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Unformatted text preview: Assignments 3/28/11 9:15 AM Student View Summary View Diagnostics View Print View withEdit Assignment
Answers Settings per Student MP_Assignment#6 [ Print ] Due: 8:00am on Monday, March 28, 2011
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy Introduction to Two Source Interference
Description: This problem provides students with a basic introduction to the interference of two identical
waves. The conditions for constructive and destructive interference are discussed in terms of phase
difference and path length difference.
Learning Goal: To gain an understanding of constructive and destructive interference.
Consider two sinusoidal waves (1 and 2) of identical wavelength , period , and maximum amplitude snapshot of one of these waves taken at a certain time is displayed in the figure below. Let .A and represent the displacement of each wave at
position at time . If these waves were to be in the same location ( ) at the same time, they would
interfere with one another. This would result in a single
wave with a displacement
given by
.
This equation states that at time the displacement of the resulting wave at position is the algebraic sum of the displacements of the waves 1
and 2 at position at time . When the maximum
displacement of the resulting wave is less than the
amplitude of the original waves, that is, when
, the waves are said to interfere destructively
because the result is smaller than either of the individual waves. Similarly, when , the waves are said to interfere constructively because the resulting wave is larger than either of the individual waves. Notice
that
.
Part A
To further explore what this equation means, consider four sets of identical waves that move in the + x
direction. A photo is taken of each wave at time and is displayed in the figures below.
Rank these sets of waves on the basis of the maximum amplitude of the wave that results from the
interference of the two waves in each set.
Rank from largest to smallest To rank items as equivalent, overlap them.
ANSWER: http://session.masteringphysics.com/myct/assignments Page 1 of 25 Assignments 3/28/11 9:15 AM View
When identical waves interfere, the amplitude of the resulting wave depends on the relative phase of
the two waves. As illustrated by the set of waves labeled A, when the peak of one wave aligns with
the peak of the second wave, the waves are in phase and produce a wave with the largest possible
amplitude. When the peak of one wave aligns with the trough of the other wave, as illustrated in Set
C, the waves are out of phase by
and produce a wave with the smallest possible amplitude, zero! Part B
Consider sets of identical waves with the following phase differences: Identify which sets will interfere constructively and which will interfere destructively.
Enter the letters of the sets corresponding to constructive interference in alphabetical order and
the letters corresponding to sets that interfere destructively in alphabetical order separated by a
comma. For example if sets A and B interfere constructively and sets C and F interfere
destructively enter AB,CF.
http://session.masteringphysics.com/myct/assignments Page 2 of 25 Assignments 3/28/11 9:15 AM ANSWER: BCEG
ADF Do you notice a pattern ? When the phase difference between two identical waves can be written as
, where
, the waves will interfere constructively. When the phase
difference can be expressed as , where , the waves will interfere destructively.
Consider what water waves look like when you throw a rock into a lake. These waves start at the point
where the rock entered the water and travel out in all directions. When viewed from above, these waves
can be drawn as shown , where the solid lines
represent wave peaks and troughs are located
halfway between adjacent peaks. Part C
Now look at the waves emitted from two identical sources (e.g., two identical rocks that fall into a lake at
the same time). The sources emit identical waves at the exact same time.
Identify whether the waves interfere constructively or
destructively at each point A to D. Hint C.1 How to approach the problem Recall that constructive interference occurs when the two waves are in phase when they interfere, so
http://session.masteringphysics.com/myct/assignments Page 3 of 25 Assignments 3/28/11 9:15 AM that the peak (or trough) of one wave aligns with the peak (or trough) of the other wave. Destructive
interference occurs when waves are
out of phase so that the peak of one wave aligns with the
trough of the other wave.
Study the picture to find where each type of interference occurs.
For points A to D enter either c for constructive or d for destructive interference. For example if
constructive interference occurs at points A, C and D, and destructive interference occurs at B,
enter cdcc.
ANSWER: ccdd Each wave travels a distance
consecutive peaks is equal to
you can see that Point B is
and or from its source to reach Point B. Since the distance between , from the picture
away from Source 1 away from Source 2. The path  length difference, , is the difference in the distance each wave travels to reach Point B:
. Part D
What are the path  length differences at Points A, C, and D (respectively, , , and )? Enter your answers numerically in terms of
separated by commas. For example, if the path length differences at Points A, C, and D are
,
, and , respectively, enter 4,.5,1. ANSWER:
http://session.masteringphysics.com/myct/assignments Page 4 of 25 Assignments 3/28/11 9:15 AM ANSWER:
, , = , , Knowing the path  length difference helps to confirm what you found in Part C. When the path  length
difference is
, where
, the waves interfere constructively. When the
path  length difference is , where , the waves interfere destructively. Part E
What are the path  length differences at Points L to P?
Enter your answers numerically in terms of
separated by commas. For example, if the path length differences at Points L, M, N, O, and P are
,
,
, , and
, respectively, enter
5,2,1.5,1,6. ANSWER: , , , , = , , , , Every point along the line connecting Points L to P corresponds to a path  length difference . This means that at every point along this line, waves from the two sources interfere constructively.
The figure below shows two other lines of constructive interference: One corresponds to a path  length
difference
, and the other corresponds to
. It should make sense that the line
halfway between the two sources corresponds to a path  length difference of zero, since any point on
this line is equally far from each source. Notice the symmetry about the
line of the
and the lines. http://session.masteringphysics.com/myct/assignments Page 5 of 25 Assignments 3/28/11 9:15 AM A similar figure can be drawn for the lines of destructive interference. Notice that the pattern of lines is
still symmetric about the line halfway between the two sources; however, the lines along which
destructive interference occurs fall midway between adjacent lines of constructive interference. Multislit Interference and Diffraction Gratings
Description: Qualitative questions about multislit interference, leading to understanding of diffraction
gratings.
Learning Goal: To understand multislit interference and how it leads to the design of diffraction gratings.
Diffraction gratings are used in modern spectrometers to separate the wavelengths of visible light. The
working of a diffraction grating may be understood through multislit interference, which can be understood as
an extension of two  slit interference. In this problem, you will follow the progression from two  slit to many slit
interference to arrive at the important equations describing diffraction gratings.
A typical diffraction grating consists of a thin, opaque object with a series of very closely spaced slits in it.
(There are also reflection gratings, which use a mirror with nonreflecting lines etched into it to provide the
same effects.) To see how a diffraction grating can separate different wavelengths within a spectrum, we will
first consider a "grating" with only two slits. http://session.masteringphysics.com/myct/assignments Page 6 of 25 Assignments 3/28/11 9:15 AM Recall that the angles for constructive interference from a pair of slits are given by the equation , where is the separation between the slits, is the wavelength of the light, and is an integer.
Part A
Consider a pair of slits separated by
maximum with micrometers. What is the angle for red light with a wavelength of to the interference nanometers? Express your answer in degrees to three significant figures.
ANSWER: = Part B
Consider the same pair of slits separated by
maximum with micrometers. What is the angle for blue light with a wavelength of to the interference nanometers? Express your answer in degrees to three significant figures.
ANSWER: = Notice that the maxima for these two wavelengths are separated by over 20 degrees. Purely on the
basis of this information, it would appear that a pair of slits would be useful for physically separating
wavelengths for spectroscopy. However, these peaks of intensity are rather wide, and so it is difficult
to resolve similar colors, such as orange and yellow. As you will see, the more slits, the better! Part C
The intensity pattern from the two slits for a single wavelength looks like the one shown on the left side of
the figure.
If another slit, separated from one of the original
slits by a distance , is added, how will the intensity
at the original peaks change?
By examining the phasors for light from the two
slits, you can determine how the new slit affects the
intensity. Phasors are vectors that correspond to the
light from one slit. The length of a phasor is
proportional to the magnitude of the electric field
from that slit, and the angle between a phasor and
the previous slit's phasor corresponds to the phase
difference between the light from the two slits.
Recall that at points of constructive interference,
light from the original two slits has a phase
difference of
, which corresponds to a complete
http://session.masteringphysics.com/myct/assignments Page 7 of 25 Assignments 3/28/11 9:15 AM revolution of one phasor relative to the first.
Notice that, as shown in the figure, undergoing a complete revolution leaves the phasor pointing in the
same direction as the phasor from the other slit. Think about the phase difference between the new slit
and the closer of the two old slits and what this implies about the direction of the phasor for the new slit.
Hint C.1 Phase differences for multiple slits The phase difference between the original two slits is not affected by the presence of a third slit.
Similarly, the phase difference between the new slit and the closest of the old slits, slit 2 in the figure,
can be determined in the same way that you would determine the phase difference for a regular two  slit
interference problem. Recall that in two  slit interference, phase shift is proportional to
. How do
and for slit 2 and the new slit compare to the values of ANSWER: and for slit 1 and slit 2 ? The peak intensity increases.
The peak intensity remains constant.
The peak intensity decreases. Part D
Are there any points between the maxima of the original two slits where light from all three slits interferes
constructively? If so, what are they ?
ANSWER: No
Yes, halfway between the original maxima
Yes, at intervals one  third of the distance between the original maxima
Yes, halfway between every second pair of maxima Since the original two slits do not interfere constructively anywhere except for the original maxima, it
would be impossible for all three slits to interfere constructively somewhere between the original
maxima. However, there are two places between the maxima where the intensity is zero, instead of
just one such minimum as in the case of just two slits. This means that the intensity dies off more
quickly as you move away from one of the peaks than it would for two slits, and then stays low until
the next peak is reached. Part E
What is the angle to the second order maximum from a diffraction grating with slits spaced light of wavelength ? Hint E.1 apart, for Equation for diffraction gratings Recall that you have shown that the maxima for a diffraction grating are the same as the maxima for a
two  slit setup. Therefore, the locations of the maxima are given by the equation
. http://session.masteringphysics.com/myct/assignments Page 8 of 25 Assignments 3/28/11 9:15 AM Express your answer in terms of and . ANSWER:
= Part F
Recall that phasor diagrams give the magnitude of the electric field, and that the intensity is related to the
electric field squared. In going from two slits to three, the amount of energy in the total interference pattern
increases by
(three slits worth of light instead of two), but the peak intensity of the interference
maxima increases by (from to , where is the electric field due to one slit). What does this suggest about the intensity maxima for three slits ?
ANSWER: They are wider than the maxima for two slits, because they contain more
energy.
They are narrower than the maxima for two slits because of conservation of
energy.
They have a more blue color than the maxima for two slits, because they have
more energy.
They are the same width and color as the maxima for two slits. Part G
By the same reasoning that worked with three slits, you can see that no matter how many slits you have,
the maxima will still fall at the same locations as the maxima for two slits. The peak intensity of the
maxima will be proportional to
, where
is the total number of slits.
The energy at the screen is roughly equal to the product of the number of maxima, the peak intensity of a
maximum, and the width of a maximum. As
increases, the number and location of the maxima will not
change, while the peak intensity of the maxima will increase proportionally to
available increases proportionally to
Hint G.1 . If the total energy , how does the width of the maxima change? Writing an equation If you take the information above and put it into an equation, you obtain , where energy at the screen, is the width of the is the number of maxima, maxima. You know that is constant, asked to determine the power of
equation remain equal as is the peak intensity, and is proportional to to which , and is proportional to is the
. You are must be be proportional so that both sides of the varies. ANSWER:
http://session.masteringphysics.com/myct/assignments Page 9 of 25 Assignments ANSWER: 3/28/11 9:15 AM The width does not change.
The width is proportional to
The width is proportional to
The width is proportional to
The width is proportional to .
.
.
. Now you can see why diffraction gratings have many thousands of slits. The more slits, the narrower
the maximum for a particular wavelength, and thus the more finely wavelengths may be separated for
analysis. Why Butterfly Wings Shimmer
Description: Basic conceptual and quantitative questions on thin film interference, followed by an
application to the iridescence of butterfly wings.
Learning Goal: To understand the concept of thin film interference and how to apply it.
Thin  film interference is a commonly observed phenomenon. It causes the bright colors in soap bubbles and
oil slicks. It also leads to the iridescent colors on many insects and bird feathers. In this problem, you will
learn how to work with thin film interference and see how it creates the dazzling display of a tropical
butterfly's wings.
When light is incident on a thin film, some of the light
will be reflected at the front surface of the film, and
the rest will be transmitted into the film. Some of the
transmitted light will be reflected from the back surface
of the film. The light reflected from the front surface
and the light reflected from the back surface will
interfere. Depending upon the thickness of the film,
this interference may be constructive or destructive.
We will be studying the interference of light normal to
the surface of the film. The figure shows the light
entering at a small angle to normal only for the
purpose of showing the incident and reflected rays. For this problem, you will only be concerned with the geometric aspects of thin film interference, so ignore
phase shifts caused by reflection from a medium with higher index of refraction. (Because of the structure
of a butterfly's wings, such phase shifts do not contribute much to what you actually see when you look at
the butterfly.)
Part A
Assume that light is incident normal to the surface of a film of thickness . How much farther does the light reflected from the back surface travel than the light reflected from the front surface?
http://session.masteringphysics.com/myct/assignments Page 10 of 25 Assignments 3/28/11 9:15 AM Express your answer in terms of . ANSWER: Part B
For constructive interference to occur, the difference between the two paths must be an integer multiple of
the wavelength of the light (as is true in any interference problem), i.e. the general criterion for
constructive interference is
,
where is a positive integer. This is usually stated in the slightly more explicit form
. Given the thickness of the film, , what is the longest wavelength that can exhibit constructive interference?
Express your answer in terms of
ANSWER: . = Part C
If you have a thin film of thickness 300 , what is the third  longest wavelength of light that exhibits constructive interference with the reflected light?
Note that this corresponds to
.
Express your answer in nanometers to three significant figures.
ANSWER: = Part D
The criterion for destructive interference is very similar to the criterion for constructive interference. For
destructive interference to occur, the difference between the two paths must be some integer number of
wavelengths plus half a wavelength:
,
or
, http://session.masteringphysics.com/myct/assignments Page 11 of 25 Assignments where 3/28/11 9:15 AM is a nonnegative integer. What is the second longest wavelength that will not be visible (i.e., will have strong destructive interference for the reflected waves) when reflected from a film of
thickness 300
?
Note that the longest wavelength corresponds to
notation used for the second longest wavelength is for destructive interference. This is why the
instead of . Express your answer in nanometers to three significant figures.
ANSWER: = The blue morpho butterfly lives in tropical rainforests and can have a wingspan greater than 15 cm. The
brilliant blue color of its wings is a result of thin film
interference. A pigment would not produce such
vibrant, pure colors. What cannot be conveyed by a
picture is that the colors vary with the viewing angle,
which causes the shimmering iridescence of the
actual butterfly.
The scales of the butterfly's wings consist of two thin
layers of keratin (a transparent substance with index
of refraction 1.54), separated by a 200gap filled
with air. Part E
What is the longest wavelength of light that will exhibit constructive interference at normal incidence? The keratin layers are thin enough that you can think of them as representing the surfaces of a 200"film" of air.
Express your answer in nanometers to two significant figures.
ANSWER: = This wavelength lies near the cutoff between visible (violet) and ultraviolet light. The shorter
wavelengths that are strongly reflected (corresponding to
in
) will not be relevant to
what humans see when they look at the butterfly.
To understand why the color changes with viewing angle, try drawing a diagram of light incident on a
thin film at a large angle. The distance the light travels within the film will be increased. However, at
large angles the light reflected from the front surface will actually have to travel farther to an observer
outside of the film than the light reflected from the back surface. The increased distance outside of
the material for the front surface reflection actually makes the net path  length difference smaller than
http://session.masteringphysics.com/myct/assignments Page 12 of 25 Assignments 3/28/11 9:15 AM it would be for normal incidence. As viewing angle increases, the largest wavelength that experiences
constructive interference gets shorter. Thus, although the butterfly is blue at normal incidence, at large
angle of incidence no particular wavelength of visible light is short enough to be strongly reflected.
Part F
The wavelength that appears in the interference equations given in Parts B and D represents the wavelength of light within the medium of the film. So far we have assumed that the medium composing the
film is air, but many thin film problems will involve films with an index of refraction different from that of air.
Suppose that the butterfly gets wet, thus filling the gaps between the keratin sheets with water (
).
What wavelength
Hint F.1 in air will be strongly reflected now? How to approach the problem In Part E, you found the wavelength that exhibits constructive interference; represents the wavelength of the light within the medium of the film (the medium being water, in this case). Now find
the wavelength of light in air that, once in water, has a wavelength of
.
Hint F.2 Relating the wavelength in air to the wavelength in water Recall that the wavelength within a medium is related to the wavelength in vacuum (or air, to the accuracy we are working with) by the equation
,
where is the index of refraction of the medium. Be careful not to confuse this with the arbitrary positive integer in the equations introduced earlier in this problem.
Express your answer in nanometers to two significant figures.
ANSWER: = Part G
Several thin films are stacked together in each butterfly wing scale. How would these multiple layers of thin
films affect the light reflected by the butterfly's wings ?
Hint G.1 A picture of the situation The figure shows a model of the layers in the
scales on the butterfly's wing. Make a similar
drawing so that you can draw in the rays of light
to help you visualize what is occurring in this
situation. Keep in mind that when light strikes a
transparent boundary, some of the light is
reflected and some is transmitted. http://session.masteringphysics.com/myct/assignments Page 13 of 25 Assignments ANSWER: 3/28/11 9:15 AM There would be no change, because all of the 400 light would be reflected by the first scale, and so the light is unaffected by the remaining scales.
More 400 light would be reflected, because some of the light transmitted through the first layer could be reflected by the second layer, light transmitted
by the second layer could be reflected by the third, etc.
Less 400 light would be reflected, because light reflected from the second layer can interfere destructively with light reflected by the first layer.
There would be no change, because the interference effects would all occur at
the first layer.
More 400 light would be reflected, because the effective gap would be three times as wide.
Less 400 light would be reflected, because the effective gap would be three times as wide.
Layering of thin films is used to make wavelength specific mirrors, used widely in laser applications,
that are far more reflective than metal mirrors. Layered films are also used to make antireflective
coatings for camera lenses. Constructive Interference
Description: Short quantitative problem on a double  slit experiment and interference pattern. Students
should know how to find the wavelength of light in different materials. Based on Young/Geller Quantitative
Analysis 26.1.
The figure shows the interference pattern obtained in
a double  slit experiment with light of wavelength . http://session.masteringphysics.com/myct/assignments Page 14 of 25 Assignments 3/28/11 9:15 AM Part A
Identify the fringe or fringes that result from the interference of two waves whose phases differ by exactly
.
Hint A.1 How to approach the problem In a double  slit experiment, when two waves reach the screen shifted by an integer number of
wavelengths, such as
or
, at that point the waves are in phase and interfere constructively.
Their amplitudes add up and the resulting band, or fringe, on the screen is a bright region where light
intensity is at a maximum. Note that several bright fringes can be seen in the interference pattern shown
in the figure. Each of them is produced by waves that are shifted by a different (integer) number of
wavelengths. Choose those that correspond to a path difference of
.
Hint A.2 Find which fringes correspond to constructive interference Which of the fringes shown in the figure correspond to constructive interference?
Hint A.2.1 Constructive interference fringes
Recall that, in a double  slit experiment, waves that interfere constructively strike the screen in regions
where light intensity is maximum, creating a series of bright bands or fringes.
ANSWER: fringe C only
fringes A, B, and C
fringes A, B, and D
fringes A, B, and E
fringes A, D, and E
fringes A, B, D, and E Which of these fringes correspond to a path difference of and which to a path difference of ? Hint A.3 Constructive interference pattern In a double  slit experiment, fringes of maximum intensity are produced by waves whose path difference
is an integer number of wavelengths, such as
,
,
,
. The bigger this number, the
farther from the central bright band is the corresponding fringe. For example, if the center of the fringe
produced by waves that are shifted by
is at 1
from the center of the central band, the center of
the fringe produced by waves shifted by will be at approximately 4 from the central band. Moreover, the fringes appear symmetrical with respect to the central band. Thus, when the center of the
fringe produced by waves that are shifted by is at 1
from the center of the central band, the
center of the fringe produced by waves shifted by http://session.masteringphysics.com/myct/assignments will be located at about the same distance from Page 15 of 25 Assignments 3/28/11 9:15 AM the center of the central band, but on the opposite side.
Enter the letter(s) indicating the fringe(s) in alphabetical order. For example, if you think that
fringes A and C are both correct, enter AC.
ANSWER: AD Part B
The same double  slit experiment is then immersed in water (with an index of refraction of 1.33) and
repeated. When in the water, what happens to the interference fringes ?
Hint B.1 How to approach the problem When in water, the wavelength of light changes. Will this affect the location of the interference fringes on
the screen ? Recall that in a double  slit experiment the positions of the centers of the bright fringes on
the screen can be expressed in terms of the difference in path length traveled by the waves.
Hint B.2 Find the wavelength of light in water Consider light with wavelength in air. What is the wavelength of the same light in water, ? Hint B.2.1 Wavelength of light in a material
The wavelength of light is different in different materials. In particular, the wavelength material of index of refraction of light in a is given by
, where is the wavelength of the same light in vacuum. This is because in any material the velocity of a wave is given by the product of frequency and wavelength,
a material is less than it is in vacuum, but its frequency . Since the speed does not change, its wavelength of light in
must be less than the wavelength of the same light in vacuum. Note that the wavelength of light in air is
essentially the same as that of light in vacuum.
Express your answer in terms of . ANSWER:
= Thus, when the experiment is immersed in water, the wavelength of light decreases. How does this
affect the location of the interference fringes ?
Hint B.3
Let Find how the centers of the fringes change with the wavelength of light
be the position of the center of the mth band (measured from the center of the central band) http://session.masteringphysics.com/myct/assignments Page 16 of 25 Assignments 3/28/11 9:15 AM when the experiment is in air, and let
be the same position when the experiment is in water.
Which of the following expressions is correct?
Hint B.3.1 Using proportional reasoning
To solve this problem use proportional reasoning to find a relation between the position of the mth
band when the experiment is in air,
, and the same position when the experiment is in water,
.
Find the simplest equation that contains these variables and other known quantities from the problem.
Write this equation twice, once to describe
and again for
.
You need to write each equation so that all the constants are on one side and your variables are on
the other. Since your variable is
in this problem, you want to write your equations in the form
.
To finish the problem you need to compare the two cases presented in the problem. For this question
you should find the ratio
. Hint B.3.2 Position of the centers of the bright fringes in a double slit experiment
In a double  slit experiment with light of wavelength
is an integer) interfere constructively and create the , waves whose path difference is (where bright band on the screen. The position of the center of the mth band (measured from the center of the central band) on the screen is given by
,
where and are, respectively, the distance from the screen to the slits and the distance between the slits.
ANSWER: As you found out, when the experiment is immersed in water, the centers of the bright fringes are
closer to the central band than when the experiment is in air. But does the location of the center of
the central band change when the wavelength of light decreases?
Hint B.4 Find how the center of the central band changes with the wavelength of light In a double slit experiment, the central fringe of the interference pattern corresponds to waves that
traveled the same distance from the source. If the wavelength of the waves changes, will the position of
the center of this fringe change? Assume the slits are horizontal so that the fringe pattern displayed on
a screen is oriented vertically..
Hint B.4.1 How to approach the question http://session.masteringphysics.com/myct/assignments Page 17 of 25 Assignments 3/28/11 9:15 AM In a double  slit experiment, the central fringe is created by waves that travel the same distance from
the slits, and its center is always located at the same distance from both slits. Thus, its position
changes only if the location of the slits changes, regardless of the properties of the light that shines
through the slits.
ANSWER: The position of the center of the central fringe will shift upward.
The position of the center of the central fringe will shift downward.
The position of the center of the central fringe will not change. ANSWER: They are more closely spaced than in air by a factor of 1.33.
They are more widely spaced than in air by a factor of 1.33.
They are spaced the same as in air.
They are shifted upward.
They are shifted downward. When the experiment is immersed in water, the wavelength of light decreases because the index of
refraction of water is higher than that of air. Since the positions of the centers of the bright bands
depend on the wavelength of light, light with a smaller wavelength will produce interference fringes
that are more closely spaced; the higher the index of refraction, the more closely spaced are the
fringes. Huygens' Principle
Description: Problem goes through the concepts of the Huygens' principle and looks at some specific
situations.
Huygens' principle (first described by the Dutch scientist Christiaan Huygens in 1678) uses geometry to
determine the shape of a wavefront at a time , given some initial wavefront at an earlier time. This can be
constructed by imagining that the initial wavefront is a source of wavelets that propagate from each point at
the speed of light. From this principle, one can determine the angles of reflection and refraction by
considering the speed of light at each point on the wavefront. In this problem, we will explore some of these
concepts.
Part A
First, let's look at a plane wave that is incident on a flat piece of material with an index of refraction . Part of the light is transmitted through the material and part of it is reflected. For the moment, let's just
look at the part that is transmitted. At time
, the wavefront is a distance away from the surface of
the material . At time , the wavefront is at the position of the material interface . Use the Huygens'
principle to determine how far into the material ( )
the wavefront has propagated by time http://session.masteringphysics.com/myct/assignments . Page 18 of 25 Assignments Hint A.1 3/28/11 9:15 AM Speed of light in a material Recall that the speed
refraction of light in a material is less than the speed is defined by the relation of light in a vacuum. The index of . Use this relation to find the speed of light in a material, which you will need to answer this part.
Express your answer in terms of the variables , , and the speed of light in a vacuum . ANSWER: http://session.masteringphysics.com/myct/assignments Page 19 of 25 Assignments 3/28/11 9:15 AM = Note that there would be no change in the direction of the wavefront at any point because the
waverfront encountered the material interface at the same time at all points. Thus all of the individual
wavefronts all propagated at the same speed, thereby maintaining the flat wavefront.
Now, instead of having a flat wavefront propagating normal to the material interface we have a flat
wavefront propagating toward the material at an angle of
relative to the axis perpendicular to the
material interface. In this part, we will look at the relative positions of a few points  A, B, and C   on the
wavefront to illustrate Huygens' principle. Point C
touches the vacuum/material interface at time
whereas point B is a distance
distance and point A is a away from the interface. Part B
What is the time
Hint B.1 it will take for point B of the wavefront to encounter the vacuum/material interface? Looking at the direction Point B propagates along a flat wavefront in the direction that's relative to the axis perpendicular to the material surface. As a result, one can use standard geometry and the fact that point B is originally
only a distance away from the surface.
Express your answer numerically to two decimal places in units of
travel a distance
ANSWER: (the time it takes light to in a vacuum). = It should be noted that the accuracy of the method increases the smaller the distance is before
making a new wavefront and redoing the wavelets. If you tried to just draw a large wavelet from point
B then it would hit the surface after only traveling a distance . This large wavelet would make it
difficult to visualize how to add up all of the other wavelets to make a coherent wavefront. As a result,
one should just draw a line perpendicular to the wavefront at point B until it hits the surface, at which
point the wavelets will change owing to the new index of refraction.
http://session.masteringphysics.com/myct/assignments Page 20 of 25 Assignments 3/28/11 9:15 AM Part C
How far did point C go into the material interface in the time that it took for point B to get to the interface? For this part we are looking for the distance traversed by the point C in the material (
Give your answer in terms of the time , , and ). . ANSWER:
= Part D
What is the new angle at which point C of the wavefront is propagating (relative to a line perpendicular to the vacuum/material interface) ? Try to use the fact that you have a spherical wavefront propagating
from
at the point where C met the vacuum/material interface until time
when the wavefront at
point B reached the interface.
Hint D.1 Geometry of the problem Huygens' principle relies on looking at each point on a wavefront as a source of circular (two dimensional) wavelets. As a result, we can look at how a wavelet propagates from where point C
touches the interface. While this wavelet is propagating in the material (at a speed less than ), point B
of the wavefront continues to propagate at speed until it too hits the interface. When this occurs, one can draw a line from the point at which B touches the interface to the tangent of the wavelet from point
C earlier. Since this line is tangent to the circle, it is perpendicular to a line from where point C touched
the interface (i.e., the center of the circular wavelet) to the point of tangency (on the circular wavelet).
From geometry and the previous part, one should be able to deduce the relevant angles. Note that the
distances from point C to points B and A are arbitrary and thus can be considered very small.
Express your answer in terms of inverse trig functions and . ANSWER:
= Problem 22.66
Description: A 0.10  mm thick piece of glass is inserted into one arm of a Michelson interferometer that is
using light of wavelength 500 nm. This causes the fringe pattern to shift by 200 fringes. (a) What is the
index of refraction of this piece of glass ?
A 0.10  mm thick piece of glass is inserted into one arm of a Michelson interferometer that is using light of
wavelength 500 nm. This causes the fringe pattern to shift by 200 fringes. http://session.masteringphysics.com/myct/assignments Page 21 of 25 Assignments 3/28/11 9:15 AM Part A
What is the index of refraction of this piece of glass ?
ANSWER: = Problem 22.54
Description: The figure shows the light intensity on a screen behind a single slit. The wavelength of the
light is 600 nm and the slit width is 0.15 mm. (a) What is the distance from the slit to the screen ?
The figure shows the light intensity on a screen
behind a single slit. The wavelength of the light is
600
and the slit width is 0.15
. Part A
What is the distance from the slit to the screen ?
Express your answer using two significant figures.
ANSWER: = Using a Michelson Interferometer
Description: A Michelson interferometer is used to find the wavelength of a laser and then the index of
refraction for an unknown fluid.
You are asked to find the index of refraction for an unknown fluid, using only a laser and a Michelson
interferometer. A Michelson interferometer consists of two arms  paths that light travels down, which end in
mirrors   attached around a beam splitter. The beam
splitter separates the incoming light into two separate
beams and then recombines them once they return http://session.masteringphysics.com/myct/assignments Page 22 of 25 Assignments 3/28/11 9:15 AM from the ends of the arms. The recombined beams
are sent to a telescope, where their interference
pattern may be observed in detail. Part A
First, you must find the wavelength of the laser. You shine the laser into the interferometer and then move
one of the mirrors until you have counted
fringes passing the crosshairs of the telescope. The
extremely accurate micrometer shows that you have moved the mirror by
wavelength
Hint A.1 millimeters. What is the of the laser ?
Relating wavelength and distance in a Michelson interferometer Since the change in path difference for a Michelson interferometer comes from both the increased
distance that light travels to the moving mirror and the distance traveled back from it, the equation for
how far the mirror has moved takes the form
, where is the distance traveled by the mirror,
is the number of fringes that have moved past the crosshairs, and is the wavelength of the light. Express your answer in nanometers, to four significant figures.
ANSWER: = Part B
You now immerse the interferometer in a tank filled with some unknown liquid and carefully align the laser
into the interferometer. You move the mirror until you count 100.0 fringes passing the crosshairs of the
telescope. The micrometer indicates that the mirror has moved
millimeters. What is the mystery
fluid?
Hint B.1 Find the index of refraction What is the index of refraction for this liquid ? Hint B.1.1 Find the wavelength of the light in the fluid
Find the wavelength
for of the light in the fluid. To do this, solve the equation from Part A . http://session.masteringphysics.com/myct/assignments Page 23 of 25 Assignments 3/28/11 9:15 AM Express your answer in nanometers, to four significant figures.
ANSWER: = Hint B.1.2 Relating wavelengths and the index of refraction
Recall that the wavelength of light in a medium is related to its wavelength relation is the index of refraction for the medium. , where in air by the Express your answer to four significant figures.
ANSWER: ANSWER: = water ( ) methanol ( ) ethanol ( ) acetone ( ) isopropyl alcohol (
saline ( ) ) Problem 22.29
Description: A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having
wavelengths 589.0 nm and 589.6 nm. The interferometer is initially set up with both arms of equal length
L_1=L_2, producing a bright spot at the center of the...
A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths
589.0 nm and 589.6 nm. The interferometer is initially set up with both arms of equal length
,
producing a bright spot at the center of the interference pattern.
Part A
How far must mirror be moved so that one wavelength has produced one more new maxima than the other wavelength?
ANSWER: = mm http://session.masteringphysics.com/myct/assignments Page 24 of 25 Assignments 3/28/11 9:15 AM Score Summary:
Your score on this assignment is 0%.
You received 0 out of a possible total of 9 points. http://session.masteringphysics.com/myct/assignments Page 25 of 25 ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.
 Fall '08
 drake

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