energy 0 A 0 kA2
2 0 q2
2C Li 2
2 x<A v = x’ kx 2
2 mv 2
2 imax = I0 0 LI 2
2 0 vmax 0 2
mvmax
2 q < Q0 i = q’ q2
2C Li 2
2 x<A v = x’ kx 2
2 mv 2
2 Q0 0 Q02
2C 0 A 0 kA2
2 0 Page 56 §25. ELECTROMAGNETIC FIELD
1. Fluctuated electric field and fluctuated magnetic field
Faraday stated that when a flux through a closed loop varies with time, it produces an induced current in
the loop.
Maxwell discovered that when a magnetic field varies with
time, it produces a rotating electric field of which the electric
field lines surround the magnetic field lines.
According to Maxwell, the induced electric field itself exists
without the help of wires. The closed loop only helps to
discover the current, as well as the rotating electric field even
when the loop is not around.
Maxwell also discovered that when an electric field varies in
time, it produces a rotating magnetic field of which the
magnetic field lines surround the electric field lines.
Maxwell’s theory is confirmed by experiments. When a capacitor is charged, there is a varying electric
field between the two bodies of the capacitor. This varying electric field produces a rotating magnetic field
like a current flow through the capacitor. The varying of electric field between two bodies of the capacitor
is equivalent to a current flowing in the wire. It’s called displacement current. With this definition, it can
be said that the current in §23, §24 is a closed current, composed of the displacement current and the
current flowing in the wire. 2. Electromagnetic field
According to Maxwell, there no electric field or magnetic field existing separately. A varying electric field
produces a varying magnetic field, and a varying magnetic field produces a varying electric field.
A permanent magnet produces a magnetic field around it. An observer moves with a closed loop can
observe the current in the loop, that means electric field and magnetic field both exist. Similarly, a moving
observer can observe the magnetic field existing with the electric field of a stationary charge.
Electric field and magnetic field is the two sides of the only field called electromagnetic field. In some
special cases, an observer can only see electric field or magnetic field. 3. Transmission of electromagnetic interaction
Suppose that there is a varying undamped electric field E1. This produces a rotating magnetic field B1
around. If the varying speed of E1 changes, B1 also varies. Hence, the varying B1 continues to produce a
varying electric field E2. This process continues. Electromagnetic field spreads out in space, far away from
O.
Questions
1. How does the electromagnetic field form according to Maxwell?
2. What is displacement current? Compare displacement current and
3. Why say that static electric field is a special case of electromagnetic field? Translated by VNNTU – Dec. 2001 Page 57 §26. ELECTROMAGNETIC WAVES
1. Electromagnetic waves
At O there is a charge oscillating periodically with frequency f in vertical direction (figure 4.4). It
produces a harmonic varying electric field E with frequency f. This electric field produces a harmonic
varying magnetic field B with frequency f. In short, at O there is a varying electromagnetic field.
Maxwell proved that this electromagnetic field spreads out in the form of wave called electromagnetic
waves. It can be said that an oscillating charge radiates electromagnetic wave. Electromagnetic wave has
electric component oscillates in vertical direction and magnetic component oscillates in horizontal
direction (figure 4.4). The frequency of electromagnetic wave is equal to that of the charge and its velocity
is equal to that of the light, c = 300000km/s. The field directions are perpendicular to each other and to
the direction of propagation. According to Maxwell’s theory, the energy of electromagnetic wave is proportional to the fourth power of
its frequency. 2. Properties of electromagnetic waves
10 years after Maxwell’s death, Hertz was the first one who generated electromagnetic waves by making
electric pulses varying very rapidly between two points connected two plates of a highvoltage capacitor.
He researched the properties of electromagnetic waves, and found that they also have similar properties
like known mechanical waves. They can reflect on the surfaces of metal, they can interfere together. When
making interference between a electromagnetic wave and its reflexive wave, Hertz managed to create
standing wave. From them, Hertz could measure the wave length λ, and the frequency f of the electrical
pulse. He found the speed of electromagnetic wave using the formula v = λf. The result was
v = c = 300000km/s = 3×108 m/s, that matches the theoretical prediction of Maxwell.
Nowadays, we have already known that electromagnetic waves have all the characteristics as mechanical
waves, but mechanical waves are transmitted in elastic environment while electromagnetic waves can
transmitted by themselves without any need of deformation of an elastic environment, therefore they can
be transmitted in vacuum. In chapter 7, we will learn that the light is also an electromagnetic wave. 3. Electromagnetic waves and wireless communication
Nowadays, electromagnetic waves are widely used in broadcast and television, as in other fields as radars,
astronomy, cybernetics... Electromagnetic waves are typical in their wave length. Between the wave length
(measured in meter) and the frequency (measured in Hertz) has the formula: λ= c 3.108
=
f
f Translated by VNNTU – Dec. 2001 Page 58 Electromagnetic oscillations with frequencies of tens to hundreds hertz radiate very weakly. Their
corresponding waves cannot be transmitted far. In wireless communication they use waves with
frequencies of thousands hertz and greater, called radio waves. Those waves are divided into some types:
Wave type Frequency Wavelength Long and very long waves 3300kHz 1001km Middle waves 0.33MHz 1000100m Short waves 330MHz 10010m Microwaves 3030000MHz 100.01m As mentioned above, the shorter the wave length is (the higher the frequency is), the greater the energy it
has. The radio waves are almost not absorbed in water. They are used to broadcast under water and rarely
used to broadcast on the ground since their energy is low and cannot go far.
Middle waves can be transmitted on the surface of the earth. During the day time, they are strongly
absorbed by the ionosphere and therefore cannot be transmitted on a long distance. In the night, the
ionosphere reflects middle waves, so they can be transmitted far. Therefore, in the night it’s clearer to
listen to the radio than during the day time.
Short waves have more energy than middle waves. They are reflected by
the ionosphere back to the earth, the earth also reflects them for the
second time, the ionosphere reflect them for the third time... (figure 4.5).
So a highpower short wave radio station can transmit wave to
everywhere on the earth.
Microwaves have the highest energy. They are not absorbed or reflected
by the ionosphere, therefore can be transmitted very far in straight line.
They are used in universal communication. Televisions use microwaves
and cannot travel far. In order to broadcast far, they have to use
intermediate repeaters or satellites to receive waves from the generator
and then broadcast back to the earth in a specific direction.
Question:
1. State the characteristics of electromagnetic waves ?
2. State all the type of electromagnetic waves?
3. Why short wave can be transmitted on a long distance ?
4. How can the television wave be propagated to the televisions that are far from the television station ? §27. TRANSMITTING AND RECEIVING ELECTROMAGNETIC WAVES
In order to broadcast television wave, we must propagate electromagnetic wave at the generator and
receive it at the receiver. The propagation process bases on the oscillation of LC circuit. The frequency of
the circuit is f = 1
2π LC . If we choose such small L and C value, we can get a very high frequency f. 1. Periodicoscillation transmitters using transistors
When an oscillating circuit operates, the stored energy in capacitors
decreases at the beginning because of thermal emission in the wires
Translated by VNNTU – Dec. 2001 Page 59 and the radiation of electromagnetic wave. As a result, oscillation becomes damped. To maintain
oscillation, a system called periodicoscillation transmitters using transistors is used.
Two bodies of capacitor C are connected to a DC circuit through a transistor T (figure 4.6). An inductor L’
is placed next to the inductor L of the oscillating circuit. Two terminals of L’ are connected to the base and
emitter. Capacitor C’ prevents DC circuit from flowing from source P to the base.
When an oscillating circuit operates, the varying magnetic flux of inductor L produces an induced current
flowing in L’. Two inductors L and L’ are placed so that when the current IC through collector increases,
base voltage is larger than emitter voltage, and there is no current flowing through transistor, and viceversa. A periodicoscillation transmitter using transistor is made so that the amount of energy provided to
the oscillating circuit in each period is equivalent to the loss of energy.
Figure 4.6 shows only the diagram of operating principle. In fact, the diagram is more complicated. 2. Open oscillation circuit. Antenna
In a LC oscillating circuit of a periodicoscillation transmitter, electromagnetic oscillation is undamped,
but no electromagnetic wave is emitted. The reason is because most of the varying magnetic field is in the
inductor, and most of the varying electric field is in the capacitor. The emitted electromagnetic field is
negligible (figure 4.7a). This is called closed oscillating circuit. If two bodies of the capacitor are not
parallel, a part of electric field in the capacitor is emitted out of the circuit, and electromagnetic wave can
go a longer way. This is called the open oscillating circuit (figure 4.7b). In case the angle between two
bodies of the capacitor is 180 degrees, the capacity of emitting is largest (figure 4.7c). In fact, a long wire
of which an inductor is in the middle is used. One end is open and the other is connected to the ground. A
wire like this is called an antenna, as shown in figure 4.7d. 3. Principles of transmitting and receiving electromagnetic waves
For emitting electromagnetic wave, a periodicoscillation transmitter is
connected to an antenna (figure 4.8). A LC oscillating circuit has an
oscillating circuit with frequency f.
The inductor L of the oscillating circuit produces an oscillating
magnetic field with frequency f in the inductor LA of the antenna. This
produces an induced electric field, making electrons in the antenna
oscillate in the direction of antenna with the same frequency. The
antenna then emits an electromagnetic field t\with frequency f. Translated by VNNTU – Dec. 2001 Page 60 To receive electromagnetic wave, an antenna is connected to an oscillating circuit (figure 4.9). Antenna
receives many waves with different frequencies at which the electrons oscillate. With the help of two
inductors L and LA, the LC circuit also oscillates at this frequency.
In the oscillating circuit, capacitor C has a variable capacitance. To receive
wave with a certain frequency f, capacitor C is adjusted so that the circuit
oscillation has a frequency of f. Then resonance occurs, and the amplitude
of this oscillation is much larger than the others. It is said that the receiver
makes a ‘wave selection’.
Questions
1. Draw diagram and explain the operation of a periodicoscillation
transmitter using transistor.
2. Describe an openoscillating circuit.
3. Describe the principle of transmitting and receiving wave. §28.  §29. A GLANCE AT RADIO TRANSMITTERS AND RECEIVERS
Electromagnetic waves with the above principle of operation can not be used, as their energy is too small.
To overcome this disadvantage, electromagnetic oscillation is amplified before being transmitted to
antenna. Electromagnetic waves of receiver are also amplified, as their energy is lost during transmission
process.
On the other hand, commonly used television waves has a frequency from 300 kHz onwards, meanwhile,
sound has a frequency less than 20 kHz, human voice has a frequency less than 1 kHz. If oscillation with
frequencies less than 20 kHz is transmitted to antenna, they can not travel for a long way.
To solve this problem, low frequency oscillation is combined with high frequency oscillation before being
transmitted to antenna. The transmitted wave is the high frequency television wave which brings along
low frequency oscillation. At the receiver, after receiving wave from antennas, low frequency is separated. 1. Principle of oscillation amplification
Normally, oscillation is made by using transistors, making the amplitude of oscillation become much
larger.
When a current flows through transistors, the emitter current is equal to the sum of the base current and
the collector current:
IE = IB + IC
For each transistor, the proportion of IC and IB has a certain value: IC
=β
IB It is said that the current is amplified β times. β is called the amplifying coefficient of transistor. The value
of β is from 20 to 500 depending on the construction of transistors.
In the diagram of oscillation amplification (figure 4.10),
the oscillation LC circuit produces an induced current
flowing in the inductor L’. It is the base current which
flows through transistor T, and oscillates with the
frequency of LC circuit. The collector current IC” has an
amplitude IC” = βIB. In the load R there is a current
oscillating with the same frequency as that of the current in Translated by VNNTU – Dec. 2001 Page 61 LC circuit, but its amplitude is β times larger. That means the oscillation of LC circuit is amplified β
times. 2. Principle of amplitude modulation
Suppose that in the oscillation LC circuit of a periodicoscillation transmitter (figure 4.6), there is a current
i1 oscillating in high frequency f maintained by a transistor. If voltage u between emitter and collector
remains unchanged, the curve demonstrating i1 oscillation is as shown in Figure 4.11a.
Let voltage u have a low frequency oscillation with frequency f’ (figure 4.11b). The varying of u leads to
the varying of oscillation amplitude of i1 with frequency f. Hence, the current itself oscillates with
frequency f (high frequency), and its amplitude oscillates with frequency f’ (low frequency). The current i1
changes into current i2 (figure 4.11c). This process is called amplitude modulation. 3. Operational principle of radio transmitters
Figure 4.12 displays the diagram of principle of radar transmitters. Sound wave strikes at the micro,
making an electric oscillation with frequency f’ (low frequency) in the micro’s circuit. This oscillation is
transmitted to the low frequency amplifier. After going out of the amplifier, the oscillation is as shown in
Figure 4.11b, then it is transmitted to the modulator. Meanwhile, a high frequency oscillation transmitter
generates an electric oscillation of which the shape is shown in Figure 4.11a. This oscillation also
transmitted to the modulator. Translated by VNNTU – Dec. 2001 Page 62 After going out of the modulator, the oscillation is as shown in Figure 4.11c.
A modulated oscillation is transmitted through the high frequency amplifier, to antenna, then antenna
produces an electromagnetic wave with frequency f’.
Amplitude modulation is the simplest way of modulating. In radar technology, frequency modulation and
phase modulation are also used. 4. Operational principle of radio receivers
Figure 4.13 shows the diagram of principle of a radio receiver. The antenna A receives radio waves from
different transmitters, these waves are called the modulated waves. When adjusting the capacitor C, we
select the wave of desired frequency f. The receiver has an opposite function to the transmitter: it separates
low frequency waves from high frequency waves. It is called the wave separation. The high frequency oscillation in the LC circuit has a very low energy. Before separating waves, this
oscillation is transmitted through the high frequency amplifier KC. After going out of the amplifier, the
current’s high frequency oscillation has a shape as shown in figure 4.11c, which is similar to the
modulated high frequency oscillation in the transmitter before being transmitted to antenna. Translated by VNNTU – Dec. 2001 Page 63 The diode D of the waveseparating circuit allows the current to go through in only one direction. After
going out of the diode, the current will have the demonstrating curve as shown in figure 4.14a if it is
transmitted directly to the load R. In the waveseparating circuit, connected in parallel to resistor R is a
capacitor C’ which plays the role of a filter. As a result, the current flowing through R is continuous and
smooth, and has an intensity oscillating with low frequency f’ (figure 4.14b). Thus, the waveseparating
circuit has separated low frequency f’ and high frequency f.
After going through the waveseparating circuit, the current is transmitted to low frequency amplifier KA,
then to the speaker L. It makes the loudspeaker’s diaphragm oscillate with frequency f’, producing a sound
wave with the same intensity as that of the sound wave transmitted to the micro.
In fact, the transmitters and receivers have more complicated circuits, with many types of resistors,
capacitors, inductors, etc, so as to eliminate sound jam, increasing sound intensity. In short, the aim is to
make the sound transmitted from the receiver is resemble to that transmitted to the micro.
Questions
1. Explain the principle of oscillation amplification using transistors.
2. Explain the principle of amplitude modulation. Where is the low frequency oscillation?
3. Draw diagram and explain the principle of operation of radio transmitter.
4. Draw diagram and explain the principle of operation of radio receiver.
SUMMARY OF CHAPTER IV
1. A closed circuit with a capacitor C and an inductor L with negligible resistor is an oscillating circuit.
After being charged, the capacitor and the inductor produce an electromagnetic oscillation. The circuit’s
oscillation and the voltage between two ends of the inductor vary periodically with its own frequency
ω0 = 1
.
LC The equation of electromagnetic oscillation is like that of mechanical oscillation. That means
electromagnetic oscillation and mechanical oscillation have similar properties.
2. A varying electric field produces a rotating magnetic field, and a varying magnetic field produces a
rotating electric field, which is like a current called displacement current. Static electric and magnetic
fields are special cases of electromagnetic field.
Electromagnetic field spreads out in space in the form of electromagnetic wave, with a velocity
c = 300,000km/s. At certain points on transmitting direction, electromagnetic wave has electric component Translated by VNNTU – Dec. 2001 Page 64 and magnetic component oscillating periodically in perpendicular direction. If a screw moves along wave
propagation, its moving direction is from vector E to vector B. These two vectors are perpendicular to
each other and to wave propagation. Unlike mechanical wave, electromagnetic wave can spread out in
vacuum, without depending on the deformation of any environment.
Electromagnetic wave has some properties like mechanical wave. They can be reflected or interfered, and
can produce a standing wave, etc. Light is also a kind of electromagnetic wave.
3. Electromagnetic wave is used widely in radar information. The smaller the wavelength is, the larger the
energy is.
To produce electromagnetic wave, a periodicoscillation transmitter is connected to an antenna. Antenna is
a kind of open oscillating circuit, it has induced oscillation with periodicoscillation transmitter, then
electromagnetic wave is radiated.
To receive electromagnetic wave, an antenna is connected to an oscillating circuit with its own adjustable
frequency.
In the radar transmitter, low frequency wave is transmitted through a micro and low frequency oscillation
is amplified before being combined with high frequency oscillation by using principle of modulation.
Modulated high frequency oscillation is amplified and transmitted to antenna to become radar wave.
In radarradio, the oscillating circuit makes wave selection. After amplifying, waveseparating circuit
separates low frequency oscillation and high frequency oscillation. Low frequency oscillation is amplified
and transmitted to the loudspeaker. the loudspeaker diaphragm oscillates with the same frequency as that
of the transmitter, and the sound transmitted to micro is echoed. Translated by VNNTU – Dec. 2001 Page 65 Part II. OPTICS Optics is the science studying lightrelated phenomena, including the transmission of light, the formation
of images, the properties of light as well as the nature of light.
Chapter V – LIGHT REFLECTION AN D REFRACTION §30. LIGHT TRANSMISSION. LIGHT REFLECTION. PLANE MIRROR
The textbook “Physics 8” has mentioned some fundamental definitions and rules of optics. Here we
remind the most important knowledge of optics in that book. 1. Light propagation
a) Light source and light object: Light sources are those which can emit light. Light objects includes light
sources and objects which are illuminated.
b) Lightbarring object – transparent object: Lightbarring objects are those which do not allow light to
pass through.
Transparent objects are those who let almost all of the light pass through. A transparent medium does not
contain any imperfections and we cannot see the trails of light propagating inside.
c) Law of light’s straight propagation
In a transparent and identical medium,
light propagates in a straight line
direction.
The law of light’s straight propagation is
used to explain some phenomena: the
appearance of black shadows and halfdark areas; solar eclipse, lunar eclipse
(figure 5.1) and it is applied to adjust a
straight line on the ground by using
stakes.
d) Ray of light. Beam of light
A ray of light is the path of light’s
propagation. In a transparent and identical
medium, rays of light are straight lines.
In order to have the concept of ‘beam of
light’, we carry the following experiment: Let
the light from a source S which is rather small
(for example from a small candle radiating
through a round hole O, bored on a lightbarring screen M) (figure 5.2).
We can see that there is a flow of light which has a conical shape, propagating through the round hole. We
call that flow of light a beam of light.
The shape of the beam of light depends on the shape of the hole. Translated by VNNTU – Dec. 2001 Page 66 We can observe the beam of light by blowing smoke into the space area behind the hole, or by using a
screen E which is parallel with the screen M, to bar the beam of light. At that time, the beam of light will
create a roundshining area on the screen.
In this beam of light, the light emitted from the source S propagates in the form of many halflines passing
through S, lying inside the cone of the beam. Therefore, we can imagine that the above beam of light
comprises countless of rays emitted from S. Sometimes, people call this beam of light ‘beam of rays of
light’ (or beam of rays).
There are many kinds of beam of rays. We only
examine there kinds of them:
+ Diverging beam of ray is the one in which the rays
of light are emitted from one point (figure 5.3) (or
the extended lines of the rays, in the opposite
direction of the light propagation’s direction,
intersect at one point).
+ Converging beam of ray is the one in which the
rays intersect at one point.
+ Parallel beam of ray is the one in which the rays are parallel with each
other.
e) Principle of light reversibility
AB is a path of light propagation (a ray of light). On this path, we can
the light travel from A to B, or from B to A (figure 5.4). let Many simple experiments can be done to check this principle. 2. Light reflection
a) Light reflection phenomenon: The phenomenon, in which a ray
changes its direction and goes back to its old medium when striking a
smooth surface, is called reflection of light.
The smooth surface can be the surface of an object or the separation
surface between two different transparent media.
In figure 5.5, line xy represents the reflecting surface (plane of
reflection); SI represents the incident ray and IR represents the reflected
ray, where I is the incident point.
Line IN perpendicular to the place of reflection is called a normal to the plane of reflection, at I.
The plane, which is formed by the incident ray SI and the normal IN is called the plane of incidence. The
"
"
angle SIN is called the angle of incidence (i); and the angle NIR is called the angle of reflection (i’).
b) Law of light reflection
 The reflected ray lies in the plane of incidence and is on
the other side compared with the incident ray.
 The angle of reflection is equal to the angle of incidence
(i’ = i) 3. Plane mirror Translated by VNNTU – Dec. 2001 Page 67 a) Plane mirror is a part of a smooth surface, reflecting almost all of the light radiating to it.
b) Properties of the image created by a plane mirror
 A point object in front of a plane mirror G is called an object with respect to the mirror (figure 5.6). S
emits a diverging beam of light to the mirror. The reflected beam of ray is also a diverging one. The
extension of the reflected rays intersect at one point S’. S’ is the image of S through the mirror G.
 The image S’ is symmetric with S through the mirror.
When we look into the mirror, we see the image S’. However, we cannot capture this image S’ on a
screen. We call S’ an virtual image.
If we put an object which has a finite size in front of a plane mirror, its image through the mirror will be
the collection of all the images of all the points on the object. This image also has the properties as stated
above. Besides, we can see that it has the same size as the object, but in general, the image is not exactly
the same as the object (it’s like the left hand and the right hand).
Questions
1. State the Law of light’s straight propagation
2. State the Principle of light reversibility of light propagation.
3. State the law of light reflection
4. State the properties of the image of a point object and of an object
which has a finite size through a plane mirror
5. Radiate a thick beam of light into a plane mirror G, which has a finite
size (figure 5.7). Draw the reflected beam of light.
6. Given a point object S and an arbitrary point M in front of a plane mirror G. a) Draw a ray starting form
S, reflecting on the mirror and than passing through M. b) Prove that among many paths from S to the
mirror G and then M, the path that the light travels is the shortest.
7. Radiate a ray SI to a plane mirror G. The reflected ray is IR. Keep the incident ray SI unchanged, rotate
the mirror G through an angle α around the axis which is perpendicular to the plane of incidence. The new
reflected ray is I’R’. Determine the angle between IR and I’R’.
Hints: 7) 2α §31. CONCAVE SPHERICAL MIRRORS
1. Definitions
The concave spherical mirror is a part of a sphere (its
shape is usually a partial sphere) capable of reflecting
light, the reflecting surface of which faces the center of
the sphere (figure 5.8).
The center of the sphere is called the center of
curvature (C). The center of the mirror surface is
called the vertex (O).
The line connecting the center of curvature and the vertex is called the main optic axis. Other lines passing
through the center are called the supplemental optic axis.
Any plane that contains the main optic axis is a straight section of the mirror. Translated by VNNTU – Dec. 2001 Page 68 The angle ϕ between two supplemental optic axis passing through the edge of the mirror and lying in a
same straight section is the opening angle of the mirror.
We only consider the propagation of the rays which lie in straight sections. 2. Reflection of a light in a concave spherical mirror
Radiate a ray of light SI, which strikes the mirror surface at I. If we want to draw the reflected rays, we
consider a very small part of the mirror around point I, a small plane mirror whose normal is the radius of
the sphere passing through I.
Hence, when we have known the incident ray, we
can draw its reflected ray: the reflected ray lies in the
plane which is formed by ray SI and point C; and is
symmetric with SI through IC (figure 5.9).
Special cases:
If the direction of the ray passes through the center
of curvature, when it strikes the mirror, the light will
be reflected back along its original path.
If the incident ray passing through the vertex, its reflected ray will be symmetric with itself through the
main optic axis. 3. Formation of images by concave spherical mirror
Put a candle in front of a concave spherical mirror, on the main optic axis and rather far away from the
mirror. Adjust a small screen along the main optic axis in order to capture the image of the candle. We
have to arrange in such a way that the screen do not block all the rays from the candle to the mirror and the
screen is always perpendicular to the main optic axis of the mirror (figure 5.10).
When the small screen is at a right position,
we will see the image of the candle on the
screen. This image’s direction is reversed and
is smaller than the object. This is a real
image.
Move the candle nearer to the mirror. We
have to move the screen further away from
the mirror in order to capture the image. This
image is larger than the previous one.
The nearer the candle is to the mirror, the
further and larger is the image.
If the candle is near to the mirror at a certain extent, we will no longer be able to capture its image on the
screen. At that time, if we look into the mirror, we will see an image whose direction is the same as the
candle and it is larger than the candle. This is a virtual image.
Theory and experiment have shown that: if we want to get a clear virtual image given by the mirror, the
following conditions (termed the paraxial conditions) must be satisfied:
a) the opening angle ϕ must be very small.
b) the angles of incidence of the rays must be very small, too, i.e. the incident rays must be approximately
parallel to the main optic axis. Translated by VNNTU – Dec. 2001 Page 69 Those angles must be as small as we can consider sine, tangent and the value of the angles measured in
terms of radian are equal. Cosines of those angles are approximately equal to 1.
When the paraxial conditions are satisfied, the image of any point object through a concave spherical
mirror will the a point (and not a trail of light). 4. Main focal point. Focal length
a) Experiment: adjust a concave spherical facing the Sun in such a way that its main optic axis passes
through the center of the Sun. Adjust a small screen, which is perpendicular to the main optic axis, we will
find a position at which there is a very small and bright image of the Sun on the screen. This image lies on
the main optic axis of the mirror.
b) The beam of light emitted from the Sun to the mirror is considered as a parallel beam of light. If the size
of the Sun’s image is neglected, we can draw a conclusion: the parallel incident beam of light, after
reflected at the concave spherical mirror, will converge at one point on the main optic axis of the mirror.
That point is called the main focal point of the mirror (point F in figure 5.11).
c) Let us find the position of F: Consider an incident ray SI of the incident beam of light which is parallel
to the main optic axis (figure 5.11). The reflected ray intersects the main optic axis at F. At I, the angle of
incidence and the angle of reflection are equal to i. The triangle IFC is an isosceles one. IC = R is the
radius of the mirror. FC = R
R
≈
2 cos i 2 According to the paraxial condition, cos i ≈1. Hence, F is the central point of line segment OC. And as a
result, all the reflected rays pass through F. In other words, the reflected beam of light converges at point
F.
According to the Principle of light reversibility, if the incident ray passes through the main focal point F,
the reflected ray will be parallel to the main optic axis.
The distance f from the vertex to the main focal point is called the focal length of the mirror: f = OF.
We have: f= R
2 (51) 5. Method to draw an object’s image obtaining from a concave spherical mirror
Different from plane mirror and depending from cases, a concave spherical mirror can give a real image
(in front of the mirror) or a virtual image (behind the mirror). The image can be larger or smaller than the
object. etc…
a) The image of a point object not lying on the
main optic axis: Assume that we have to draw
the image of a point A on an object AB (figure
5.12), A emits a diverging beam of light to the
mirror.
b) If the reflected beam of light converges at a
point A’, A’ is the real image of A. If the
reflected beam of light diverges, the point A’ at
which all the extended rays, behind the mirror,
will be the virtual image of A. Translated by VNNTU – Dec. 2001 Page 70 To determine the position of A’, we only need to draw the paths of any two out of the four following
principal rays emitted from A.
 The ray passing through the center of curvature of the mirror (or a ray whose extension passes through
the center of curvature). This ray, on striking the mirror, will reflect backwards, through the center of
curvature.
 The ray which is parallel to the main optic axis. After reflected back, this ray will pass through the main
focal point.
 The ray passing through the main focal point (or a ray whose extension passes through the main focal
point). After reflected back, this ray will become parallel to the main optic axis.
 The ray passing through the vertex. After reflected back, this ray will follow the direction which is
symmetric with the incident ray through the main optic axis.
b) The image of an object given by the concave spherical mirror: The image of an object is the collection
of the images of all the points on the object. If the object is an arrow, which is perpendicular to the main
optic axis, the image will has an erect shape and is perpendicular to the main optic axis, too. Hence, to
draw the image A’B’ of the object AB above (B lies on the main optic axis), we will draw the image A’ of
A, and then project A’B’ perpendicularly to the main optic axis. As a result, we will get the image A’B’ of
AB.
By drawing the image of an object, we can see that:
 When the object lies outside the focal length of
the concave spherical mirror, its image is virtual
and has an opposite direction with the object itself.
 When the object is inside the range OF of the
mirror, its image is real, and has the same direction
with the object (figure 5.12b).
 When the object lies exactly at the focal point of
the mirror, its image is at infinity and we cannot
capture the its image.
Questions
1. What is a concave spherical mirror? What is the main focal point of the mirror?
2. Prove the expression: f = R
2 3. State the method to draw the image of a point object and an erect object given by the concave spherical
mirror.
4. A small arrow AB, which is perpendicular to the main optic axis
of a concave spherical mirror, is put in front of the mirror, and its
distance from the vertex is d. Draw the image of AB in the
following cases: a) d = 2R ; b) d = R and c) d = R/4, where R is the
radius of the mirror.
5. In figure 5.13, line xy is the main optic axis of a concave spherical mirror and A is a point object in
front of the mirror. A’ is the image of A. By using drawing method, determine the position of the vertex,
the center of curvature and the main focal point of the mirror. Translated by VNNTU – Dec. 2001 Page 71 §32. CONVEX SPHERICAL MIRRORS.
CONVEX SPHERICAL MIRRORS CONVEX SPHERICAL MIRROR EQUATIONS. APPLICATIONS OF 1. Convex spherical mirror
a) Convex spherical mirror is a spherical mirror whose center lies behind the mirror. An incident beam of
light, which is parallel to the main optic axis of the mirror, after reflected back, will become a diverging
beam of light. The extensions of the reflected rays will intersect at one point F which lies on the main
optic axis. This point F is a main focal point of the mirror, and it is a virtual focal point.
The main focal point is also lies at the central point of
the line connecting the vertex and the center of
curvature (figure 5.14).
The method to draw the image given by a convex
spherical mirror is exactly the same as the one for
drawing the image given by a concave spherical mirror.
We can see that the image of an object, given by a
convex spherical mirror is always a virtual one, with
the same direction as the object and smaller than the
object (figure 5.15).
b) Field of vision of a convex spherical mirror: Assume that there is a point object M in front of a convex
spherical mirror. The diverging beam of light emitting from M, coming to the mirror M, after reflected
back will be come another diverging beam of light, whose extensions will pass through the image M’ of
M, behind the mirror (figure 5.16).
If the shape of the mirror surface is a circle, the
reflected rays will lie inside a cone, whose top is M’
and the cone’s surface will lean on the edge of the
mirror.
Now, we replace point object M by putting our eyes
there and look into the mirror. Apply the Principle of
light reversibility, we can see that: every point object A
lying inside the cone stated above, can emit a ray of
light (or a thin beam of light) which, after reflected at the
mirror, will goes into our eyes, as long as the extension
lines of the incident rays must pass through M’. And our
eyes will see the image of A in the mirror.
The space area in front of the mirror, restricted by this
cone, is called the field of vision of the mirror. For those
objects which are outside this field of vision, the
observer cannot see their images in the mirror.
It can be proved that the field of vision of a convex
spherical mirror is always larger then the one of a plane
mirror with the same surface size and the same position
of the observer’s eyes. 2. Convex spherical mirror equations
a) Sign rules: Assume that we have a small arrow AB which is perpendicular to the main optic axis of a
concave spherical mirror, in front of the mirror. Point A lies on the main optic axis (figure 5.17). Its image
Translated by VNNTU – Dec. 2001 Page 72 is a small arrow A’B’, perpendicular to the main optic axis, with A’ lying on the main optic axis. Let’s
find an equation which can determine the position of the image A’B’.
b) The relative positions of the object, the image and the main focal point with respect to the vertex are
determined by line sections d = OA ; d’ = OA' and f = OF . We follow the sign rules: if the object AB is
a real one, d has a positive value (d > 0); if the image A’B’ is a real one, d’>0; and if A’B’ is a virtual
image, d’<0; if the focal point F is a real one (concave spherical mirror), f>0; if the focal point F is a
virtual one (convex spherical mirror), f<0.
Let h be the height of the object: h = AB and h’
be the height of the image: h’ = A' B ' . We
establish a convention that if the image is reverse
with respect to the object, h’ has the opposite sign
with respect to h; and on the other hand, if the
image is in the same direction with respect to the
object, h and h’ have the same signs.
b) Spherical mirror equations: Consider two similar triangles OA’B’ and OAB (figure 5.17), we have: OA' A' B'
=
OA
AB (1) Consider two similar triangles CA’B’ and CAB, we have: CA' A' B'
=
CA
AB
Compare (1) and (2), we have: (2) OA' CA'
=
OA CA As OA = d, OA’ = d’ so CA’ = OC – OA’ = 2f – d’, CA’ = OA – OC = d – 2f.
Substitute the values into (3), we have: 2 f − d' d'
or 2dd’ = 2df + 2d’f
=
d −2f
d
Divide both sides by 2dd’f, we get 111
+=
d d' f (52) This equation still can be applied for convex spherical mirror and for other image formations.
c) The lateral magnification: The lateral magnification is the ratio between the height of the image and the
height of the object. k= A' B'
AB Follow the above steps: k=− d'
d Translated by VNNTU – Dec. 2001 (53) Page 73 k>0 in the case where the image has the same direction with the object and k<0 in the case where the
image is reverse with respect to the object. The absolute of k gives the relative magnification of the image
with respect to the object. 3. Applications of convex spherical mirrors
a) Concave spherical mirror: concave spherical mirror has many applications
 In a solar furnace, people use concave spherical mirrors with large surface to focus the energy of the Sun
into the focal points of the mirrors. At those points, people put the devices which use solar energy (gas
ovens, furnaces, etc.)
 In reflex astronomical telescopes, people use a very large concave spherical mirror to create, at the focal
point, the real images of the heavenly bodies which they want to research (the Sun, the Moon, the stars,
etc.)
In some projection lamps, to create a parallel beam of light which is
strong enough to travel far, people use a concave spherical mirror
put the light source at its focal point. and In order to focus most of the energy of a parallel beam of light, the
mirror must have a large opening angle. Concave spherical mirror
doesn’t satisfy this requirement because it is limited by the paraxial
condition. So, in the headlight of the cars, motorbikes, people have
replace the concave spherical mirror by a paraboloid mirror (figure
5.18). to  In some kinds of lamp or flash light, the light from the those devices are emitted in all directions.
Therefore, we can make use only a part of it. In order to enhance the useful light, we sometimes use an
eyeshade whose shape is like a part of a sphere and we put the filament at the center of it. The rays of
light when reaching the eyeshade will be reflected backwards. Hence, the amount of light which is useful
will increase substantially.
In flash lights, people sometimes plate a thin layer of silver at the inner surface of the lights.
 Concave spherical mirrors are also used in hospitals and or help actors to do the makeup …
b) Convex spherical mirror: convex spherical mirrors are used in the rear view mirrors of cars and
motorbikes …
Questions
1. What is a convex spherical mirror? State the properties of the image of an object through a convex
spherical mirror.
2. State the definition of the field of vision of a convex spherical mirror.
3. Prove the convex spherical mirror equation.
4. State the applications of convex spherical mirror.
5. An short arrow object AB, perpendicular to the main optic axis of a convex spherical mirror, is put in
front of the mirror, 50cm away from it. The mirror’s radius is 1m. Determine the position, properties and
the lateral magnification of the image. Draw the image to the right scale the.
6. An object AB is put perpendicularly to the main optic axis of a concave spherical mirror and 20cm
away from the mirror. We observe a virtual image, which is three times larger than AB. Determine the
focal length of the mirror. Draw picture to demonstrate. Translated by VNNTU – Dec. 2001 Page 74 7. An object AB is put perpendicularly to the main optic axis of a concave spherical mirror. A lies on the
main optic axis. Let O, C and F be the vertex, the center of curvature and the main focal point of the
mirror. Draw the image of AB in the following cases: a) A lies outside OC; b) A lies at C; c) A lies inside
CF; d) A lies inside FO.
In each case, specify:
 The relative position of the image A’B’ (A’B’ lies inside or outside which range).
 The relative size of the image (larger or smaller than the object).
 The property of the image.
Hints: 5) d’ = 25cm, k = ½. 6) 30cm. §33. LIGHT REFRACTION
1. Light refraction phenomenon
a) Experiment: Radiating a thin parallel beam of light (considered
as a ray of light) to the interface between the air and the water
contained in a small upstanding glass basin. The incident ray of
light SI inclines above the interface (figure 5.19). We will see
there is a part of the beam goes into the water; however, at the
point of incidence I, the ray is bent. This phenomenon is called
light refraction. The ray of light traveling in the water is called
refracted ray.
We can easily observe the ways that the incident ray and the
refracted ray travel by letting them skim over a whitepainted
board.
b) The phenomenon in which the light traveling across the interface between two transparent media is
broken (changing directions abruptly) at the interface is called light refraction. 2. The law of light refraction
a) Experiments:
Experiment 1. In the above experiment, we put the board perpendicularly to the water surface and change
the angle of incidence in such a way that the incident ray skims over the surface of the board. We will see
the refracted ray IK also skim over the surface of the board. Hence, the refracted ray and the incident ray
always lie in a same plane, which is perpendicular to the interface. This plane is exactly the plane of
incidence. It contains the incident ray SI and the normal IN of the interface at the point of incidence I.
Experiment 2. This experiment aims to determine the positions of the refracted ray in the plane of
incidence. In order to achieve that, people measure the angle of incidence (the angle between the incident
ray and the normal) and the angle of refraction (the angle between the refracted ray and the normal)
correctly and find the relationship between these two angles.
This experiment was done a long time ago. At first, people thought that the angle of refraction is
proportional to the angle of incidence. However, experiments showed that this is only true with small
angles of incidence and totally wrong in the case of large angles of incidence. Until 1621, Snell (a Dutch
scientist), through experiments, discovered that sine of the angle of refraction is proportional to sine of the
angle of incidence. Some time thereafter, Descartes (a French scientist) proved this result by theory and
state it under a law. Translated by VNNTU – Dec. 2001 Page 75 The experiment examining the relation between sin i and sin r is described in the experiment exercise.
b) Law of light refraction:
+ The refracted ray of light lines in the plane of incidence and is on the other side of the normal with
respect to the incident ray of light.
+ For any pair of identical transparent media, the ratio between sine of the angle of incidence (sin i) and
sine of the angle of refraction (sin r) is always a constant. This constant depends on the nature of the two
media and is called the relative index of refraction of the medium containing the refracted ray (medium 2)
with respect to the one containing the incident ray (medium 1); its notation is n21 sin i
= n21
sin r (54) Example: when the light travels from air to water: sin i
= nwaterair = 1.333 ≈ 4/3
sin r
When light travels from air to glass sin i
= nglassair = 1.5 ≈ 3/2
sin r
+ If n21 > 1, the angle of refraction is smaller than the angle of incidence. It is said that the medium 2 is
more refringent than the medium 1.
+ If n21 < 1, the angle of refraction is larger than the angle of incidence. It is said that the medium 2 is less
refringent than the medium 1.
+ If i = 0, r = 0: the ray of light which is perpendicular to the interface will travel directly.
+ If the incident ray follows the direction of KI (f. 5.19), the refracted ray will follow the direction of IS
(due to the Principle of light reversibility). Hence, we get:
n21 = 1
n12 (55) 3. Index of refraction (refractive index)
The absolute index of refraction of a medium is its relative index of refraction with respect to vacuum.
Below is the table of the absolute index of refractions of some solid, liquid and gas media(*).
Normal glass 1.52 Water 1.333 Crown glass 1.51 Wine Flint glass 1.65 Benzin Salt 1.54 Glycerin 1.47 Diamond 2.42 Sulfur carbon 1.63 Ice 1.31 Air 1.000293 1.3 CO2 1.00045 1.5 Hydrogen 1.00014 (*) In fact, the index of refraction depends also on the color of light (see Chapter VII). The values given in this table
are the indexes of refraction of media to the yellow light of sodium gas. Translated by VNNTU – Dec. 2001 Page 76 Due to the fact that the index of refraction of the air is approximately equal to 1, when we don’t need high
accuracy, we can assume that the index of reflection of a material with respective to the air is equal to its
absolute index of refraction.
Between the relative index of refraction n21 of medium 2 with respect to medium 1, and their absolute
indexes of refraction, there is a following relationship:
n21 = n2
n1 (56) Apart from that, based on the theory about the wave nature of the light proposed by Huyghen, people have
proved that:
The absolute indexes of refraction of the transparent media are inversely proportional to the traveling
velocity of the light in those media.
n 2 v1
=
n1 v 2 (57) In the expression (57), if medium 1 is vacuum, we have n1 = 1 and v1 = 3x108 m/s.
We get the result:
n2 = c
c
or v2 =
v2
n2 (58) Because the speed of the light in any medium is smaller than the speed of the light traveling in vacuum.
Therefore, the absolute index of refractions of all the medium is always larger than 1.
The expression (58) shows that: the absolute index of refraction of a transparent medium lets us know the
speed of the light in that medium is how many times smaller than the one in vacuum.
Questions
1. What is light refraction ? State the law of light refraction.
2. What is the relative index of refraction? What is the absolute index of refraction? State the relation
between those two kinds of index of refraction.
3. State the relation between the absolute index of refraction and the speed of light propagation. As a
result, state the meaning of the absolute index of refraction of a medium.
4. Radiating a ray of light from air into a medium whose index of refraction is n. Find the formula for
determining the angle of incidence in the case that the refracted ray is perpendicular to the reflected ray.
5. Radiating a ray of light from water to air. Determine the angle of refraction, given the angle of
incidence: a) 30o ; b) 45o ; c) 60o. The index of refraction of water is 4/3.
6. A stake is pitched vertically in a wide pool of water with horizontal bottom. The part of the stake
emerging upon the water is 0.6m long. The shadow of the stake on the water surface is 0.8m long; and the
shadow on the bottom of the pool is 1.7m long. Find the depth of the
pool, given the index of refraction of water is 4/3.
Hints: 5) a) 41o50’; b) 70o30’ ; c) no refraction ray. 6) 1.2m. §34. TOTAL INTERNAL REFLECTION
1. Total internal reflection
Translated by VNNTU – Dec. 2001 Page 77 Radiating a thin parallel beam of light (considered as a ray of light SH) from air to water in a direction
which is perpendicular to the water surface. The water is contained inside a small glass basin, which is
vertical. At the bottom of the basin, we put a plane mirror G which is inclined. The degree of inclination of
the mirror can be changed (figure 5.20). The ray SH strike the surface of the mirror at I. It is reflected back
to the water surface at J. There, part of the light is reflected (ray JR) and the other part is refracted into the
air (ray JK).
We can observe the path of IJ, JR and JK by letting them skim over a whitepainted board. When the
degree of inclination of the mirror is increased, then angle of incidence of the incident ray IJ at the
interface also increases. As a result,
+ When the angle of incidence is small, the refracted ray JK is very bright while the reflected ray JR is
very dim.
+ When the angle of incidence i increases, the angle of refraction r also increases. However, i is always
larger than r. At the same time, we can see that the refracted ray JK gets brighter and brighter while the
reflected ray JR is dimmer and dimmer.
+ When the angle of incidence increases to some extent (at which we call critical angle: igh), r = 90o)
At that instant of time, the refracted ray skims over the water surface and very dim, while the reflected ray
is very bright.
+ If we continue increasing in such a way that i > igh, their will be no longer refracted ray. All of the
incident rays are reflected back. Therefore, the reflected ray is as bright as the incident ray. That is total
internal reflection. 2. Conditions to achieve total internal reflection
Firstly, total internal reflection only occurs at the interface between two transparent media when light
travels from the more refringent medium (with larger index of refraction) to the less refringent medium
(with smaller index of refraction).
For the example, total internal reflection can happen when light travels from water to air, from glass to air,
from glass to water etc. 3. Critical angle
When total internal reflection hasn’t happed, we have:
n
sin i
= n21 = 2 , where n2 < n1
sin r
n1 When there starts happening total internal reflection: i = igh and r = 90o.
sin igh = n2
n1 (59) If light travels from a medium (water, glass …) to air, n2 = 1 and
sin igh = 1
n1 (510) For example: water (n ≈ 4/3) ⇒ igh ≈ 48o30’ Translated by VNNTU – Dec. 2001 Page 78 4. Applications of total internal reflection
a) Total internal reflection prism: Total internal reflection prism is an object made of glass and is like a
vertical prism with crosssection being an isosceles triangle ABC (figure 5.21).
There are two ways using total internal reflection prism
+ Radiate the incident ray perpendicularly to the sidesurface (AB for example) of the prism. The ray traveling
inside the prism will be totally reflected back at the
inclined side BC of the prism. (The critical angle of glass
is about 41o50’) (figure 5.21a)
+ Radiate light perpendicularly to inclined surface BC of
the prism. The ray will be totally reflected back twice on
the side surfaces (figure 5.21b)
Total internal reflection prisms are used to substitute spherical mirrors in some optical instruments such as
binoculars, periscopes etc. because they have an advantage, that is we don’t need a silverplate and the
ratio of the reflected light is very large.
b) Optical illusion: that is the phenomenon happening in the atmosphere due to the total internal reflection
of the rays at the interface between the layer of cold air (large index of refraction) and the layer of hot air
(small index of reflection) (see figure 5.22). c) Optical fiber: Total internal reflection is applied in optical fibers. Optical fibers are cylindrical
transparent, flexible strings with a polished surface. A ray goes into the string at one end, will be
continuously reflected many times, and then comes out at the other end. Therefore, optical fiber acts like a
“lightconduction string”. Optical fibers have many applications in science and modern technology, as
well as in medicine.
Questions
1. What is total internal reflection? State the conditions for this phenomenon to occur.
2. State and analyze some examples of application which applies total internal reflection.
3. A small lightbulb S lies at the bottom of a small pool which is 20 cm depth. A piece of wood, is put
floating on the water surface such that no light from the lightbulb will come out from the water surface.
Find the position, the shape and the smallest size of the lightbulb. Given the index of refraction of water
is 4/3.
4. Given a total internal reflection prism, with the index of refraction n = 2 . A ray of light, which lies in
the crosssectional area ABC of the prism and is parallel to inclined surface BC, strikes the side surface
AB at I, near the top B. Draw the path of the ray.
Hints: 3) The piece of wood is round, its center lies on the vertical line passing through S, and its radius is
22.7 cm. Translated by VNNTU – Dec. 2001 Page 79 §35. PRISM
1. Definition
A prism is an object made of transparent materials (glass, quartz, water, etc.). It has the shape like a prism,
with the cross section of a triangle.
The two surfaces of the prism, which we used, are polished and called
two sidesurfaces (the two surfaces ABB’A’ and ACC’A in figure 5.23).
The other surface BCC’B’ is called the bottomsurface of the prism. This
surface might not be used, hence it is harsh or blackcoated.
The angle A created by the two sidesurfaces is called the apex angle of
the prism.
The intersection AA’ of the two sidesurfaces is called the edge of the
prism.
A plane P perpendicular to the edge will cut the prism through a crosssection (A1B1C1).
We only consider the rays of light, when passing through the prism, lie in a certain cross section.
Let n be the relative index of refraction of the prism material to the medium in which the prism is located.
In short, we usually call n the index of refraction of the prism. 2. Path of a monochromatic ray through a prism. Angle of deviation.
In fact, the indexes of refraction of a prism to the light with different colors are different. Here, we only
consider the path of a monochromatic ray through a prism. For this kind of ray, the prism has a fixed index
of refraction. We examine the case where n>1 and the incident ray
going up from the bottom of the prism (figure 5.24).
A very important characteristic of the light propagation through a
prism (with n>1) is that: after passing through the prism, the direction
of the emergent ray deviates to the bottom of the prism with respect
to the incident ray. This is because when refracted at I, the refracted
ray IJ has deviated to the bottom BC (figure 5.24); next, when
refracted at J, the emergent ray JR is deviated to the bottom, too.
We call the angle of deviation D is the angle through which we have
to rotate the incident ray SI so that it coincides with the direction of emergent ray JR (figure 5.24). 3. Prism equations
We can determine the angle of deviation D of the emergent ray, if we know the angle of incident I1 of the
incident ray SI, the apex angle A and the index of refraction n of the prism. It’s true that we can prove the
following equations with ease:
sin i1 = n sin r1
sin i2 = n sin r2 (511) A = r1 + r2
D = i1 + i2  A 4. Minimum deviation angle Translated by VNNTU – Dec. 2001 Page 80 With a certain prism, the angle of deviation only depends on the angle of incidence i1 of the incident ray.
When we vary i1, D also varies, but it will vary about a minimum value Dmin. This can also be seen clearly
in the following experiment (figure 5.25). Put a glass
prism on a rotating table, such that the edge of the
prism lies along the axis of the rotating table.
Radiate a thin monochromatic parallel beam of light SA
to the edge of the prism such that a part of it does not
pass through the prism. This parts gives a trail of light
H on the screen E, which is perpendicular to SA. Part of
the light, when passing through the prism, deviates to
the bottom of the prism and produces a trail of light M
on the screen. The angle HAM is exactly the angle of
deviation D of the emergent ray. Slowly rotating the
rotating table in the direction of the arrow, we will see that the trail H is stable, while the trail M moves
closer to H (D decreases). After some time, the trail stops and then reverses, becomes further away from H
(D increases).
Based on the prism equations, people can prove that when the angle of deviation D achieves the minimum
value Dmin, the angle of emergence is equal to the angle of incidence, i2 = i1. At that time, we also get r2 =
r1, it means the triangles NIJ and DIJ in figure 5.24 are isosceles ones. The bisector planes of the apex
angle becomes a symmetric plane of the ray of light’s path through the lens.
This result is easy to understand, because the equation (511) shows that: D equally depends on i1 and i2.
Therefore, the minimum of D corresponds to the value of i1, will also correspond to the same value of i2.
From the equations (511), with i2 = i1, r2 = r1, we can find the equation for the minimum angle of
deviation:
sin D min + A
A
= n sin
2
2 (512) The equation (512) shows that Dmin depends only on A and n. This is a very important characteristic of a
prism.
If we can measure Dmin and A, we will be able to determine n. This is the basics of the measurements of
the indexes of refractions of the solids and liquids by using goniometers.
Questions
1. What is a prism? When a ray of light goes through a prism with index of refraction n>1, how does the
emergent ray travels?
2. Write down and prove the equations to calculate the angle of deviation D and the minimum angle of
deviation Dmin of a ray passing through a prism.
3. Given a prism with the apex angle of A = 60o and the index of refraction n = 2 . Radiating a ray of
light, lying in a cross section of the prism, to a sidesurface of the prism with the angle of incidence i1 =
45o.
a) Calculate the angle of deviation of the ray of light.
b) If we increase or decrease the angle of incidence by a few degrees, how will the angle of deviation
change? Why?
4. Given a prism with the index of refraction n = 3 and its cross section is an equilateral triangle.
Radiate a ray of light, lying in a prism’s cross section, to its sidesurface. Translated by VNNTU – Dec. 2001 Page 81 a) Calculate the angle of incidence and the angle of deviation of the ray for the case of minimum angle of
deviation. b) Draw the path of the ray of light when the incident ray is perpendicular to the sidesurface of
the prism.
5. Given a wedge prism, with the apex angle 6o (can be considered a small angle) and the index of
refraction 1.6. Radiate a ray of light to its sidesurface with a small angle of incidence. Find the equations
and calculate the angle of deviation of the emergent ray.
Hints: 3) a) 30o, b) The angle of deviation will increase; 4) a) i1 = 60o; Dmin = 60o;
5) D = A(n1) = 3o36’. §36. THIN LENSES
1. Definition
A lens is a transparent medium bound by two curvature surfaces, usually two spherical surfaces. One of
the two can be a plane surface (figure 5.26).
A thin lens is a lens whose the distance between the two vertexes (O1 and O2) is small compared with the
two radii R1 and R2 of the spherical surfaces: O1O2 << R1, R2.
We only examine the case of thin lenses
with spherical surfaces, in an identical
medium and let n be the relative index of
refraction of the lens with respect to that
medium (assume n >1).
Based on the shape and the effect of lenses,
people classify them into two categories:
converging lenses (also called thinedge
lenses, see figure 5.26a) and diverging lenses (also called thickedge lenses, see figure 5.26b).
The line connecting the two centers of the two partial sphere is called the main optic axis of the lens. In
case there is one spherical surface and one plane surface, the main optic axis is the line passing through
the center of the sphere and perpendicular to the plane surface. 2. Main focal point. Optical center. Focal length
a) Put a thinedge lens facing the Sun in such a way that its main optic axis passes through the Sun. Adjust
a screen behind the lens, make sure that it is always perpendicular to the main optic axis. We will find a
particular position on the main optic axis, at which we can capture the image of the Sun., which is very
small and bright. If we neglect the size of the image, we can consider that the beam of light from the Sun
is parallel to the main optic axis of the lens, and after passing through the lens, that beam converges at one
point (point F’ in figure 5.27a). This point is one of the main focal points of the lens. For that reason, thinedge lenses are also called
converging lenses.
Translated by VNNTU – Dec. 2001 Page 82 Hence: if the incident rays are parallel to the main optic axis of a converging lens, their emergent rays
will intersect the main optic axis at the main focal point (F’) of the lens.
Based on the Principle of light reversibility, if the incident light passes through the main focal point of a
lens, the emergent ray will be parallel to the main optic axis.
We can explain this phenomenon as follows: imagine that we can divide the lens into plenty of small
parts; each part can be considered a prism with a small apex angle (figure 5.27a). The nearer this system of
prism to the edge, the larger the apex angle. Therefore, when passing through the prisms, the rays of light
will be bent downwards to the main optic axis. The near is a ray to the edge, the more it is bent down. As a
result, the rays of light will converge at the same point. In general physics courses, will we prove that the
rays of light intersect at one point.
For the case of thickedge lenses (diverging lenses), when the incident beam of ray are parallel to the main
optic axis, the emergent rays will diverge and their extensions will cut the main optic axis at one point F’,
called main focal point of the diverging lens (figure 5.27b). We call this focal point a virtual focal point.
We also explain the refractions of the incident rays passing through the diverging lenses in a similar
manner with the converging lenses. (see figure 5.27b).
b) The middle part of the lens, between the two vertexes of the two partial sphere can be considered as a
transparent layer with two parallel faces. The two vertexes can be considered coincident at that point. The
rays passing through this point will not change its direction. This point is call the optical center of the
lens and its notation is O.
Those lines which passes through the optical center O and does not coincide with the main optic axis are
called supplemental optic axes.
Experiments showed that each thinedge lens has two main focal points lying on both sides of the optical
center.
Their notations are F and F’. One is called the object focal point (F) and the other is called the image
focal point (F’).
Image focal point is the point through which the emergent rays will pass, if the incident rays are parallel to
the main optic axis. Object focal point (F) is such a point that if the incident ray passes through this point,
the emergent ray will be parallel to the main optic axis. It is clear that the classification is based on the
direction of the incident rays.
The distance f from the optical center to the main focal points is the focal length of the lens.
f = OF = OF’ 3. Supplemental focal points. Focal plane
a) Experiments showed that if the incident beam of light is parallel to one of the supplemental optic axis of
a converging lens, the emergent beam will converge at one point F’1 on that supplemental optic axis. F’1 is
one of the supplemental focal points of the converging lens (figure 5.28a).
b) For diverging lenses, when the incident beam of light is parallel to one of the supplemental optic axis,
the emergent beam will diverge and their extensions will intersect at one point F’1 on that supplemental
axis. F’1 is one of the supplemental focal points of the diverging lens (figure 5.28b). Translated by VNNTU – Dec. 2001 Page 83 b) The are numerous of supplemental focal points: all supplemental focal points lie in a same plane
perpendicular to the main optic axis at the main focal point. That plane is called the focal plane of the
lens. Each lens has two focal planes lying on both sides of the optical center (figure 5.28). 4. Lens power
The power of a lens, denoted by D, is the reciprocal of its focal length:
D= 1
f (513) The unit for lens power is the diopter. The power as well as the focus of a converging lens are positive,
whereas those for a diverging lens are negative.
For thin lenses, the power can be calculated as follows
D= 1
1
1
= (n  1) (
+
)
f
R1 R 2 (514) where n is the relative index of refraction of the lens material to the medium in which the lens is located;
R1 and R2 are the radii of curvature of the lens boundaries. We establish a convention that the radius of a
convex spherical surface is positive, of a concave spherical surface is negative and of a plane is infinity.
Concave spherical surface is the surface whose center is on the same side with the lens material with
respect to is vertex.
Questions
1. What is a lens? What are the optical center, the main focal points, the supplemental focal points, the
focal plane of a lens?
2. Explain the effect of a converging lens to a beam of light parallel to the main optic axis.
3. What is the power of a lens? State the unit of lens power and the equation to determine the power of a
thin lens.
4. Calculate the focal lengths of the lenses with following power lenses: +0.5diopter, +1diopter, +5diopter,
4diopter, 2diopter, 0.4diopter. Classify which lenses are converging ones and which lenses are
diverging ones.
5. A lens with two similar surfaces has a power lens +2diopter and a index of refraction 1.5. Calculate it
focal length and the radii of its two surfaces.
6. A glass lens (n = 1.5) located in air, with power lens +1diopter. Calculate its focal length when it is
dipped into water. Give the index of refraction of water 4/3.
Hints: 4) 2m; 1m; 0.2m; =0.25m ;0.5m; 2.5m; 5) f = 50cm; R1 = R2 = 50 cm; 6) 4m. Translated by VNNTU – Dec. 2001 Page 84 §37. IMAGE OF AN OBJECT THROUGH LENSES. LENSES EQUATIONS
1. Observing an object’s image through a lens
a) Diverging lenses: put an object in front of a diverging lens (nearsighted spectacles for example). Put
the eyes behind the lens to observe. We will see there is a revert and smaller image in front of the lens.
This is a virtual image. When we move the object (nearer to the lens or further away from it), the
properties of the image stated above do not change.
b) Converging lenses: Put a light object (a candle for example) in front of a converging lens (a farsighted
spectacle or a spectacle for old people, for instant). After the lens, we put a screen perpendicularly to the
main optic axis to capture the image (figure 5.29). We can see that if the object is fairly far away from the les, and if we put the screen at an appropriate
position, we will be able to capture a reversed real image which is smaller than the object and similar to it.
 Move the object nearer to the lens, to be able to capture the image, we have to move the screen further
away from the lens. The image is still a real one, reversed and similar to the object. However, it is bigger
than the previous one.
 When the object is rather near the lens, the image become a virtual one, and we cannot capture it on the
screen. At that time, if we look through the mirror, we will see that there is a revert image and it is larger
than the object.
Experiments have shown that if we want the lenses to produce sharp images, absolutely similar to the
object, we have to satisfy some conditions which are similar to the paraxial conditions for spherical
mirrors. 2. Method to draw an object’s image through a lens
a) Assume there is a point object in front of a lens and it emits a diverging beam of light to the lens. If the
emergent beam of light is a converging one S’, S’ is the real
image of S through the lens.
If the emergent beam of light is a diverging one, and their
extensions intersect at one point S’, S’ is the virtual image of S
through the lens.
The image of an object is the collection of all the images of all
the points belonging to the object.
b) Draw the image of a point object outside the main optic
axis: Assume that we have to draw the image B’ of a point B
lying outside the main optic axis of a lens. As B’ is the
Translated by VNNTU – Dec. 2001 Page 85 intersection of all the emergent rays (or their extensions), to draw B’, we just need to draw the paths of
any two principal rays among the incident beam of light (figure 5.30, a and b).
In Geometrical Optics figures, for convention, people use they drawing symbols to represent converging
and diverging lenses as in figures 5.30 a and b.
We can draw 2 out of 3 following principal rays:
 Ray BO, passing through the optical center O of the lens. This ray travels in a straight direction.
 Ray BI, parallel to the main optic axis of the lens. This ray, when coming out of the lens, will pass
through the image focal point F’ of the lens (or its extension will pass through F’).
 Ray BF, passing through the image focal point F (or its extension passing through F’’). This ray, when
coming out of the lens, will become parallel to the main optic axis.
These kinds of rays (or there extensions) will intersect at B’, the image of B.
c) Draw the image of an object which is like an arrow AB perpendicular to the main optic axis; and A is
on the main optic axis. The image A’B’ of AB is also an arrow, perpendicular to the main optic axis, and
A also lies on the main optic axis.
To draw the image A’B’, we only need to draw the image B’ and project B’A’ down to the main optic
axis. 3. Lens equation
a) Sign rules: We use the sign convention as follow:
For real object (in front of the lens), the distance from lens to the object has positive value (f.5.30a and b)
d = OA > 0
For real image (behind the lens), the distance from the lens to the image has a positive value (f. 5.30a)
d = OA' > 0
 For virtual image (in front of the lens), the distance from the lens to the image has a negative value (f.
5.30b)
d = OA' < 0
Note that the positive direction of d and d’ are opposite to each other.
+ The focal length of a converging lens is positive (f = OF = OF ' > 0), whereas the focal length of a
diverging lens is negative (f <0).
+ If the image A’B’ has the same direction with the object AB, A' B ' and AB have the same sign.
Otherwise, they have opposite signs.
b) Lens equation: Let’s find an equation in order to determine the position of the image A’B’ of the object
AB.
Consider two similar triangles OA’B’ and OAB (figure 5.30a), we have:
OA ' A ' B '
=
OA
AB (1) Consider two similar triangles F’A’B’ and F’OI, we have: Translated by VNNTU – Dec. 2001 Page 86 F'A ' A 'B' A 'B'
=
=
OF
OI
AB
Compare (1) and (2), we have: OA ' F ' A '
=
OA
OF (2)
(3) Since OA’ = d’, OA = d so FA’ = OA’ – OF’ = d’ – f (where OF’ = f). Substitute the values into (3):
d ' d '− f
=
or dd’ = df + df
d
f
Divide both sides by dd’f, we get:
1
1
1
+
=
d
d'
f (515) In the case for figure (5.30b), we have OA = d ; OA’ = d’.
Consider similar triangles O’A’B’ and OAB ; F’A’B’ and F’OI, we still have
OA ' A ' B '
=
OA
AB (4) F'A ' A 'B'
=
OF
AB (5) Compare (4) and (5): OA ' F ' A '
=
where F’A’ = OF’ – OA’.
OA
OF Substitute OF’=f; OA’ = d’ we get
 d'
−f + d '
=
or d’f =  df + dd’
d
−f Divide both sides by dd’f, we have: 1
1
1
+
= , this is exactly the equation (515).
d
d'
f This equation is applied for all image formations through lenses. 4. Lateral magnification
Lateral magnification of the image is the ratio between the height of the image (measured in the direction
perpendicular to the main optic axis) and the height of the object.
k= A 'B'
AB From the above formulas, we have:
k= d'
d If k >0, A’B’ has the same direction with AB; if k <0, AB has an opposite direction with AB. The absolute
value of k represents the relative height of the image compared with the object.
Questions
1. Explain the method to draw the image of a point object through a converging lens. Translated by VNNTU – Dec. 2001 Page 87 2. Explain the method to draw the image of a point object through a diverging lens.
3. Prove the equation which determines the position of an image through a lens.
4. What is the lateral magnification of the image? Prove the equation determining the lateral magnification
of an image through a lens.
5. Put a lens 20cm away from a bookleaf. Looking through the lens, we see the image of the lines of
words with the same direction and its height is half of the lines of words’ height. What kind of lens is that?
Calculate the focal length of the lens. Draw figures.
6. Put an object 12cm away from a lens, we get an image that is three times higher than the object.
Calculate the focal length of the lens. Draw figures.
7. Put an object AB parallel to a screen and 90 cm away from the curtain (L=90cm). After that, put a
converging lens between the object and the screen such that its main optic axis passes through A and is
perpendicular to AB. Moving the lens along the main optic axis, we find that there are two positions at
which there are clear images of the object on the screen. These two positions are 30 cm apart (l = 30cm).
Calculate the focal length f of the lens.
Hints: 5) Diverging lens, f = 20cm; 6) If the image is real: f=9cm; if the image is virtual, f=18cm;
7) f = 20 cm.
SUMMARY OF CHAPTER V
1. Law of light’s straight propagation: In a transparent and identical medium, light propagates in a straight
line direction.
2. Principle of light reversibility:
On a light propagation path, light can travel from any direction.
3.Law of light reflection:
a) The reflected ray lies in the plane of incidence and is on the other side compared with the incident ray.
b) The angle of reflection is equal to the angle of incidence (i’ = i)
Then plane of incidence is the one formed by the incident ray and the normal to the plane of reflection at
the incident point.
4. Law of light refraction:
+ The refracted ray of light lines in the plane of incidence and is on the other side of the normal with
respect to the incident ray of light.
+ For any pair of identical transparent media, the ratio between sine of the angle of incidence (sin i) and
sine of the angle of refraction (sin r) is always a constant. This constant depends on the nature of the two
media and is called the relative index of refraction of the medium containing the refracted ray (medium 2)
with respect to the one containing the incident ray (medium 1); its notation is n21
sin i
= n21
sin r
5. The relation equation between the index of refraction and the speed of light propagation:
n 2 v1
c
=
and n =
v
n1 v 2 Translated by VNNTU – Dec. 2001 Page 88 6. The conditions to achieve total internal reflection for a ray of light at the interface between two media:
a) The ray of light must be from medium with higher index of refraction n1 to the medium with smaller
index of refraction n2.
b) The angle of incidence must be larger than the critical angle.
7. Equations to calculate the critical angle:
sin igh = n2
n1 If light is from a medium with the index of refraction n to air:
sin igh = 1
n 8. Prism equations
sin i1 = n sin r1 ; sin i2 = n sin r2
r1 + r2 = A; D = i1 + i2  A
9. The angle of deviation is smaller (D = Dmin) when: i1 = i2 (at as result, r1 = r2)
sin D min + A
A
= n sin
2
2 10. The equation to determine the position an object’s image through a spherical mirror or a lens:
111
+=
d d' f
Real object, in front of the lens or in front of the spherical mirror, d>0.
Real image, behind the lens or in front of the spherical mirror, d’>0.
Virtual image, in front of the lens or behind spherical mirror, d’<0.
Converging lenses and concave spherical mirrors have positive focal lengths f>0.
Diverging lenses and convex spherical mirrors have negative focal lengths f<0.
11. The equations to calculate the lateral magnification of an image through a mirror or a lens:
k= A'B'
AB =− d'
d k>0, the image has the same direction with the object; k<0, the image has an opposite direction with the
object.
12. The equation to calculate the power of a lens
D= 1
1
1
= (n − 1)( +
)
f
R1 R 2 The unit of lens power is diopter (corresponding to the length unit  meter). Converging lenses have D>0;
diverging lense have D<0.
A convex spherical surface has positive radius, a concave spherical surface has negative radius, and a
plane has a radius which equal to ∞. Translated by VNNTU – Dec. 2001 Page 89 13 .The equation to calculate the focal length of spherical mirrors:
f= R
2 14. To draw the image of a point object through a spherical mirror or a lens, we can use any two out of the
four following principal rays emitted from A.
 The ray passing through the center of curvature of the mirror (or passing through the optical center of the
lens).
 The ray which is parallel to the main optic axis.
 The ray passing through the focal point of the mirror (or the object focal point of the lens)
 The ray passing through the vertex. Translated by VNNTU – Dec. 2001 Page 90 Chapter VI – THE HUMAN EYE AND OPTICAL INSTRUMENTS §38. CAMERA AND THE HUMAN EYE
1. Camera
a) Camera is an optical instrument for
obtaining the real image of an object on a
film frame. The image is smaller than the
object.
b) Structure: the main part of camera is a
converging lens (or a system of
converging lenses with positive focal
length) called the lens. In the normal
cameras, the lens has a focal length of
approximately 10 centimeters. Lens is
attached in front of dark room and film is
attached tightly into the opposite side,
inside the dark room (figure 6.1b).
The lenstofilm distance is changeable.
Closely attached to lens (or between the lenses) is a shutter, at the center of which there is one small
circular hole which diameter is adjustable. This one is used to adjust the light rays lighting up to film.
Besides there is an aperture M which stands in front of film and prevents the lights from continuously
shining the film. This aperture opens only within a very short period of time decided by user by pressing
button.
c) Adjusting a camera: In order for the image of
desired object to appear clearly on the film, we
change the distance d’ from lens to film by
getting lens closer to or farther from film. To
detect the clearness of image on film, we use
one glass available in camera.
Furthermore, depending on the intensity of light,
we have to choose the appropriate time of taking
photograph and the size of a hole on a shutter. 2. The human eye
a) In term of geometric optic, the human eye likes a camera. It has a function of creating a real image
which is smaller than object on the layer of cells sensitive to lights, from which nervous signals are caused
and transferred to brain. However, the optical system of eye is much more complicated than that one of
camera.
b) Structure: The main part of the eyes is a converging lens
which is translucent and soft. It’s called crystalline lens (5)
(figure 6.2). A curving degree if two sides of the crystalline
lens is adjustable subject to the stretching of ciliary’s muscle. Translated by VNNTU – Dec. 2001 Page 91 In front of the crystalline lens is a translucent liquid with a refraction index n ≈ 1.333 called the aqueous
humor (2).
Behind the crystalline is another translucent liquid which discount n ≈ 1.333 called the vitreous humor
(6).
The outer face of eye is a translucent, thin and rather hard layer called the cornea (1).
The inner side of eye which is opposite to crystalline lens is called the retina (7). It plays a role of the
screen at which there are sensitive receptors located at the beginning of optic nerves.
On the retina, there is a small yellow area, very sensitive to light, located near the crossing point V of the
main axis and the retina. This area is called the fovea.
A little under fovea is the blind point M which is absolutely nonsensitive to light, because at that point
there is no optic nerve.
Closely to the front side of crystalline lens is there a nontranslucent layer, black (or blue, brown) called
the iris (3).
At the center of the iris there is a small circular hole called the pupil (4). Depending on the intensity of
coming lights, a diameter of pupil will automatically change to adjust the lights shining to retina.
One very important characteristic in term of structure of human eye is that: the curving degree (and
therefore, the focal length) of crystalline is adjustable. Whereas, the distance from the optical center of
crystalline lens to the retina (d’ = OV) is always fixed (d’≈ 2,2cm).
c) The accommodation – The far and near points:
When our eyes see any objects, on the retina appears a real, oppositelydirected and very small image of
that object.
When we get an object closer to our eyes (d decreases), if the focal length of crystalline lens is unchanged
image of object will reach behind retina (d’ increases). In order for image to be on retina (i.e. d’ is
unchanged), focal length f of crystalline must decrease. The ciliary’s muscle supporting the crystalline
lens has to shrink to makes crystalline lens fill out. In contrast, when the object becomes farther from eyes,
in order for image to be on the retina focal length of crystalline lens must increase. In this case the ciliary’s
muscle has to stretch that makes crystalline lens flatter.
The change in curving degree of eyes (and therefore, the change in focal length or the strength of lens) so
as for the desired image of an object to appear clearly on the retina is called the accommodation.
The farthest point on the main axis of eye at which a given object is still observed clearly is called the far
point (CV).
In the case of nondefect eye, a far point is infinite. When seeing object put at the far point, eye do not
need to regulate. The strength of crystalline lens is smallest; its focal length is the largest and its focal
point is located exactly on the retina: fmax = OV.
Hence, a nondefect eye is the eye, with no regulation, has its focal point located on the retina.
The closest point on the main axis of eye at which a given object is still observed clearly is called the near
point (CC) of the eye. It is due to the fact that the crystalline lens only fills out to a fixed limit, and
therefore the focal length of crystalline lens can only decrease to a minimum value. At that time, the
distance from the object, whose image is still clearly on the retina, to eye is the smallest. It is called the
smallest clearobservation distance and denoted by the letter D. If the object reach closer the crystalline
does not have ability to have image appear clearly on the retina any longer. Translated by VNNTU – Dec. 2001 Page 92 With the youth who has not defect in eyes, the near point is from 10cm to 20cm. The older the person is,
the farther the near point is.
When observing an object put at the near point, our eyes have to regulate most strongly. The crystalline
lens fill out at maximum, hence, our eyes soon become tiring.
To observe for a long time and clearly (reading, writing, seeing an object through an optical instrument,
etc.), we often put an object (or an image need to be observed) far from eyes a distance which is a bit
longer than that from eye to the near point. That distance is approximately 25cm.
Distance from the near point CC to the far point CV is called the clearobservation limit of eye.
d) Sighted angle and productivity of separation:
The sighted angle of an object AB in a form of stick perpendicular to the main axis of eye, is an angle
caused by two light rays coming from two endpoints A and B of the object through the optical center of
AB
eye (figure 6.3): tgα =
.
l
To distinguish two points A and B, not only this two points
lie within the clearobservation limit but also the sighted angle
AB is large enough. When the line AB becomes shorter, the
sighted angle of AB decreases, two images A’ and B’ reach
closer to each other. When two images A’, B’ lie on the same
sensitive receptor, it is impossible to distinguish points A and must
of B. Therefore, we call the productivity of separation as the smallest sighted angle αmin between two points A
and B that our eyes can still distinguish them. In that case, two images A’ and B’ lie on two very close
sensitive receptors.
The productivity of separation depends on each eye. The statistical measurement shows that:
αmin ≈ 1’ ≈ 1
rad.
3500 e) The effect of saving image on the retina:
After switching off the stimulus light to retina, it takes a period of time of nearly 0.1 second for a retina to
recover as before. In this time, our sense of light still exists and observer still sees an image of object. It is
the effect of saving image on the retina.
This effect is used in movies. We do not have films run continuously before the lens of projector but let
each one stop before the lens about 0.04 second. After that, there is a propeller coming to hide this lens
and film is replaced by the other one, so on and so forth. Hence, the images we observe on the screen
move continuously.
The similar technique is also used in television.
Questions
1. State the structure and adjusting way of camera.
2. State the structure of human eye in term of geometric optic.
3. State the conception about regulation of eye as well as the near point and the far point.
4. State the conception about sighted angle and productivity of separation of eye. Translated by VNNTU – Dec. 2001 Page 93 5. The lens of one camera has a focal length f =10cm. This camera is used to take
photograph of one man who is 1.6m high and stands 5m away from camera.
Calculate the length of image on film and the lenstofilm distance.
6. Using one camera which lens has focal length f = 10cm to take photograph of a
picture with the size of 1m×0.6m onto a film with the size of 24mm x 36mm. Find
the smallest distance from picture to lens such that we can get image of the whole
picture on film. Also compute the magnification of image in that case.
7. Draw two short lines 1mm away from each other on the paper (figure 6.4). Take this paper further and
further from your eyes until you see two lines seem to be one line. Find approximately the distance from
eye to paper and deduce the productivity of separation of your eyes.
Hints: 5) 3.26cm and 10.2cm ; 6) ≈ 2.9m and 0.036 §39. EYE’S DEFECTS AND CORRECTING METHODS
1. Nearsightedness (myopia)
a) A nearsighted eye is the eye whose focal point lies before the retina when there is no accommodation
(figure 6.5).
It means that the maximum value of focal length is smaller than the distance from optical center to retina.
fmax < OV
b) The far point (CV) of the eye lies in a short distance from eye (usually less
than 2m depending on the degree of nearsightedness). The nearsighted eye
can not observe any object far away from it.
When observing an object at the far point, the nearsighted eye does not need
to regulate. In that case, the strength of crystalline lens is
minimum and the focal length is maximum.
The near point (CC) of nearsighted eye lies very closely to it.
c) Correcting nearsightedness means making the nearsighted
eye be able to clearly observe objects far away. In order to do
that, the nearsighted eye must be worn a diverging lens
(assuming this lens is very close to the eye) so that the images
of object at infinity through lens appear at the far point. The
eye will see such images clearly without the accommodation.
On the other hand, because the images of object at infinity will appear on the focal surface of lens, we
deduce that the far point must lie on the focal surface (figure 6.6) and the focal length is equal to distance
from optical center to the far point.
fk = OCV (62)
(minus sign corresponds to diverging lens).
When we wear glasses, a new far point is also farther from the eye.
To try out the glasses, nearsighted man has to wear one after another the diverging lens which absolute values of
strength of lens (D) increase, and observe the lines of characters from small to big on the board which is about 5m
away from eyes. If we could comfortably see the line with sign 10/10, we chose the appropriate glasses. At that time,
we actually can observe the far objects without regulating. Translated by VNNTU – Dec. 2001 Page 94 2. Farsightedness (hyperopia)
Farsighted eye is the eye which focal point lies behind the
retina when there is no regulation (fmax > OV) (figure 6.7).
Farsighted eye has to regulate in order to observe objects at
infinity. But as it regulates most strongly, it only may observe
objects which are relatively far way. In the other words, in
comparison to normal eye, the near point CC of farsighted one
farther (OCC > 25cm). lies Correcting farsightedness is making farsighted man be able to observe clearly objects at infinity without
regulating. So this person has to wear converging lens with a suitable strength of lens.
However, because in fact it is difficult to do such thing, we simply let farsighted eye wear one converging
lens such that we can see near objects as normally. Images of objects next to eye produced by lens will
appear within the clearlyobserving limit.
For instance, in order to observe clearly the object which is 23cm away from the eyes farsighted man
must choose lens such that image of this object through his eyes appear at the near point CC. This image is
virtual, farther from lens than its object (since OCC > 25cm), hence the lens to wear should be converging
lens, i.e. the farsighted lens is a converging lens.
To detect this kind of glasses, farsighted man must wear glasses and read the book pages which are at
normal distance from his eyes such that he can read comfortably.
Questions
1.State the characteristics of nearsighted eye and the correcting method.
2. State the characteristics of farsighted eye and the correcting method.
3. One nearsighted person has a far point of 50cm away from eyes and a near point of 12.5cm.
a) Find the strength of lens to be worn.
b) When wearing this lens how far is the nearest point that person can see from the eyes? Optical center of
lens is also that one of eye.
4. One farsighted person observe clearly the nearest point of 40cm away from eyes.
a) Find the strength of lens to be worn such that he/she is able to see the nearest point of 25cm away from
eyes. Given that the lens is worn closely to eyes.
b) If he/she wears a lens of 1dp, how far is the nearest point that person can see from the eyes?
Hints: 3) a) –2diopters; b) 16.7cm; 4) a) 1.5diopters; b) ~29cm. §40. MAGNIFYING GLASS
1. Definition
Assume that we have to observe a very small object as a straight line AB and that even if that object is put
at a near point the observing angle is still very small. To increase this angle, an object has to be got closer
to the eyes. But at that time our eyes may not see object clearly since object is outside the clearlyobserving limit.
To get our goal, we use one converging lens and put object AB within the distance from object focal point
to optical center of lens such that there is a virtual image A’B’ which is greater and farther from lens than Translated by VNNTU – Dec. 2001 Page 95 an object. The closer to a focal point AB is, the greater and farther from lens A’B’ is. Therefore, we could
adjust the lens such that an image A’B’ lies within a clearlyobserving limit. Eyes are put behind lens and
observe an image A’B’ with an observing angle that is much greater that productivity of separation. The
above lens is called the magnifying glass (or the magnifier).
Hence, magnifying lens is an optical instrument aiding the eyes in observing the small objects. It has a
function of increasing imageobserving angle by producing a virtual image which is greater than object
and lies within the clearlyobserving limit.
The simplest magnifying glass is the converging lens with a small focal length. 2. Near point and infinite point
To observe a small object through magnifying glass, we
have to put the object within the distance from object to
optical center of lens to get a virtual image. Our eyes are
put behind the glass to observe that virtual image. We
also must adjust the position of glass or object so that a
virtual image appears within the clearlyobserving limit
CC  CV (figure 6.8).
If we adjust the glass so that the image A’B’ appears at
the near point, this way is called viewing (observing) at
the near point.
Normally, to reduce eyetiring observer adjust so that image of object lies at the far point CV. Because in
case of normal eyes the far point is at infinity, this observing method is called viewing (observing) at
infinity. 3. Angular magnification
a) It is defined that the angular magnification of an optical instrument aiding the eye is the scale between
imagesighted angle of an object through that instrument (α) and objectsighted angle when this object is
observed directly at the near point of the eyes (α0)
G= α
α0 (63) Because the sighted angles α and α0 both are very small, we usually substitute values of them by the
values of their tangents:
G= tgα
tgα 0 As shown in figure 6.9, we have tgα0 = (64)
AB
D (65) where D is the smallest clearlyobserving distance of our eyes (D = OCC).
In case of a magnifying glass (figure 6.8) given that l is the
distance from eyes to glass and d’ is the distance from the
image A’B’ to glass (d’ < 0), we have
tgα = A 'B'
A 'B' D
D
tgα
and G =
.
=k
=
d' +l
AB d ' + l
d' +l
tgα 0 Translated by VNNTU – Dec. 2001 Page 96 where k is the magnifying degree of the image.
The value of angular magnification G of a magnifying glass depends on the observer’s eyes (D) and the
observing method (k, d’ and l).
Generally, the value of angular magnification G is not the
same as value of the magnifying degree k.
When the observer views at the near point, d’ + l = D and
G = k.
When observing at infinity, an observed object is put at
object focal point of magnifying glass, the image A’B’ is at
infinity, the rays leaving a glass are parallel. Hence the
imagesighted angle of A’B’ is always α no matter what
the position of our eyes (figure 6.10).
In this case, we have
tgα =
and AB
AB
=
OF
f G∞ = D
f (66) To have a great G∞ then f must be small. The method of observing at infinity not only helps the eye avoid
adjusting but also makes the angular magnification of glass independent on the position of the eye.
When we observe at infinity, a conception about magnifying degree of image is not important.
For convenience, we usually choose D = 0.25m. Thus the value of G∞ will be
G∞ = 0.25
(m)
f (67) With the widelyused glasses, this value varies from 2.5 to 25. It is often shown on the rim of glass. For
example, X 2.5, X 5, etc.
Questions
1. What’s magnifying glass? State structure and objectobserving method of magnifying glass.
2. State conception about angular magnification of an optical instrument aiding eyes.
3. State conception about the observation at the near point as well as the far point. Demonstrate all the
formula of angular magnification in these cases.
4. Use a 10dp lens as magnifying glass:
a) Compute angular magnification when observing at infinity.
b) Compute angular magnification and magnifying degree of image when observer views at the near point.
The smallest clearlyobserving distance is 25cm.
5. One nearsighted person whose distance from his face to the near point is 10cm and to the far point is
50cm, observes a small object through a 10dp magnifying glass. Eyes are behind this glass.
a) How long is the distance from object to a glass? (object is put in front of glass)
b) Find the angular magnification corresponding to that person and the magnifying degree of image in the
following cases: that person views at the far point; that person views at the near point. Translated by VNNTU – Dec. 2001 Page 97 Hint: 4) a) 2.5; b) G = k = 3.5; 5) a) 5cm ≤ d ≤ 8.3cm; b) kV = 6; GV = 1.2; kC = GC = 2. §41. MICROSCOPE AND TELESCOPE
1. Microscope
a) Microscope is an optical instrument helping eye increase the imageobserving angle of very small
objects. Its angular magnification is much greater than the one of magnifying glass.
b) Structure: Microscope has two main parts which are objective and eyepiece (figure 6.11).
Objective O1 is a converging lens with very short focal
length. It’s used to produce a real and very large image of the
observed object.
Eyepiece O2 is also a converging lens with very short focal
length, used as one magnifying glass to observe the above
image(*).
Two lenses are attached to two heads of one cylinder tube
that the main axes of them coincide with each others and the
distance between them is unchanged. real
such Besides, there is a lightcollecting part used to light up the
observed object. The simple lightcollecting part may be a
concave mirror G.
Observing method: an object AB (normally a specimen) is put outside but very close to focal point of an
objective. Through objective, we get the real image A1B1 which is k1 times greater than object.
We have to adjust microscope such that image A1B1 lies within the distance from object focal point F2 to
optical center O2 of eyepiece. Through eyepiece, we have the last virtual image A2B2 that is very great and
in a contrary direction to an object AB (figure 6.12).
The observer’s eyes are put behind the eyepiece to view an image A2B2. Optical center O of the eyes is
considered to coincide with optical center O2 of eyepiece. Translated by VNNTU – Dec. 2001 Page 98 To see clearly an image A2B2, observer must adjust microscope such that an image A2B2 lies within the
clearlyobserving limit of eyes. To adjust microscope, we change the distance d1 between object and
objective by getting the whole tube higher or lower. At that time, the distance d’2 from the last image A2B2
to the eyepiece, i.e. to the eyes, also changes.
Usually, to avoid tiring our eyes, we adjust to observe an image A2B2 at infinity. In this case, an image
A1B1 lies at the object focal point F2 of eyepiece (figure 6.13).
d) Angular magnification of microscope: Let’s compute the angular magnification G∞ of microscope in
the case of observing at infinity. Basing on figure 6.13, we have:
tgα = A1 B1 A1 B1
=
O 2 F2
f2 and tgα0 is still computed by the formula (65).
Therefore, angular magnification of microscope will be calculated by:
G∞ =
or AB D
tgα
= 1 1×
AB f 2
tgα 0 G∞ = k1.G2 (68) The angular magnification G∞ of the microscope in the case of observing at infinity is equal to the product
of magnifying degree k1 of an image A1B1 through eyepiece and angular magnification G2 of eyepiece.
This two values is usually shown in the rims of eyepiece and objective.
Now we consider two similar triangles A1B1F’1 and O1IF’1 as shown in figure 6.13. We have
A1 B1
AB
F' F
δ
= 1 1= 1 1 =
AB
O1 I
O1 F '1
f1
where δ = F’1F2. The distance δ from image focal point of eyepiece to object focal point of objective is
called the optic length of microscope.
As a result, we have G∞ = δD
.
f1 f 2 (69) We usually take D = 25cm.
To have great angular magnification, focal points f1 and f2 of objective and eyepiece must be small. The
angular magnification of microscope is normally not greater than 1500 to 2000. 2. Telescope
Telescope is an optical instrument helping eyes increase the imageobserving angle of objects which are
very far (planets) (figure 6.14a). (*) Normally, the objective and eyepiece are a complex system which plays the role of a converging lens. Translated by VNNTU – Dec. 2001 Page 99 Telescope has two main parts which are objective and eyepiece.
The objective is a converging lens with long focal length; the eyepiece is a converging lens with short
focal length. Two lenses are attached coaxially at two heads of one cylinder tube. Distance between them
is adjustable.
An object AB (may be a diameter of the moon) considered at infinity through the objective gives a real
image A1B1 at the image focal surface F’1 of objective (figure 6.14b). The eyepiece is used as a
magnifying glass to observe an image A1B1. The last image A2B2 is virtual image. An observer puts eyes
behind the eyepiece and observes image A2B2. We have to adjust telescope (change distance O1O2
between eyepiece and objective) such that an image A2B2 lies within the clearlyobserving limit of our
eyes. In the case of observing at infinity, observer adjusts to have A2B2 at infinity. At that time, image A1B1 is at
the object focal surface F2 of the eyepiece. So the image focal point F1 of the objective will coincide with
the object focal point F’2 of the eyepiece (figure 6.15). Hence, the last imageobserving angle through the
"
"
telescope is angle A O B (α = A O B ); and the objectobserving angle of AB (when the telescope is
1 2 1 1 2 1 AB
AB
"
"
not used) is equal to an angle A1O1 B1 (α0 = A1O1 B1 ). So tgα = 1 1 and tgα0 = 1 1 .
f2
f1 Therefore, the angular magnification of telescope in the case of observing at infinity is given by:
G∞ = f
tgα
=1
f2
tgα 0 Translated by VNNTU – Dec. 2001 (610) Page 100 Beside this type of telescope, there are many types we do not consider here. However their basic operation
is similar.
Questions
1. State the use and structure of microscope.
2. State the observing method and angular magnification of microscope.
3. State the use, structure and angular magnification of telescope.
4. The objective of one microscope has focal length f1 = 1cm; eyepiece has focal length f2 = 4cm. Two
lenses are 17cm away from each other.
a) Calculate the angular magnification of microscope in the case of observing at infinity. Given
D = 25cm. b) Calculate the angular magnification and magnifying degree of microscope in the case of
observing at the near point.
5. The objective of one microscope has focal length f1 = 1cm; eyepiece has focal length f2 = 4cm. Optic
length of microscope is 15cm. Observer has the near point of 20cm away from eyes and the far point at
infinity. How long is the distance in which an object is put before objective?
6. The objective of one school telescope has focal length of 1.2m. The eyepiece is a converging lens which
focal length is 4cm.
a) Find the distance between two lenses and angular magnification of telescope in the case of observing at
infinity. b) One student uses the above telescope to observe the moon. His far point is 50cm away from his
eyes. Find the distance between two lenses and angular magnification of telescope when that student
observes without eyeregulating.
Hints: 4) a) G∞ = 75; b) kC = GC = 91; 5) 1.064cm ≤ d1 ≤ 1.067cm; 6) a) O1O2 = 124cm; G∞ = 30;
b) O1O2 = 123.7cm; G = 32.4
SUMMARY OF CHAPTER VI
1. Camera
Objective has a fixed focal length (f ~ 10cm).
Distance d’ from lens to film is adjustable.
2. Human eye
a) Normal eye
 Without regulation, focal point lies on the retina: f max = OV
 Far point CV is at infinity.
 Near point CC is 1520cm away from eye when young and go farther when older.
b) Nearsighted eye
 Without regulation, focal point lies before the retina: f max < OV
 Far point CV is at the definite distance from eye.
 Near point is very close to eye.
 To correct, the nearsighted person has to wear diverging lens such that he/she can observe clearly
objects at infinity without eyeregulating: f max =  OCV Translated by VNNTU – Dec. 2001 Page 101 c) Farsighted eye
 Without regulation, focal point lies behind the retina: f max > OV
 Has to regulate to observe objects a infinity.
 Near point CC is too far from eye in comparison to normal eye.
 To correct, the farsighted person has to wear converging lens such that he/she can observe clearly
objects at infinity without eyeregulating.
3. Magnifying glass
 Magnifying glass is a converging lens with short focal length, used to observe very small objects with
angular magnification that is not great.
 Angular magnification
G= image  observing angle through glass
tgα
α
=
≈
α 0 tgα 0
object  observing angle, at near po int  Angular magnification of magnifying glass in case of observing at infinity: G∞ = D
.
f D is usually chosen as 0.25m.
4. Microscope
 Both the objective and eyepiece are converging lenses with very short focal length, coaxially situated and
in a constant distance to each other.
 The angular magnification G∞ of the microscope in the case of observing at infinity is equal to the
product of magnifying degree k1 of an image A1B1 through eyepiece and angular magnification G2 of
eyepiece.
G∞ = k1.G2 = δD
.
f1 f 2 δ = F1F’2 is the optic length of the microscope.
5. Telescope:
 The objective is a converging lens with long focal length; the eyepiece is a converging lens with short
focal length. Two lenses are attached coaxially at two heads of one cylinder tube. Distance between them
is adjustable.
 In the case of observing at infinity, the image focal point of objective coincides with the object focal
f
point of eyepiece. The angular magnification of telescope in this case is G∞ = 1 .
f2 Translated by VNNTU – Dec. 2001 Page 102 Chapter VII – THE WAVENATURE OF LIGHT
In previous chapter we has discussed the transmission of light, light beam and the creation of image in
optical equipment. In chapter VII and VIII we will discuss the phenomena related to the nature of light. §42. LIGHT DISPERSION PHENOMENON
1. Experiment on light dispersion phenomenon
This experiment was carried out first by Newton in 1672.
Using a screen, in which there is a small split A (figure
7.1) to portion out a natural light beam (white light) in
the form of a thin light band. Guiding the light band
through a prism whose sides are all in parallel with the
split A. Behind the prism we put another screen to catch
the image of the light beam. In the screen, we can
observe a range of rainbow colors varying from red to
violet, in which the red are less distorted and the violet
are the most distorted in the light beam.
Thus, going through a prism, the white beam has not only been refracted towards the bottom of the prism
but also has been separated out to become different light beams with colors. This phenomenon is called
the dispersion of light phenomenon.
The range of colors that we observed in the final screen is called the spectrum of white (natural) light. In
its spectrum there are 7 main colors named: red, orange, yellow, green, blue, indigo, violet. In fact, there
are not only these 7 colors but a number of different colors, changing continuously from one color to
another. 2. Experiment on monochromatic light
This experiment has also been carried out the first time by Newton. In the final screen (figure 7.1), there
has been created one more slit which is in parallel with the original split A and was placed so that it stays
in the place of one specific color in the spectrum (say blue). Behind the screen B, we place another screen
C, which again portion out only a certain small part of the beam (figure 7.2). This arrangement will make
sure that the light beam coming out of screen C is only blue. Let the light beam going through another
prism and observe the result at the forth screen E, we can only observed 1 thin and blue band.
Thus, a blue light beam going through prism
is still a blue light beam, which means the
beam has not been dispersed. We called
such a beam as monochromatic light beam.
Doing the same experiments with different
colors in the spectrum, we also have
received the same result.
Thus, in conclusion, monochromatic light
beam are the light beams that will not be dispersed when going through a prism. Each corresponding color
is called monochromatic color. Translated by VNNTU – Dec. 2001 Page 103 3. Synthesizing white light
In the above experiments, we have been separating the different monochromatic light beam from one
natural white light beam. However, the question is that: Is this possible to synthesize these collection of
monochromatic light beam into one white light beam?
Newton has also been the first one to carried out the experiment of synthesizing white light beam. Below
is the description of one of his experiments.
Letting a white light beam through a small hole placed in front of the focal point of a convergence lens so
that a real image will be created somewhere behind the lens. Let the out going beam going through a prism
(figure 7.3). The white light beam will be dispersed into different color light beams. Using another
convergent lens, placing behind the prism and be fore the prism's new image, and observe the new image
in a screen putting behind the second lens, we'll be able to see back a point of white color situated in the
screen. The experiment has proved that the synthesizing the full spectrum of colors will give us the
original white color light beam. Thus, white color light beam is the combination of various monochromatic color lights ranging from red to
violet. 4. Dependence of the index of refraction of a transparent medium on the color of the light
We have observed that, when light beam is going through a prism, the coming out beam will be refracted
towards the bottom of the prism. The more the index of refraction of the prism material, the more the
refraction angle will be.
The experiment in figure 7.1 give us the conclusion that: while going through the prism, the refraction
angle of different monochromatic colors is different: the red is refracted lest and the violet is refracted the
most.
Thus the index of refraction of the material to different monochromatic light is different. And that the
index of refraction to the red color is least and to the violet color is the most. This characteristic holds true
with every transparent material.
Below are some value of the index of refraction of several material to different colors:
Red ray
Yellow ray
Green ray
Violet ray Flint glass
1.6444
1.6499
1.6657
1.6852 Crown glass
1.5145
1.5170
1.5230
1.5381 Water
1.3311
1.3330
1.3371
1.3428 The dispersion phenomena is often used in spectrometer equipments to separate a complex  combination
of colors  into a collection of monochromatic light beams. Translated by VNNTU – Dec. 2001 Page 104 Several phenomena in the nature like rainbow, halo, etc. can also be explained by the dispersion rule.
Questions
1. What is the dispersion? Describe the Newton's experiment on dispersion
2. What is the monochromatic color? Describe the experiment.
3. What is the white light? Describe the experiment.
4. Describe the dependence of the index of refraction of a medium on the color of the light. §43. LIGHT INTERFERENCE PHENOMENON
1. Young's experiment on light interference phenomenon
The first experiment on light interference of two light sources was carried out by Young in 1802.
Light from a lamp D goes through a onecolor filter F (say red, for example) and towards a small split S
placed on a screen M (figure 7.4). S becomes a split source of monochromatic light beam.
The monochromatic light beam going out of S will continue to spreading through to two small splits S1,
S2, which are in parallel to each other and stay very closely in one screen M12. S1, S2 are both in parallel
with S.
After M12, we place our eyes so
that we can see both the light
beams coming out from S1 and
S2, adjusting our eyes towards
the splits, we can see a small
range of light, in which there
are bright (red) bands and dark
bands situating alternatively
and evenly.*
The phenomenon of having these bright bands and dark bands is called the light interference
phenomenon.
If we do not use the filter M but letting the white color light entering the splits, there will be white color
band in the center of the bright range, in its two sides will be a rainbow range of color bands with violet
bands inside and red bands at the outside. 2. Explanation of the phenomenon
The observation of different dark and bright bands locating alternatively can only be explained by the
interference of two waves. The bright bands are where the two waves having constructive interference
while the dark bands are where the destructive interference occurs. We call these bright and dark bands as
interference fringes.
If we admit that the light having a wave nature, we can explain the observation as follows:
The light beam coming from the lamp D to the
split S make S becomes a lighting wave source.
* In fact, to make the experiment to become objective, the images of these bands should be put on a screen behind the
splits. However, since the bands are very close and the light intensity is weak, the observation would be very
difficult. Translated by VNNTU – Dec. 2001 Page 105 The waves spreading towards the splits S1 and S2 making this to become two coordinating light sources. In
the final screen, there will be an overlap of the two sources thus creating an interference, resulting bright
and dark fringes (figure 7.5).
The reasons that an interference can be done is that the two sources S1 and S2 are coordinative to each
other, which means:
+ The light waves that emitted from the two sources have the same frequency.
+ Since the distance between the S1, S2 with S has been fixed that the phase difference between S1 and S2
is not changed over time.
If we used a white color light beam instead of a monochromatic light then the systems of interference
fringes of different color ray do not coincide with one others, except at the center of the screen (where a
white color band, namely the central white fringe, can be seen). In the two sides of the band, the
interference fringes do not coincide and create rainbow range of colors at each side.
We can better understand the idea if we use other filter instead of the red
one (green, yellow, etc.). In that case, we can also see the bright and dark
bands but they stays at different places in comparison with the original
(red) case. In details, the interference fringes distance in the red light
colors is greater than that of the yellow color and so on.
Sometime when we observe the reflection light from a soapy or oil
bubbles, we also see beautiful colors appearing. This is also resulted from
the light interference phenomenon. The two sources of interference are:
reflecting sources at the up surface of the film. The other is the light
which after refraction at the upper surface, get reflected at the lower
surface, that comes out and interference with the previous one (figure
7.6). 3. Conclusion
The phenomena of light interference is an important experimental evidence that helps to conclude the
wave nature light.
Recently, based on a number of other important evidence, especially on the Maxwell's electromagnetic
theory, that scientists have been proved successfully that light is a kind of electromagnetic wave.
Questions
1. Describe the experiment on the interference of light?
2. Explain the experiment's result and produces necessary conclusion. §44. MEASURING THE WAVELENGTH OF THE LIGHT. THE WAVELENGTH AND THE COLOR OF THE LIGHT 1. Interference fringe distance
a) The placement of the interference fringes in the
Young’s experiment: Suppose that we got the
interference fringes in the Young’s experiment on the
screen E placed in parallel with screen M. The
interference fringes having the forms of: lines of bright Translated by VNNTU – Dec. 2001 Page 106 and dark bands situated in parallel with the splits S1 and S2, and placed alternatively and evenly.
In figure 7.7, the drawing plane is the plane that perpendicular to the splits and screens. I is the center of
S1S2 and IO is perpendicular to S1S2, the placement of the bands (both dark and bright) is represented by
OA = x.
In point A there is a bright fringe when two light waves coming from two sources S1 and S2 to A are in
phase and constructive to each other. This condition will be satisfied if the difference in traveling length of
light waves from two sources S1 and S2 to A equals an integral multiple of the light wavelength:*
S2A  S1A = kλ
r2  r1 = kλ or (71) where r1 = S1A; r2 = S2A and k is an integer (k = 0; ±1; ±2; etc.).
"
"
"
Denoted H1H2 as the projection of S1S2 on AI. Denoted S1S2 = a; α = OIA ; α1 = IAS1 ; α2 = IAS2 , we
have
H1A = r1cosα1 = IA – IH1 = IA – (asinα)/2
H2A = r2cosα2 = IA + IH2 = IA + (asinα)/2
Since α1 and α2 are rather small, we have r1cosα1 ≈ r1 and r2cosα2 ≈ r2. Thus r2  r1 = asinα.
Moreover, since α is also quite small we can approximate sinα ≈ tgα = x/D, where D = IO, and formula
x
= kλ.
(71) becomes r2  r1 = a
D
Thus, the positions of bright fringes on the screen can be determined by the correlation
x=k λD
a (72) where k = 0; ±1; ±2; etc.
At point O (x=0), with k = 0 we have a bright fringe, namely the central bright fringe. In two sides of the
central bright fringe in turn are firstlevel bright fringes (with k = ±1), secondlevel bright fringes (with
k = ±1), etc. These bright fringes are equally distributed.
Between two successive bright fringes are one dark fringe.
X=(2k+1) λD/2a
b) The fringe interval: The fringe interval is the distance between two successive bright (or dark) fringes.
The fringe interval between a klevel bright fringe and a (k+1) level bright fringe is
i = xk+1 – xk = (k+1)
i= λD
λD
k
a
a λD
a (73) For example, if a = 0.35mm, D = 1mm and λ = 0.7µm(*) (the red ray) then i = 2mm.
* Two sources, S1 and S2, are considered independent to each other. (*) 1µm = 106m Translated by VNNTU – Dec. 2001 Page 107 2. The wavelength and the color of the light
a) Measuring the wavelength by the interference method:
It can be measured accurately the distance D from two splits S1, S2 to the screen E with an accuracy in
millimeters. Furthermore, the distance a between two splits S1 and S2, and the fringe interval i can also be
measured by using appropriate equipment (microscope, magnifier).
With these information in hand, we can easily calculate the light wavelength by using formula (73). It is
the principle of measuring a light wavelength by the interference method.
b) Wavelength and the color of the light: While measuring the wavelength of different monochromatic
light by the interference method, it is realized that each monochromatic light has a specific wavelength.
For example:
+ The red light at the start of the spectrum has λ=0.760um
+ The violet light at the end of the spectrum has λ=0.400um
+ The yellow light emitted by a sodium fluorescent lamp has λ=0.589um
Thus, the monochromatic light is the light with a specific wavelength. The color corresponding to this light
is called the monochromatic color or the spectrum color.
In fact, monochromatic lights with wavelengths close to each other can be considered as having the same
color. Therefore the continuous spectrum has been classified into different color ranges:
Color range From To Red
Yellow and orange
Green
Blue – indigo
Violet 0.760µm
0.640µm
0.580µm
0.495µm
0.440µm 0.640µm
0.580µm
0.495µm
0.440µm
0.400µm Besides these colors, there are colors which are combinations of these monochromatic colors with
different ratios.
Questions
1. Describe the method to measure a light wavelength based on the interference phenomenon.
2. Describe the relationship between a light’s color and its wavelength
3. In Young experiment, let’s a = 0.3mm; D = 1m and the fringe interval i = 2mm, determine the light
wavelength λ.
a) Calculate the interval between the firstlevel bright fringe of red color (λR = 0.76µm) and that of violet
color (λV = 0.400µm); b) Calculate the interval between the secondlevel bright fringe of red color and that
of violet color.
Hints: 3) 6×107m; 4)a) 2.4mm; b) 4.8mm. §45. SPECTROMETER. CONTINUOUS SPECTRUM
1. Relation between the index of refraction of a medium and the wavelength of the light
In the dispersion experiment, we have seen that the index of refraction of a certain material but to different
monochromatic light is different. On the other hand, we have also known that each monochromatic light
Translated by VNNTU – Dec. 2001 Page 108 has a certain wavelength. Thus, the relative indexes of refraction of a specific transparent medium to
different monochromatic lights depends on their light wavelength.
When measuring the index of refraction of different material to different monochromatic lights, it can be
observed that the index of refraction to those lights that have a longer wavelength is smaller than that to
those lights which have shorter wavelength. For example with water we have:
Light color Wavelength (µm) Index of refraction Red
Yellow
Blue
Violet 0.6563
0.5893
0.4861
0.4047 1.3311
1.3330
1.3371
1.3428 2. Spectrometer
One of the most important applications of the dispersion phenomena is its application in analyzing lights
in spectrometer.
Spectrometer is an optical equipment that is used to analyze and separate a complex light beam into
different components of monochromatic lights. In other words, it can be used to recognize the components
of a light source.
A spectrometer has 3 main components:
+ Collimator: which helps to create a beam of parallel lights. It has a small split S staying at the focal point
of a convergence lens L (figure 7.8). The light beam created from the source J has been adjusted to come
to S. the outcome beam after L is a corresponding parallel beam
+ Prism P has the duty of transforming a parallel light beam coming out from L to become a system of
parallel monochromatic light beams.
+ Image room: include a convergence lens L2 that serves to converge the systems of outcoming beams to
a systems of converged lines which can be observed at the screen S placed at the focal point of the second
lens L2. If the source J has created a number of light colors with different wavelength: λ1, λ2, λ3 … then on the
screen F we can see the same number of lines: S1, S2, S3 … Each of the line represent a monochromatic
component of the emitted light beam.
The combination of lines is called the spectrum of the source J. 3. Continuous spectrums:
If J is a lamp with a hot burned filament, than in the screen we can observe a range of color lines, varying
from red to violet. This is called the continuous spectrum of the lamp.
Objects of liquid, solid or gas, being heated off will emit a continuous spectrum of light. The Sun is a big
gas with high mass being heated and emit lights. The spectrum of the sunlight is a continuous spectrum. In Translated by VNNTU – Dec. 2001 Page 109 a continuous spectrum the different color lines situated so closely that it can be seen as if there is no gap
between them, and thus they are a continuous spectrum of light.
One of the important aspect of a continuous spectrum is that it does not depend on the structure of the light
source, it only depends on the temperature of the source. This means that one iron bar and a porcelain bar
being heated to the same temperature will create two spectrum which are almost the same.
In a temperature of 500oC, an object starts to emit lights in the red zone, however the amplitude of the
light emitted is still very weak. The result is that we hardly can see the lights and the object is still dark.
The more temperature of the object, the light emitting of it moving more towards those having a smaller
wavelength in the spectrum. The filament of electric lamps, being heated to around 2500K emit quite a
number of lights in the visible area and thus can possible to create a full range of spectrum. The light of
this lamp emitted can be considered as white color light.
The temperature at the surface of the sun is around 6000K, and the strongest light emitted by the sun is at
around violet color (wavelength = 0.47µm). Again the light emitted by the sun is white color lights.
In the sky, sometime we can see the start emitting the blue light. This means that the temperature at the
surface of such a star is much greater than that of the Sun.
Scientists often make use of this characteristic to find the temperature of a certain heated object like: the
sun, the stars, the fire in a tunnel, etc. To measure the temperature, the spectrum of the source in question
is being compared to that of a known temperature source.
Questions
1. Describe the relationship between the index of refraction of a certain material to a certain light.
2. What is a spectrometer? Describe a spectrometer’s design.
3. What is continuous spectrum. What are the conditions that an object is emitting a continuous spectrum.
Describe the characteristics and the applications of continuous spectrum. §46. LINE SPECTRUM
1. Emission line spectrum
Directing the light beam radiated from a fluorescent lamp (that contains gaseous mercury or hydro, or
sodium, etc.) towards the split of a spectrometer, we can observe on the screen of the spectrometer the
emission spectrum of the fluorescent. The spectrum is a systems of separate color lines and is called the
line spectrum.
Emission line spectra are created by lowpressure gases being stimulated. The stimulation can be done by
heating up the gas, or by creating an electric spark going through the gas.
Experiments has showed that: different atom will have very different number and types of spectrum lines.
Also the place and the ratio of the lines are different from one atom to another.
For example: the spectrum of a sodium fluorescent have two yellow lines placing very closely to each
other with its wavelength of 0.5890µm and 0.5896µm. While the spectrum of a hydro fluorescent have 4
lines: from red (0.6563µm) to blue (0.4861µm) to indigo (04340µm) and lastly violet (0.4102µm).
Thus, each atom in its evaporated gas form in lowpressure will have its specific line spectrum, that
characteristic for that atom. Translated by VNNTU – Dec. 2001 Page 110 2. Absorption line spectrum
a) Directing a white color light beam created by a incandescent filament lamp towards the split of a
spectrometer (figure 7.9) we can observed a full continuous spectrum in the screen of the spectrometer. If
in the course of the light beam, we placed a fluorescent evaporated sodium lamp in, there will appears in
the spectrum two dark lines situated closely to each other. These two dark lines situated right in the place
of the two bight lines in the emission line spectrum of sodium as mentioned earlier. The new spectrum
obtained is called the absorption spectrum of sodium. If we substitute the sodium fluorescent with a potassium fluorescent then in the final spectrum there will
be dark lines appeared right in the place of the absorption spectrum of potassium. This is also called the
absorption spectrum of potassium.
The spectrum of the sun that we obtained on the earth is an absorption spectrum. The surface of the sun
creates a full range continuous spectrum. Lights going through the sun’s atmosphere to the earth give us
the absorption spectrum of the sun’s atmosphere.
The condition to obtain an absorption spectrum is that the temperature of the absorptive gas should be
lower than that of the light source that emits the continuous spectrum.
b) The chromoinversion phenomenon: There is a special phenomenon that create a special link between
the emission spectrum and the absorption spectrum of one element: it is the chromoinversion phenomena.
Suppose that the fluorescent has been heated to a certain degree that it can emit lights, but still lower than
the temperature of the white light source. The absorption spectrum of this fluorescent is recorded on the
screen of the spectrometer.
Then, we suddenly remove the white light source. In the screen of the spectrometer we can now see the
colorful background of the continuous spectrum has gone away, and the dark absorption lines now become
the bright emission lines of the emission spectrum. This is called the chromoinversion phenomena of
spectrum lines.
For example: in the absorption spectrum of sodium there is a double dark line situated in the same place of
two yellow lines (0.5890µm and 0.5896µm) of the emission spectrum of sodium.
Thus, in a certain degree, when a fluorescent can emitting which monochromatic lights, will also has the
ability to absorb these monochromatic lights.
c) The absorption spectrum of a certain atom is also characterized to that specific atom. Thus we can based
on the absorption spectrum, we can detect the presence of a certain atom in a mix or compound. This is
called the absorptive spectroscopic analysis approach.
Thanks to this approach that scientist can detect the present of helium in the sun before they even find it in
our earth. Besides, people can detect the present of a number of atom in the surface of the sun including:
hydro, sodium, calcium, iron, etc. Translated by VNNTU – Dec. 2001 Page 111 3. The spectroscopic analysis approach and its advantages
The approach to detect the structure of a mix or compound by analyzing spectrum is called the
spectroscopic analysis approach.
In a qualitative study, scientist only tries to detect the present of a certain atom in a sample under study.
The spectroscopic analysis approach has been proved to be simpler and faster than chemical analysis
approaches.
In a quantitative study, scientist also try to find out the concentration of a certain substance in a mix or
compound. The spectroscopic analysis approach is very sensitive. It can even detect of a very small
concentration of a substance in a mix or compound (usually is around 0.002%)
Thanks to the spectroscope approach, scientists can know the structure and the temperature of a very
distant away object like the sun and stars.
Questions
1. What is the emission spectrum? In what condition an emission spectrum can be obtained? Describe the
characteristics and applications of the emission spectrum.
2. What is an absorption spectrum in what condition an absorption spectrum can be obtained? Describe the
characteristic and applications of the absorption spectrum.
3. Describe the chromoinversion phenomena of spectrum lines.
4. What is the spectroscopic analysis approach? Describe the advantages of the approach. §47. INFRARED AND ULTRAVIOLET RAYS
1. Experiments to discover infrared and ultraviolet rays
Directing the light beams of an arc J to the split of a spectrometer, in the screen of the spectrometer we can
observe a spectrum of the light beam. In the screen, make a new split F so that a certain monochromatic
light beam can be extracted. Direct this light beam to a weld of a sensitive thermocouple battery. The other
weld of the battery is kept at a fixed temperature (figure 7.10). We can see that the electroscope G in the circuits of the thermal battery showed a current going through it.
This has proved that the light beam has a thermal effects, which heated the wield of the thermal battery
and created current in the circuit.
Adjust the screen so that the split F go through a full range of the spectrum. We can observe that the hand
of the electric meter always shows a current, through the exact amount of the number is varied. Thus,
thermal effects of different monochromatic lights are different.
If we continued to adjust the split F out of the visible spectrum, we still see the electrometer’s hand
showing a current run in the circuit. This has proved that, besides the type of visible light, there are still
lights (or so called radiations) invisible. Translated by VNNTU – Dec. 2001 Page 112 2. The infrared ray
Infrared rays are radiations that is not visible and having its wavelength greater than that of the red light
(0.75µm). Infrared has its nature as a electromagnetic wave. It can be radiated from heated objects.
Objects that are heated at small temperature can only emit infrared lights. For example: human body at its
normal temperature of 37oC only radiate infrared lights, within which the strongest light is at around 9µm
in wavelength. Another object at about 500oC is starting to emit red lights, but still its strongest lights are
infrared having its wavelength at 3.7µm.
In the Sun light, there are about 50% energy of the combined light beams are in the form of infrared. The
infrared source usually are the heated filament having its power from 250W to about 1000W. The
temperature of a redlight filament is about 2000oC.
The outstanding characteristics of infrared lights are their thermal effects. Besides, infrared light also has
its effects on some special kind of films which is called infrared films. If we have taken picture of the
clouds by infrared films we can see the clouds to comes up very clear. This is because that the clouds
which have a lot of water in it, more or less will absorb different types of infrared radiations.
The most important applications of infrared is used for drying and warming purposes. In industrial
applications, infrared are used to dry painted objects (like cases of car or fridge) or to dry fruits. In medical
applications, infrared are used to warm up outer skins to help better movements of blood. 3. The ultraviolet ray
Ultraviolet rays are invisible radiations that have wavelengths shorter than that of the violet light (0.4µm).
The ultraviolet ray is an electromagnetic wave in nature.
The Sun is a strong source of ultraviolet rays. Around 9% power of the sunlight beam is from ultraviolet
rays. Electric arcs are also sources of ultraviolet rays. In hospitals, mercury fluorescent lights are used to
radiate ultraviolet rays. Besides, objects at a temperature higher than 3000oC also radiate ultraviolet rays.
The ultraviolet ray has very strong effects on films. It can also makes some material emit lights. It can
ionize gas. And it can helps to makes several photochemical reaction to happen… ultraviolet also has
some biological effects.
In industrial applications, ultraviolet can be used to find small crack in a structure. To do so, they drop
some special smooth eradicative ponder into the structure’s surface. The ponder will then goes into the
crack on the surface. When we bring the surface into the light of an infrared light the crack will bright up.
In medical, ultraviolet also helps to cure rickety.
Questions
1. Describe the experiment that helps to find out infrared and ultraviolet rays.
2. What is the infrared ray? What are the sources of infrared rays? In what conditions does an object
radiate infrared ray? Describe some applications of infrared rays.
3. What is the ultraviolet ray? What are the sources of ultraviolet rays, and in what conditions the
ultraviolet rays are radiated? Describe effects and applications of ultraviolet rays. Translated by VNNTU – Dec. 2001 Page 113 §48. XRAYS
1. Xray tube
In 1895, the famous German physic scientist Roentgen has discovered that when a cathode beam
(electronic beam) crash into a iron bar with high atom density weight (platinum, wolfram, etc.), then there
will be an invisible radiation that can goes through glasses and can stimulate some material and can
blacken the films. Such a radiation is called Roentgen ray, or in short, is called X ray.
A simple Roentgen tube is simply a tube that direct
electronic beam, in which there will be another electrode
that have high atom density weight and hard to melt
material to block the electronic beam. This electrode is
called the anticathode AK and is usually connected to
anode (figure 7.11). The atmosphere in the tube is usually
kept at 103 mmHg, and the voltage difference between
anode and cathode is about several tens thousands of volts.
Since when activated, the anticathode is usually being
heated very rapidly, thus inside modern X ray tube, a water
current is often inserted inside the anticathode. Besides, to
increase the electronic beam in the tube, the cathode is
usually heated off when used. 2. The nature of Xrays
When it was first discovered, the X ray was mistakenly believed that having a photon nature. However,
when inserted into a magnetic or electric field the X ray does not change its direction. Thus the X ray does
not bring with it electricity.
Later on, scientists have been proved that X ray is, in fact an electromagnetic wave that have its
wavelength shorter than even that of the ultraviolet lights. Scientists have been tried to measure the
wavelength of X ray and have find out that its wavelength is in the range of 1012 (hard X ray) to around
108m (soft X ray).
The procedure that created an X ray has been explained as follows:
The electrons in the cathode beam was accelerated in the electric field, and thus having gained a big
kinetic energy in their movement. When they come to “opposite cathode” they crash with electron in the
metal bar and make interaction with atoms of the metal. In this interaction there will be an electromagnetic
wave with very short wavelength that we often called the “block radiation”. This radiation is what we have
known as X ray.
Most of the kinetic that the electron have before crashing with the metal
bars become the internal energy of the metal bars and heathen it up. The
rest of the energy becomes the power of the X ray. 3. Properties and uses of Xrays
X ray has several characteristics and applications:
The most outstanding characteristics of the X ray is the ability to go
through. It can goes through most of untransparent materials like:
paper, wood, … It can goes through metal, though it has some difficulty
in doing so. The more massive the metal, the more difficult the X ray to
go through. For example: the X ray can goes quite easy through an
Translated by VNNTU – Dec. 2001 Page 114 aluminum bar with thickness of several centimeters, but it is blocked by a layer of lead with thickness of
some millimeters. Thus lead is often used to block X rays.
Thanks to the ability to goes through things, X rays is widely used in medical to do roentgenizations.
(figure 7.12). In industrial applications, x rays is used to detect the faulty in casting products.
X rays has strong effects on films, thus is often used to “X ray”.
X rays can also stimulate some material to emit lights. Fluorescent monitor is a screen that have been
covered by layer of platinocyanic barium. This screen emit green light under the effects of X rays.
X rays can also ionize gases. Scientists has been making used of this characteristic to create equipment
that can measure the density of X rays.
X rays also have physiological effects. It can kills cells and bacteria. Thus X rays can be used to cure
shallow cancer cases, and also cure some other diseases. 4. Electromagnetic waves scale
X rays, ultraviolet, visible lights and infrared as well as radio waves have the same nature as
electromagnetic waves.
The main difference between them is their wavelength. The following table will give such a view.
Electromagnetic types Wavelength (m) X rays 10 12 to 10 9 Ultra violet 10 9 to 4*10 7 Visible lights 4*10 7 to 7.5* 10 7 Infra red 7.5*10 7 to 10 3 Radio waves Greater than 10 3 Besides, in the process of disintegration of an atom, we can see another type of wave length that have a
very short wavelength and is called gamma waves.
In fact, there is no clear boundary between scales.
However, since the different wavelength that each type of waves have, their corresponding properties are
different:
• Those waves that have short wavelength (like gamma or X rays) will have very good going
through ability. Have strong effects on films, can easily stimulate material, and ionize gases. • For those waves that have longer wavelength, we can more easily to observe their interferences. The way to create and receive different types of waves is also different (figure 7.13). Translated by VNNTU – Dec. 2001 Page 115 Questions
1. What is X ray? Describe the design of a X ray tube
2. Show the characteristics and applications of X rays
3. Show the general conclusions on the scale of electromagnetic waves
SUMMARY OF CHAPTER VII
1. When a whitecolor light going through a prism it will be dispensed into different monochromatic light
components, in which the violet light is refracted the most and the red light is refracted the least.
+ The reasons for the dispersion of light is the dependence of the index of refraction on a light’s
wavelength. The longer the wavelength of a light, the smaller its corresponding index of refraction.
+ Monochromatic lights are lights that have only one colors, and that it will not be dispersed after going
through a prism. It has a fixed wavelength
+ Whitecolor light is the combination of a number monochromatic light.
+ The dispersion phenomena is used in spectrometer to analyze the structure and design of a light source.
2. The light interference is happened when two coordinating lights overlap.
+ The two coordinating light wave are created by separating one light beam into two portions and letting
them overlap (Young’s experiment).
+ Interference fringes are the dark and bright bands staying alternatively and evenly. Their basic distance
λD
.
is i =
a Translated by VNNTU – Dec. 2001 Page 116 3. There are three types of spectrum: continuous spectrum, line emission spectrum, line absorption
spectrum.
+ The continuous spectrum only depends on the temperature of the light source.
+ Line absorption and emission spectrums of different elements are different. They are used in the
spectroscopic analysis approach.
+ The dark bands in absorption spectrum situated right in the place of bright bands in emission spectrum.
4. Infra red, visible lights, ultra violet lights and X rays are all electromagnetic waves, but with different
wavelengths. Thus they have different properties and applications. The first three types can be radiated
from heated objects, but the X rays can only created by using X ray tubes, where there is a crash between
the electron beam and the anticathode. Translated by VNNTU – Dec. 2001 Page 117 Chapter VIII – LIGHT QUANTUM
In this chapter, we will investigate some phenomena that cannot be explained by the wave property of
light. These phenomena however can be explained by the new hypothesis: quantum hypothesis. The born
of quantum hypothesis helps us to understand about the light, alone and the microworld, generally, more
deeply. §49. THE PHOTOELECTRIC EFFECT
1. Hertz’s experiment
In 1887, Hertz, a German scientist, did the following experiment: exposed a negativelycharged zinc (Zn)
plate, attached to a electroscope, to a light beam from an electric arc (figure 8.1). He observed that the two
copper leaves of the electroscope closed; this means that the leaves lost its negative electrons. If the zinc
plate was positively charged, then nothing happened.
This phenomenon also happened when the zinc piece was replaced by a piece of copper (Cu),
aluminum(Al), silver (Ag) or nickel (Ni).
If a noncolor glass piece was used, then the photoelectric phenomenon did not happened. We know that
glass strongly absorbs the ultraviolet rays.
Many similar experiments also came to the conclusion
that: when a metal piece is exposed to a suitable light
beam (having short wavelength), the electrons of the
surface of the metal piece will be emitted. This is the
photoelectric effect.
The emitted electrons are called photoelectric
electrons.
Actually, when an ultraviolet beam is directed into the
positivelycharged zinc plate, there are still electrons
emitted. However, they are pulled back immediately,
so the electric charge of the zinc plate is considered
unchanged. 2. The experiment with a photocell
a) Photocell is a small vacuum tube in which there are two
electrodes: anode A and cathode K. An anode is a metal wire
ring. A cathode has a hemispherical shape; it is made from
metal (which we need to examine) coated on the interior side
of the cell (figure 8.2).
+ Light emitted from an electric arc, goes through a glass
filter F to obtain a certain monochromatic light, and then to
cathode K.
+ Setting up a electric field between anode and cathode by
using a set of batteries E. Voltage difference U between A
and K is changeable (in magnitude and sign) by changing the
position of the plug C on the source. Translated by VNNTU – Dec. 2001 Page 118 A voltmeter V is used to measure the potential difference U and a sensitive milliampmeter G is used to
measure the current intensity flowing through the photoelectric cell. The internal resistances of batteries
are negligible compared to the resistance of photocell.
+ When the cathode is illuminated with light having short wavelength, the circuit has a current that is
called photoelectric current.
In the photoelectric cell, the photoelectric current has the direction from cathode to anode under the effect
of electric field between anode and cathode.
b) Investigating the dependency of photoelectric phenomenon on the wavelength of trigger light (which
impinges on the cathode), it is found that: to each metal used for cathode, the photoelectric phenomenon
happens only if the trigger light has a wavelength at least smaller than some limit λ0. If the trigger light’s
wavelength is larger than the limit, then no matter how strong the light beam is, the photoelectric
phenomenon will also not happen.
c) After illuminating the cathode to cause the photoelectric
phenomenon, they examined the dependency of photoelectric current
intensity I on the voltage difference UAK between anode and cathode.
The result was illustrated by curve 1 on figure 8.3. This graph is
called the characteristic voltamp line of photoelectric cell.
Initially, increasing UAK makes the photoelectric current to increase.
As UAK is at a certain value, then the photoelectric current intensity
acquires a saturated value Ibh.
After that, the value of photoelectric current will not change no
matter how UAK is increased.
d) Investigating the dependency of the intensity of saturated photoelectric current Ibh on the intensity of the
trigger light beam(*), it is found that Ibh is proportional to that intensity (curve 2 in figure 8.3 corresponds
to the case when trigger light intensity is increased by 1.5 times).
e) To completely annul the photoelectric current, it is necessary to apply a certain negative potential
difference Uh (Uh = UAK<0) between anode A and cathode K. Uh is called the stopping voltage. The value
of Uh corresponds to the intersection point of the characteristic voltamp line of photocell and the Uaxis.
It is seen from the experiment that: values of stopping voltage Uh corresponding to each metal used for
cathode definitely does not depend on the intensity of the trigger light but only depends on the wavelength
of that trigger light beam. If the two monochromatic light beams 1 and 2 have the same wavelength, then
characteristic voltamp lines 1 and 2 will intersect the Uaxis at the same point Uh (figure 8.3).
Questions
1. State out the Hertz’s experiment to find out the photoelectric phenomenon and give the definition of this
phenomenon.
2. State out the experiment with photocell and give the results of that experiment.
3. Intensity of the optimum photoelectric current is 40µm. Find the number of electron emitted from
cathode of the photocell in every minute.
4. Explain why increasing voltage difference between anode and cathode of the photocell to a certain
value, then the intensity of the photoelectric current acquires an optimum value.
(*) We call the intensity of a light beam at a point is a quantity measured by amount of energy transmitted from that
light beam to a unit charge, positioned perpendicular to the light beam passing through that point, in a time unit. Translated by VNNTU – Dec. 2001 Page 119 Hints: 3) 25.1013 electrons/s. §50. THE QUANTUM HYPOTHESIS AND PHOTOELECTRIC LAWS
1. Photoelectric laws
a) The first photoelectric law: to each metal used for cathode, there is a certain threshold wavelength λ0,
which is called photoelectric threshold. Photoelectric phenomenon only happens when wavelength λ of the
trigger light is smaller than the photoelectric threshold (λ<λ0).
Table of photoelectric thresholds λ0 of some metals (µm)
Silver 0.26 Calcium 0.45 Copper 0.30 Nitrogen 0.5 Zinc 0.35 Potassium 0.55 Aluminum 0.36 Cesium 0.66 From the above table, it is seen that the visible light can only cause photoelectric phenomenon with
calcium and alkaline metals.
b) The second photoelectric law: for trigger light having a wavelength satisfied the first photoelectric law,
the intensity of the saturated photoelectric current is proportional to the intensity of the trigger light beam.
c) The third photoelectric law: the existence of stopping voltage difference Uh proved that the
photoelectric electrons emitted from the metal surface have initial velocity v0. The pulling electric field’s
strength is to an extent at which even the electron having highest initial velocity v0max also cannot fly to
anode. At that time, the photoelectric current is cancelled out completely and the work of pulling electric
field is equal to initial maximum kinetic energy of the photoelectric electron.
2
mv0 max
eU h =
2 From the experimental values of Uh showed in last lesson, we come to the following law:
The initial maximum kinetic energy of photoelectric electrons does not depend on the intensity of the
trigger light beam, but only depend on the wavelength of the trigger light and the nature of metals used for
cathode. 2. The quantum hypothesis
a) The photoelectric laws contradict to the wave property of light. Indeed, from wave theory, when light
arrives at the cathode surface, the fluctuating electric field in light wave will make the electrons in metal
vibrate. The larger the intensity of trigger light beam is, the stronger the electric field is and it will make
the electrons vibrate more. To a certain extent, the electrons will be released and create the photoelectric
current. Therefore, any light beam can also cause the photoelectric phenomenon, as long as its intensity if
strong enough and the initial kinetic energy of photoelectric electrons must depend on the intensity of the
trigger light beam.
b) The photoelectric laws can only be explained if we assume a new hypothesis called the quantum
hypothesis, initiated by Planck, a German scientist, in 1900.
According to the quantum hypothesis, atoms or molecules of matter do not absorb or emit light
continuously, but in separated and discrete packets. Each of those packets carries a definite amount of Translated by VNNTU – Dec. 2001 Page 120 energy, having magnitude equal to ε = hf, in which, f is the frequency of light emitted, and h is a constant
called Planck’s constant.
h = 6.625×1034Js
Each of those packets is called a light quantum.
For example, with a violet ray λ = 0.4×106m then f = c
3.108
=
= 7.5×1014Hz.
λ
0.4x10−6 The light quantum of this violet ray has a value of ε = hf = 6.62×1034 × 7.5×1014 = 5.965×1019J.
It is seen that the light quantum is very small, even a weak light beam also contain a larger amount of light
quanta. Therefore, we feel that light beam is continuous.
While light is transmitted, light quanta are not changed, and independent of the distance to light source,
even though the light source is a star, which is millions of light years away from us. 3. Explaining photoelectric laws by using the quantum hypothesis
Albert Einstein, a German scientist, is the first person who used the quantum hypothesis to explain the
photoelectric laws. He considered a light beam as a collection of particles and called each particle a
photon. Each photon corresponds to a light quantum.
Einstein suggested that there is a complete absorption of impinging photons in the photoelectric
phenomenon. Each absorbed photon will transmit all its energy to an electron. For the electrons lying on
the surface of metal, this energy is used for two purposes:
+ Supply that electron a work A to overcome the binding forces in the crystal and escape. This work is
called work function.
+ Supply that electron an initial kinetic energy. Compared with the initial kinetic energy obtained by
electrons on the inner layers, this energy is maximum: hf = A + 2
mvo max
2 (81) This is the Einstein’s formula on photoelectric phenomenon.
For the electrons in inner layers of the metal surface, they collide with ions of metal and lost part of its
energy. Therefore, their initial kinetic energy is smaller than forementioned maximum one.
The formula (81) says that initial maximum kinetic energy of photoelectric electrons depends only on
frequency f (or wavelength λ) of the trigger light and nature of metals used for cathode (K), not on
intensity of the trigger light beam. That is the main content of the third photoelectric law.
The formula (81) also indicates that: if energy of photon is smaller than work function A, then it can not
make the electron escape from cathode and the photoelectric phenomenon will not happen.
We have
Let hf ≥ A or h c
hc
≥ A; λ ≤
λ
A hc
= λ0, we have λ ≤ λ0
A (82) λ0 is photoelectric threshold of metal. Inequality (82) indicates the first photoelectric law.
Finally, we explain the second photoelectric law as below: Translated by VNNTU – Dec. 2001 Page 121 For the light beams capable of causing the photoelectric phenomenon, the number of photoelectric
electrons emitted from cathode in a time unit is proportional to number of electrons striking on the metal
surface in that period of time. On the other hand, this number of electrons is again proportional to intensity
of light beam; and intensity of the optimum photoelectric current is proportional to number of
photoelectric electrons ejected from cathode in a time nut. Therefore, intensity of the optimum
photoelectric current is proportional to intensity of the trigger light beam. 4. Waveparticle duality of the light
In the chapter VII, we showed that: visible light as well as infrared, ultraviolet, Roughen rays all are
electromagnetic wave having different wavelengths. It is said that they all have electromagnetic natures.
Here, we learned that light has particle property (quantum property). So light has both wave property and
particle property. It is said that: light has a waveparticle duality.
The shorter wavelengths the electromagnetic waves have, the stronger energy they contain.
Experimentally, the clearer their particle property is exhibited, the less clear their wave property is shown.
It is possible to consider the following effects as expressions of wave property: penetrating ability,
photoelectric effect, ionization effect, and luminescent effect.
In contrast, the longer wavelengths the electromagnetic waves have, the less energy they contain.
Experimentally, the more difficult their particle property is exhibited, the easier their wave property is
shown. It is easily to see interfering, dispersing phenomena of those waves.
The simultaneous existence of two opposite properties (wave and particle) in a thing (light) is a clear
illustration for the philosophically theoretical point about the united dialectics of opposite sides.
Questions
1. State three photoelectric laws and clearly point out the contradiction of those laws with the wave
property of light.
2. State the sketchy content of quantum hypothesis and apply that hypothesis to explain the photoelectric
laws.
3. Based on the table of photoelectric threshold values of some metals given in the text, calculate the work
functions of electrons from surfaces of the following metals: silver (Ag), copper (Cu), zinc (Zn),
aluminum(Al), calcium(Ca), sodium (Na), potassium (K), cesium (Cs).
4. Photoelectric threshold of sodium is 0.50µm. An infrared ray of wavelength 0.25µm passes to sodium.
Calculate the initial maximum kinetic energy and the initial maximum velocity of photoelectric electrons.
5. The cathode of a photocell is made of cesium (Cs) which has a photoelectric threshold of 0.66µm. The
cathode is illuminated with an infrared light of wavelength 0.33µm. Calculate the required stopping
voltage between anode and cathode of the photocell in order to completely annul the photoelectric current.
Hints: 3) Silver 7.64x1019J; Calcium 4,41x1019J; Copper 6.62x1019J; Sodium 3.97x1019J;
Zinc 5.67x1019J; Potassium 3.61x1019J; Aluminum 5.52x1019J; Cesium 3.01x1019J;
4) 3.97x1019J; 9.34x105m/s; 5) UAK= 1.88V Translated by VNNTU – Dec. 2001 Page 122 §51. LIGHT DEPENDANT RESISTOR AND PHOTOELECTRIC BATTERY
1. The photoconduction phenomenon
Some semiconductors are insulated when they are not illuminated and become electroconductive when
illuminated. The resistancedecreasing phenomenon of semiconductors when being illuminated is called
the photoconduction phenomenon.
In photoconduction phenomenon, when suitable light illuminates cathode of the photoelectric cell,
electrons will be ejected out of the cathode. Therefore, this phenomenon is also called the outside
photoelectric phenomenon.
In the photoconduction phenomenon, each photon of trigger light when being absorbed will release a
binding electron so that this electron will become a free electron moving in that semiconductor block.
These electrons will become conducting electrons. Besides, each released binding electron will leave a
hole having positive charge. These holes can also move freely from one lattice knot to another and take
part in electroconducting process.
The phenomenon in which the binding electrons are released to become conducting electrons is called the
inside photoelectric phenomenon.
Because the energy required to release a binding electron, to become a conducting electron, is not large,
the photon need not to posses large energy to cause the photoconducting phenomenon. Many
photoconductors work under infrared light. For example, CdS has a photoconductive threshold of 0.9µm.
We understand that the photoconductive threshold of a material is the longest wavelength of light, which
is capable of causing the photoconductive phenomenon in that material. This is an advantage of the
photoconductive phenomenon over the photoelectric phenomenon. 2. Light dependant resistor (LDR)
Composition: A LDR consists of a layer of semiconductor (ex, cadmium sulfur CdS) (1) coated on an
insulated plastic plate (2). Two electrodes (3) and (4) are attached to that semiconductor layer (figure 8.4).
Connecting a source of few volts to the LDR through a milliampmeter.
We found that when the LDR was put in the dark, there was no current in
the circuit. When the LDR was illuminated with light having wavelength
shorter than the photoconductive threshold of the LDR, current appeared in
the circuit.
The resistance of LDR decreases tremendously when being illuminated by
the above light. Measuring the resistance of the CdS LDR, it was seen that:
when not being illuminated, its resistor is about 3×106Ω; when being
illuminated, the resistor is only about 20Ω.
Nowadays, LDR is used to replace the photocells in most of automatic
control circuits. Let take an example: the automatic openclose street lights circuit (figure 8.5). Translated by VNNTU – Dec. 2001 Page 123 A light dependent resistor (LDR) is connected between the base B and the emitter E of a transistor T (say, npn).
The LDR divides the voltage difference with a resistor R1 (about 10kΩ) that is connected between the collector C
and the base B. A source of 6V is used to create either a voltage difference UBE or collector current Ic. The current Ic
flows through an electric magnet of an electromagnetic relay to turn streetlights on or off. The electric magnet is
connected in the collector circuit of the transistor.
In daylight, when the striking light on the LDR is strong enough, its resistor becomes very small compared to R1. The
voltage difference UBE is also very small. The base current and, hence, the collector current will be zero. The electric
magnet will not work.
At night, when the striking light on the LDR is weak enough, its resistor will become sufficiently large. The voltage
difference UBE increases. When UBE gets to a certain value (about 0.7V), there exists a base current (about 0.3mA)
and, hence, the collector current Ic (about 60mA). The Ic current flows through the electric magnet and induce it to
attack the power handle and close the circuit to turn on the streetlights. 3. The photoelectric battery
Photoelectric battery is an electric source in which photo energy is converted directly to electricity. The
battery works based on the inside photoelectric phenomenon occurring inside a semiconductor.
Let examine a simple photoelectric battery, Cu2O battery (figure 8.6).
The battery has a copper electrode. A layer of Cu2O is coated on this
copper plate. A very thin layer of gold is also sprayed over the Cu2O
layer to form a second electrode. The gold layer is so thin that light
can pass through it. On the interface between Cu2O and Cu, there is a
formation of a layer having special usage: it only allows electrons go
through it in the direction from Cu2O to Cu.
When shining a light beam having suitable wavelength onto the surface of Cu2O layer, the light will
release binding electrons in Cu2O to become conducting electrons. Part of these electrons is dispersed to
the Cu electrode. The Cu electrode receives electrons and become negatively charged. Cu2O is positively
charged. An electromotive force is formed between two electrodes of the battery.
If we connect the two electrodes together by conducting wire and through a galvanometer, we find that
there is a current flowing from Cu2 to Cu.
Beside Cu2O photoelectric batteries, there are many other kinds of photoelectric battery.
Nowadays, photoelectric battery has many applications. Solar batteries in pocket calculators, on artificial
satellites etc. are all photoelectric batteries.
Questions
1. What is the photoconducting phenomenon? State the composition and working mechanism of a light
dependent resistor (LDR).
2. State the composition and working mechanism of a Cu2O photoelectric battery. Translated by VNNTU – Dec. 2001 Page 124 §52. OPTICAL PHENOMENA RELATING TO THE QUANTUM PROPERTY OF THE LIGHT
1. The luminescence
An infrared light having wavelength λ is shined onto a bottle containing alcoholfluorescence solution.
This solution will emit light green light having wavelength λ’ (λ’ > λ). That is the fluorescent
phenomenon.
The same phenomenon also occurs with gases. The fluorescent light is gone nearly immediately after the
trigger light is off.
The mechanism of fluorescent phenomenon is as follows: when each fluorescent molecule absorbs an
ultraviolet photon having energy hf, it jumps to stimulated state. This molecule stays in the stimulated
state for a very short period of time. During this period, a part of its energy is lost due to collisions with
other molecules. Eventually, it returns to the original state and emits another photon having smaller energy
c
c
hf ’:
hf ‘ < hf or h
<h
λ'
λ
Hence λ’ > λ..
When an ultraviolet light is shined onto the zinc sulfur (ZnS) crystals doped with a small amount of
copper (Cu) and cobalt (Co), the zinc sulfur crystal will give out green light. The lighting up of crystals
when being triggered by suitable light is called the phosphorescence. Phosphorescent light can last for a
very long time after the trigger light is off.
The fluorescence of gases and liquids together with the phosphorescence of rigid are commonly called the
luminescence. The luminescence is often called the cold glow to differentiate with the glow of objects
when being heated. The mechanism of phosphorescence is also different with that of fluorescence.
Instead of using ultraviolet rays, rays having short wavelengths are also used to trigger the luminescence.
For example, Roughen rays (in Xray photographing), γ rays and electron beams (in the electron tube of
television) etc. to stimulate the luminescence of television screen.
The luminescence phenomenon of rigid is used in fluorescent lights. The light consists of a glass
cylindrical tube containing mercury gas at low pressure. Two electrodes at the two ends of the tube having
a hairlike shape are heated up. The interior side of the tube is coated with a layer of luminescent powder.
Light given out from mercury gas contains lots of ultraviolet rays. These rays will trigger the luminescent
powder layer coated on the interior side of the light to emit visible light. Consuming the same power, the
fluorescent lights are much brighter than incandescent bulbs. 2. Photochemical reactions
Photochemical reactions are chemical the reactions which occur under the effect of light.
There are many kinds of photochemical reactions: disintegrating reactions, synthetic reactions,
polymerization reactions, etc.
Example 1: Photosynthesis
Photosynthesis is an important photochemical reaction for the existence of verdure, so it affects
tremendously people’s lives. In this phenomenon, under influence of ultraviolet photons, verdure absorbs
and disintegrates CO2 of the air to produce nutrients like glucose, cellulose, starch etc.
Example 2: Disintegrating reaction of AgBr
This reaction is the foundation of photography. Being influenced by photons of trigger light, some AgBr
molecules in AgBr grains are disintegrated into silver and bromine: Translated by VNNTU – Dec. 2001 Page 125 AgBr + hf → Ag + Br
The lightstimulated molecules disintegrated in photographing process will become shoots from which all
AgBr grains will be chemically disintegrated in the photodeveloping substances. Since the AgBr grains
do not contain molecules disintegrated under influence of light, the photodeveloping substances do not
affect them.
Example 3: Synthetic reaction of HCl
If a container holding a solution of chlorine and hydrogen is illuminated with shortwavelength light, then
chlorine will combine with hydrogen to form HCl. The reaction takes place so strong that it cause
explosion and destroy the container. It is assumed that this reaction takes place in two continuous
processes:
 A primary process: Cl2 + hf → Cl + Cl  Secondary processes:
Cl + H2 → HCl + H
H + Cl2 → HCl + Cl etc.
Questions
1. What is the luminescence? Differentiate the fluorescence and the phosphorescence. Explain why the
luminescent light has longer wavelength than the trigger light does.
3. Describe the composition and action of a fluorescent light.
4. What are the photochemical reactions? Give few examples of them. §53. APPLICATION OF THE QUANTUM HYPOTHESIS TO HYDROGEN ATOM
1. The Bohr model of the atom
a) The axiom of stationary states In 1911, after many careful experimental works, Rutherford initiated the planet model of atoms (refer to
the textbook Chemistry 10). However, this model has a problem that it cannot explain the firmness of
atoms and the formation of line spectra of atoms.
To overcome the above problems, in 1912, the Danish scientist Bohr applied the quantum hypothesis’s
concept to explain phenomena of atomic system. He proposed the following two hypotheses (considered
as two axioms in mathematic)
a) The axiom of stationary states: atoms only exist in states having certain energy levels,
called the stationary states. Atoms do not radiate in the stationary states.
Energy of atoms in stationary states includes kinetic energy of electrons and their potential energy, with
respect to the nucleus. To calculate the energy of electrons, Bohr still used the planet model of atoms*.
b) The axiom of energy emission and absorption of atoms: the smaller energy a stationary state has, the
firmer it is.
The higher energy a stationary state has, the less firm it is. Therefore, when an atom has a high energy
stationary state, it always tends to change to a smaller energy stationary state. * Energy of atomic system is also considered as that of electrons Translated by VNNTU – Dec. 2001 Page 126 When an atom changes from a stationary state of energy Em to a stationary state of En (Em > En), the atom
emits a photon having energy equal to the difference Em – En.
ε = hfmn = Em – En
where fmn is the frequency of light wave associated with that photon.
In contrast, if an atom staying in a low energy stationary
state En absorbs a photon having energy hfmn equal
exactly the difference Em – En, then it will jump to the
stationary having larger energy Em. (figure 8.7).
A very important consequence deduced from the above
two axioms is that: in the stationary states of atoms,
electrons only move around the nucleus in the orbits,
having definite radii, called stationary orbits.
Bohr showed that: for the hydrogen atom, radii of stationary orbits increase proportionally to the square of
consecutive nature numbers:
Radius: r0 4r0 9r0 16r0 25r0 36 r0 Orbit’s name: K L M N O P where ro = 5.3×1011 m. ro is called the Bohr radius.
An orbit having large radius corresponds to high energy, and small radius to small energy. 2. Using the Bohr model to explain hydrogenous line spectrum
One of the important successes of Bohr atomic model is that it fully explained for formation of line spectra
of hydrogen.
It is observed that the lines in radiation spectrum of hydrogen arrange into definite and completely
separated series.
In the ultraviolet region, there is one series called Lyman series(*).
The second series is called Balmer series. Part of this series lies in the ultraviolet region, other part lies in
the visible region. The latter part has 4 lines: red line Hα (λα = 0.6563µm), blue line Hβ
(λβ = 0.4861µm), indigo line Hγ (λγ = 0.4340µm) and violet line Hδ (λδ = 0.4102µm).
In the infrared region, there is one series called Paschen series.
Firstly, let explain the formation of line spectra.
In normal states (primary states), the hydrogen atom has lowest energy, so electrons move in orbit K.
When an atom receives trigger energy, it is excited to higher energy orbits: L, M, N, O, P, etc.
The atom remains in the excited state for a very short period of time (about 108s). Then, the electron
returns to some inner orbit and releases a photon.
Every time an electron changes from a highenergy orbit to a lowenergy orbit, it emits a photon having
energy equal to the energy difference between those two orbits
hf = Ehigh – Elow
Each photon having frequency f corresponds to a monochromatic wave light of wavelength λ.
(*) The lines in this series are obtained by using photographing method Translated by VNNTU – Dec. 2001 Page 127 λ= c
f Each monochromatic wave light produces a spectrum line having certain color. Hence, the spectrum is a
line spectrum.
The formation of series is explained as below:
The lines in Lyman series are produced when electrons return from outer orbits to orbit K : L → K;
M → K; N → K; O → K; and P → K.
The lines in Paschen series are produced when
electrons return from outer orbits to orbit L.
Line Hα corresponds to the change M → L
Line Hβ “ “ N→L Line Hγ “ “ O→L Line Hδ “ “ P→L The lines in Paschen series are produced when
electrons return from outer orbits to orbit M.
If we represent every energy level associated with a
stationary orbit by a horizontal line, then we have a
map of energy levels as in figure 8.8.
In that map, an arrow, directed downward, represents
a change. Below the map, there are spectrum lines
corresponding to that change.
The successful explanation for spectrum rules of hydrogen indicates that atomic system (microsystem)
obeys rules, which are different from those of classical physics, quantum rules.
Questions
1. State the two axioms of Bohr’s atomic structure and their consequence.
2. Use the Bohr model of atoms to explain the formation of line spectrum of the hydrogen atom.
3. Given the wavelengths of the four lines in Balmer series: red line Hα = 0.6563µm, blue line
Hβ = 0.4861µm, indigo line Hγ = 0.4340µm and purple line Hδ = 0.4102µm. Calculate wavelengths of
three spectrum lines of Paschen series in infrared region.
Hints: 3) .0939µm; 1.2811µm and 1.8744µm
SUMMARY OF CHAPTER VIII
1. The photoelectric phenomenon, luminescence of materials, photochemical reactions and formation of
line spectrum are phenomena involving the quantum property (particle property) of light.
2. Quantum hypothesis: atoms or molecules of matter do not absorb or radiate light continuously but in
separated and discrete packets. Every that packet carries a definite amount of energy equal to ε = hf, where
f is the frequency of light emitted by the atom, h is the Planck constant: h = 6.62×1034Js; every that small
light packet is called a light quantum or photon.
3. The photoelectric phenomenon is the phenomenon in which electrons are ejected from a metal surface,
when being illuminated with light having suitable wavelength. Translated by VNNTU – Dec. 2001 Page 128 The first photoelectric law: for every metal, the photoelectric phenomenon only takes place when the
wavelength of the trigger light is smaller than the photoelectric threshold λo of that metal (λ ≤ λo).
This law indicates the condition: hf ≥ A
where A is the work function of that metal.
The second photoelectric law: intensity of the optimum photoelectric current is proportional to intensity of
the trigger light beam.
The third photoelectric law: initial maximum kinetic energy of photoelectric electrons emitted from metal
is independent of intensity of the trigger light beam, and depends only on the wavelength of the trigger
light and nature of the metal used for cathode.
This law indicates the condition: hf = 2
mv 0 max
+A
2 (The Einstein’s formula of photoelectric phenomenon)
+ Photoelectric phenomenon is applied in photoelectric cells, tools that used to convert light signals to
electric signals.
4. The photoconducting phenomenon is the phenomenon in which a semiconductor’s resistor is reduced
tremendously when being illuminated.
In the photoconducting phenomenon, light releases binding electrons to become conducting electrons:
that is the inside photoconducting phenomenon. The photoconducting phenomenon is used in light
dependent resistors (LDR); the inside photoconducting phenomenon finds its application in photoelectric
cells.
5. The two axioms of Bohr:
+ Atoms only exist in states having definite energy, called the stationary states. In stationary states, atoms
do not radiate.
+ When each atom changes from a stationary state of energy Em to a stationary state of energy En, it will
radiate (or absorb) a photon carrying energy: ε = hfmn = Em – En
+ The important consequences deduced from the two axioms of Bohr are:
 In stationary states, electrons in atoms only move around a nucleus on the orbits, having definite radii,
called the stationary orbits.
 The Bohr model of atoms explained successfully the line spectrum formation of the hydrogen atom. Translated by VNNTU – Dec. 2001 Page 129 Part III. NUCLEAR PHYSICS Chapter IX – BASIC KNOWLEDGE ON THE ATOMIC NUCLEUS §54. STRUCTURE OF THE NUCLEUS. THE UNIT FOR ATOMIC
1. Structure of the nucleus
Rutherford’s experiments showed that, even an atom is very small (with a diameter of about 109m) it still
has a complex structure with a particle, namely the nucleus, at the centre and electrons surrounding. The
diameter of the nucleus is hundreds thousand times less than that of the atom, only about 1014 – 1015m.
The nucleus also has its own structure. Radioactivity and nuclear reaction phenomena showed that the
nucleus is composed of smaller particles called nucleons. Two types of nucleons are the proton (p) which
carries a positive unit charge +e, and the neutron (n) which has neutral charge.
If an element has an order number Z in the periodic table of the elements (Z is called the atomic number),
its atom has Z electrons in its outer shell, and its nucleus has Z protons and N neutrons. The electron shell
has a charge of –Ze, the nucleus has a charge of +Ze, thus the whole atom is neutral in charge. The sum of
Z + N, denoted as A, is called the mass number.
For example:
 hydrogen atom is the first atom in the periodic table, Z = 1, it has 1 electron in its outer shell, its nucleus
has 1 proton and 0 neutrons; its mass number A = 1.
 carbon atom with Z = 6 has 6 electrons, its nucleus has 6 protons and 6 neutrons, mass number A = 12.
 sodium atom with Z = 11 has 11 electrons, its nucleus has 11 protons and 12 neutrons, mass number A =
11 + 12 = 23.
An atom or its nucleus is represented by writing, on the left of the chemical symbol, the atomic number Z
(subscript) and the mass number A (superscript). So the above atoms can be represented as 1 H , 12 C ,
1
6
23
11 Na . Sometimes we can omit the subscript Z because the chemical symbol already determines Z. We
also can use the form C12, Na23, U235, etc. 2. Nuclear forces
Protons carry positive charge so the repulse each other. However, nucleus is still stable because between
nucleons (including both protons and neutrons), there exists a very strong attractive force called nuclear
force. Nuclear force is the strongest among the known forces, but it comes to effect only when the distance
between the two nucleons is less than or equal to the nuclear size, i.e. the range of nuclear force is about
1015m. 3. Isotopes
The atoms H, C and Na mentioned above are the most common in nature. There are other hydrogen atoms
which also have Z = 1, but its nucleus has either one or two neutrons.
The atoms of which nucleus has the same number of protons Z but different number of neutrons N (and so
different mass number A = Z + N) are called isotopes (share the same position in the periodic table). Translated by VNNTU – Dec. 2001 Page 130 Hydro has 3 isotopes: ordinary Hydro ( 1 H ), heavy hydro or deuterium ( 2 H or D), and super heavy hydro
1
1
or tritium ( 3 H or T).
1
Deuterium combines with oxygen and becomes heavy water D2O which is a key material for atomic
technology.
Most of the elements comprise of many isotopes.
For example:
 carbon has 4 isotopes with the number of neutrons ranging from 5 to 8 (A from 11 to 14), 2 of which are
3
stable ( 12 C and 16 C ). The natural abundance of the 12 C isotope is about 99%.
6
6
 tin (Z = 50) comprises of 10 isotopes with mass number from 112 to 122, and percentage from 0.4% to
33%. 4. The unified atomic mass unit
In atomic and nuclear physics the unified mass unit is used, called atomic mass, denoted as u, which is
equal to one twelfth of the mass of the most common isotope 12C. So sometimes this unit is also called the
Carbon unit.
The isotope 12C has 12 nucleons thus the mass of each nucleons is approximately u.
The exact mass of Proton is mp = 1.007276u, and neutron is mn = 1.008665u.
Comparing with the mass of electron mc = 0.000549u, we can see that the most of the atomic mass is
contained in the nucleus. Nuclear substance has very high mass density, about hundreds million tons/cm3.
The Avogadro number NA is the number of atoms in 12 grams of 12C.
NA = 6.022×1023/mol
So u = =1.66058×1027kg
An atom with mass number A has the mass of about A in u unit, because its nucleus has A nucleons. Mol
is a quantitative unit in SI system (the old name of mol is atoms gram or molecules gram). A mol of a
substance has NA molecules of that substance, or NA atoms if the substance is an element.
The atomic mass table in the reference books provides the mass of atoms in u unit, but taking into
consideration the percentage of isotopes in nature. For example C = 12.011, not C = 12 because in nature
carbon has 1% of isotope C13 which is heavier than C12.
For instance, looking up in the table we see He = 5, 1 mol of Helium has NA atoms and mass is NA4u.
But NAu = 1gram. So m = 4gr, that means the mass of 1 mol of an elemental substance in gram unit has
same value as in the atomic mass table.
Questions
1. What is the structure of nucleus?
2. Write the notation of the atoms whose nucleus comprises: 2p and 1n, 2p and 2n, 3p and 4n, 7p and 7n.
3. What are isotopes? Give examples.
4. Explain the structure of the nucleus of the following atoms: 16
8 O , 17 O , 18 O ,
8
8 235
92 U, 238
92 U 3
5. What is the atomic mass unit? Compare the mass of the following nuclei: D, T and 2 He . Translated by VNNTU – Dec. 2001 Page 131 6. Calculate: the number of atoms in 1gram of helium; the number of molecules in 1 gram of oxygen; the
number of oxygen atoms in 1 gram of carbonic.
Given He = 4.003, O = 15.999, C = 12.011.
3
Hints: 5) T is almost equal to 2 He , and is one and a half times of D. 6) 1.50×1023; 188×1020; 274×1020 §55. RADIOACTIVITY
During the middle age, the metallurgists put a lot of efforts looking for a method to change one element to
another, especially to gold. They were not successful then due to the lack of scientific and technological
bases. However, their hope is not totally vain. In fact, a nucleus is not invariable, but it can change to
another nucleus spontaneously (natural decay) or in nucleus reaction (artificial). 1. Radioactivity
Radioactivity is a phenomenon when a nucleus spontaneously emits
radiations, namely radioactive rays, and changes into another nucleus.
The radiation is invisible to the eyes but it causes physical and chemical
effect like ionizing the environment, darkening photographic plate,
causing chemical reaction, etc. If we direct the radiation to go through
the electric field between two plates of the capacitor (figure 9.1), we
can determine the nature of the radiation emitted by various radioactive
substances. There are three types of radiation.
a) Alpha rays: The first radioactive substance discovered by the French
scientist Becquerel was uranium (Z = 92), the radiation deflected
toward the negative plate of the capacitor, called alpha ray, denoted as
α. Figure 9.1 After more careful study, the alpha ray is then known to comprise of nuclei of the atom 4 He with two
2
positive unit charges, called alpha particles. The alpha particle is emitted from the nuclei with velocity of
about 107m/s, ionizing the environment and gradually losing its energy. The alpha particle can only travel
up to 8cm in the air and cannot penetrate through a thin glass.
b) Beta rays: two types: the more common one is β, in which emitted particles are electrons, so the β ray
is deflected toward the positive plate of the capacitor (figure 9.1). The deflection is more than α ray,
because the mass of electron is much smaller that that of alpha particle. The isotope 14 C is the radioactive
6
carbon, emits β.
Another type is β+, which is less common. The β+ particle is also called positive electrons or positrons
1
because it has the same mass as electron but carry a positive unit charge. The isotope 16 C is also the
radioactive carbon but emits β+.
The β particles are emitted with very high velocity, which can be near to the velocity of light.
The β ray also ionizes the environment, but weaker than the α ray, so range of the β ray is longer, maybe
up to a few hundred meters in the air.
c) Gamma rays: denoted γ, they are electromagnetic waves with very short wave length (less than
0.01nm), and also photon particles with very high energy. They are not deflected in electric field, has a
high penetrating power (it can penetrate through a lead wall of decimeters thick), and is very dangerous to
human. All radiations possess energy (e.g. the kinetic energy of particles, the energy of electromagnetic
waves), so the radioactivity emits energy, a part of this energy becomes heat energy that heat up the
container of radioactive substance.
Translated by VNNTU – Dec. 2001 Page 132 2. The radioactive decay law
The radioactivity phenomenon is caused by the internal structure of nuclei themselves, and totally
independent from external impacts. When the nuclei are in different conditions such as different
compound, different pressure and different temperature, the radioactive substances still decay without
being affected. They emit radioactive rays and change into other substances, obeying the following law
which is called the radioactive decay law.
Each of the radioactive substance is characterized by
a time T called the halflife so that after every period
of length T, half of a given number of the atoms of that
substance change into another substance.
Thus after the time T, 2T, 3T,…, kT (k is a positive
integer), the number of atoms in the radioactive
substance is N0/2, N0/4, N0/8..N0/2k, where N0 is the
initial number of atoms. We can use mathematics to
prove that the number of atoms N or the mass m of the
substance is an exponential function of time with
negative power (see figure 9.2). Figure 9.2 N = N0eλt (91) λt (92) m = m0e where m0 is the initial mass.
The coefficient λ is called the decay constant related to the halflife. Given t = T then m = m0/2, and
m
replace it into (92) we get: 0 = m0eλT. Thus:
2
T= Ln2 0.693
=
λ
λ (93) The radioactive substances have very different values of halflife. Uranium has T = 4.5×109 years, which
is a very slow decay process, so they still exist on earth until now. Radium has T = 106s, they changes into
another substance right after they are born.
The activity H of an amount of a radioactive substance is the quantity for radioactive power, measured by
the number of decays in 1 second. The unit of activity is becquerel (Bq), equal to 1 decay/ 1 second.
Another applicable unit is curie (Ci).
1 Ci = 3.7x1010Bq, it is the approximate the activity of 1g of radium, one of the firstly found radioactive
substances.
The activity H reduces with respect to time with the same law as the number of atoms N(t).
−dN(t)
= λN0eλt = λN(t), which means that the activity is equal to the multiplication of the
dt
number of atoms and the decay constant.
Really, H(t) = Thus if H0 is the initial activity, H0 = λM0, we have the law of reduction of activity:
H(t) = H0eλt (94) Questions
1. What is the radioactivity? List the radioactive rays and their natures. Translated by VNNTU – Dec. 2001 Page 133 2. What is the halflife of an radioactive substance? Write the equation describe radioactivity law.
3. Radioactive iodine
weeks? 131
53 I has a halflife of 8 days. If the initial amount is 100g, how many left after 8 4. The age of the earth is about 5×109 years. Assume that the uranium exists from the very beginning of
the earth. If the initial amount of uranium is 2.72kg, how many still left nowadays? The halflife of
Uranium is 4.5×109 years.
5. The radioactive polonium 210Po has the halflife of 138 days. Calculate the amount of Po which has the
activity of 1Ci.
Hints: 3)0.78g; 4) 1.26kg; 5) 0.222mg §56. NUCLEAR REACTIONS
1. Nuclear reactions
Nuclear reaction is the interactivity between two nuclei which then changes them into the other particles.
For example, two nuclei A and B interact with each other and become nuclei C and D. The equation of
this reaction is written in the form:
A+B→C+D (95) The newly formed particles can be simpler than nuclei, i.e. the primary particles such as nucleons,
electrons, photons, etc.
In nature, the atmosphere of the earth is bombarded by rays from the universe which always cause nuclear
reactions.
A particular case of nuclear reaction is radioactivity: the left side of the equation (95*) there is only one
particle A, called the parent nucleus
A→B+C (95*) If B is the new nucleus it is called the daughter nucleus. C will be alpha or beta particle. 2. Conservation laws in nuclear reactions
The following laws of conservation have been verified of their correctness with all nuclear reactions.
a) Conservation of the number of nucleons (mass number A): Proton can change into neutron and vice
versa, but the number of nucleons on the left hand and right hand of the equation (96) are always equal.
Conservation of the number of nucleons is also conservation of mass number A.
b) Conservation of charge (or atomic number Z): The nuclei during the reaction only interact with each
other, so they form a closed system, whose charge is isolated from outside. As we know, the charge in a
closed system is constant, which means the algebraic sum of the electric charges is constant. The sum of
charge (sum of atomic number Z) of the particles on the left and right hand of the equation is therefore
always equal.
c) Conservation of energy and conservation of linear momentum of the system of particles in the reaction:
the two laws of conservation observed in the macroscopic world are still applicable to the microscopic
world, which means they must be obeyed by a closed system of atoms, nuclei, etc.
In our scope we will just consider the first two laws a and b.
Please note that there is no law of conservation of the mass of the system. Translated by VNNTU – Dec. 2001 Page 134 3. Application of conservative laws to radioactivity. Transmutation rules
a) αdecay: Assume that the parent nucleus X has the atomic number Z and mass number A, denoted A X ,
Z
emits the α particle, which is the 4 He particle, and change into the daughter nucleus Y which has atomic
2
number Z’ and mass number A’, denoted as
A
Z X → 4 He +
2 A'
Z' Y . We have the equation: A'
Z' Y (96) From the law of conservation of mass number we have A’ = A – 4; from the law of conservation of charge
we have Z’ = Z – 2. It means that the daughter nucleus stands two cells before the parent nucleus in the
periodic table, and the mass number of the daughter nucleus is 4 units smaller.
b) βdecay: We have the equation:
A
Z X→ 0−
−1 e+ A
Z +1 Y (97) So the daughter nucleus is one cell after the parent nucleus and has the same mass number.
Studying the radioactivity β of bismuth, it can be seen that if the equation of the reaction is just (97)
210
83 Bi → 0−
−1 e+ 210
84 Po (polonium) then the energy is not conserved. Believing in the correctness of the principle of conservation of energy in
the microscopic world, in 1933 the Switzerland scientist Paoli proposed a hypothesis that there exists in β
decay a third particle, namely neutrino (symbol ν). Twenty years later, experiment verified this
hypothesis. The neutrino is electrically neutral, has little or no rest mass, and moving at velocity of light,
so they almost never interact with other matter and hardly detected.
The nature of β decay is that inside the nucleus, one neutron changes into one proton and one electron and
one neutron: n → p + e + ν.
c) β +decay:
We have the equation:
A
Z X→ 0+
−1 e+ A
Z −1 Y (98) In the periodic table the daughter nucleus stands one cell before the parent nucleus and has the same mass
number. The nature of this β+ decay is the change of proton into one neutron and one positron and one
neutron:
p → n + e+ + ν
d) γdecay: The daughter nucleus is born in the excited state and will change from the higher energy level
E2 to the lower energy level E1, simultaneously emits photon at frequency f determined by the relation: E2
 E1 = hf.
So the γ decay is the accompanying radiation with α and β decay. There is no change in the nucleus in the
γ decay.
A radioactive substance has only one of the radiations: α, β or β+, and maybe accompanied by γ radiation.
In figure 9.1, inside the container there are different radioactive substances, but they all maybe originate
from the same substance.
The nucleus also has discrete energy levels, similar to the energy levels of electron studied before, but the
distance between its levels is millions time larger, so photon γ emitted from the nucleus has very high
energy (with very high frequency f or very short wavelength) Translated by VNNTU – Dec. 2001 Page 135 Questions
1. State and explain: a) Laws of conservation in radioactivity. b) Transmutation rules in radioactivity.
2. Determine particle x in the following reactions:
19
9 F +p→ 25
12 Mg + x → 16
8 O +x
21
11 Na + α 3. Uranium decays into radium in the following chain of radioactivity:
238
92 β− α β− α α U → Th → Pa → I → Th → Ra Write the full chain (determine Z and A of the nuclei).
4. The above chain continues until the daughter nucleus becomes the stable isotope of
after how many α and β decays will 238
92 U become 206
82 206
82 Pb (lead). So Pb ? Hints: 4) 8 and 6. §57. ARTIFICIAL NUCLEAR REACTIONS. APPLICATIONS OF ISOTOPES
1. Artificial nuclear reactions
In nature there are nuclear reactions, which are natural nuclear reactions. Besides, human can also create
nuclear reactions, called artificial nuclear reactions, by using small particles to bombard the target nucleus.
The first artificial nuclear reaction was done by Rutherford in 1919. He used radioactive polonium 210
which emitted α particles, to bombard nitrogen. As a result, nitrogen changed to oxygen and proton:
4
2 He + 14
7 N→ 17
8 O + 1H
1 In 1934, Pierre and Marie Curie used α particles to bombard an aluminum sheet and got the following
equation:
4
2 He + 37
13 Al → 30
15 1
P + 0n The interesting thing is that the phosphor nuclei born from the experiment are not stable but have β+
radioactive attribute:
30
15 The 30
15 P→ 30
14 1
Si + 0 e+ P atom is called artificial radioactive isotope because it does not exist in nature. The natural 31
phosphor is the stable isotope P31 (i.e. 15 P ). The number of natural radioactive isotopes is only about 325,
but by using particleaccelerators, more than 1500 artificial radioactive isotopes can be created. Also by
using this method, they can extend the Mendeleyev periodic table and created the elements which stand
behind uranium with Z > 92. All of those elements are radioactive, and the most important one is
plutonium, Z = 94, used as nuclear fuel. 2. Particle accelerators
Using α particle to bombard only results in a small number of nuclear reactions as the particle’s velocity is
not high enough to overcome the repulsive Coulomb force of the nuclei which have more protons. We
need the particles which have velocity high enough to approach nucleus within the effective range of
nucleus force. Thus in nuclear physics there is a need of creating the devices which can accelerate charged Translated by VNNTU – Dec. 2001 Page 136 particles such as protons, α, ion, etc. Those devices are called particle accelerators. There are different
kinds of accelerators.
Cyclotron is the first evermade particle accelerator (1932). It comprises of two Dshape boxes put in
vacuum (figure 9.3). There is a magnetic field B perpendicular to the boxes. The Lorenz force causes the
charged particles (emitted from a source at the centre of the device) to move in circular motion inside the
boxes with radius:
R= mv
qB where M, q and v are the mass, charge and velocity of the
particle respectively.
We can prove (exercise 5) that the rotational frequency of the
particle is independent of its orbital radius or its velocity.
Between the two boxes there is an AC potential which has the
same frequency with the particle’s rotational frequency, so
every time the particle goes through the gap between the two
boxes, it is accelerated. Thus the radius R increases, and the
of the particle is in spiral form. At the end of the path, when
particles achieve a certain kinetic energy, they are directed to
bombard the target. orbit
Figure 9.3 Cyclotron can accelerate proton to the kinetic energy of tens million MeV. Above this limit, due to the
relative effect, the synchronism between the AC potential and the particle’s rotation ceases to exist, so
other types of accelerators must be used such as: Synchrocyclotron, Synchrophasotron, etc.
The modern particle accelerators are huge devices. The accelerator in Geneva, which has the radius of the
vacuumed box of 1.1km, can accelerate proton to the kinetic energy of 400GiV. Recently people has
managed to make an accelerator which have two clusters of particles accelerated separately which then
collide with each other and create the relative kinetic energy of about 1TeV (Tetra = a thousand billion).
With that device they even expect to break nucleon to study their internal structure.
3. Applications of radioactive isotopes
The natural and artificial radioactive isotopes have many applications in sciences as well as in daily life.
60
Cobalt 27 Co radiates the γ ray which has high penetration ability, and can be used to detect the defect in
machines and devices (same as using xray to get image of human organs), to preserve foods (because the
γ ray can free bacteria), to cure human from cancer, etc. To observe the movement of phosphate fertilizer
in a tree, people mix some phosphate P32 with normal phosphate P31. In the sense of plants physiology,
the two isotopes are totally similar because of their same electronics outer cell, but P32 is radioactive β,
so we can easily detect their movement, which is also the movement of the whole phosphate. That is the
method of marking atoms being used widely in science.
Archeology has a very accurate method to determine the age of organic samples, namely the carbon dating
method. Carbon has 4 isotopes: C12 (most common), C13 (stable), C14 (radioactive β) and C11
(radioactive β+). C14 is created in the atmosphere and then infiltrating into everything on Earth. Its halflife is 5600 years. The decay is balanced with the creation., so for tens thousand year the ratio of C13 in
the atmosphere remains unchanged. In every 1012 carbon atoms there is one C14 atom. While a plant is
still alive, it still exchanging with the environment thus still keep the above ratio of carbon. However, after
it dies, there is no more exchanging with environment, C14 only decays without being compensated back
so the ratio is reduced. After 5600 years only half is left, and the activity H reduces accordingly. We can
measure this activity to determine how much time has lapsed from the time when the plant died. For Translated by VNNTU – Dec. 2001 Page 137 example, measure the activity H of an ancient Egypt wooden plate, we have H = 0.15Bq. With the same
amount (mass) of wood taken from a living tree, we measure its activity and have H0 = 0.25Bq. As H =
0.0693
H0eλt with λ =
, we can deduce t = 4100 years. That is the age of the plate.
5600
Animals take plants for food so the ratio C14/C in their body also reduces as above after they die. Thus we
can apply the same method to determine the age of animal bones in archeological sites.
Questions
1. What are artificial radioactive isotopes?
2. What are the use of particle accelerators? Explain how cyclotron operates.
3. What is the marking atoms method?
4. Prove the formula 9.9.
5. If a particle (mass m, charge q) rotates many times on a circular orbit which is perpendicular to the
magnetic field with induction B, the rotating frequency is called cyclotron frequency. Calculate this
frequency: Does it depend on the orbit radius and the velocity of the particle?
6. Calculate the age of an ancient wooden statue knowing that its activity β is 0.77 that of a chunk of
wood, which has the same mass and has just been cut down.
Hints: 5) f = qB
; 6) ≈ 2,100 years.
2πm §58. EINSTEIN’S RELATION BETWEEN MASS AND ENERGY 1. Einstein’s axioms
According to the principle of Galilean relativity, all mechanics phenomena are the same in all inertial
systems of reference (systems that moves with constant velocity with respect to the Earth), or in other
words, all the laws of mechanics have the same mathematics forms in those systems. But how about
electromagnetic phenomena? Light is an electromagnetic wave. From the old point of view, light
propagates in a special medium which is absolutely at rest, namely ether. The Earth is in relative motion
with ether. Thus according to the addition law for velocities from classical physics, the speed of light
measured by the observer (standing on the earth) when the light traverses in opposite direction with the
rotation of the earth must be larger than that when the light traverses in the same direction with that
rotation. However, the Michelson’s experiment performed in 1881 with very high accuracy shows that the
speed of light, in any inertial system of reference, always has the same value.
In 1905 the Germany scientist Einstein (1879, 1955) abandoned the hypothesis of ether and proposed a
new theory which is called the Principle of Relativity. This theory has two parts. The first part which
considers only the inertial systems of reference, called the narrower theory of relativity, had been
completed. The second part considers any arbitrary system of reference, called the general theory of
relativity, was still being developed. Einstein proposed two axioms, called the axioms of the narrower
theory of relativity.
Axiom 1 generalizes the Galilean principle of relativity to all physics phenomena.
All the law of physics are the same in all inertial reference frames. In other words, their mathematics
forms are the same in all inertial reference frames.
Axiom 2.The speed of light in a vacuum has the same value c in all inertial reference frames, independent
of the speed of the light source or the observer.
Translated by VNNTU – Dec. 2001 Page 138 c = 299,792,458m/s ≈ 300,000km/s
Theory and practice also proved that: the speed of light c in a vacuum is the limit speed, no physical
objects can have velocity higher than c.
The second axiom simply denies the classical physics. If a source of light, which is traveling at the speed
of v relative to the observer, emits in the same direction the light at speed c, according to the classical
physics the speed of light observed by the observer should be (c + v). But according to the second axiom,
the speed is still c! Based on the 2 axioms, Einstein constructed the new mechanics called the relativity
mechanics, which has some rather bizarre consequences to classical mechanics, but they have been
verified in practice. Those results are only different from the classical results when the velocity is very
high, comparable to the speed of light. If the velocity is small , the equations in relativity mechanics can
be simplified to the equations in classical mechanics, so classical mechanics is only the special case of
relativity mechanics. In every day life we only deal with very small velocities compared to c, so classical
mechanics is still applicable. But in research of nuclear physics, astronautics, designing particle
accelerators, etc. we have to use the narrower theory of relativity (see exercise 5).
The quantum theory and the relativity theory have become two fundamental bases of the modern physics.
2. Einstein’s relation between mass and energy
Einstein’s theory of relativity stated a very important equation which relates the mass and energy of a
particle (or system of particles):
If the mass of a particle is m then the object has the energy M that proportionate with m, called the rest
energy:
E = mc2 (910) where c is the speed of light in vacuum.
From this relationship¸1 gram of any substance possesses a very large amount of energy, equivalent to 25
million kWh.
According to the theory of relativity, the rest energy can change to the usual energy, i.e. kinetic energy,
and vice versa. When the rest energy increases or decreases, the rest also increases or decreases obeying
the equation (910).
In classical physics, the energy and mass of the isolated system is conserved. But from the theory of
relativity, mass, and hence rest energy, may change; there is only the conservation law of total energy,
comprising of the usual energies and the rest energy.
From equation (910), in nuclear physics the mass of the particle is not only measured in kg unit, but also
in the energy unit divided by c2, i.e. eV/c2 or MeV/c2:
1 MeV
1.6022x10−13 J
=
= 1.7827×1030kg
c2
(2.99792x108 )2 (m / s) 2 and viceversa, 1 kg = 0.561x1030 MeV/c2.
For example, the mass of electron is me = 9.1095×1031kg = 0.511 MeV/c2.
Questions
1. State Einstein axioms
2. State Einstein’s relation between mass and energy. Show that the unit on both sides are the same.
3. Verify that every gram of any substance has the amount of rest energy equal to 25 million kWh.
4. Change to MeV/c2: atomic unit mass u, proton’s and neutron’s mass.
Translated by VNNTU – Dec. 2001 Page 139 5. The relativity effect is only considerable when the at velocity > 0.4c. Do we need to use the relativity
mechanics in the following situation
 the jet flying at 2500km/h;
 the spaceship flying at 360,000 km/h;
 proton moving inside the particle accelerator on a circular orbit of which radius is 100m, at a frequency
of 3×105 revolution per second.
Calculate the ratio v/c in each case.
6. Prove that the momentum of particle can be measured in the unit MeV/c. 1 SI unit of momentum is
equivalent to how many of this unit?
Hints: 4) 931MeV/c2, mp = 938.3MeV/c2, mn = 939.6MeV/c2; 5) 2.3×106; 3.3×104; 0.63 > 0.4;
6) 1kgms1 = 1.9×1021MeV/c. §59. THE LOSS OF MASS. NUCLEAR ENERGY
1. The loss of mass and binding energy
Assuming that Z protons and N neutrons at first are separated and at rest.
Their total mass is : m0 = Zmp + Nmn, where mp and mn are mass of protons and neutrons respectively.
When the nuclear force binds the nucleons together to form a nucleus with mass m, the interesting thing is
that m less than mo (there is no more conservation law of mass). But according to the principle of
relativity, the total energy of initial nucleons system is E0 = m0c2, and the energy of the bound system is E
= mc2 < E0. Energy must be conserved, thus there must be an amount of energy
ΔE = E0 – E = (m0  m)c2 released.
The difference ∆m = m0 – m is called the loss of mass, the equivalent energy ΔE = (m0 – m)c2 is called the
binding energy because of the reasons which will be explained later.
The energy is released in the form of kinetic energy of the nucleus or energy of a γ ray.
Contrarily, to break the nucleus of mass m to nucleons with total mass m0 > m, we have to provide an
amount of energy ΔE = (m0 – m)c2 to overcome the nucleus force. The higher of ΔE, the stronger of the
binding between nucleons, the higher energy we need to break the bond, so ΔE is called the binding
energy. The binding energy per nucleons is ΔE/A.
The higher of the binding energy per nucleons that a nucleus has, the more stable it is. 2. Exothermic and endothermic nuclear reactions
The loss of mass of each nucleus is followed by the nonconservation of
mass in nuclear reaction. Consider the following nuclear reaction:
A+B→C+D
And suppose that the A and B are at rest. The number of nucleons is
unchanged in the reaction, but A, B, C, D have different loss of mass so
the total mass Mo of the nuclei A + B can be different from the total
mass M of the nuclei C + D.
Figure 9.4
Assuming M < M0. The initial system has the rest energy E0 = M0c2,
2
after the reaction the rest energy E = Mc . The total energy must be conserved, so the reaction must
release an amount of energy ΔE = (M0 – M)c2, in the form of kinetic energy of C and D particles, or
Translated by VNNTU – Dec. 2001 Page 140 photon γ. M is smaller than M0 because the created nuclei have higher loss of mass compared with the
initial nuclei, which means more stable (figure 9.4).
In short, a reaction in which the newly created particles have total mass smaller than that of the initial
particles, which means more stable, is an exothermic reaction.
In contrast, if M > M0, and E > E0, the reaction cannot happen by itself,
but it requires an amount of energy W to be provided to its particles A
and B, maybe in the form of kinetic energy of A (bombard B by A).
W is larger than ΔE = E – E0, because the created particles have kinetic
energy Wd
W = ΔE + Wd (figure 9.5)
So a reaction in which the created particles have total mass larger than
that of the initial particles (thus less stable) is an endothermic reaction. Figure 9.5 Sometimes it is said that in exothermic reactions, the loss of mass (M0 – M) “changes” into the released energy (M0 M)c2 = ΔE, and in endothermic reactions, a part of the delivered energy ΔE “changes” into the increase in mass: ∆E
= M  M0
C2
This way of saying helps us to memorize the result of the reaction easier, but in fact it is incorrect because mass and
energy are two different quantities, they are in proportion, but cannot change into each other. 3. Two exothermic nuclear reactions
There might be two nuclear reactions which emits energy, namely nuclear energy.
a. When a heavy nucleus like uranium, plutonium, etc. absorbs a neutron and splits into two roughly equal
pieces. This phenomenon is named nuclear fission.
b. When two small nuclei like hydrogen, helium combine and build a larger nucleus. This phenomenon is
named nuclear fusion.
These two nuclear reactions will be discussed in details in next chapters.
Questions
1. What are the loss of mass and binding energy of nucleus? How are they related to the stability of
nucleus?
2. What are the conditions for a reaction to be exothermic or endothermic? Why the natural decay must be
exothermic?
3. What types of reactions that can be exothermic?
∇ For the following exercises, it is given: u =1.66x1027kg; mp = 1.0073u; mn = 1.0087u;
NA = 6.02×1023mol1. 4. Deuterium nucleus ( 2 H or D) has the mass of 2.013i. Calculate its binding energy.
1
5. The α particle has the mass of 4.0015u. Calculate the amount of energy released when 1 mol of helium
is created.
6. Consider the nuclear reaction in which aluminum is bombarded by α particles:
27
13 Al + α → 30
15 Translated by VNNTU – Dec. 2001 P +n Page 141 Knowing the masses of the nuclei are mAl = 26.974u; mp = 29.970u; mα = 4.0015u, calculate the minimum
energy of α particles required for the reaction to happen. Neglect the kinetic energy of the created
particles.
Hints: 4) 2.3Mev; 5) 2.7×1012J; 6) 3Mev. §60. NUCLEAR FISSION. NUCLEAR REACTION PLANTS
Nuclear fission is one of the two exothermic nuclear reactions observed before the second World War by
Otto Hahn in 1938. During the World War, some top secret researches were carried out to make use of
those energy for the purpose of war (making nuclear bomb). After the World War, those energy have been
used for peaceful purposes (like producing electricity in nuclear reaction plants). 1. Chain nuclear reactions
Nuclear fission occurs when a heavy nucleus absorbs a neutron and then splits (fissions) into two smaller
nuclei. Slow neutrons, of which the kinetic energy is equivalent to the average kinetic energy of thermal
motion (less than 0.1eV), are more easily absorbed than fast neutrons.
Among the isotopes that are fissionable when absorbing slow neutron, we will focus our interest on the
natural isotope uranium 235 and the artificial isotope plutonium 239. Natural uranium is the combination
of 3 isotopes, in which U 238 accounts for 99.27%, U 235: 0.72% and U 234: 0.01%, Among which,
isotope U235 is easiest to fissions. The fission equation of U235 is
235
92 1
U + 0n → 236
92 U → AX +
Z A'
Z' 1
X ' + k 0 n + 200MeV X and X’ are the average nuclei, with the mass number
from 80 to 160. The process results in the production
of k = 2 or 3 (on average 2.5) neutrons, and release the
energy of about 200MeV in the form of kinetic energy
of the particles. They are the two very important
points.
1) A large number of the production neutrons are lost
because of some reasons (leak out of the uranium core,
absorbed by other nuclei,,,) But after every fission,
there still exists s neutrons averagely, and s > 1, then
those s neutrons collide with the nuclei of other U235
Figure 9.6
and causing s other fissions, produce s2 neutrons and
34
then s , s , etc. neutrons. The number of fissions increases sharply within a very short period: we have a
chain nuclear reaction; s is called the neutron reproduction constant K.
Figure 9.6 illustrates the case when s = 2.
2) Each fission only releases an amount of 200MeV = 3.2x1011J of energy, but 1 gram of U 235 contains
2.5x1021 nuclei, so the total energy is very high, equal 8x1010J and equivalent to 220000kWh. This fission
energy is called incorrectly as nuclear energy.
When s < 1: the reactor is said to be supercritical: the chain nuclear reaction is uncontrollable, the
released energy is very devastating. This case is used in making nuclear bomb.
When s = 1: the reactor is said to be critical: the chain reaction is selfsustained without increasing, the
energy released is unchanged and controllable. That is the case of nuclear reactors in the nuclear reaction
electric plants.
When s < 1: the reactor is subcritical and the reaction dies out, the chain reaction does not occur.
Translated by VNNTU – Dec. 2001 Page 142 The percentage between the leaking out neutrons (in proportion with the outer surface area of the uranium
core), and the production neutrons (in proportion with the volume of the core) is smaller if the mass of the
uranium is larger. This mass must reach a minimum value, namely the critical mass m, in order to have s
≥1. The nuclear bomb in Hiroshima in 1945 containing pure uranium 235 of mass mh ≈ 50kg. At first, the
uranium was divided to two separated cores, each has the mass less than mh so that it did not explode.
Combining the two cores, the mass exceeded mh hence the bomb exploded.
The isotope U238 is also a nuclear material because when it is bombarded by a neutron, it changes to
Pu239 which is a fission substance
238
92 1
U + 0n → 239
92 β− U→ 239
93 β− Np → 239
94 Np The process to purify U235 is expensive, therefore in nuclear reactors it is widely used the enriched
natural uranium in which the percentage of U235 is increased to several or tens. The critical mass is
from tons to tens of tons depending on the enrichment level. 2. Nuclear reaction plants
The main part of a nuclear reaction plant is
a nuclear reactor, where the chain fission
reaction is controlled at the critical level
(figure 9.7).
Inside the reactor there are pieces of alloy of
enriched uranium (A) which are placed in
the moderator substance B (heavy water
D2O, beryllium, etc). Neutrons produced
from the fission are fast neutrons; they
collided with the nuclei of the moderator,
gradually losing kinetic energy and become
slow neutrons which are easily absorbed by
Figure 9.7
uranium. The reactor also have control rods
C made of materials, such as cadmium, that
are efficient in absorbing neutrons. When those rods are adjusted to the lower position, the reproduction
constant s decreases, when they are adjusted higher, s increases. When the reactor is running, the rods are
adjusted automatically so that s = 1. The fission reaction release energy in the form of kinetic energy of
nuclei fragments and other particles, which then becomes heat energy of the reactor. The heat energy is
then carried out by the heat carrying substance, usually a liquid, that runs through the reactor, and then
provides heat for the steam engine D. The steam runs the turbine as in usual normal thermoelectricity
power plant. Although both of the plants use the collected heat to run the turbine, the nuclear reaction
plant must have a special design, because there must be safety measures such as steel and concrete wall to
resist the radioactive rays that are harmful to human, and other special devices to keep the reactor from
being supercritical.
Many nuclear reaction plants have been built in developed countries and provided considerable amount of
electricity: more than 35% total amount of electricity produced annually in France, Swede,. 30% in Japan,
12% in the United States, 7% in the former Soviet Union, etc. However, the breakdown of Chernobyl
nuclear power plant made some countries to reconsider in building nuclear power plants.
In Vietnam there is a small nuclear reactor in Da Lat, used for research purpose and to produce radioactive
isotopes (power of 500kW, with 89 pieces of alloys of enriched uranium (35% U235)). Translated by VNNTU – Dec. 2001 Page 143 Nuclear reactors have also been placed on ships, submarines. Once refueled, they can run continuously in
several years. Currently people are doing research on reducing the weight of the reactors so that they can
be put on airplanes.
Questions
1. What is nuclear fission and its characteristics?
2. What is chain nuclear reaction? Under which conditions will it happen?
35
1
95
3. One fission reaction of uranium 235 is 292 U + 0 n → 42 Mo + 139 La +2n + 7 e − , where Mo is
57
molybdenum, La is lanthanum. Knowing that the nucleus mass mU = 234.99u; mMo = 94.88u; mLa =
138.87u. Ignoring the mass of electrons. a) Calculate in MeV the energy released by one fission.
b) U235 can split in different ways, if we take the result from part a, to be the average value of energy
released in one fission, then what is the total energy released by 1 gram of U235? Calculate the equivalent
required amount of gasoline, knowing that the thermal efficiency of gasoline is 46x106 J/kg.
Hints: 3) 215MeV; 88×109 J; 1.9 tons. §61. THERMONUCLEAR REACTION
Another exothermic nuclear reaction is the combination of two light nuclei to form a heavier nucleus. For
examples, the reaction when isotopes of hydrogen (deuterium or tritium) combine.
2
1 3
1
H + 2 H → 2 He + 0 n + 3.25MeV
1 2
1 1
H + 3 H → 4 He + 0 n + 17.6MeV
1
2 Although one fusion reaction relation releases less energy than a fission reaction, but with respect to the
mass of fuel, fusion reaction releases more power.
However, it is very hard for fusion reaction to occur because of the repulsive force between positive
charge nuclei. It has been calculated that only when hydrogen is raised to the temperature of 50 – 100
million degrees then the nuclei have enough kinetic energy to overcome the repulsive Coulomb force, go
very near to each other and hence combined by the nucleus force. Because the combination reaction only
occurs at extremely high temperature, it is called the thermonuclear reaction.
Long ago, people had been trying to discover the source of energy of the sun and other stars. The sun
endlessly releases a very large amount of energy, with radiation power up to 3.8x1026W. Nowadays it is
believed that the fusion reaction is the source of that energy , because there exists a very high temperature
in the sun’s interior, which allows the reactions to occur.
An American scientist, Bethe, proposed a chain of combination reaction , namely the carbonnitrogen
cycle, which includes 6 continuous reactions, with carbon and nitrogen as catalysts. But in the end the
whole cycle can be shorten to the creation of helium nucleus from hydrogen nuclei.
4 1 H → 4 He + 2 0 e + + 26.8MeV
1
2
1
The whole cycle takes tens million year to complete, but each reaction occurs continuously, and this cycle
provide part of the energy for the sun (besides other cycles). The sun is losing its energy due to radiation,
hence according to Einstein’s relation, its mass reduces incessantly. However, the mass of the sun is very
large, so the reduction is only considerable after a few million years. Translated by VNNTU – Dec. 2001 Page 144 People now are able to create thermonuclear fusion reaction in an uncontrollable manner; that is the
explosion of hydrogen bomb (or fusion bomb). The bomb contains a mixture of deuterium and tritium,
and a nuclear bomb as the detonator. First the nuclear bomb explodes, generate the temperature of
hundreds million degree, and then the fusion reaction occurs, its energy added to the fission energy, so
hydrogen bomb has extremely devastating potential (equivalent to tens million tons of the usual explosive
TNT).
One of the most important target of the modern physics is to do the fusion reaction in a controllable
manner., with small amount of fuel, and releases limitary energy which can be used for peaceful purposes.
If it can be done, human will no more worry about the inadequacy of energy because fusion fuel is almost
unlimited: in the water from rivers or oceans there always exists 0.015% heavy water D2O from which
deuterium can be extracted. Tritium can be taken from lithium 6 Li . Ecologically, fusion reaction is also
3
“cleaner” than fission reaction, because there are less radiations or radioactive disposals which pollute the
environment.
The developed countries are putting much efforts in the search for controllable fusion reaction, but there
are still a lot of difficulties to overcome.
Questions
1. What is the nuclei combination reaction? Why it is also called thermonuclear fusion reaction?
2. Where does the sun’s energy come from?
3. Have people managed to create fusion reaction?
4. Consider the combination reaction: D + D → T + p. Knowing the mass of the nuclei:
mD = 2.0136u; mT = 3.016u; mP = 1.0073u.
a) Calculate the energy released by one reaction
b) Calculate the energy that can be generated from 1kg water if all the extracted deuterium as nuclear fuel.
How much gasoline required to create that amount of energy (thermal efficiency of gasoline is
46x106J/kg)?
Hints: 4) a) 3.7MeV; b) 2.7×109J; 58kg. SUPPLEMENTAL READING: PRIMARY PARTICLES 1. Properties of the primary particles:
During the study of the structure of matter, there are smaller and smaller particles discovered: molecule,
atom, nucleus, electron, nucleon, etc. Conventionally, the particles that are smaller than nucleus are called
primary particles. Thus electron, nucleon, are primary particles. The primary particles are not the
elementary ones that construct matter but rather, they are just the current limitation of the experimental
discovery of small particles. There are theoretical bases to assert that, for example, nucleon has a very
complicated structure.
Primary particles have the following main properties:
a) Mass: Only photon is massless (however they still have momentum, a quantum property of photons).
Neutrinos also have mass approximate 0.
b) Charge: equal to ±e. e is the unit charge (exceptionally, some equal to ±2e). Those neutral in charge
like neutrons, neutrinos, etc. do not have charge. Translated by VNNTU – Dec. 2001 Page 145 c) Spin is another quantum property which is similar to angular momentum of the selfspin of the particle
d) Magnetic moment is a quantum property of magnetism.
e) If the particle is unstable, the halflife T is then another property. For example: neutrons outside of
nucleus are unstable, with T = 638s. 2. Antiparticles. Antimatter.
Each particle, except photon, always has its antiparticle; they have the same mass, spin and halflife (if
they are unstable) but opposite charge and magnetic moment.
A particle and its antiparticle can annihilate each other and produce photons, that is the pairannihilation
phenomenon. For example electron and positron:
e + e+ → γ
Under certain conditions, there may be a process in which a photon changes into an particle and an
antiparticle, which is called the pair production. For example:
γ → e + e+
Antiparticles can form antimatter. For example, the hydrogen antiatom consists of one positron rotating
around one antiproton. Our galaxy is mainly composed of matter, thus antiparticles are rare. But in the
universe, there may exist some antigalaxies which is composed of antimatters. Some discovered “points”
in the universe, namely quasars, very far from us, are emitting energy which is tens time stronger than that
from the strongest galaxies. There is hypothesis that quasar is the result of a pairannihilation between a
galaxy and an antigalaxy. 3. Fundamental interactions. Classification of primary particles.
The primary particles can be divided into two categories: substance particles and field particles. In other
words, matter has two forms: substance and field. Substance particles are the usual particles, which
interact with each other through fields. For example the charged particles like electron, proton interacts
with each other through the electromagnetic field. Fields are both wavelike and particlelike (the quantum
property of field). For example, the electromagnetic field is both an electromagnetic wave as well as a
stream of photons. The photon is hence the substance particle of electromagnetic field.
There are four fundamental interactions between particles, in the form of four types of field. Besides
electromagnetic interaction, we have also studied the gravitational interaction between the particles
which have mass (according to Newton’s universal law of gravity, or more precisely, Einstein’s general
theory of relativity). We have also mentioned the interaction between nucleons (nucleus force), which is
called the strong interaction.
The interaction which make particles, except photons, to decay, is called the weak interaction.
The strength of the each interaction is specified by a coefficient α, namely relative strength of
interactions. The following table shows those relative strength in the descending order. There is also the
corresponding range of force and field particle.
Interaction
1) Strong
2) Electromagnetic Relative strength Range of Force (m) Mediating field Particle 1 1015 Gluon (presumed) ∞ Photon 10 2 10 3) Weak 10 4) Gravitational 1038 Translated by VNNTU – Dec. 2001 10 18 ∞ W±, Z0
Graviton (presumed) Page 146 Based on the properties of the particles, especially mass, we can divide the substance particles into two
smaller categories:
a) Hadrons, meaning big particles, include baryons (meaning heavy particles) and mesons (average
particles). Baryons include nucleons (protons and neutrons) and hyperons (super heavy particles).
b) Leptons, light (or small) particles, include electrons and neutrinos. 4. Quarks
Including the antiparticles, the total number of discovered primary particles so far is nearly 400! The
physicists are trying to reduce this number, by looking for the smaller particles that build primary
particles. In 1961, an American physicist GellMann proposed a hypothesis that hadrons are composed of
smaller particles named quarks. There are 6 types of quarks, denoted as u, d, s, c, b, t, and 6
corresponding antiquarks. For example protons are composed of 3 quarks (u, u, d). Oddly, the charge of
quarks is ±e/3 or ±2e/3, which goes against the perception that e is the unit charge, or the smallest charge.
Several experiments have been conducted to discover quarks but so far they are still unsuccessful, and it is
concluded that quarks do not exist independently, but only inside hadrons. However, the hypothesis of
quarks has achieved some great successes, such as the deduction of existence of the Ω  particle. This
particle was discovered in 1964 with all properties as expected.
Until now, most of the physicists have agreed with the hypothesis of quarks, and it can be said that the real
primary particles comprises only the 6 quarks, leptons (electrons and neutrinos), and the field particles.
However, the discovery of the construct of matter is an endless practice, as Lenin wrote in 1908: “electron
is as endless as atoms, nature is endless…”. SUMMARY OF CHAPTER IX
1. Nucleus is composed of nucleons: protons which carry a positive charge unit, and neutrons which are
neutral in charge, combine with each other by a very strong but short effective distance nucleus force.
A nucleus which has atomic number Z contains Z protons and N neutron: A = Z +N is called mass
number. The atoms of which nucleus has the same number of protons Z but different number of neutrons
N are called isotopes.
The atomic mass u is equal to one twelfth of the mass of the isotope 12C; u is approximate the mass of a
nucleon, thus a nucleus of mass number A will have a mass approximate to Au.
2. Nucleus spontaneously emits radiation and changes into another nucleus. There are different types of
radiation: α, β, β+, γ. The α particle is the nucleus of helium 4 He .
2
β particle is the electron, denoted as e, β+ is a positive electron or positron e+.
γ ray is an electromagnetic wave with very short wavelength.
The halflife T of an radioactive substance is the time after which the number of atoms in the given
amount of that substance is equal to a half of the initial number of atoms No. The number of atoms N
reduces exponentially with respect to t: N = N0eλt.
The decay constant λ is adversely proportion to the halflife: λ = Translated by VNNTU – Dec. 2001 0.693
T Page 147 In α decay, the daughter nucleus stands two cells before the parent nucleus in the periodic table; in β  or
β+decay it stands one cell before or after. There is no change in the nucleus in the γ decay, but there is
shifting to lower energy level.
3. Nuclear reaction is the interactivity between nuclei which then changes them into the other particles;
they can be nature (e.g. the decay) or artificial.
In a nuclear reaction, the following quantities are conserved: number of nucleons, charge, energy and
momentum . Mass may or may not be conserved.
4. Classical mechanics is only applicable when the velocity v of the objects are small compared with the
speed of light in vacuum (c ≈ 300000km/s)
If v > 0.4 c, the narrower theory of relativity must be used, and relativity mechanics in particular. The
theory is based on two axioms of Einstein:
a) All the law of physics are the same in all inertial reference frames.
b) The speed of light in a vacuum has the same value c in all inertial reference frames, c is the limit of all
physics speed.
One result of the narrower theory of relativity I s the relation between mass m and energy E of an object:
E = mc2.
5. The mass of the nucleus which is composed of some nucleons is less than the total mass of the
nucleons. The difference ∆m is called the loss of mass. The equivalent released energy ΔE = ∆mc2 is
called the binding energy because it is the energy required to break the nucleus to nucleons. The higher of
ΔE, the more stable the nucleus is.
In the nuclear reaction, the total mass of the initial particles Mo may be different from the total mass of the
created particles M. If Mo > M the reaction is exothermic.
If Mo < M , the reaction only occurs if the initial particles are provided energy.
6. There are two types of exothermic nuclear reaction, namely nuclear fuel.
a) Nuclear fission occurs when a heavy nucleus absorbs a neutron and then splits into two smaller nuclei,
with 23 neutrons. If the fission has a chain property, it releases a very large amount of energy. The
reaction is controlled in the nuclear reactor.
b) Two light nuclei such as isotopes of hydrogen (deuterium or tritium) combine to form a heavier
nucleus. This reaction only happens in extremely high temperature, so it is called thermonuclear fission
reaction. This reaction has only been made in an uncontrollable manner (hydrogen bomb).
Question
63
1. The isotope 29 Cu has the radius of 4.8 fermi (1 fermi = 1015m). Calculate the mass density of the
nucleus of the isotope and compare that with the mass density of Cu 8.9 g/cm3. Calculate the charge
density of that nucleus. Taking the unit charge e = 1.6×1019C. 2. Radius R of a nucleus increases with mass number A following an approximate relation R = R0A1/3,
with R0 = 1.2 fermi.
a) Compare the radius of nucleus 1 H and
1 238
93 U. b) Prove that the mass density of the nucleus approximates a constant value. Translated by VNNTU – Dec. 2001 Page 148 3. A radio active substance with the halflife T = 10s initially has the activity H0 = 2×107Bq. Calculate the
decay constant λ, the initial number of atoms, the number of atoms left and the activity after 30 seconds.
4. Uranium U238 changes to lead after a chain of α and β decay:
238
92 U→ 206
83 Pb + 8α + 6e The halflife of this conversion is 4.6×109 years. Assuming initially the rock contained only uranium, no
m(U)
lead. If currently the ratio between uranium and lead in that stone is
=37, how old is that rock?
m(Pb)
5. The particular binding energy is the binding energy of 1 nucleon.
a) Calculate the binding energy of an α particle, knowing the mass ma = 4.0015u; mp = 1.0073u;
mn = 1.0087u.
b) Calculate the energy released when 1 gram of helium is created.
Taking NA = 6.022×1023 mol1; He = 4.003
6. The fission of uranium U235 can be in many ways. One of the possible reaction is:
235
92 1
U + 0n → 140
58 Ce + 93
41 Nb + 3n + 7 e− The particular binding energy of U235 is 7.7MeV, of Ce140 is 8.43MeV, of Nb93 is 8.7Mev. Calculate
the energy released in this fission.
(Ce is cerium used to make sparking stone. Nb is niobium used to create retard alloy. The particular
binding energy is the energy calculated for 1 nucleon).
7. The sun has the mass of 2×1030 kg and radiation power of 3.8×1026W
a) After every second how much weight does the Sun lose?
b) If the radiation power is unchanged, then one billion year later, what is the percentage between the lost
weight and the current weight?
c) Assuming that every second in the Sun there are 200 million tons of helium being created as the result
of the carbonnitrogen cycle. What is the percentage of the contribution of this cycle to the total radiation
power?
Hints: 1) 2.26×1014g/cm3; 1019C/cm3; 3) λ = 0.0693s1; N0 = 2.9×108; N = N0/8; H = H0/8; 4) 2×108 years;
5) 7.1MeV; 68×1010J; 6) 182.6MeV; 7) a) 4.2 million tons; b) 0.007%; c) ≈34%. Translated by VNNTU – Dec. 2001 Page 149 Part IV. PRACTICAL EXPERIMENTAL EXERCISES EXPERIMENTAL EXERCISE 1 – CLARIFICATION OF THE LAW ON THE SIMPLE PENDULUM’S
OSCILLATION. DETERMINATION OF THE GRAVITY ACCELERATION.
I  Purpose
+ Using experiments to clarify the formula for determining the oscillation period of a simple pendulum in
l
.
the condition of small amplitude: T = 2π
g
+ From which, determine the gravity acceleration g at the experimental location.
II  Theoretical preparation
1) Is it possible to determine the oscillation period T of a simple pendulum with an error ∆T ≤ 0.02s by
using a clock/watch with a second hand? If yes, describe the procedure.
2) Describe experimental procedures to clarify that the oscillation period T is proportional to l. 3) When using a simple pendulum to determine the gravity acceleration g based on the formula g
∆g
∆π ∆l
∆T
l
=2
+ +2
. To have the value of g with an
= 4π2 2 , the following relative error occurs:
π
g
l
T
T
error not larger than 5%, how should the experimental conditions be selected?
III  List of required accessories
1) A small ball (or a similar mass) connected to a 1m string to make a simple pendulum; 2) a 500mm rule;
3) a rid to hang the pendulum; 4) a clock/watch with a second hand.
IV  Experimental procedures
1) Clarifying the formula for determining the oscillation period of a simple pendulum in the condition of
small amplitude:
a) Hang a pendulum with length l1 = 80cm to the experimental rid. The value of l1 is measured from the
hanging point to the centre point of the ball, with an error of 1mm. Lift the ball to make a slope angle
of 7o to the vertical (about 10cm from the balance position), then drop the ball. Measure the time
interval t1 in which the pendulum carries out 50 complete oscillations. Calculate the period
T1 = t1 50 . Calculate the absolute error ∆T1 = ∆t1 50 .
Repeat the above procedure with a slope angle smaller than 7o. Measure the time interval t '1 in which
the pendulum carries out 40 complete oscillations. Calculate the period T '1 = t '1 50 . Calculate the
absolute error ∆T '1 = ∆t '1 50 .
Compare T1 and T '1 . Draw appropriate conclusions.
b) Repeat the above procedure with different pendulum lengths: l2 = 60cm, l3 = 40cm and arbitrary l4.
2 T l
l
Determine the corresponding values T2, T3 and T4. Compare the following ratios: 2 and 2 ; 3
l1 T1 l1
2 2 T l
T and 3 ; 4 and 4 . Draw appropriate conclusions. T1 l1 T1 Translated by VNNTU – Dec. 2001 Page 150 2) Determining the gravity acceleration g at the experimental location:
a) Based on measured values l1, T1:
 Calculate g1 = 4π2
 Calculate l1
= ...
2
T1 ∆g1
∆π ∆l1
∆T
=2
+
+ 2 1 = ...
π
l1
g1
T1  Calculate ∆g1 = g1 ∆g1
= ...
g1 Note: If π = 3.142 then ∆T
∆π
2
∆l
≈
is quite smaller than 1 and 2 1 , and therefore ignorable.
T1
π 3.142
l1  Write out the result obtained: g1 = … ± … (m/s2)
∆g 2
and ∆g2.
g2
Write out the result obtained: g2 = … ± … (m/s2)
∆g1
∆g 2
and
.
Compare g1 and g2, and compare
g2
g1 b) Based on measured values l2, T2 to calculate g2, c) Which result should be selected? Why?
To increase the accuracy of measuring g, what should you do?
V  Form of the experimental report
1) Purpose
2) Results
A – Measuring results:
l ± ∆l t ± ∆t T ± ∆T Setup 1
Setup 2
Setup 3
Setup 4
B – Conclusions:
a) Conclusion from the comparison between T1 and T2 (setup 1 and 2):
b) Calculation results from setup 1 to 4:
l3
=
l1 l2
=
l1
2 T2 = T1 l4
=
l1
2 T3 = T1 2 T4 = T1 The corresponding conclusion drawn:
C – Determining the gravity acceleration: Translated by VNNTU – Dec. 2001 Page 151 l1
= ... (m/s2)
2
T1
∆g1
∆π ∆l1
∆T
=2
+
+ 2 1 = ...
π
l1
g1
T1 a) g1 = 4π2 ∆g1 = g1 ∆g1
= ... (m/s2)
g1 Result: g1 = ± (m/s2) ± (m/s2) b) Similarly for g2.
Result: g2 = c) The result ……………….. should be selected since ………….
To increase the accuracy of measuring g, we can ……………….. EXPERIMENTAL EXERCISE 2 – DETERMINATION OF SOUND WAVELENGTHS AND
FREQUENCIES
I  Purpose
+ To make the resonance between the oscillation of air inside a tube and the oscillation of a tuning fork.
+ Based on the conditions of resonance (standing wave) to determine the sound wavelength of the tuning
fork’s oscillation, from which determine the sound frequency.
II  Theoretical preparation
1) Present the equation for calculating wavelength λ of a sound wave with frequency f that travels with
speed v in the air in temperature t.
Hint: In the air with temperature t, the sound travels with a speed v = v0 1 + 0.0037t m/s. In this
experimental exercise the value of sound speed is approximate v = 332 1 + 0.004t ± 3 m/s.
2) Figure 1a describes the standing wave phenomenon inside a tube when a tuning fork that is placed in
one end oscillates.
It shows that the resonance happens when this end corresponds to a standing wave’s node, i.e. when the
λ
λ
λ
lengths of air columns are l = ; l’ = 3 ; l” = 5 ; ... From this characteristic, introduce a method to
4
4
4
determine the sound wavelength and frequency. Figure 1a Translated by VNNTU – Dec. 2001 Page 152 III  List of required accessories
1) A cylinder tube with length 60cm, diameter 4 – 5cm, having a piston that is movable along the tube’s
axis; 2) a tuning fork la; 3) a rubber hammer; 4) a ruler; 5) a thermometer.
IV  Experimental procedures
1) Put the piston inside the tube and move it to the middle of the tube. Place the tube horizontally on the
experimental table. Knock on the tuning fork using the hammer to, then bring the fork near the open end
of the tube so that both two ends of the fork lie on the tube’s axis (figure 1b).
Move the piston from the open end toward the middle of the tube. Note the position of piston where the
sound heard is loudest (corresponding to the resonant position). Measure the length of air column l (from
the open end of the tube to the noted point).
Repeat this procedure 5 times, take the average value of results to have the value of l and the deviation ∆l. Figure 1b
2) Move the piston toward the far end of the tube. Note the position of piston where the sound heard is
loudest. Determine the length l’ from the open end of the tube to the noted point following the procedure
similar to that one stated in (1).
3) The difference between l and l’ equals the distance between two successive nodes of the standing wave,
i.e. equals a half of the sound wavelength λ:
l’ – l = λ
2 Calculate the average value of λ. Determine the absolute error ∆λ and the relative error ∆λ
.
λ 4) Measure the indoor temperature by using the thermometer. Calculate the sound speed in air:
v = 332 1 + 0.004t (m/s)
5) Determine the sound frequency f =
Determine v (m/s)
(Hz).
λ (m) ∆f ∆v ∆λ
∆v
1
=
+
(approximately
=
)
f
v
λ
v 100 Determine ∆f and write out the result f = … ± … (Hz)
V  Form of the experimental report
1) Purpose
2) Results
a – Determine the length of air column corresponding to first resonance:
l1 (cm) l2 (cm) Translated by VNNTU – Dec. 2001 l3 (cm) l4 (cm) l5 (cm) Page 153 l= l1 + l2 + l3 + l4 + l5
= … (cm)
5 ∆l = (lmax − l ) + (l − l min ) l max − lmin
=
= … (cm)
2
2 Note: lmax and lmin are the maximum and minimum values of l.
b – Determine the length of air column corresponding to second resonance:
l’2 (cm) l’1 (cm)
l' = l’3 (cm) l’4 (cm) l’5 (cm) l '1 + l '2 + l '3 + l'4 + l '5
= …. (cm)
5 ∆ l' = (l'max − l ') + (l ' − l'min ) l'max − l'min
=
= …. (cm)
2
2 c – Determining the sound wavelength λ:
λ = 2 (l’ – l) = …. ; ∆λ = ∆l’ + ∆l = …. ; So λ = .… ± …. (m) and ∆λ
= ….
λ d – The sound speed in air corresponding to the air temperature t = 20oC:
v = 332 1 + 0.004t = …. (m/s)
The sound frequency produced by the tuning fork:
f= v
= …. (Hz)
λ ∆f ∆v ∆λ
3
=
+
=
+ ... =
f
v
λ 332
∆f = f ∆f
= …. (Hz)
f So f = .… ± …. (Hz) EXPERIMENTAL EXERCISE 3 – THE ALTERNATING CURRENT CIRCUIT WITH R, L, C
I  Purpose
+ To qualitatively investigate the effect of objects having either a resistor R, a capacitor C, or an inductor
L to an AC circuit.
+ To observe the phenomenon of resonance in an AC circuit with R, L and C connected in series.
II  Theoretical preparation
Present the equation for calculating
1) the resistance of a circuit with only a resistor R Translated by VNNTU – Dec. 2001 Page 154 2) the capacitive impedance of a circuit with only a capacitor C
3) the inductive impedance of a circuit with only an inductor L.
4) the total impedance of a circuit having R, L and C connected in series.
Prove that if L =
in series is I = 1
1
or C = 2 2 then the maximum current in a circuit having R, L, C connected
4π f C
4π f L
22 U
(where f is the frequency of the alternating current).
R III  List of required accessories
1) A 6V DC source and an 6V AV source; 2) a resistor R ≈ 15 – 20Ω; 3) a coil (750 – 1000 closely packed
turns of wire) with ferrite core; 4) a capacitor C = 20µF; 5) a capacitor C’ = 50µF; 6) a flashlamp bulb
(6V – 0.1A); 7) an electric switch; 8) a set of 6 wires.
IV  Experimental procedures
1) a) Connect resistor R in series with lamp D and switch K to the DC
source of a voltage U = 6V (figure 2). Observe the lamp’s luminosity when
close switch K.
b) Repeat step (1a) with the AC source of an effective voltage U = 6V.
Compare the lamp’s luminosity in two cases (the resister is kept unchanged). Figure 2 c) Draw corresponding conclusions.
2) a) Connect coil L in series with lamp D and switch K to the DC source
of a voltage U = 6V (figure 3). Observe the lamp’s luminosity when close
switch K.
b) Slowly withdraw the ferrite core from coil L. Remark changes of the
lamp’s luminosity.
c) Repeat step (2a) with the AC source of an effective voltage U = 6V.
Compare the lamp’s luminosity in cases (2a) and (2c). Figure 3 d) Again, slowly withdraw the ferrite core from coil L. Remark changes in the lamp’s luminosity.
c) Draw corresponding conclusions.
3) a) Connect capacitor C in series with lamp D and switch K to the DC
source of a voltage U = 6V (figure 4). Observe the lamp’s luminosity when
close switch K.
b) Repeat step (3a) with the AC source of an effective voltage U = 6V.
Observe the luminosity of lamp D. Figure 4 c) Connect capacitor C’ in parallel with capacitor C in (3b). Observe the luminosity of lamp D.
d) Draw corresponding conclusions.
4) a) Connect resistor R, coil L, capacitor C, lamp D and switch K in series to the AC source of a voltage
U = 6V (figure 5). Close switch K, slowly withdraw the ferrite core from coil L, observe the lamp’s
luminosity.
b) Determine the position of the ferrite core so that the luminosity of lamp D is maximum. Explain the
phenomenon. Translated by VNNTU – Dec. 2001 Page 155 c) Determine the inductance of coil L when the ferrite core is in the position so that the lamp is lightest,
knowing that C = 20µF and the frequency of the AC source is f = 50Hz.
d) Repeat step (4a) with capacitor C is replaced by capacitor C’. Determine the position of the ferrite core
corresponding to the resonance of the circuit. Determine the inductance of coil L in this case.
V  Form of the experimental report
1) Purpose
2) Results
a) The lamp’s luminosity in (1a) is (greater than), (equal to), (less than) that in (1b). This means ....
b) When the ferrite core is withdrawn as in (2a), the lamp’s luminosity (increases), (keeps unchanged),
(decreases). This means ....
When the ferrite core is withdrawn as in (2c), the lamp’s luminosity (increases), (keeps unchanged),
(decreases). This means ....
c) The lamp in (3a) is (on) (off). This means ....
The lamp in (3b) is (on) (off). This means ....
The lamp’s luminosity in (3c) is (greater than), (equal to), (less than) that in (3cb). This means ....
d) When the ferrite core is withdrawn, the lamp’s luminosity (increases), (keeps unchanged), (decreases).
In the case of resonance, the lamp’s luminosity is ....
The inductance L0 of the coil is calculated by the following equation: L0 = ................
With f = 50Hz and C = .... then L0 = ....
With f = 50Hz and C’ = .... then L0 = .... EXPERIMENTAL EXERCISE 4 – THE REFRACTION INDEX OF GLASS
I  Purpose
To determine the refraction index of glass.
II  Theoretical preparation
1) Present the law of refraction and write out the equation to calculate the refraction index of a substance
(in comparison with that of air).
2) Since the errors in measuring angles of incidence and of reflection are quite large, to determine the
sin α
can be
value of refraction index n precisely the ratio
sin γ
SH
determined by using the ratio
as in figure 6. If taking
S'H '
SI = IS’ (S is an arbitrary point in the incident ray), then
n= sin α
SH / SI
SH
=
=
.
sin γ
S'H '/ IS'
S'H ' Explain why we should select SI ≥ 60mm and α ≥ 30o ? Translated by VNNTU – Dec. 2001 Figure 6 Page 156 3) Given that 3 needles and a block of glass are available, describe the method to determine the refracted
ray in glass, with the incident ray is predetermined.
III  List of required accessories
1) A prism of glass with the main section is rectangle (trapezium, triangle, semicircular); 2) 3 needles; 3) a
ruler in millimeter scale; 4) pencils and compasses.
IV  Experimental procedures
Determine the refraction index of glass:
a) Draw in a blank paper a circle with center point I, a horizontal diameter PQ = 120mm and a vertical
diameter MN. Attach this paper to a notebook (or book) on the horizontal surface of a table. Lay the prism
on the paper so that MN is perpendicular to one of its vertical faces at I. Draw the boundary ABCD of the
prism on the paper (figure 7).
b) Pitch the 1st needle vertically next to the prism’s face at point
I. Pitch the 2nd needle vertically at point S on the circle and away
from MN a distance SH = 40mm (figure 7).
c) Look on these two needles through the prism from side CD.
Pitch the 3rd needle vertically at a certain point I’ next to face CD
of the prism so that this needle hides the images of two other
needles (figure 7).
d) Take the prism away and draw lines SI (incident ray) and II’
(refracted line). Extend II’ until it cuts the circle at S’. Draw line
S’H’ perpendicular to MN, measure the length of S’H’.
e) Calculate n of glass by the following equation: n = SH
= ....
S'H ' Figure 7 f) Repeat steps (2b) to (2e) with different positions of S (3 times), find corresponding positions of S’ and
the value of reflection index of glass.
g) Calculate the average value of the obtained results and the averaged absolute error of the above
mentioned method.
V  Form of the experimental report
This report should write on the same paper to be used in the experiment.
1) Purpose
2) Results
a) Determine the (relative) refraction index of glass.
Based on the figure obtained from the experiment, we have
Experimental order
1
2
3
n + n 2 + n3
n= 1
=
3
∆n = SH (mm) S’H’ (mm) n = SH/S’H’
n1 =
n2 =
n3 = (n max − n) + (n − n min ) n max − n min
=
=
2
2 Translated by VNNTU – Dec. 2001 Page 157 So n = .… ± ….
b) Supplementary questions
Do you know any other methods to determine the refraction index of glass?
Describe the method you think the best one. EXPERIMENTAL EXERCISE 5 – OBSERVATION OF LIGHT DISPERSION AND INTERFERENCE
PHENOMENA
I  Purpose
To create and observe:
a) the dispersion phenomenon of the white light
b) the interference phenomenon of the white light through Young’s splits.
II  Theoretical preparation
1) In which way the dispersion phenomenon of the white light can be created? Why is it necessary to use a
source of narrow white beam to create a colorful spectrum?
2) Sketch the principle diagram of Young’s experiment to observe the light dispersion phenomenon.
Present necessary conditions to see bright fringes come between dark ones. Figure
III  List of required accessories8a Figure 8b 1) A single split: selfmade from a piece of board (50mm×50mm) with a rectangle hole (5mm×20mm) in
the middle which is sheltered by two halves of blade to make a split of 0.5mm to 1mm (figure 8a);
2) Young’s double split: made following the way described above, but between two halves of blade we
stretch a thin copper wire to create two parallel slips; 3) A galvanized steel shaft; 4) a chromatic filter; 5) a
plate of diameter 20cm full of water and a handy mirror; 6) a prism of glass.
IV  Experimental procedures
1) Observe the light dispersion phenomenon:
a) Place the handy mirror in the plate containing water with a small slope to make a prism of water. Direct
the mirror to an open window, and find an appropriate observing position to see the image of the
window’s horizontal frame. Describe and explain the phenomenon observed. Translated by VNNTU – Dec. 2001 Page 158 b) Take the prism of glass by its sides and direct it to the open window. Adjust the slope of the prism until
having a clear image of the window bars. Describe and explain the phenomenon observed.
2) Observe the light resonance phenomenon:
a) Put the galvanized shaft vertically on the table surface so that it is lightened by the sunlight (or the light
of a lamp). Observe the reflection light through Young’s double split (in vertical direction) in a distance of
1m far from the shaft (figure 9). Describe the image obtained.
How is the resonant image changed when:
 increase the distance from the shaft to Young’s split;
 increase the distance from the eyes to Young’s split;
 put a chromatic filter before (or after) Young’s split.
b) Put the single split horizontally on the window so that it is lightened by the sunlight. Observe this
lightened split through Young’s double split (in horizontal direction) in a distance of about 1m far from
the shaft (figure 10). Describe the phenomenon and answer similar questions as in (2a). Figure 9 Figure 10 Note: Try to observe the single split which is lightened by a fluorescent lamp or a candle if available.
V  Form of the experimental report
1) Purpose
2) Results
A  Observe the light dispersion phenomenon:
a) Describe and explain the phenomenon observed through the prism of water and the plane mirror.
b) Describe and explain the phenomenon observed through the prism of glass.
B  Observe the light resonance phenomenon:
Based on experimental results:
a) answer questions in (2a);
b) answer questions in (2b). Translated by VNNTU – Dec. 2001 Page 159 COMBINED EXPERIMENTAL EXERCISES
EXPERIMENTAL EXERCISE A – DETERMINATION OF CAPACITANCE AND INDUCTANCE (2
SECTIONS)
I  Purpose
To apply a combination of knowledge and experimental skills to determine using one of different
methods:
1) the capacitance C of a capacitor; 2) the inductance L of a coil with ferrite core.
II  Theoretical preparation
A) Methods to determine the capacitance C of a capacitor
1) Connect the capacitor to an AC source of effective voltage U. If the effective current through the
U
1
=
capacitor has a value of I, then the impendence of the capacitor is ZC =
I 2πfC
Measure U, I and with the frequency f known, we have C = I
2πfU 2) Connect the capacitor C and a resistor R in series to an AC source. Use an AC voltmeter (RV >> R) to
measure the effective voltage UC of the capacitor and the effective voltage UR of the resistor. Since
U C ZC I ZC
1
, so if we select a predetermined resistance R and with f = 50Hz we have C =
=
=
=
UR
RI
R 2πfCR
UR I
U C 2πfR
B) Methods to determine the inductance L of a coil
1) If a coil is connected to an AC source of effective current I, since the coil has both resistance r and
U
impendence of inductor ZL = 2πfL so the total impendence Z =
= r 2 + 4π2 f 2 L2
I
Z2 = r 2 + 4π2 f 2 L2 = U2
1
and hence L =
2
I
2πf U2
U
U
then L =
.
− r 2 . If r <<
2
I
I
2πfI 2) Connect the inductor L and a capacitor C in series to an AC source of a constant effective voltage U.
When the capacitance C changes, the effective current in the circuit LC also changes.
2 1 r + ωL − = r. The effective current in the circuit
Cω 1
1
reaches the maximum value (resonance). So L =
=
2
Cω
C4π2 f 2
1
, the total impendence is Z =
When ωL =
Cω 2 III  List of required accessories
1) Capacitors 5µF (2 pcs.); 2) capacitors 50µF (2 pcs.); 3) the capacitor of unknown capacitance CX; 4) a
coil (750 – 1000 closely packed turns of wires) with a fixed ferrite core; 5) a set of AC and DC sources
with voltages can be changed from 4V to 10V; 6) a multifunction electric meter POLYTEST (or an AC
voltmeter 10V, a DC voltmeter 3V, a DC ammeter 100mA and an AC ammeter 100mA); 7) a coaltype
resistor 100Ω; 8) a coaltype resistor 50Ω; 9) an electric switch; 10) a set of wires.
Translated by VNNTU – Dec. 2001 Page 160 IV  Experimental procedures
A) Determine the capacitance C of the capacitor
1) Method I: a) Connect the capacitor CX to the AC circuit of frequency
50Hz as described in figure 11. f= b) Adjust the variable resistor, measure the effective voltage U by using
AC voltmeter, measure the corresponding effective current I by using
AC ammeter. the
the Figure 11 c) Record the data obtained and calculate CX:
Experimental order
U (V)
I (A)
ZC = U/I (Ω)
1
6
2
8
3
9
4
10
Calculate the average capacitance CX = ................ (F)
∆CX = CX = 1/(2πfU) (F) CX max − CX max
= ........... (F)
2 2 Method II: a) Connect the capacitor CX in series with the coaltype
resistor R1 = 50Ω to the AC circuit of frequency f = 50Hz as
described in figure 12. Measure the effective voltages UC of the
capacitor and UR of the resistor by using the AC voltmeter. So CX = UR I
.
U C 2πfR Figure 12 b) Repeat step (2a) with R1 is replaced by R2 = 100Ω and R3 = 150Ω (R3 is made by connecting R1 and R2
in series) in turn in order to determine the corresponding values of CX.
c) Record the data obtained and calculate CX:
Experimental order
1
2
3 R (Ω)
50
100
150 UR (V) CX = UR/(UC2πfR) (F) UC (V) Calculate the average capacitance CX = ................ (F) and ∆CX = CX max − CX max
= ........... (F)
2 ∆C X
∆C X
depends mostly on the ratio
, about 10%. If the coal resistors have a high
CX
CX
accuracy so the relative error is rather small). (The relative error B) Methods to determine the inductance L of a coil
1) Method 1: a) Determine the resistance r of the coil by using a DC
source (about 3V), a DC voltmeter (3V scale), a DC ammeter (100mA
scale).
b) Connect the coil to the AC circuit of frequency f = 50Hz as described
in figure 13. Measure the effective voltage U by using the AC voltmeter Translated by VNNTU – Dec. 2001 Figure 13 Page 161 (10V scale) and measure the effective current I by using the AC ammeter (100mA scale) in
correspondence with different resistance of the variable resistor.
c) Record the data obtained and calculate the inductance L:
Experimental order U (V) 1
2
3
4 Z = U/I (Ω) I (A) L= 6
8
9
10 Calculate the average inductance L = ................ (H) and ∆L = 1
2πf Z2 − r 2 (H) L max − L min
= ........... (H)
2 2 Method II: a) Connect the coil in series with the capacitor C1 = 10µF and an AC ammeter to the AC
circuit of voltage U = 6V and frequency f = 50Hz as described in figure 14. Record the value of current I
in this circuit.
b) Replace C1 by the following sets of capacitors in turn:
C2 = 25µF (two 50µF capacitors connected in series)
C3 = 30µF (C2 and another 5µF capacitor connected in parallel)
C4 = 35µF (C2 and two other 5µF capacitor connected in parallel)
C5 = 50µF Figure 14 C6 = 55µF (C5 and another 5µF capacitor connected in parallel)
C7 = 60µF (C5 and two other 5µF capacitors connected in series)
C8 = 100µF (two 50µF capacitors connected in parallel)
Write down the values of current I in each case to the following table:
Capacitance C (µF) 10 25 30 35 50 55 60 100 Current I (mA)
c) Based on this table, determine the capacitance value needed to have the resonance in the LC circuit.
1
∆C
∆L
Then calculate the inductance L of the coil: L =
. Determine
, from which calculate
and
22
4π f C
C
L
∆L.
V  Form of the experimental report
1) Purpose
2) Results
A  Determine the capacitance CX
a) Draw the circuit diagram in figure 11 and complete the table of results obtained in (1c)
Result: CX = ................ ± ................ (F)
b) Draw the circuit diagram in figure 12 and complete the table of results obtained in (2c)
Result: CX = ................ ± ................ (F) Translated by VNNTU – Dec. 2001 Page 162 B  Determine the inductance L
a) Draw the circuit diagram in figure 13 and complete the table of results obtained in (1c)
Result: L = ................ ± ................ (H)
b) Draw the circuit diagram in figure 14 and complete the table of results obtained in (2b)
Result: L = ................ ± ................ (H)
C  Questions
*1) Do you know any other methods to determine C or L? Describe the method you think the most reliable
one.
*2) Determine the coefficient cosϕ of the part MN in experiment (A.2) and in experiment (B.1). EXPERIMENTAL EXERCISE B – CHARACTERISTICS AND APPLICATIONS OF TRANSISTORS
(2 SECTIONS)
I  Purpose
+ To experimentally investigate characteristics of transistors
+ Assembly some simple circuits using transistors.
II  Theoretical preparation
1) Each transistor has 3 poles: E (emitter), B (base), C (collector); thus we have to connect these poles
precisely to the circuit as required.
Connect a transistor to the circuit as described in figure 15a, when we adjust the variable resistor to change
the base current Ib, the collector current Ic will change following the characteristic curve
Ic = f(Ib) as in figure 15b. From this curve we can see that: Figure 15a Figure 15b a) when Ib is in the range of 0 to Ibs, Ic is (linear) proportional to Ib. The constant coefficient β = Ic
is
Ib called the coefficient of current amplification.
b) when Ib is greater than Ibs then Ic is maximum (saturation value), and is constant to any Ib > Ibs.
2) Some applications of transistor Translated by VNNTU – Dec. 2001 Page 163 a) Choose a resistor for assembly in the base circuit so that Ib > Ibs. When the base circuit is open, Ib = 0
and Ic = 0. When it is close, Ic is at its saturation value and is large enough to make lamps, bells or
electromagnetic relays which are assembled in the collector circuit operate. In this mod, the transistor
plays a role as an electriccontrolled switch or an electronic switch. It is widely used in alarming systems,
liquid level indicating systems, photoelectric counters, etc.
b) Choose a resistor for assembly in the base circuit so that 0 < Ib < Ibs, the transistor will work in the
proportional amplifying mode. In this mode, Ic can be 30 to 300 times larger than Ib, and this intensity can
satisfy certain requirements of the collector circuit.
c) Choose a resistor for assembly in the base circuit so that Ib ≈ Ibs/2, and connect a capacitor C to this
circuit so that in the base circuit there is a periodically
fluctuated current (ib) of an amplitude Ab < Ibs/2. In
this case, in the collector circuit we will have a the
periodically fluctuated current ic of an amplitude Ac =
β.Ab (figure 16).
This characteristic of transistor is usually used to
amplify high frequency currents such as those in
radios, amplifiers, etc.
d) Transistor are also used for SHM generators (see
§27 of this book).
III  List of required accessories
1) Transistors (2 pcs.); 2) a coaltype variable resistor
100kΩ; 3) coaltype resistors: 50Ω, 2kΩ, 5kΩ, 5kΩ,
50kΩ, 270kΩ; 4) capacitors: 0.1µF, 0.1µF; 5) 2 coils
Figure 16
with ferrite core; 6) a microphone and an earphone; 7) a lamp: 4V – 0.05A; 8) a battery set 4.5V; 9) a set
of wires; 10) a cup of salt solution.
IV  Experimental procedures
1) Investigate the characteristics of transistor:
a) Assemble the circuit as described in figure 15a. The transistor and ammeters should be connected
precisely. Before closing switch K, the variable resistor’s contact should be set in the position nearest to
the base pole of transistor.
b) Close switch K, adjust the variable resistor to increase the base current Ib from 0 to about 3mA. Record
the values of Ic corresponding to Ib on the following table:
Ib (mA)
Ic (mA) 0 0.20 0.40 0.60 0.80 1.0 1.5 2.0 2.5 3.0 c) Based on these results, determine the range of Ib in which this transistor operates in the saturation mode
and in the proportional amplifying mode.
d) Determine the current amplifying coefficient β = Ic/Ib of this
transistor.
2) Assemble a shortcircuit alarming device to measure the liquid
level inside a tank:
Assemble the circuit as described in figure 17. Pour the salt
solution into the cup until the liquid level reaches the wire end A.
Observe and explain the phenomenon. Translated by VNNTU – Dec. 2001 Figure 17 Page 164 What will happen if the liquid is kerosene? Explain why.
3) Assemble a hearing machine for deaf men:
a) Assemble the circuit as described in figure 18.
Wear the earphone, and talk onto the microphone to
hear the voice.
Explain the operation of this circuit. Explain the
roles of resistors 270kΩ and 50kΩ, and the roles of
capacitors 0.1µF.
b) Take microphone A out of MN and connect it
between PQ. Talk to microphone A, how does the
voice received in the earphone B change? Explain why. Figure 18 3) Assemble a simple SHM generator:
a) Assemble the circuit as described in figure 19. Close switch K, adjust the ferrite cores of two coils,
compare changes of sound produced in the earphone.
Explain the operation of this circuit.
b) Reduce the number of wire turns of coil L1, or replace
capacitor C1 by a capacitor 0.2µF (two capacitors 0.1µF
connected in parallel). How does the sound received in the
earphone B change? Explain why.
L1 or Note: if no sound is heard, try to exchange two ends of coil
L2.
V  Form of the experimental report Figure 19 1) Purpose
2) Results
a) Rewrite the table of results in (1b).
The transistor operates in the saturation mode when ....
The transistor operates in the proportional amplifying mode when ....
The coefficient of current amplification is β = ....
b) When the level of salt solution reaches the wire end A, the lamp .... because ....
If the liquid is kerosene then .... because ....
c) There is a voice in the earphone B because ....
Resistor 270kΩ plays a role of .....
Resistor 50kΩ plays a role of .....
Capacitor 0.1µF plays a role of .....
When the microphone is connected to PQ, the voice heard in the earphone B is .... because ....
d) The SHM generator can operate because ....
When reduce the number of wire turns of coil L1, the sound produced by the generator is ... because ...
When increase the capacitance of C1, the sound produced by the generator is ... because ...
Translated by VNNTU – Dec. 2001 Page 165 EXPERIMENTAL EXERCISE C – DETERMINATION OF FOCAL LENGTH OF LENSES (2
SECTIONS)
I  Purpose
To determine the focal length of positive and negative lenses in various ways, and to assess the accuracy
of experimental results.
II  Theoretical preparation
Let’s the distance between the object and the lens is d, the distance between the lens and the image of
111
object is d’, we can determine the local length f of the lens by using the equation = + .
f d d'
The value of f can be experimentally determined by using the following methods:
1) For each positive lens of a constant focal length f, if d increases then d’ decreases. If d is very large then
d’ is negligible in comparison with d, thus we can approximate f = d’. Hence, to determine f we only have
to measure the distance d’ between the lens and the screen at which we have the clearest image of an
object shining from far.
2) If lay an object in a distance d far from the positive lens, and move the screen to a position at which we
dd '
, where d’ is the
can have the clearest image of object, then the focal length can be calculated by f =
d + d'
distance between the lens and the screen.
∆f ∆d ∆d ' ∆(d + d ')
=
+
−
, we can see that when d
f
d
d'
d + d'
∆d
∆d '
∆f
small then d’ is small and
is large. The result is that
is large.
is selected rather large to have
d
d'
f
∆f
Therefore, to reduce
we have to try through experiments so that the difference between d and d’ is
f
rather small. Note: From the formula to calculate the relative error 3) Especially if we can adjust through experiments to have d0 = d0’ then
have the focal length f = 12
2
=
= ' , and therefore we
f d0 d0 d0
d'
∆d 0
∆d '0
= 0 and ∆f =
=
. The error ∆f of this method is rather small, i.e.
2
2
2
2 the accuracy is high.
4) The negative lens only gives us the virtual image of the object. Since it is difficult to determine the
position of the virtual image accurately by experiments, the focal length is usually determined by joining a
negative lens and a positive lens. Through this joined lens, an object at position S will have its real image
on a screen at S’ as in figure 20. After that we take out the negative lens and move the object from S to a
position S1 that gives the clearest image on the screen at S’. Position S1 is actually the position of the
virtual image of the object in S through the negative lens, i.e. SO1 = d and O1S1 = d’. Thus the focal length
dd '
.
of the negative lens can be calculated by f’ =
d + d' Translated by VNNTU – Dec. 2001 Page 166 Due to the fact that the image of S through the negative lens is a virtual one at S1 so d’ has a minus sign (), and since  d  >  d’ then f’ has a minus sign ().
III  List of required accessories
1) Positive lenses (from +5diopter to +10diopter);
2) Negative lenses (from 10diopter to 12diopter);
3) a flashlamp bulb 2.5V attached on a base; 4) a
screen; 5) a straight wooden bar of length 1m; 6) a
500mm ruler; 7) a 3V source; 8) a switch; 9) a set
of wires.
IV  Experimental procedures
A  Determine the focal length of a positive lens: Figure 20 1) Direct the lens to an open window looking into the landscape outside (about 30 – 40m away). Move the
screen behind the lens to find a position at which the real image of a certain object over the window (the
roof of a house, the top of a tree, etc.) is clearest. Measure the distance d’ from the lens to the screen, we
have f = d’.
Repeat this procedure several times with different objects. Calculate the average value of f and the error
∆f.
2) Arrange the lamp, the positive lens and the screen straight along the wooden bar as described in figure
21. Figure 21
First, locate the lamp in a distance d = 450mmm ± 1mm far from the lens, switch on the lamp and move
the screen to a position at which the image of the filament is clearest. Measure the distance from the lens
to the screen we have d’ = ....
Repeat this procedure several times with d = constant we realize that the value of d’ is in a range of two
extreme values d’max and d’min. From which we can calculate the extreme values of the focal length by
these formulae
fmax = d max .d 'max
d .d '
and fmin = min min
d max + d 'max
d min + d 'min Thus we can infer that f = f max + f min
f −f
and ∆f = max min
2
2 Result: f = ............... ± ................. (mm)
3) Locate the positive lens in the middle of the wooden bar. Move both the lamp and the screen
symmetrically over the lens until the image of the filament is clearest. Measure the distance from the lens
to the lamp (d0) and the distance from the screen to the lens (d0’) and adjust to have d0 = d0’ corresponding
to the clearest image of the filament. Translated by VNNTU – Dec. 2001 Page 167 From which we infer that the focal length f = d0
d'
(or f = 0 ). Then determine ∆d0 and ∆f.
2
2 Result: f = ............... ± ................. (mm)
4) Compare the results obtained from 3 different methods described above. Which result is the most
reliable one? Explain why.
B  Determine the focal length of a negative lens:
1) Arrange the lamp, the negative lens, the positive lens and the screen straight along the wooden bar as
described in figure 22. Switch on the lamp, move the positive lens until the image of the filament on the
screen is clearest. Figure 22
2) Measure the distance SO1 = d from the lens to the negative lens as in figure 20. Keep the positions of
the positive lens and of the screen. Mark on the wooden bar the position O1 of negative lens. Take out the
negative lens and move the lamp nearer to the positive lens until the image of the filament on the screen is
clearest. At that time, the lamp is in the position S1 of the virtual image through the negative lens in the
experiment described in figure 22.
The distance O1S1 is actually the distance d’ from the lens to the virtual image of a lamp located at a point
S which has SO1 = d. Measure the distance O1S1 = d’.
3) Calculate the focal length of the negative lens by f’ =
image at S1 is a virtual one. dd '
, in which d’ has a minus sign since the
d + d' 4) Repeat this experiment with some values of d = SO1 so that these values are nearly the same. Determine
the value of f’ in each case, calculate the average value f ' and the averaged absolute error ∆f ' .
C – Supplementary questions
Do you know any other methods to determine the focal length of lenses? Describe the method you think
the best one, and implement this method if applicable.
V  Form of the experimental report
1) Purpose
2) Results
A  Determine the focal length of a positive lens:
When the object is very far, we have d ' = '
'
d1 + d '2 + d 3 + d '4
d ' − d 'min
= .... and ∆d ' = max
.
4
2 Thus f = ............... ± ................. (mm)
2) From experiment A.2 we have dmax = .............; dmin = ...............; d’max = .............; d’min = ...............
Therefore fmax = .............; fmin = .............; ∆f = .............
Translated by VNNTU – Dec. 2001 Page 168 So f = ............... ± ................. (mm)
3) From experiment A.3 we have d0 = ............. ± ...............; d’0 = ............. ± ...............
So f = ............... ± ................. (mm)
4) The result determined by method ............ is the most reliable one because ...............
B  Determine the focal length of a negative lens:
No.
f'= Distance d (mm) Distance d’ (mm) Focal length f (mm) f1 + f 2 + f 3
f' −f'
= ............... (mm) and ∆f ' = max min = ............... (mm)
3
2 So f’ = ............... ± ................. (mm)
C  Determine the focal length of a lens by other methods Translated by VNNTU – Dec. 2001 Page 169 ...
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This note was uploaded on 12/09/2011 for the course ELECTRICAL 101 taught by Professor Huanhoang during the Fall '11 term at National University of Singapore.
 Fall '11
 HuanHoang

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