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Chapter 37 Questions: 1, 3, 4, 5, 6, 8, 9, 10, 11, 13, 14, 16 Problems: 1, 3, 9, 11, 15, 17, 25, 31, 36, 41, 49, 51, 55, 57, 61 Questions 1. The maximum intensity in a di¤raction grating occurs at d sin = m ; where is the wavelength. As increases must also increase. Thus a di¤raction grating bends red light more than blue. 3. Yes, as long as the di¤erence in length was less than the coherence length of the light. 4. It becomes too thin for any constructive interference, see example 37-4. 5. Eyeglasses are too thick. This means that the order of interference is so large that all interference e¤ects are smeared out. 6. If your …ngers are close enough together it is possible to secondary di¤raction peaks typical of single-slit di¤raction. 8. The wavelength of light is much shorter than the wavelength of sound. Hence di¤ractions e¤ects allow us to hear around corners, but not see around them. 9. The EM …elds from the two slits of the monochromatic light wave that add. It is the square sum of these …elds that gives the intensity and the cross term that results from this sum gives rise to interference. 10. It is the …elds that add not the intensity. Thus the intensity is proportional to I / (E1 + E2 )2 = E12 + 2E1 E2 + E22 ; I = I1 + I2 + 2 hE1 E2 i : For incoherent sources the time average of E1 E2 vanishes. Thus we can simply add the intensities. 11. The widths of the peaks due to constructive interference are much narrower for a grating versus two slits. This increase in resolution enables a di¤raction grating to resolve interference peaks between two slightly di¤erent wavelenghts. 13. For a broad slit, the intensity pattern is tightly centered about the centerline. As the slit narrows this intensity pattern broadens and it is possible to observe small secondary peaks in the intensity on the wings of this pattern. Continuing to narrow the slit results in a very broad intensity pattern and the secondary peaks occur at greater separation.. 1 14. In the center both wavelengths of light form a constructive intensity peak. For higher order peaks, the red light occurs at larger angles than the blue, see …gure 37-7. 16. You need several di¤erent orientations in order to see the lattice spacings in di¤erent directions through the crystal (Not all crystals are simple cubic.) Problems 1. The constructive interference pattern is determined from the criterion d sin = m : 10 2 ; we …nd For m = 1; d = 15 m, and sin = :071=2:2 = 3:227 = d sin = 15 10 6 3:227 10 2 = 484nm. 3. (a) The constructive interference pattern is determined from the criterion d sin = m : For adjacent …gures at small angles this expression reduces to = 5mm=L; = 633 10 6 mm, and d = :12mm, we have 633 10 9 = 12 10 55 10 L 3 !L= 60 10 633 10 =d : For 8 9 L = 95cm (b) The fringe spacing, `; is proportional to the wavelength, hence `=5 480 = 3:8mm 633 9. The constructive interference pattern is determined from the criterion d sin = m : If instead the path di¤erence is 1=4 of a wavelength versus a full wavelength then sin 1 1 600 10 6 = = 6 10 4 ; 4d 4 :25 180 ' 6 10 4 rad = 6 10 4 = :034 : = 2 11. The constructive interference pattern is determined from the criterion d sin = m : (a) The height of the m fringe is (L is the distance to the screen) ym = L tan ym = L q d2 m =L sin cos m m m sin m sin2 1 m = Lq m =d 1 (m )2 =d2 (m )2 For the case d = 6:5 m, fringe is y2 = 1:7 q = Lp = :633 m, and L = 1:7m the height to the m = 2 2 (6:5 633 10 6 )2 9 10 (2 = 33:76cm 633 10 9 )2 The height to the m = 1 fringe is y1 = 1:7 q (6:5 633 10 10 6 )2 9 = 16:63cm (633 The separation between these two fringes, y = 33:76 10 9 )2 y; is 16:63 = 17:13cm (b) The height to the m = 4 fringe is y4 = 1:7 q 4 (6:5 633 10 6 )2 9 10 (4 = 71:90cm 633 10 9 )2 The height to the m = 3 fringe is y4 = 1:7 q 3 (6:5 10 633 6 )2 9 10 (3 = 51:93cm 633 The separation between these two fringes, y = 71:09 10 9 )2 y; is 51:93 = 19:97cm 3 15. (a) The maxima occur at the same locations as that for two-slit di¤raction. Hence sin 1 sin 2 :633 10 6 = :0844 ! 1 = 4:8 d 7:5 10 6 2 2 :633 10 6 = = = :169 ! 2 = 9:7 d 7:5 10 6 = = (b) The minima occur when d sin = m : N For a grating with 5 slits (N = 5) the third minima occurs when m = 3: The wavelength is = 633nm and the slit spacing is d = 7:5 m: Hence the angular location for the third minima is sin = 3 3 = 5:064 5d 10 2 ! 3 = 2:9 : There are only 4 minima between maxima. Hence the 6th minima occurs when m = N + 2; sin 6 = 1+ 2 5 d = :1182 ! 6 = 6:79 17. The criterion for constructive interference is the same for a di¤raction grating as it is for twin slits. For 3000 lines/cm, the slit spacing is d = 10 2 m=3000 = 1=3 10 5 m. Thus (a) sin 1 1 520 10 9 = :1560, d 10 5 = :1566rad = 8:975 : = = 3 (b) For the …fth order fringe sin 5 5 520 10 d 10 5 = :8947rad = 51:3 : = = 5 3 9 = :78; 25. The angular location for the 6th order 484:3nm is found from 6 5 7 sin = 484:3nm = 581:2nm = 415:1nm: d d d 4 Hence it could be a 5th order 581nm line or a 7th order 415nm line. 31. The criterion for constructive interference in a thin …lm is 2d = (m + 1=2) n; where n is the wavelength of the light inside the …lm. Light with a wavelength of = 550nm will have a wavelength of n = =n = 550nm=1:33 = 413:5nm, inside the soap …lm. The minimum thickness for constructive interference occurs at m = 0; hence dmin = n =4 = 103:4nm: 36. For the oil …lm ‡oating on water the …rst interface (air to oil) there is a phase shift during the re‡ection. At the second interface (oil to water) that is an additional phase shift. These will cancel each other out. Hence the condition for constructive interference is 2d = m For green light, = m =1:25: oil = 500nm; the minimum thickness is d= 2:5 = 200nm: For the soap …lm suspended in air there is only one the condition for constructive interference is 2d = (m + 1=2) soap phase shift. Hence = (m + 1=2) =1:333: The minimum thickness (m = 0) is d= 4 (4=3) = 93:75nm 41. The criterion for constructive interference in a thin …lm is again 2d = (m + 1=2) 5 n; only now n = 1: The maximum thickness is d = :065mm = 65 From the expression just above we have m = 2d= 1=2 = 130 103 =550 = 236:4 103 nm. :5 = 235:9: So the largest m is m = 235: Counting the fringe for m = 0 results in 236 observable fringes. 49. The intensity pattern for di¤raction from a single slit of width a is given by 2 sin (( a sin ) = ) : S = So ( a sin ) = The …rst minima occurs when the argument of the sine function is equal to : Thus = a sin : When this occurs at =2, the sine function is unity, and = a: 51. The wavelength for a 29M Hz radio wave is = 10:34m. The intensity pattern for di¤raction from a single slit of width a is given by S = So sin (( a sin ) = ) ( a sin ) = 2 : For a slit spacing of a = 45m, the …rst minima occurs at ( a sin ) = = sin = ; 10:34 = :23; a 45: :232rad = 13:3 : = = The full angular width is = 26:6 : 55. For a circular aperture the Rayleigh criteria is 1:22 : D For 5cm resolution from a distance of 100km with a wavelength of 550nm, the diameter of the lens must be sin D = 1:22 550 min = 10 9 = :05=105 = 1:22 110 10 2 = = 1:342m. 57. If the radius of the spot is given by r; then the subtended angle for a spot at the focal length is simply r=f: From the Rayleigh criteria the minimum angle is 1:22 r = : min = D f 6 Since f =D = 1:4, and the wavelength of light is r = 1:22 = 580nm, we have 1:4 = 990nm = :99 m. Thus the diameter of the spot is d = 1:98 m. 61. For a circular aperture the Rayleigh criteria is min = 1:22 : D Now one arcsec is 1=3600 of a degree or 4:85 10 6 rad. The diameter of an aperature for this minimum angle for = 550nm is D= 1:22 550 10 4:85 10 6 9 = :138m = 13:8cm. 7