Unformatted text preview: Phasors This is vector technique for adding large numbers of wave disturbances of angular frequency ( ) but with different phases . A phasor is a vector of magnitude E0 rotating with an angular frequency, . E Eo t The rules for using phasors are as follows Construct a series of phasors representing the functions to be added. Draw them end to end maintaining the proper phase relations between adjacent phasors. Construct the vector sum of this array. The length of the vector sum gives the amplitude of the resulting phasor. The angle between the vector sum and the first phasor is the phase of the resultant with respect to this first phasor. The projection of this vectorsum on the vertical axis gives the time variation of the resultant wave. A sinusoidal wave disturbance can be considered as a projection of this phasor on the vertical axis such that E = E0 sin( t) Tuesday, 12 August 2008 1 Tuesday, 12 August 2008 2 Using Young’s two slit experiment as an example E1 = E0sin t and E2 = E0sin( t+ ) i.e. E=2E0cos sin( t+ ) with = /2 The amplitude of the interference is 2E0cos The intensity is the square of the electric vector so Eo E2=4E2cos2 0 I=4I0cos2 How is the phase difference where the maxima are? 2 E2 E1 Eo t Eo E2 E1 t Eo 2 that tells The magnitude of the sum is 2E0cos while the time varying component is sin( t+ ). The sum of the interior angles is equal to the exterior angle . i.e. = /2. related to the angle Tuesday, 12 August 2008 3 Tuesday, 12 August 2008 4 phase difference = 2 path difference The path difference is dsin so = 2 d sin I=4I0cos2 and = 2 d sin 2 Coherence The interference pattern produced by Young's experiment only works because the waves at both the slits are in phase. This is because the waves originate from a single slit. The waves are said to be completely coherent. If we replaced the two slits with lamps then we destroy the interference pattern because the lamps emit light with a random phase with respect to each other and so the sources in this case are completely incoherent. When two incoherent sources combine we add intensities while if we have coherent sources then we add the amplitudes before squaring to find the combined intensity. Tuesday, 12 August 2008 5 Tuesday, 12 August 2008 6 Which of the following pairs of light sources are coherent: (a) two candles; (b) one point source and its image in a plane mirror; (c) two pinholes uniformly illuminated by the same point source; (d) two headlights of a car; (e) two images of a point source due to reflection from the front and back surfaces of a soap film. Huygens’ Principle All points on a wave front serve as a point sources of spherical secondary wavelets. After a time t, the new position of the wavefront will be that of a surface tangent to these secondary wavelets. t t=0 ct Tuesday, 12 August 2008 7 Tuesday, 12 August 2008 8 Diffraction Huygens principle implies that rays do not travel in straight lines from a source. If a wave encounters a barrier that has an opening of dimensions similar to the wavelength, the wave will flare out into the region beyond the barrier. Light from one part of the slit will interfere with light from another part of the slit giving rise to diffraction. 10x10 3 5 0 5 10 3.0 x/L 1.5
10 Tuesday, 12 August 2008 9 Tuesday, 12 August 2008 10 Physics Demonstrations in Light (2.4) Fraunhofer and Fresnel Diffraction ©Physics Curriculum & Instruction (www.physicscurriculum.com)
Tuesday, 12 August 2008 11 Tuesday, 12 August 2008 12 12 Diffraction from a single slit Using Huygens principle we treat all points within the slit as point sources. Using the phasor technique to add vectorially the individual amplitudes Ao of the sources we can find the resultant amplitude A at P. Am is the amplitude opposite the slit where =0 A = 2r sin 2 D P = Am r slit width a asin D>>a
13 14 Tuesday, 12 August 2008 13 Tuesday, 12 August 2008 14 Combining the last two equations A=2 Am sin 2 At the screen I=Im sin
2 == 2 a sin I=Im sin 2 Phase difference = 2 path difference = 2 asin == 2 a sin It is easy to work out where the minima are in the fringe pattern. When sin =0 then I=0 (except when =0) i.e. =m for m=1,2,3, .... m= a sin asin =m for m=1,2,3 . 16 Tuesday, 12 August 2008 15 Tuesday, 12 August 2008 16 a sin =m for m = 1,2,3 ... If a is small compared to large and so the pattern spreads out. If a= then = 90˚ when m = 1. For a >> then 0 i.e geometric optics. then is Diffraction from a circular aperture When projecting the image of a distant star on a screen using a converging lens we find that the image has a diffraction pattern I=Im sin 2 == 2 a sin sin =1.22 Minima at asin =m for m=1,2,3 d The first minimum for the diffraction pattern of a circular aperture of diameter d is given by
Tuesday, 12 August 2008 17 Tuesday, 12 August 2008 18 The fact that lens images are diffraction patterns is important when we wish to distinguish two distant objects whose separation is small. d L D d 20 Tuesday, 12 August 2008 19 Tuesday, 12 August 2008 20 The Rayleigh criterion for resolving two objects is that two objects are resolved if the central maximum of the diffraction pattern of one source falls on the first minimum of the diffraction pattern of the other source. d or for small angles
R=1.22 R=sin 1 1.22 Tipler Problem 33:11 d where is the angular separation between the two objects. For objects that are far away = D/L where D is the separation of the objects and L is the distance to the objects. So D/L=1.22 /d to just resolve the two objects. True or false: (a) When waves interfere destructively, the energy is converted into heat energy. (b) Interference is observed only for waves from coherent sources. (c) In the Fraunhofer diffraction pattern for a single slit, the narrower the slit, the wider the central maximum of the diffraction pattern. (d) A circular aperture can produce both a Fraunhofer and a Fresnel diffraction pattern. (e) The ability to resolve two point sources depends on the wavelength of the light. 21 Tuesday, 12 August 2008 21 Tuesday, 12 August 2008 22 Physics Demonstrations in Light (2.7)
(a) False. When destructive interference of light waves occurs, the energy is no longer distributed evenly. For example, light from a twoslit device forms a pattern with very bright and very dark parts. There is practically no energy at the dark fringes and a great deal of energy at the bright fringe. The total energy over the entire pattern equals the energy from one slit plus the energy from the second slit. Interference redistributes the energy. ( b ) True (c) True ( d ) True (e) Tr u e 23 ©Physics Curriculum & Instruction (www.physicscurriculum.com)
23 Tuesday, 12 August 2008 24 Tuesday, 12 August 2008 Problem 33:58 The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 6 mm. (a) Using light with a wavelength of 500 nm, how far could you be from this tile and still resolve these holes? The diameter of the pupil of your eye is about 5 mm. (b) Could you resolve these holes better with red light or with violet light? Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to the right showing the overlapping diffraction patterns to express L in terms of , x, and the diameter D of your pupil. (a) Referring to the diagram, relate c, L, and x :
For circular apertures, Rayleigh’s criterion is: c x provided L
= 1.22 << 1 c D Equate these two expressions to obtain: x = 1.22 L D
26 25 Tuesday, 12 August 2008 25 Tuesday, 12 August 2008 26 Solve for L : L= xD 1.22 Diffraction Grating A multiple slit device is called a diffraction grating. Substitute numerical values and evaluate L : L= (6 mm)(5 mm ) = 1.22(500 nm ) 49.2 m P (b ) Because L is inversely proportional to , the holes can be resolved better with violet light which has a shorter wavelength. d dsin With many slits (e.g.10,000 per centimetre) we get intense, sharp maxima.
27 28 Tuesday, 12 August 2008 27 Tuesday, 12 August 2008 28 Because the maxima positions depend on the wavelength of light (dsin =m ) the maxima are different for different wavelengths of light. We can use diffraction gratings to view say the spectrum from a sodium lamp. The lamp which looks to be yellow tinted white light is actually made up of discrete wavelengths of red, yellow, green and blue. Problem 33:8 Equation 332, d sin = m , and Equation 3311, a sin = m , are sometimes confused. For each equation, deﬁne the symbols and explain the equation’s application.
Determine the Concept Equation 332 expresses the condition for an intensity maximum in twoslit interference. Here d is the slit separation, the wavelength of the light, m an integer, and the angle at which the interference maximum appears. This spectrum is repeated for all the orders (m=1,2,3) of the diffraction grating Equation 3311 expresses the condition for the first minimum in singleslit diffraction. Here a is the width of the slit, the wavelength of the light, and the angle at which the first minimum appears, assuming m = 1 . 29 30 Tuesday, 12 August 2008 29 Tuesday, 12 August 2008 30 Problem 33:39 Measuring the distance to the moon (lunar imaging) is routinely done by firing shortpulse lasers and measuring the time it takes for the pulses to reflect back from the moon. A pulse is fired from the earth; to send it out, the pulse is expanded so that it fills the aperture of a 6indiameter telescope, (a) Assuming the only thing spreading the beam out to be diffraction, how large will the beam be when it reaches the moon, 382,000 km away? (b) The pulse is reflected off a reflecting mirror left by the Apollo 11 astronauts. If the diameter of the mirror is 20 in, how large will the beam be when it gets back to the earth? (c) What fraction of the power of the beam is reflected back to the earth? (d) If the beam is refocused on return by the same 6 in telescope, what fraction of the original beam energy is recaptured? Ignore any atmospheric losses. Picture the Problem The diagram shows the beam expanding as it travels to the moon and that portion of it that is reflected from the mirror on the moon expanding as it returns to earth. We can express the diameter of the beam at the moon as the product of the beam divergence angle and the distance to the moon and use the equation describing diffraction at a circular aperture to find the beam divergence angle. We can follow this same procedure to find the diameter of the beam when it gets back to the earth. In Parts (c) and (d) we can use the dependence of the power in a beam on its crosssectional area to find the fraction of the power of the beam that is reflected back to earth and the fraction of the original beam energy that is recaptured upon return to earth. 32 Tuesday, 12 August 2008 31 Tuesday, 12 August 2008 32 (a) Relate the diameter D of the beam at the moon to the distance to the moon L and the beam divergence angle : The angle subtended by the first diffraction minimum is related to the wavelength of the light and the diameter of the telescope opening dtelescope by: Because << 1, sin and: D L
Substitute numerical values and evaluate D : sin = 1.22 d telescope D = 3.82 108 m ( ) 1.22(500 nm ) = 1.53 km 2.54 cm 1m 6 in in 102 cm 1.22 d telescope Substitute for obtain: in equation (1) to D= 1.22 L d telescope Out by a factor of 2 ?? 33 Tuesday, 12 August 2008 33 Tuesday, 12 August 2008 34 Hollograms Physics Demonstrations in Light (2.8) ©Physics Curriculum & Instruction (www.physicscurriculum.com)
Tuesday, 12 August 2008 35 Tuesday, 12 August 2008 36 ...
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This note was uploaded on 08/20/2010 for the course PHYS 142 taught by Professor Xxx during the One '09 term at University of Wollongong, Australia.
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