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1059 Chapter 33 Interference and Diffraction Conceptual Problems 1 • A phase difference due to path-length difference is observed for monochromatic visible light. Which phase difference requires the least (minimum) path length difference? (a) 90o (b) 180o (c) 270o (d) the answer depends on the wavelength of the light. Determine the Concept The phase difference δ due to a path difference Δr are related according to λπδ rΔ=2 . Therefore, the least path length difference corresponds to the smallest phase difference. ( )a is correct. 2 • 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. Determine the Concept Coherent sources have a constant phase difference. The pairs of light sources that satisfy this criterion are (b), (c), and (e). 3 • [SSM] The spacing between Newton’s rings decreases rapidly as the diameter of the rings increases. Explain qualitatively why this occurs. Determine the Concept The thickness of the air space between the flat glass and the lens is approximately proportional to the square of d, the diameter of the ring. Consequently, the separation between adjacent rings is proportional to 1/d. 4 • If the angle of a wedge-shaped air film, such as that in Example 32-2 is too large, fringes are not observed. Why? Determine the Concept There are two possible reasons that fringes might not be observed. (1) The distance between adjacent fringes is so small that the fringes are not resolved by the eye. (2) Twice the thickness of the air space is greater than the coherence length of the light. If this is the case, fringes would be observed in the region close to the point where the thickness of the air space approaches zero. 5 • Why must a film that is used to observe interference colors be thin? Determine the Concept Colors are observed when the light reflected off the front and back surfaces of the film interfere destructively for some wavelengths and Chapter 33 1060 constructively for other wavelengths. For this interference to occur, the phase difference between the light reflected off the front and back surfaces of the film must be constant. This means that twice the thickness of the film must be less than the coherence length of the light. The film is called a thin film if twice its thickness is less than the coherence length of the light. 6 • A loop of wire is dipped in soapy water and held up so that the soap film is vertical. (a) Viewed by reflection with white light, the top of the film appears black. Explain why. (b) Below the black region are colored bands. Is the first band red or violet? (a) The phase change due to reflection from the front surface of the film is 180°; the phase change due to reflection from the back surface of the film is 0°. As the film thins toward the top, the phase change due to the path length difference between the two reflected waves (the phase difference associated with the film’s thickness) becomes negligible and the two reflected waves interfere destructively. (b) The first constructive interference will arise when twice the thickness of the film is equal to half the wavelength of the color with the shortest wavelength. Therefore, the first band will be violet (shortest visible wavelength). 7 • [SSM] A two-slit interference pattern is formed using monochromatic laser light with a wavelength of 640 nm. At the second maximum from the central maximum, what is the path-length difference between the light coming from each of the slits? (a) 640 nm (b) 320 nm (c) 960 nm (d) 1280 nm. Determine the Concept For constructive interference, the path difference is an integer multiple of λ; that is, λmr =Δ . For m = 2, ( )nm 6402=Δr . ( )d is correct. 8 • A two-slit interference pattern is formed using monochromatic laser light with a wavelength of 640 nm. At the first minimum from the central maximum, what is the path-length difference between the light coming from each of the slits? (a) 640 nm (b) 320 nm (c) 960 nm (d) 1280 nm. Determine the Concept For destructive interference, the path difference is an odd-integer multiple of λ21 ; that is ( ) ...,5,3,1,21 ==Δ mmr λ . For the first minimum, m = 1 and ( ) nm 320nm 64021 ==Δr . ( )b is correct. 9 • A two-slit interference pattern is formed using monochromatic laser light with a wavelength of 450 nm. What happens to the distance between the first maximum and the central maximum as the two slits are moved closer together? (a) The distance increases. (b) The distance decreases. (c) The distance remains the same. Interference and Diffraction 1061 Determine the Concept The relationship between the slit separation d and the angular position θm of each maximum is given by ...,2,1,0,sin == mmd m λθ (Equation 33-2). Because d and sinθm are inversely proportional for a given wavelength and interference maximum (value of m), decreasing d increases sinθm and θm. ( )a is correct. 10 • A two-slit interference pattern is formed using two different monochromatic lasers, one green and one red. Which color light has its first maximum closer to the central maximum? (a) Green, (b) red, (c) both maxima are in the same location. Determine the Concept The relationship between the slit separation d, the angular position θm of each maximum, and the wavelength of the light illuminating the slits is given by ...,2,1,0 ,sin == mmd m λθ (Equation 33-2). Because λ and sinθm are directly proportional for a given interference maximum (value of m) and the wavelength of green light is shorter than the wavelength of red light, ( )a is correct. 11 • A single slit diffraction pattern is formed using monochromatic laser light with a wavelength of 450 nm. What happens to the distance between the first maximum and the central maximum as the slit is made narrower? (a) The distance increases. (b) The distance decreases. (c) The distance remains the same. Determine the Concept The relationship between the slit width a, the angular position θm of each maximum, and the wavelength of the light illuminating the slit is given by ...,3,2,1 ,sin == mma m λθ (Equation 33-11). Because a and sinθm are inversely proportional for a given diffraction maximum (value of m), narrowing the slit increases mθsin and θm. ( )a is correct. 12 • Equation 33-2 which is d sin θm = mλ, and Equation 33-11, which is a sin θm = mλ, are sometimes confused. For each equation, define the symbols and explain the equation’s application. Determine the Concept Equation 33-2 expresses the condition for an intensity maximum in two-slit interference. Here d is the slit separation, λ the wavelength of the light, m an integer, and θm the angle at which the interference maximum appears. Equation 33-11 expresses the condition for an intensity minimum in single-slit diffraction. Here a is the width of the slit, λ the wavelength of the light, and θ m the angle at which the minimum appears, and m is a nonzero integer. Chapter 33 1062 13 • When a diffraction grating is illuminated by white light, the first-order maximum of green light (a) is closer to the central maximum than the first-order maximum of red light. (b) is closer to the central maximum than the first-order maximum of blue light. (c) overlaps the second-order maximum of red light. (d) overlaps the second-order maximum of blue light. Picture the Problem We can solve λθ md =sin for θ with m = 1 to express the location of the first-order maximum as a function of the wavelength of the light.The interference maxima in a diffraction pattern are at angles θ given by: λθ md =sin where d is the separation of the slits and m = 0, 1, 2, … Solve for the angular location θ1 of the first-order maximum : ⎟⎠ ⎞⎜⎝ ⎛= − d λθ 11 sin Because λgreen light < λred light: light redlightgreen θθ < and )(a is correct. 14 • A double-slit interference experiment is set up in a chamber that can be evacuated. Using light from a helium-neon laser, an interference pattern is observed when the chamber is open to air. As the chamber is evacuated, one will note that (a) the interference fringes remain fixed. (b) the interference fringes move closer together. (c) the interference fringes move farther apart. (d) the interference fringes disappear completely. Determine the Concept The distance on the screen to mth bright fringe is given by: d Lmy nm λ= where L is the distance from the slits to the screen, nλ is the wavelength of the light in a medium whose index of refraction is n, and d is the separation of the slits. The separation of the interference fringes is given by: ( ) d L d Lm d Lmyy nnnmm λλλ =−+=−+ 11 Because the index of refraction of a vacuum is slightly less than the index of refraction of air, the removal of air increases nλ and, hence, ym − ym−1. )(c is correct. Interference and Diffraction 1063 15 • [SSM] True or false: (a) When waves interfere destructively, the energy is converted into heat energy. (b) Interference patterns are observed only if the relative phases of the waves that superimpose remain constant. (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 diffraction pattern and a Fresnel diffraction pattern. (e) The ability to resolve two point sources depends on the wavelength of the light. (a) False. When destructive interference of light waves occurs, the energy is no longer distributed evenly. For example, light from a two-slit 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 re-distributes the energy. (b) True. (c) True. The width of the central maximum in the diffraction pattern is given by a m m λθ 1sin −= where a is the width of the slit. Hence, the narrower the slit, the wider the central maximum of the diffraction pattern. (d) True. (e) True. The critical angle for the resolution of two sources is directly proportional to the wavelength of the light emitted by the sources ( D λα 22.1c = ). 16 • You observe two very closely-spaced sources of white light through a circular opening using various filters. Which color filter is most likely to prevent your resolving the images on your retinas as coming from two distinct sources? (a) red (b) yellow (c) green (d) blue (e) The filter choice is irrelevant. Chapter 33 1064 Determine the Concept The condition for the resolution of the two sources is given by Rayleigh’s criterion: Dc λα 22.1= (Equation 33-25), where αc is the critical angular separation and D is the diameter of the aperture. The larger the critical angle required for resolution, the less likely it is that you can resolve the sources as being two distinct sources. Because αc and λ are directly proportional, the filter that passes the shorter wavelength light would be most likely to resolve the sources and the longer wavelength (red) would be most likely to prevent resolving the sources. ( )a is correct. 17 •• Explain why the ability to distinguish the two headlights of an oncoming car, at a given distance, is easier for the human eye at night than during daylight hours. Assume the headlights of the oncoming car are on during both daytime and nighttime hours. Determine the Concept The condition for the resolution of the two sources is given by Rayleigh’s criterion: Dc λα 22.1= (Equation 33-25), where αc is the critical angular separation, D is the diameter of the aperture, and λ is the wavelength of the light coming from the objects, in this case headlights, to be resolved. Because the diameter of the pupils of your eyes are larger at night, the critical angle is smaller at night, which means that at night you can resolve the light as coming from two distinct sources when they are at a greater distance. Estimation and Approximation 18 • It is claimed that the Great Wall of China is the only man made object that can be seen from space with the naked eye. Check to see if this claim is true, based on the resolving power of the human eye. Assume the observers are in low- Earth orbit that has an altitude of about 250 km. Picture the Problem We’ll assume that the diameter of the pupil of the eye is 5.0 mm and use the best-case scenario (the minimum resolvable width varies directly with the wavelength of the light reflecting from the object) that the wavelength of light is 400 nm (the lower limit for the human eye). Then we can use the expression for the minimum angular separation of two objects than can be resolved by the eye and the relationship between this angle and the width of an object and the distance from which it is viewed to support the claim. Relate the width w of an object that can be seen at an altitude h to the critical angular separation αc: h w=ctanα ⇒ ctanαhw = Interference and Diffraction 1065 The minimum angular separation αc of two point objects that can just be resolved by an eye depends on the diameter D of the eye and the wavelength λ of light: D λα 22.1c = Substitute for αc in the expression for w to obtain: ⎟⎠ ⎞⎜⎝ ⎛= D hw λ22.1tan Substitute numerical values and evaluate wmin for an altitude of 250 km: ( ) m24 mm0.5 nm40022.1tankm250min ≈⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛=w This claim is probably false. Because the minimum width that is resolvable from low-Earth orbit (250 km) is 24 m and the width of the Great Wall is 5 to 8 m high and 5 m wide, so this claim is likely false. However, it is easily seen using binoculars, and pictures can be taken of it using a camera. This is because both binoculars and cameras have apertures that are larger than the pupil of the human eye. (The Chinese astronaut Yang Liwei reported that he was not able to see the wall with the naked eye during the first Chinese manned space flight in 2003.) 19 •• [SSM] (a) Estimate how close an approaching car at night on a flat, straight stretch of highway must be before its headlights can be distinguished from the single headlight of a motorcycle. (b) Estimate how far ahead of you a car is if its two red taillights merge to look as if they were one. Picture the Problem Assume a separation of 1.5 m between typical automobile headlights and tail lights, a nighttime pupil diameter of 5.0 mm, 550 nm for the wavelength of the light (as an average) emitted by the headlights, 640 nm for red taillights, and apply the Rayleigh criterion. (a) The Rayleigh criterion is given by Equation 33-25: Dc λα 22.1= where D is the separation of the headlights (or tail lights). The critical angular separation is also given by: L d=α where d is the separation of head lights (or tail lights) and L is the distance to approaching or receding automobile. Chapter 33 1066 Equate these expressions for αc to obtain: DL d λ22.1= ⇒ λ22.1DdL = Substitute numerical values and evaluate L: ( )( ) ( ) km 11nm 55022.1 m 5.1mm 0.5 ≈=L (b) For red light: ( )( ) ( ) km .69nm 64022.1 m 5.1mm 0.5 ≈=L 20 •• A small loudspeaker is located at a large distance to the east from you. The loudspeaker is driven by a sinusoidal current whose frequency can be varied. Estimate the lowest frequency for which your ears would receive the sound waves exactly out of phase when you are facing north. Picture the Problem If your ears receive the sound exactly out of phase, the waves arriving at your ear that is farthest from the speaker must be traveling one- half wavelength farther than the waves arriving at your ear that is nearest the speaker. The lowest frequency corresponds to the longest wavelength. Assume that the speaker and your ears are on the same line and let the distance between your ears be about 20 cm. Take the speed of sound in air to be 343 m/s. The frequency received by your ears is given by: λ vf = Let Δr be the distance between your ears (the path difference) to obtain: λ21=Δr ⇒ rΔ= 2λ Substituting for λ yields: r vf Δ= 2 Substitute numerical values and evaluate f: ( ) kHz 86.0cm 202 m/s 343 ==f or between 0.80 and 0.90 kHz. 21 •• [SSM] Estimate the maximum distance a binary star system could be resolvable by the human eye. Assume the two stars are about fifty times further apart than Earth and Sun are. Neglect atmospheric effects. (A test similar to this ″eye test″ was used in ancient Rome to test for eyesight acuity before entering the army. A normal eye could just barely resolve two well-known close-together stars in the sky. Anyone who could not tell there were two stars was rejected. In this case, the stars were not a binary system, but the principle is the same.) Interference and Diffraction 1067 Picture the Problem Assume that the diameter of a pupil at night is 5.0 mm and that the wavelength of light is in the middle of the visible spectrum at about 550 nm. We can use the Rayleigh criterion for the separation of two sources and the geometry of the Earth-binary star system to derive an expression for the distance to the binary stars. If the distance between the binary stars is represented by d and the Earth-star distance by L, then their angular separation is given by: L d=α The critical angular separation of the two sources is given by the Rayleigh criterion: Dc λα 22.1= For α = αc: DL d λ22.1= ⇒ λ22.1 DdL = Substitute numerical values and evaluate L: ( )( )( ) ( ) y 5.9 m 109.461 y 1km 1059.5 nm 55022.1 m 105.150mm 0.5 15 13 11 ⋅ × ⋅××≈ ×= c c L Phase Difference and Coherence 22 • Light of wavelength 500 nm is incident normally on a film of water 1.00 μm thick. (a) What is the wavelength of the light in the water? (b) How many wavelengths are contained in the distance 2t, where t is the thickness of the film? (c) What is the phase difference between the wave reflected from the top of the air–water interface and the wave reflected from the bottom of the water–air interface in the region where the two reflected waves superpose? Picture the Problem The wavelength of light in a medium whose index of refraction is n is the ratio of the wavelength of the light in air divided by n. The number of wavelengths of light contained in a given distance is the ratio of the distance to the wavelength of light in the given medium. The difference in phase between the two waves is the sum of a π phase shift in the reflected wave and a phase shift due to the additional distance traveled by the wave reflected from the bottom of the water−air interface. (a) Express the wavelength of light in water in terms of the wavelength of light in air: nm376 1.33 nm500 water air water === n λλ Chapter 33 1068 (b) Relate the number of wavelengths N to the thickness t of the film and the wavelength of light in water: ( ) 32.5 nm376 m 00.122 water === μλ tN (c) Express the phase difference as the sum of the phase shift due to reflection and the phase shift due to the additional distance traveled by the wave reflected from the bottom of the water−air interface: Nt πππλπ δδδ 222 water traveleddistance additionalreflection +=+= += Substitute for N and evaluate δ: ( ) rad6.11 rad64.11rad32.52rad π πππδ = =+= or, subtracting 11.64π rad from 12π rad, rad 1.1rad4.0 == πδ 23 •• [SSM] Two coherent microwave sources both produce waves of wavelength 1.50 cm. The sources are located in the z = 0 plane, one at x = 0, y = 15.0 cm and the other at x = 3.00 cm, y = 14.0 cm. If the sources are in phase, find the difference in phase between these two waves for a receiver located at the origin. Picture the Problem The difference in phase depends on the path difference according to πλδ 2 rΔ= . The path difference is the difference in the distances of (0, 15.0 cm) and (3.00 cm, 14.0 cm) from the origin. Relate a path difference Δr to a phase shift δ: πλδ 2 rΔ= The path difference Δr is: ( ) ( ) cm682.0 cm0.14cm3.00cm0.15 22 = +−=rΔ Substitute numerical values and evaluate δ: rad .922cm50.1 cm682.0 ≈⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= πδ Interference in Thin Films 24 • A wedge-shaped film of air is made by placing a small slip of paper between the edges of two flat plates of glass. Light of wavelength 700 nm is Interference and Diffraction 1069 incident normally on the glass plates, and interference fringes are observed by reflection. (a) Is the first fringe near the point of contact of the plates dark or bright? Why? (b) If there are five dark fringes per centimeter, what is the angle of the wedge? Picture the Problem Because the mth fringe occurs when the path difference 2t equals m wavelengths, we can express the additional distance traveled by the light in air as an mλ. The thickness of the wedge, in turn, is related to the angle of the wedge and the distance from its vertex to the mth fringe. (a) The first fringe is dark because the phase difference due to reflection by the bottom surface of the top plate and the top surface of the bottom plate is 180° (b) The mth fringe occurs when the path difference 2t equals m wavelengths: λmt =2 Relate the thickness of the air wedge to the angle of the wedge: x t=θ ⇒ θxt = where we’ve used a small-angle approximation to replace an arc length by the length of a chord. Substitute for t to obtain: λθ mx =2 ⇒ λλθ x m x m 2 1 2 == Substitute numerical values and evaluate θ : ( ) rad1075.1nm700cm 5 2 1 4−×=⎟⎠ ⎞⎜⎝ ⎛=θ 25 •• [SSM] The diameters of fine fibers can be accurately measured using interference patterns. Two optically flat pieces of glass of length L are arranged with the wire between them, as shown in Figure 33-40. The setup is illuminated by monochromatic light, and the resulting interference fringes are observed. Suppose that L is 20.0 cm and that yellow sodium light (wavelength of 590 nm) is used for illumination. If 19 bright fringes are seen along this 20.0-cm distance, what are the limits on the diameter of the wire? Hint: The nineteenth fringe might not be right at the end, but you do not see a twentieth fringe at all. Picture the Problem The condition that one sees m fringes requires that the path difference between light reflected from the bottom surface of the top slide and the top surface of the bottom slide is an integer multiple of a wavelength of the light. Chapter 33 1070 The mth fringe occurs when thepath difference 2d equals m wavelengths: λmd =2 ⇒ 2 λmd = Because the nineteenth (but not the twentieth) bright fringe can be seen, the limits on d must be: ( ) ( ) 22 2 1 2 1 λλ +<<− mdm where m = 19 Substitute numerical values to obtain: ( ) ( ) 2 nm59019 2 nm59019 2121 +<<− d or m8.5m5.5 μμ << d 26 •• Light that has a wavelength equal to 600 nm is used to illuminate two glass plates at normal incidence. The plates are 22 cm in length, touch at one end, and are separated at the other end by a wire that has a radius of 0.025 mm. How many bright fringes appear along the total length of the plates? Picture the Problem The light reflected from the top surface of the bottom plate (wave 2 in the diagram) is phase shifted relative to the light reflected from the bottom surface of the top plate (wave 1 in the diagram). This phase difference is the sum of a phase shift of π (equivalent to a λ/2 path difference) resulting from reflection plus a phase shift due to the additional distance traveled. t 1 2 wire glass plate glass plate Relate the extra distance traveled by wave 2 to the distance equivalent to the phase change due to reflection and to the condition for constructive interference: ...,3,2,2 21 λλλλ =+t or ...,,,2 252321 λλλ=t and ( )λ212 += mt where m = 0, 1, 2, …, 0 ≤ t ≤ 2r and λ is the wavelength of light in air. Interference and Diffraction 1071 Solving for m gives: m = 2tλ − 1 2 where m = 0, 1, 2, …and 0 ≤ t ≤ 2r Solve for the highest value of m: mmax = int 2 2r( ) λ − 1 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = int 4r λ − 1 2 ⎛ ⎝⎜ ⎞ ⎠⎟ where r is the radius of the wire. Substitute numerical values and evaluate m: mmax = int 4 0.025mm( ) 600nm − 1 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = int 166.2( )= 166 Because we start counting from m = 0, the number of bright fringes is mmax + 1. N = mmax + 1 = 167 27 •• A thin film having an index of refraction of 1.50 is surrounded by air. It is illuminated normally by white light. Analysis of the reflected light shows that the wavelengths 360, 450, and 602 nm are the only missing wavelengths in or near the visible portion of the spectrum. That is, for these wavelengths, there is destructive interference. (a) What is the thickness of the film? (b) What visible wavelengths are brightest in the reflected interference pattern? (c) If this film were resting on glass with an index of refraction of 1.60, what wavelengths in the visible spectrum would be missing from the reflected light? Picture the Problem (a) We can use the condition for destructive interference in a thin film to find the thickness of the film. (b) and (c) Once we’ve found the thickness of the film, we can use the condition for constructive interference to find the wavelengths in the visible portion of the spectrum that will be brightest in the reflected interference pattern and the condition for destructive interference to find the wavelengths of light missing from the reflected light when the film is placed on glass with an index of refraction greater than that of the film. Chapter 33 1072 (a) Express the condition for destructive interference in the thin film: ...,,,2 25232121 ''''t λλλλ =+ or ...,3,2,2 '''t λλλ= or n m'mt λλ ==2 (1) where m = 1, 2, 3, … and λ′ is the wavelength of the light in the film. Solving for λ yields: m nt2=λ Substitute for the missing wavelengths to obtain: m nt2nm450 = and 1 2nm360 += m nt Divide the first of these equations by the second and simplify to obtain: m m m nt m nt 1 1 2 2 nm360 nm450 += + = Solving for m gives: nm450for 4 == λm Solve equation (1) for t to obtain: n mt 2 λ= Substitute numerical values and evaluate t: ( ) ( ) nm60050.12 nm4504 ==t (b) The condition for constructive interference in the thin film is: ,...3,2,2 21 ''''t λλλλ =+ or ( ) 'm'''t λλλλ 21252321 ,...,,2 +== where λ′ is the wavelength of light in the oil and m = 0, 1, 2, … Substitute for λ′ to obtain: ( ) n mt λ212 += ⇒ 2 1 2 += m ntλ where n is the index of refraction of the film. Substitute numerical values and simplify to obtain: ( )( ) 2 1 2 1 nm1800nm60050.12 +=+= mmλ Interference and Diffraction 1073 Substitute numerical values for m and evaluate λ to obtain the following table: m 0 1 2 3 4 5 λ (nm) 3600 1200 720 514 400 327 From the table, we see that the only wavelengths in the visible spectrum are 720 nm, 514 nm, and 400 nm. (c) Because the index of refraction of the glass is greater than that of the film, the light reflected from the film-glass interface will be shifted by λ21 (as is the wave reflected from the top surface) and the condition for destructive interference becomes: ...,,,2 252321 '''t λλλ= or ( ) n mt λ212 += where n is the index of refraction of the film and m = 0, 1, 2, … Solving for λ yields: 2 1 2 += m ntλ Substitute numerical values and simplify to obtain: ( )( ) 2 1 2 1 nm1800nm6005.12 +=+= mmλ Substitute numerical values for m and evaluate λ to obtain the following table: m 0 1 2 3 4 5 λ (nm) 3600 1200 720 514 400 327 From the table we see that the missing wavelengths in the visible spectrum are 720 nm, 514 nm, and 400 nm. 28 •• A drop of oil (refractive index of 1.22) floats on water (refractive index of 1.33). When reflected light is observed from above, as shown in Figure 33-41 what is the thickness of the drop at the point where the second red fringe, counting from the edge of the drop, is observed? Assume red light has a wavelength of 650 nm. Picture the Problem Because there is a λ21 phase change due to reflection at both the air-oil and oil-water interfaces, the condition for constructive interference is that twice the thickness of the oil film equal an integer multiple of the wavelength of light in the film. Chapter 33 1074 The condition for constructive interference is: ,...3,2,2 '''t λλλ= or 'mt λ=2 (1) where λ′ is the wavelength of light in the oil and m = 1, 2, 3, … Substitute for λ′ to obtain: n mt λ=2 ⇒ n mt 2 λ= Substitute numerical values and evaluate t: ( )( ) ( ) nm53322.12 nm6502 ==t 29 •• [SSM] A film of oil that has an index of refraction of 1.45 rests on an optically flat piece of glass with an index of refraction of 1.60. When illuminated by white light at normal incidence, light of wavelengths 690 nm and 460 nm is predominant in the reflected light. Determine the thickness of the oil film. Picture the Problem Because there is a λ21 phase change due to reflection at both the air-oil and oil-glass interfaces, the condition for constructive interference is that twice the thickness of the oil film equal an integer multiple of the wavelength of light in the film. Express the condition for constructive interference: 'm'''t λλλλ == ,...3,2,2 (1) where λ′ is the wavelength of light in the oil and m = 0, 1, 2, … Substitute for λ′ to obtain: n mt λ=2 ⇒ m nt2=λ where n is the index of refraction of the oil. Substitute for the predominant wavelengths to obtain: m nt2nm690 = and 1 2nm460 += m nt Divide the first of these equations by the second and simplify to obtain: m m m nt m nt 1 1 2 2 nm460 nm690 += + = ⇒ 2=m Solve equation (1) for t: n mt 2 λ=Interference and Diffraction 1075 Substitute numerical values and evaluate t: ( )( ) ( ) nm47645.12 nm6902 ==t 30 •• A film of oil that has an index of refraction 1.45 floats on water. When illuminated with white light at normal incidence, light of wavelengths 700 nm and 500 nm is predominant in the reflected light. Determine the thickness of the oil film. Picture the Problem Because the index of refraction of air is less than that of the oil, there is a phase shift of π rad ( λ21 ) in the light reflected at the air-oil interface. Because the index of refraction of the oil is greater than that of the glass, there is no phase shift in the light reflected from the oil-glass interface. We can use the condition for constructive interference to determine m for λ = 700 nm and then use this value in our equation describing constructive interference to find the thickness t of the oil film. Express the condition for constructive interference between the waves reflected from the air- oil interface and the oil-glass interface: ,...3,2,2 21 ''''t λλλλ =+ or ( ) 'm'''t λλλλ 21252321 ,...,,2 +== (1) where λ′ is the wavelength of light in the oil and m = 0, 1, 2, … Substitute for 'λ and solve for λ to obtain: 21 2 += m ntλ Substitute the predominant wavelengths to obtain: 2 1 2nm700 += m nt and 2 3 2nm500 += m nt Divide the first of these equations by the second to obtain: 21 2 3 2 3 2 1 2 2 nm500 nm700 + += + += m m m nt m nt ⇒ 2=m Solve equation (1) for t: ( ) n mt 22 1 λ+= Substitute numerical values and evaluate t: ( ) ( ) nm60345.12 nm7002 21 =+=t Chapter 33 1076 Newton’s Rings 31 •• [SSM] A Newton’s ring apparatus consists of a plano-convex glass lens with radius of curvature R that rests on a flat glass plate, as shown in Figure 33-42. The thin film is air of variable thickness. The apparatus is illuminated from above by light from a sodium lamp that has a wavelength of 590 nm. The pattern is viewed by reflected light. (a) Show that for a thickness t the condition for a bright (constructive) interference ring is ( )λ212 += mt where m = 0, 1, 2, . . . (b) Show that for t << R, the radius r of a fringe is related to t by r = 2tR . (c) For a radius of curvature of 10.0 m and a lens diameter of 4.00 cm. How many bright fringes would you see in the reflected light? (d) What would be the diameter of the sixth bright fringe? (e) If the glass used in the apparatus has an index of refraction n = 1.50 and water replaces the air between the two pieces of glass, explain qualitatively the changes that will take place in the bright-fringe pattern. Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is air. A phase change of 180° ( λ21 ) occurs at the top of the flat glass plate. We can use the condition for constructive interference to derive the result given in (a) and use the geometry of the lens on the plate to obtain the result given in (b). We can then use these results in the remaining parts of the problem. (a) The condition for constructive interference is: ,...3,2,2 21 λλλλ =+t or ( )λλλλ 21252321 ,...,,2 +== mt where λ is the wavelength of light in air and m = 0, 1, 2, … Solving for t yields: ( ) ...,2,1,0,221 =+= mmt λ (1) (b) From Figure 33-42 we have: ( ) 222 RtRr =−+ or 2222 2 tRtRrR +−+= For t << R we can neglect the last term to obtain: RtRrR 2222 −+≈ ⇒ Rtr 2= (2) (c) Square equation (2) and substitute for t from equation (1) to obtain: ( ) λRmr 212 += ⇒ 2 12 −= λR rm Interference and Diffraction 1077 Substitute numerical values and evaluate m: ( ) ( )( ) fringes.bright 68be will thereso and 67 2 1 nm590m0.10 cm00.2 2 =−=m (d) The diameter of the mth fringe is: ( ) λRmrD 2122 +== Noting that m = 5 for the sixth fringe, substitute numerical values and evaluate D: ( )( )( ) cm14.1 nm590m0.1052 21 = +=D (e) The wavelength of the light in the film becomes λair/n = 444 nm. The separation between fringes is reduced (the fringes would become more closely spaced.) and the number of fringes that will be seen is increased by a factor of 1.33. 32 •• A plano-convex glass lens of radius of curvature 2.00 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 520-nm wavelength. The indexes of refraction of the lens and plate are 1.60. Determine the radii of the first and second bright fringe from the center in the reflected light. Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is air. A phase change of 180° ( λ21 ) occurs at the top of the flat glass plate. We can use the condition for constructive interference and the results from Problem 31(b) to determine the radii of the first and second bright fringes in the reflected light. The condition for constructive interference is: ,...3,2,2 21 λλλλ =+t or ( )λλλλ 21252321 ,...,,2 +== mt where λ is the wavelength of light in air and m = 0, 1, 2, … Solving for t gives: ( ) ...,2,1,0 , 22 1 =+= mmt λ In Problem 31(b) it was shown that: tRr 2= Substitute for t to obtain: ( ) Rmr λ21+= Chapter 33 1078 The first fringe corresponds to m = 0: ( )( ) mm721.0m00.2nm52021 ==r The second fringe corresponds to m = 1: ( )( ) mm25.1m00.2nm52023 ==r 33 ••• Suppose that before the lens of Problem 32 is placed on the plate, a film of oil of refractive index 1.82 is deposited on the plate. What will then be the radii of the first and second bright fringes? Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is oil. A phase change of 180° ( λ21 ) occurs at lens-oil interface. We can use the condition for constructive interference and the results from Problem 31(b) to determine the radii of the first and second bright fringes in the reflected light. The condition for constructive interference is: ,...3,2,2 21 ''''t λλλλ =+ or ( ) 'm'''t λλλλ 21252321 ,...,,2 +== where λ′ is the wavelength of light in the oil and m = 0, 1, 2, … Substitute for λ′ and solve for t: ( ) ...,2,1,0, 22 1 =+= m n mt λ where λ is the wavelength of light in air. In Problem 31(b) it was shown that: tRr 2= Substitute for t to obtain: ( ) n Rmr λ21+= The first fringe corresponds to m = 0: ( )( ) mm535.0 82.1 m00.2nm520 2 1 ==r The second fringe corresponds to m = 1: ( )( ) mm926.0 82.1 m00.2nm520 2 3 ==r Two-Slit Interference Patterns 34 • Two narrow slits separated by 1.00 mm are illuminated by light of wavelength 600 nm, and the interference pattern is viewed on a screen 2.00 m Interference and Diffraction 1079 away. Calculate the number of bright fringes per centimeter on the screen in the region near the center fringe. Picture the Problem The number of bright fringes per unit distance is the reciprocal of the separation of the fringes. We can use the expression for the distance on the screen to the mth fringe to find the separation of the fringes. Express the number N of bright fringes per centimeter in terms of the separation of the fringes: y N Δ= 1 (1) Express the distance on the screen to the mth and (m + 1)st bright fringe: dLmym λ= and ( ) d Lmym λ11 +=+ Subtract the first of these equations from the second to obtain: d Ly λ=Δ Substitute in equation (1) to obtain: L dN λ= Substitute numerical values and evaluate N: ( )( ) 1cm33.8m00.2nm600 mm00.1 −==N 35 • [SSM] Using a conventional two-slit apparatus with light of wavelength 589 nm, 28 bright fringes per centimeter are observed near the center of a screen 3.00 m away. What is the slit separation? Picture the Problem We can use the expression for the distance on the screen to the mth and (m + 1)st bright fringes to obtain an expression for the separation Δy of the fringes as a function of the separation of the slits d. Because the number of bright fringes per unit length N is the reciprocal of Δy, we can find d from N, λ, and L. Express the distance on the screen to the mth and (m + 1)st bright fringe: d Lmym λ= and ( ) d Lmym λ11 +=+ Subtract the first of these equations from the second to obtain: d Ly λ=Δ ⇒ y Ld Δ= λ Because the number of fringes per unit length N is the reciprocal of Δy: LNd λ= Chapter 33 1080 Substitute numerical values and evaluate d: ( )( )( ) mm95.4 m00.3nm589cm28 1 = = −d 36 • Light of wavelength 633 nm from a helium–neon laser is shone normally on a plane containing two slits. The first interference maximum is 82 cm from the central maximum on a screen 12 m away. (a) Find the separation of the slits. (b) How many interference maxima is it, in principle, possible to observe? Picture the Problem We can use the geometry of the setup, represented to the right, to find the separation of the slits. To find the number of interference maxima that, in principle, can be observed, we can apply the equation describing two-slit interference maxima and require that sinθ ≤ 1. d cm 821 =y m 12=L θ θ1 λ 0 Because d << L, we can approximate sinθ1 as: d λθ ≈1sin ⇒ 1sinθ λ≈d (1) From the right triangle whose sides are L and y1 we have: ( ) ( ) 06817.0cm82m12 cm82sin 221 = + =θ Substitute numerical values in equation (1) and evaluate d: m3.9m29.9 06817.0 nm633 μμ ==≈d (b) The equation describing two-slit interference maxima is: ...,2,1,0sin == m,md λθ Because sinθ ≤ 1 determines the maximum number of interference fringes that can be seen: λmaxmd = ⇒ λ dm =max Substitute numerical values and evaluate mmax: 14 nm633 m29.9 max == μm because m must be an integer. Because there are 14 fringes on either side of the central maximum: ( ) 29114212 max =+=+= mN Interference and Diffraction 1081 37 •• Two narrow slits are separated by a distance d. Their interference pattern is to be observed on a screen a large distance L away. (a) Calculate the spacing between successive maxima near the center fringe for light of wavelength 500 nm, when L is 1.00 m and d is 1.00 cm. (b) Would you expect to be able to observe the interference of light on the screen for this situation? (c) How close together should the slits be placed for the maxima to be separated by 1.00 mm for this wavelength and screen distance? Picture the Problem We can use the equation for the distance on a screen to the mth bright fringe to derive an expression for the spacing of the maxima on the screen. In (c) we can use this same relationship to express the slit separation d. (a) Express the distance on the screen to the mth and (m + 1)st bright fringe: d Lmym λ= and ( ) d Lmym λ11 +=+ Subtract the first of these equations from the second to obtain: d Ly λ=Δ (1) Substitute numerical values and evaluate Δy: ( )( ) m0.50 cm00.1 m00.1nm500 μ==Δy (b) According to the Raleigh criterion you could resolve them, but not by much. (c) Solve equation (1) for d to obtain: y Ld Δ= λ Substitute numerical values and evaluate d: ( )( ) mm500.0 mm00.1 m00.1nm500 ==d 38 •• Light is incident at an angle φ with the normal to a vertical plane containing two slits of separation d (Figure 33-43). Show that the interference maxima are located at angles θm given by sin θm + sin φ = mλ/d. Picture the Problem Let the separation of the slits be d. We can find the total path difference when the light is incident at an angle φ and set this result equal to an integer multiple of the wavelength of the light to obtain the given equation. Express the total path difference: msinsin θφ dd +=Δl The condition for constructive interference is: λm=Δl where m is an integer. Chapter 33 1082 Substitute to obtain: λθφ mdd =+ msinsin Divide both sides of the equation by d to obtain: d mλθφ =+ msinsin 39 •• [SSM] White light falls at an angle of 30º to the normal of a plane containing a pair of slits separated by 2.50 μm. What visible wavelengths give a bright interference maximum in the transmitted light in the direction normal to the plane? (See Problem 38.) Picture the Problem Let the separation of the slits be d. We can find the total path difference when the light is incident at an angle φ and set this result equal to an integer multiple of the wavelength of the light to relate the angle of incidence on the slits to the direction of the transmitted light and its wavelength. Express the total path difference: θφ sinsin dd +=Δl The condition for constructive interference is: λm=Δl where m is an integer. Substitute to obtain: λθφ mdd =+ sinsin Divide both sides of the equation by d to obtain: d mλθφ =+ sinsin Set θ = 0 and solve for λ: m d φλ sin= Substitute numerical values and simplify to obtain: ( ) mm m25.130sinm50.2 μμλ =°= Evaluate λ for positive integral values of m: m λ (nm) 1 1250 2 625 3 417 4 313 From the table we can see that 625 nm and 417 nm are in the visible portion of the electromagnetic spectrum. Interference and Diffraction 1083 40 •• Two small loudspeakers are separated by 5.0 cm, as shown in Figure 33-44. The speakers are driven in phase with a sine wave signal of frequency 10 kHz. A small microphone is placed a distance 1.00 m away from the speakers on the axis running through the middle of the two speakers, and the microphone is then moved perpendicular to the axis. Where does the microphone record the first minimum and the first maximum of the interference pattern from the speakers? The speed of sound in air is 343 m/s. Picture the Problem The diagram shows the two speakers, S1 and S2, the central-bright image and the first-order image to the left of the central-bright image. The distance y is measured from the center of the central-bright image. We can apply the conditions for constructive and destructive interference from two sources and use the geometry of the speakers and microphone to find the distance to the first interference minimum and the distance to the first interference maximum. S S21 θ d L y Relate the distance Δy to the first minimum from the center of the central maximum to θ and the distance L from the speakers to the plane of the microphone: L y=θtan ⇒ θtanLy = (1) Interference minima occur where: ( )λθ 21sin += md where m = 0, 1, 2, 3, … Solve for θ to obtain: ( ) ⎥⎦ ⎤⎢⎣ ⎡ += − d m λθ 211sin Relate the wavelength λ of the sound waves to the speed of sound v and the frequency f of the sound: f v=λ Substitute for λ in the expression for θ to obtain:( ) ⎥⎦ ⎤⎢⎣ ⎡ += − df vm 211sinθ Substituting for θ in equation (1) yields: ( ) ⎭⎬ ⎫ ⎩⎨ ⎧ ⎥⎦ ⎤⎢⎣ ⎡ += − df vmLy 2 1 1sintan (2) Chapter 33 1084 Noting that the first minimum corresponds to m = 0, substitute numerical values and evaluate Δy: ( ) ( )( )( )( ) cm37kHz10cm0.5 m/s343 sintanm00.1 2 1 1 min1st =⎭⎬ ⎫ ⎩⎨ ⎧ ⎥⎦ ⎤⎢⎣ ⎡= −y The maxima occur where: λθ md =sin where m = 1, 2, 3, … For diffraction maxima, equation (2) becomes: ⎭ ⎬⎫⎩⎨ ⎧ ⎥⎦ ⎤⎢⎣ ⎡= − af mvLy 1sintan Noting that the first maximum corresponds to m = 1, substitute numerical values and evaluate y: ( ) ( )( )( )( ) cm94kHz10cm0.5 m/s3431sintanm00.1 1max1st =⎭⎬ ⎫ ⎩⎨ ⎧ ⎥⎦ ⎤⎢⎣ ⎡= −y Diffraction Pattern of a Single Slit 41 • Light that has a 600-nm wavelength is incident on a long narrow slit. Find the angle of the first diffraction minimum if the width of the slit is (a) 1.0 mm, (b) 0.10 mm, and (c) 0.010 mm. Picture the Problem We can use the expression locating the first zeroes in the intensity to find the angles at which these zeroes occur as a function of the slit width a. The first zeroes in the intensity occur at angles given by: a λθ =sin ⇒ ⎟⎠ ⎞⎜⎝ ⎛= − a λθ 1sin (a) For a = 1.0 mm: mrad60.0mm0.1 nm600sin 1 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= −θ (b) For a = 0.10 mm: mrad0.6mm10.0 nm600sin 1 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= −θ (c) For a = 0.010 mm: mrad60mm010.0 nm600sin 1 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= −θ Interference and Diffraction 1085 42 • Plane microwaves are incident on the thin metal sheet that has a long, narrow slit of width 5.0 cm in it. The microwave radiation strikes the sheet at normal incidence. The first diffraction minimum is observed at θ = 37º. What is the wavelength of the microwaves? Picture the Problem We can use the expression locating the first zeroes in the intensity to find the wavelength of the radiation as a function of the angle at which the first diffraction minimum is observed and the width of the plate. The first zeroes in the intensity occur at angles given by: a λθ =sin ⇒ θλ sina= Substitute numerical values and evaluate λ: ( ) cm0.337sincm0.5 =°=λ 43 ••• [SSM] Measuring the distance to the moon (lunar ranging) is routinely done by firing short-pulse lasers and measuring the time it takes for the pulses to reflect back from the moon. A pulse is fired from Earth. To send the pulse out, the pulse is expanded so that it fills the aperture of a 6.00-in-diameter telescope. Assuming the only thing spreading the beam out is diffraction and that the light wavelength is 500 nm, how large will the beam be when it reaches the Moon, 3.82 × 105 km away? 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. L Ddtelescope dmirror Relate the diameter D of the beam when it reaches the moon to the distance to the moon L and the beam divergence angle θ : LD θ≈ (1) Chapter 33 1086 The angle θ subtended by the first diffraction minimum is related to the wavelength λ of the light and the diameter of the telescope opening dtelescope by: telescope 22.1sin d λθ = Because θ << 1, sinθ ≈ θ and: telescope 22.1 d λθ ≈ Substitute for θ in equation (1) to obtain: telescope 22.1 d LD λ= Substitute numerical values and evaluate D: ( ) ( ) km53.1 cm10 m1 in cm2.54in00.6 nm50022.1m1082.3 2 8 = ⎥⎥ ⎥⎥ ⎦ ⎤ ⎢⎢ ⎢⎢ ⎣ ⎡ ×× ×=D Interference-Diffraction Pattern of Two Slits 44 • How many interference maxima will be contained in the central diffraction maximum in the interference–diffraction pattern of two slits if the separation of the slits is exactly 5 times their width? How many will there be if the slit separation is an integral multiple of the slit width (that is d = na) for any value of n? Picture the Problem We need to find the value of m for which the mth interference maximum coincides with the first diffraction minimum. Then there will be 12 −= mN fringes in the central maximum. The number of fringes N in the central maximum is: 12 −= mN (1) Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating: a λθ =1sin Express the angle θm corresponding to the mth interference maxima in terms of the separation d of the slits: d m m λθ =sin Interference and Diffraction 1087 Because we require that θ1 = θm, we can equate these expressions to obtain: ad m λλ = ⇒ a dm = Substituting for d and simplifying yields: 55 == a am Substitute for m in equation (1) to obtain: ( ) 9152 =−=N If d = na: n a na a dm === and 12 −= nN 45 •• [SSM] A two-slit Fraunhofer interference–diffraction pattern is observed using light that has a wavelength equal to 500 nm. The slits have a separation of 0.100 mm and an unknown width. (a) Find the width if the fifth interference maximum is at the same angle as the first diffraction minimum. (b) For that case, how many bright interference fringes will be seen in the central diffraction maximum? Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the fifth interference maximum occurs to find a. We can then find the number of bright interference fringes seen in the central diffraction maximum using .12 −= mN (a) Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating: a λθ =1sin Express the angle θ5 corresponding to the mth fifth interference maxima maximum in terms of the separation d of the slits: d λθ 5sin 5 = Because we require that θ1 = θm5, we can equate these expressions to obtain: ad λλ =5 ⇒ 5 da = Substituting the numerical value of d yields: m0.20 5 mm100.0 μ==a (b) Because m = 5: ( ) 915212 =−=−= mN Chapter 33 1088 46 •• A two-slit Fraunhofer interference–diffraction pattern is observed using light that has a wavelength equal to 700 nm. The slits have widths of 0.010 mm and are separated by 0.20 mm. How many bright fringes will be seen in the central diffraction maximum? Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the mth interference maximum occurs to find m. We can then find the number of bright interference fringes seen in the central diffraction maximum using .12 −= mN The number of fringes N in the central maximum is: 12 −= mN (1) Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating: a λθ =1sin Express the angle θm corresponding to the mth interference maxima in terms of the separation d of the slits: d m m λθ =sin Because we require that θ1 = θm, we can equate these expressions to obtain: ad m λλ = ⇒ a dm = Substitute for m in equation (1) to obtain: 12 −= a dN Substitute numerical values and evaluate N:( ) 391 mm010.0 mm20.02 =−=N 47 •• Suppose that the central diffraction maximum for two slits has 17 interference fringes for some wavelength of light. How many interference fringes would you expect in the diffraction maximum adjacent to one side of the central diffraction maximum? Picture the Problem There are 8 interference fringes on each side of the central maximum. The secondary diffraction maximum is half as wide as the central one. It follows that it will contain 8 interference maxima. Interference and Diffraction 1089 48 •• Light that has a wavelength equal to 550 nm illuminates two slits that both have widths equal to 0.030 mm and separations equal to 0.15 mm. (a) How many interference maxima fall within the full width of the central diffraction maximum? (b) What is the ratio of the intensity of the third interference maximum to one side of the center interference maximum to the intensity of the center interference maximum? Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the mth interference maximum occurs to find m. We can then find the number of bright interference fringes seen in the central diffraction maximum using .12 −= mN In (b) we can use the expression relating the intensity in a single-slit diffraction pattern to phase constant θλ πφ sin2 a= to find the ratio of the intensity of the third interference maximum to one side of the center interference maximum. (a) The number of fringes N in the central maximum is: 12 −= mN (1) Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating: a λθ =1sin Express the angle θm corresponding to the mth interference maxima in terms of the separation d of the slits: d m m λθ =sin Because we require that θ1 = θm, we can equate these expressions to obtain: ad m λλ = ⇒ a dm = Substitute in equation (1) to obtain: 12 −= a dN Substitute numerical values and evaluate N: ( ) 91 mm030.0 mm15.02 =−=N (b) Express the intensity for a single- slit diffraction pattern as a function of the phase difference φ: 2 2 1 2 1 0 sin ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= φ φII (2) where θλ πφ sin2 a= Chapter 33 1090 For m = 3: d λθ 3sin 3 = and ⎟⎠ ⎞⎜⎝ ⎛=⎟⎠ ⎞⎜⎝ ⎛== d a d aa πλλ πθλ πφ 632sin2 3 Substitute numerical values and evaluate φ: 5 6 mm15.0 mm030.06 ππφ =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= Solve equation (2) for the ratio of I3 to I0: 2 2 1 2 1 0 3 sin ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= φ φ I I Substitute numerical values and evaluate I3/I0: 25.0 5 6 2 1 5 6 2 1sin 2 0 3 = ⎥⎥ ⎥⎥ ⎦ ⎤ ⎢⎢ ⎢⎢ ⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛ ⎟⎠ ⎞⎜⎝ ⎛ = π π I I Using Phasors to Add Harmonic Waves 49 • [SSM] Find the resultant of the two waves whose electric fields at a given location vary with time as follows: iE ˆsin2 01 tA ω= r and ( )iE ˆsin3 2302 πω += tAr . Picture the Problem Chose the coordinate system shown in the phasor diagram. We can use the standard methods of vector addition to find the resultant of the two waves. y x δ 1E r 2E r E r The resultant of the two waves is of the form: ( )iEE δω += tsinrr (1) The magnitude of E r is: ( ) ( ) 02020 6.332 AAA =+=Er The phase angle δ is: rad 98.0 2 3tan 0 01 −=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −= − A Aδ Interference and Diffraction 1091 Substitute for E r and δ in equation (1) to obtain: ( )iE rad 98.0sin6.3 0 −= tA ωr 50 • Find the resultant of the two waves whose electric fields at a given location vary with time as follows: iE ˆsin4 01 tA ω= r and ( )iE ˆsin3 6102 πω += tAr . Picture the Problem Chose the coordinate system shown in the phasor diagram. We can use the standard methods of vector addition to find the resultant of the two waves. y x δ °60 1E r 2E r E r The resultant of the two waves is of the form: ( )iEE δω += tsinrr (1) Apply the law of cosines to the magnitudes of the scalars to obtain: °−+= 120cos2 1 22 1 2 22 EEEEE rrrrr or °−+= 120cos2 1 22 1 22 EEEEE rrrrr Substitute for 1E r and 2E r and evaluate E r to obtain: ( ) ( ) ( )( ) 0002020 08.6120cos34234 AAAAA =°−+=Er Applying the law of sines yields: EE rr °= 120sinsin 2 δ Solve for δ to obtain: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ °= − E E r r 120sin sin 21δ Substitute numerical values and evaluate δ: rad 43.008.6 120sin3sin 0 01 =⎥⎦ ⎤⎢⎣ ⎡ °= − A Aδ Chapter 33 1092 Substitute for E r and δ in equation (1) to obtain: ( )iE rad 43.0sin1.6 0 += tA ωr Remarks: We could have used the law of cosines to find R and the law of sines to find δ. 51 •• Monochromatic light is incident on a sheet with a long narrow slit (Figure 33-45). Let I0 be the intensity at the central maximum of the diffraction pattern on a distant screen, and let I be the intensity at the second intensity maximum from the central intensity maximum. The distance from this second intensity maximum to the far edge of the slit is longer than the distance from the second intensity maximum to the near edge of the slit by approximately 2.5 wavelengths. What is the ratio if I to I0? Picture the Problem We can evaluate the expression for the intensity for a single-slit diffraction pattern at the second secondary maximum to express I2 in terms of I0. The intensity at the second secondary maximum is given by: ⇒⎥⎦ ⎤⎢⎣ ⎡= 2 2 1 2 1 0 sin φ φ II 2 2 1 2 1 0 sin ⎥⎦ ⎤⎢⎣ ⎡= φ φ I I where θλ πφ sin2 a= At this second secondary maximum: λθ 2 5sin =a and πλλ πφ 5 2 52 =⎟⎠ ⎞⎜⎝ ⎛= Substitute for φ and evaluate 0I I : 0162.0 2 5 2 5sin 2 0 = ⎥⎥ ⎥⎥ ⎦ ⎤ ⎢⎢ ⎢⎢ ⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛ = π π I I 52 •• Monochromatic light is incident on a sheet that has three long narrow parallel equally spaced slits a distance d apart. (a) Show that the positions of the interference minima on a screen a large distance L away from the sheet that has the three equally spaced slits (spacing d, with d >> λ) are given approximately by dLmym 3λ= where m = 1, 2, 4, 5, 7, 8, 10, . . . that is, m is not a multiple of 3. (b) For a screen distance of 1.00 m, a light wavelength of 500 nm, and a source spacing of 0.100 mm, calculate the width of the principal interference maxima (the distance between successive minima) for three sources. Interference and Diffraction 1093 Picture the Problem We can use phasor concepts to find the phase angle δ in terms of the number of phasors N (three in this problem) forming a closed polygon of N sides at the minima and then use this information to express the path difference Δr for each of these locations. Applying a small angle approximation, we can obtain an expression for y that we can evaluate for enough of the path differences to establish the pattern given in the problem statement. (a) Express the phase angle δ in terms of the number of phasors N forming a closed polygon of N sides: ⎟⎠ ⎞⎜⎝ ⎛= N m πδ 2 where m = 1, 2, 3, 4, 5, 6, ,7, … For three equally spaced sources, the phase angle is: ⎟⎠ ⎞⎜⎝ ⎛= 3 2πδ m Express the path difference corresponding to this phase angle toobtain: 32 λδπ λ mr =⎟⎠ ⎞⎜⎝ ⎛=Δ (1) Interference maxima occur for: m = 3, 6, 9, 12, … Interference minima occur for: m = 1, 2, 4, 5, 7, 8, … (Note that m is not a multiple of 3.) Express the path difference Δr in terms of sinθ and the separation d of the slits: θsindr =Δ or, provided the small angle approximation is valid, L ydr =Δ ⇒ r d Ly Δ= Substituting for Δr from equation (1) yields: ...,8,7,5,4,2,1 , 3min == m d Lmy λ (b) For L = 1.00 m, λ = 500 nm, and d = 0.100 mm: ( )( ) ( ) mm33.3mm0.1003 m00.1nm50022 min ==y 53 •• [SSM] Monochromatic light is incident on a sheet that has four long narrow parallel equally spaced slits a distance d apart. (a) Show that the positions of the interference minima on a screen a large distance L away from four equally spaced sources (spacing d, with d >> λ) are given approximately by dLmym 4λ= where m = 1, 2, 3, 5, 6, 7, 9, 10, . . . that is, m is not a multiple of 4. (b) For a screen distance of 2.00 m, light wavelength of 600 nm, and a source Chapter 33 1094 spacing of 0.100 mm, calculate the width of the principal interference maxima (the distance between successive minima) for four sources. Compare this width with that for two sources with the same spacing. Picture the Problem We can use phasor concepts to find the phase angle δ in terms of the number of phasors N (four in this problem) forming a closed polygon of N sides at the minima and then use this information to express the path difference Δr for each of these locations. Applying a small angle approximation, we can obtain an expression for y that we can evaluate for enough of the path differences to establish the pattern given in the problem statement. (a) Express the phase angle δ in terms of the number of phasors N forming a closed polygon of N sides: ⎟⎠ ⎞⎜⎝ ⎛= N m πδ 2 where m = 1, 2, 3, 4, 5, 6, ,7, … For four equally spaced sources, the phase angle is: ⎟⎠ ⎞⎜⎝ ⎛= 2 πδ m Express the path difference corresponding to this phase angle to obtain: 42 λδπ λ mr =⎟⎠ ⎞⎜⎝ ⎛=Δ (1) Interference maximum occur for: m = 3, 6, 9, 12, … Interference minima occur for: m = 1, 2, 4, 5, 7, 8, … (Note that m is not a multiple of 3.) Express the path difference Δr in terms of sinθ and the separation d of the slits: θsindr =Δ or, provided the small angle approximation is valid, L ydr =Δ ⇒ r d Ly Δ= Substituting for Δr from equation (1) yields: ,...9,7,6,5,3,2,1 , 4min == m d Lmy λ (b) For L = 2.00 m, λ = 600 nm, d = 0.100 mm, and n = 1: ( )( ) ( ) mm00.6mm0.1004 m00.2nm60022 min ==y For two slits: ( ) d Lmy λ21min 22 += Interference and Diffraction 1095 For L = 2.00 m, λ = 600 nm, d = 0.100 mm, and m = 0: ( )( ) mm0.12 mm0.100 m00.2nm6002 min ==y The width for four sources is half the width for two sources. 54 •• Light of wavelength 480 nm falls normally on four slits. Each slit is 2.00 μm wide and the center-to-center separation between it and the next slit is 6.00 μm. (a) Find the angular width of the of the central intensity maximum of the single-slit diffraction pattern on a distant screen. (b) Find the angular position of all interference intensity maxima that lie inside the central diffraction maximum. (c) Find the angular width of the central interference. That is, find the angle between the first intensity minima on either side of the central intensity maximum. (d) Sketch the relative intensity as a function of the sine of the angle. Picture the Problem We can use aλθ =sin to find the first zeros in the intensity pattern. The four-slit interference maxima occur at angles given by ,md λθ =sin where m = 0,1,2, … . In (c) we can use the result of Problem 53 to find the angular spread between the central interference maximum and the first interference minimum on either side of it. In (d) we’ll use a phasor diagram for a four-slit grating to find the resultant amplitude at a given point in the intensity pattern as a function of the phase constant δ, that, in turn, is a function of the angle θ that determines the location of a point in the interference pattern. (a) The first zeros in the intensity occur at angles given by: a λθ =sin ⇒ ⎟⎠ ⎞⎜⎝ ⎛= − a λθ 1sin Substitute numerical values and evaluate θ : mrad242m00.2 nm480sin 1 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= − μθ (b) The four-slit interference maxima occur at angles given by: λθ md =sin ⇒ ⎥⎦ ⎤⎢⎣ ⎡= − d m m λθ 1sin where m = 0, 1, 2, 3, … Substitute numerical values to obtain: ( ) ( )m m m 0800.0sin m00.6 nm480sin 1 1 − − = ⎥⎦ ⎤⎢⎣ ⎡= μθ Chapter 33 1096 Evaluate θm for m = 0, 1, 2, and 3: ( )[ ] 00800.00sin 10 == −θ ( )[ ] mrad1.800800.01sin 11 ±== −θ ( )[ ] mrad1610800.02sin 12 ±== −θ ( )[ ] rad.24200800.03sin 13 ±== −θ where θ3 will not be seen as it coincides with the first minimum in the diffraction pattern. (c) From Problem 53: d n 4min λθ = For n = 1: ( ) mrad20m00.64 nm480 min == μθ (d) Use the phasor method to show the superposition of four waves of the same amplitude A0 and constant phase difference .sin 2 θλ πδ d= A0 A0 A A0 A0 f α δ δ δ δ δ " " δ ' φ φ α Express A in terms of δ ′ and δ ′′: ( )'A''AA δδ coscos2 00 += (1) Because the sum of the external angles of a polygon equals 2π: πδα 232 =+ Interference and Diffraction 1097 Examining the phasor diagram we see that: πδα =+ '' Eliminate α and solve for ''δ to obtain: δδ 23='' Because the sum of the internal angles of a polygon of n sides is (n − 2)π : πδφ 323 =+ '' From the definition of a straight angle we have: πδδφ =+− ' Eliminate φ between these equations to obtain: δδ 21=' Substitute for ''δ and 'δ in equation (1) to obtain: ( )δδ 21230 coscos2 += AA Because the intensity is proportional to the square of the amplitude of the resultant wave: ( )221230 coscos4 δδ += II The following graph of I/I0 as a function of sinθ was plotted using a spreadsheet program. The diffraction envelope was plotted using ,sin4 2 2 1 2 1 2 0 ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= φ φ I I where .sin2 θλ πφ a= Note the excellent agreement with the results calculated in (a), (b) and (c). Chapter 33 1098 -2 0 2 4 6 8 10 12 14 16 18 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 sin(theta) I /I 0 intensity diffraction envelope 55 ••• [SSM] Three slits, each separated from its neighbor by 60.0 μm, are illuminated at the central intensity maximum by a coherent light source of wavelength 550 nm. The slits are extremely narrow. A screen is located 2.50 m from the slits. The intensity is 50.0 mW/m2. Consider a location 1.72 cm from the central maximum. (a) Draw a phasor diagram suitable for the addition of the three harmonic waves at that location. (b) From the phasor diagram, calculate the intensity of light at that location. Picture the Problem We can find the phase constant δ from the geometry of the diagram to the right. Using the value of δ found in this fashion we can express the intensity at the point 1.72 cm from the centerline in terms of the intensity on the centerline. On the centerline, the amplitude of the resultant wave is 3 times that of each individual wave and the intensity is 9 times that of each source acting separately.m 50.2=L cm .721y θ (a) Express δ for the adjacent slits: θλ πδ sin2 d= For θ << 1, θθθ ≈≈ tansin : L y=≈ θθ tansin Interference and Diffraction 1099 Substitute to obtain: L dy λ πδ 2= Substitute numerical values and evaluate δ : ( )( ) ( )( ) °== = 270rad 2 3 m50.2nm550 cm72.1m0.602 π μπδ The three phasors, 270° apart, are shown in the diagram to the right. Note that they form three sides of a square. Consequently, their sum, shown as the resultant R, equals the magnitude of one of the phasors. δ δ A A A 0 0 0 0R = A (b) Express the intensity at the point 1.72 cm from the centerline: 2RI ∝ Because I0 ∝ 9R2: 2 2 0 9R R I I = ⇒ 9 0II = Substitute for I0 and evaluate I: 22 mW/m56.5 9 mW/m0.50 ==I 56 ••• In single-slit Fraunhofer diffraction, the intensity pattern (Figure 33- 11) consists of a broad central maximum with a sequence of secondary maxima to either side of the central maximum. This intensity is given by 2 2 1 2 1 0 sin ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= φ φ II , where φ is the phase difference between the wavelets arriving from the opposite edges of the slits. Calculate the values of φ for the first three secondary maxima to one side of the central maximum by finding the values of φ for which dI/dφ is equal to zero. Check your results by comparing your answers with approximate values for φ of 3π, 5π and 7π. (That these values for φ are approximately correct at the secondary intensity maxima is discussed in the discussion surrounding Figure 33-27.) Chapter 33 1100 Picture the Problem We can use the phasor diagram shown in Figure 33-26 to determine the first three values of φ that produce secondary intensity maxima. Setting the derivative of Equation 33-19 equal to zero will yield a transcendental equation whose roots are the values of φ corresponding to the intensity maxima in the diffraction pattern. Referring to Figure 33-26 we see that the first subsidiary maximum occurs where: πφ 3= An intensity minimum occurs where: πφ 4= Another intensity maximum occurs where: πφ 5= Thus, secondary intensity maxima occur where: ( ) ...,3,2,112 =+= n,n πφ and the first three secondary intensity maxima are at πππφ 7 and ,5,3= The intensity in the single-slit diffraction pattern is given by: 2 2 1 2 1 0 sin ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= φ φ II Set the derivative of this expression equal to zero for extreme values (relative minima and maxima): ( ) extremafor 0 sincossin 2 2 2 1 2 1 2 1 2 1 4 1 2 1 2 1 0 =⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ − ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= φ φφφ φ φ φ Id dI Simplify to obtain the transcendental equation: φφ 2121tan = Solve this equation numerically (use the ″Solver″ function of your calculator or trial-and-error methods) to obtain: πππφ 6.94 and 92.4 86.2 ,,= At the three intensity minima φ = 2π, 4π, and 6π , and at the three intensity maxima φ = 2.86π, 4.92π, and 6.94π. At the intensity maxima φ ≈ 3π, 5π, and 7π. Interference and Diffraction 1101 Remarks: Note that our results in (b) are smaller than the approximate values found in (a) by 4.9%, 1.6%, and 0.86% and that the agreement improves as n increases. Diffraction and Resolution 57 • [SSM] Light that has a wavelength equal to 700 nm is incident on a pinhole of diameter 0.100 mm. (a) What is the angle between the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern? (b) What is the distance between the central maximum and the first diffraction minimum on a screen 8.00 m away? Picture the Problem We can use D λθ 22.1= to find the angle between the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern and the diagram to the right to find the distance between the central maximum and the first diffraction minimum on a screen 8 m away from the pinhole. L miny θ D (a) The angle between the central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern is given by: D λθ 22.1= Substitute numerical values and evaluate θ : mrad54.8mm100.0 nm70022.1 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛=θ (b) Referring to the diagram, we see that: θtanmin Ly = Substitute numerical values and evaluate ymin: ( ) ( ) cm83.6 mrad54.8tanm00.8min = =y 58 • Two sources of light that both have wavelengths equal to 700 nm are 10.0 m away from the pinhole of Problem 57. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh’s criterion? Chapter 33 1102 Picture the Problem We can apply Rayleigh’s criterion to the overlapping diffraction patterns and to the diameter D of the pinhole to obtain an expression that we can solve for Δy. Pinhole α α Δy L c c Rayleigh’s criterion is satisfied provided: D λα 22.1c = Relate αc to the separation Δy of the light sources: L yΔ≈cα provided αc << 1. Equate these expressions to obtain: DL y λ22.1=Δ ⇒ D Ly λ22.1=Δ Substitute numerical values and evaluate Δy: ( )( ) cm54.8 mm100.0 m0.10nm70022.1 ==Δy 59 • Two sources of light that both have wavelengths of 700 nm are separated by a horizontal distance x. They are 5.00 m from a vertical slit of width 0.500 mm. What is the smallest value of x for which the diffraction pattern of the sources can be resolved by Rayleigh’s criterion? Picture the Problem We can use Rayleigh’s criterion for slits and the geometry of the diagram to the right showing the overlapping diffraction patterns to express x in terms of λ, L, and the width a of the slit. c c Lα α x Interference and Diffraction 1103 Referring to the diagram, relate αc, L, and x: L x≈cα For slits, Rayleigh’s criterion is: a λα =c Equate these two expressions to obtain: aL x λ= ⇒ a Lx λ= Substitute numerical values and evaluate x: ( )( ) mm00.7 mm500.0 m00.5nm700 ==x 60 •• The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 6.0 mm. (a) Using light that has a wavelength of 500 nm, how far could you be from this tile and still resolve these holes? Assume the diameter of the pupil of your eye is about 5.0 mm. (b) Could you resolve these holes better using red light or using violet light? Explain your answer. 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 λ, x, and the diameter D of your pupil. Your pupil α α x L c c (a) Referring to the diagram, relate αc, L, and x: L x≈cα provided α << 1 For circular apertures, Rayleigh’s criterion is: D λα 22.1c = Equate these two expressions to obtain: DL x λ22.1= ⇒ λ22.1 xDL = Chapter 33 1104 Substitute numerical values and evaluate L: ( )( ) ( ) m49nm50022.1 mm0.5mm0.6 ==L (b) Because L is inversely proportional to λ, the holes can be resolved better with violet light which has a shorter wavelength. The critical angle for resolution is proportional to the wavelength. Thus, the shorter the wavelength the farther away you can be and still resolve the two images. 61 •• [SSM] The telescope on Mount Palomar has a diameter of 200 in. Suppose a double star were 4.00 light-years away.
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