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f5ch36 - CHAPTER 36 INTERFERENCE 983 CHAPTER 36 Answer to...

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Unformatted text preview: CHAPTER 36 INTERFERENCE 983 CHAPTER 36 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (a) top; (b) bright intermediate illumination (phase dierence is 2:1 wavelengths (a) 3, 3; (b) 2:5, 2:5 a and d tie (amplitude of resultant wave is 4E0 ), then b and c tie (amplitude of resultant wave is 2E0 ) (a) 1 and 4; (b) 1 and 4 b (least n), c, a Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. (a) peak; (b) valley (a) 300 nm; (b) exactly out of phase (a) 2d; (b) (odd number)=2 (c) (a) intermediate closer to maximum, m = 2; (b) minimum, m = 3; (c) intermediate closer to maximum, m = 2; (d) maximum, m = 1 (a) increase; (b) 1 (a){(c) decrease; (d) blue down (a) and (c) tie, then (b) and (d) tie (zero) (a) maximum; (b) minimum; (c) alternates The moving milk particles in the partially coherent sunlight produce eeting speckle. a, c, b d (a) third longest; (b) 2 (a) 0:5 wavelength; (b) 1 wavelength 984 CHAPTER 36 INTERFERENCE 16. 17. 18. 19. (a) no; (b) 2(0) = 0; (c) 2L bright L1 = =4n1 and L2 = =4n2 all Solutions to Exercises & Problems (a) (b) (c) 1E c :00 108 f = = 3589 10 m/s = 5:09 1014 Hz : 9m v nm 0 = f = c=n = n = 58952 = 388 nm : f 1: v = 0 f = (388 10 9 m)(5:09 1014 Hz) = 1:97 108 m/s : 2E 1 v = vs vd = c n s = (3:00 108 m/s) 1 1:77 nd 1 1 = 4:55 107 m/s : 2:42 3E i' i A' r' r B A B' Refer to the ray diagram shown above. Consider a wave front AA0 of the incident wave. Suppose the time it takes for the wave at point A0 to travel to point B 0 is t, then A0 B 0 = vt, CHAPTER 36 INTERFERENCE 985 where v is the speed of light in the medium. Meanwhile the wave which was re ected at point A must have reached point B , where AB = vt. Thus AB cos = AB 0 = A B0 = cos ; AB or = . This is equivalent to the law of re ection. 4E 0 0 1 92 108 n = v = 3::00 108 m/s = 1:56 : c m/s The index of refraction of fused quartz at = 550 nm is about 1:459, which is obtained from Fig. 34-19. So 5E c 108 v = n = 3:00 :459 m/s = 2:06 108 m/s : 1 Vmin = c=n = (3:00 108 m/s)=1:54 = 1:95 108 m/s: The travel time would be greater when the tube is lled with air, as the speed of light v in air is less than that in vacuum. The time dierence is 7E 6E d t = d v = d 1 n1 c c air 3m 1 1:609 10 = 3:00 108 m/s 1 1:00029 = 1:55 10 9 s : 8P Refer to Fig. 18-22. Suppose that the stick starts from point S1 at t = 0 and reaches point S at time t, then S1 S = vt. Meanwhile the wave front produced by the stick at t = 0 at point S1 has grown to a circular cross section of radius r = ut. Thus a conical wave front is set up, with v sin = SrS = ut = u : vt 1 986 CHAPTER 36 INTERFERENCE Apply the law of refraction: sin = sin 30 = vs =vd . So 9P = sin 1 vs sin 30 = sin vd 1 (3:0 m/s) sin 30 = 22 : 4:0 m/s The angle of incidence is gradually reduced due to refraction, such as shown in the calculation above (from 30 to 22 ). Eventually after many refractions will be virtually zero. This is why most waves come in normal to a shore. Set up a coordinate system as shown. Let A = (0; Ay ) and B = (Bx ; By ). If OC = x then CD = Bx x. The time t it takes for light to travel form point A to B is 10P y A t = tAC + tCB = p A2 + x2 x c=n1 q + 2 (Bx x)2 + By 1 c=n2 : O x C D n1 n2 x To minimize t, set dt=dx = 0: dt c dx =n1 p 2 A2 + x2 x x ! 2 5 n2 4 q Bx x 2 (Bx x)2 + By =n1 sin 1 n2 sin 2 = 0 : 3 B Thus n1 sin 1 = n2 sin 2 . (a) The time t2 it takes for pulse 2 to travel through the plastic is 11P L L L L t2 = c=1:55 + c=1:70 + c=1:60 + c=1:45 = 6:30L : c Similarly for pulse 1 L L t1 = c=2L59 + c=1:65 + c=1:50 = 6:33L : 1: c CHAPTER 36 INTERFERENCE 987 So pulse 2 travels through the plastic in less time. (b) t = t2 t1 = 6:33L 6:30L = 0:03L : c c c 12P 2 Now 1 = air =n1 , where air is the wavelength in air and n1 is the index of refraction of the glass. Similarly 2 = air =n2 , where n2 is the index of refraction of the plastic. This means that the phase dierence is 1 2 = (2=air )(n1 n2 )L. The value of L that makes this 5:65 rad is 9 m) 1 L = (2(n 2 )n air = 5:65(400 10:50) = 3:60 10 6 m : 2(1:60 1 1 2) (b) 5:65 rad is less than 2 rad (= 6:28 rad), the phase dierence for completely constructive interference, and greater than rad (= 3:14 rad), the phase dierence for completely (a) Take the phases of both waves to be zero at the front surfaces of the layers. The phase of the rst wave at the back surface of the glass is given by 1 = k1 L !t, where k1 (= 2=1 ) is the angular wave number and 1 is the wavelength in glass. Similarly, the phase of the second wave at the back surface of the plastic is given by 2 = k2 L !t, where k2 (= 2=2 ) is the angular wave number and 2 is the wavelength in plastic. The angular frequencies are the same since the waves have the same wavelength in air and the frequency of a wave does not change when the wave enters another medium. The phase dierence is 1 1 L: = (k k )L = 2 1 2 1 2 1 destructive interference. The interference is therefore intermediate, neither completely constructive nor completely destructive. It is, however, closer to completely constructive than to completely destructive. Use (a) (b) (c) 13P L = !t = 2 v T 2 L = 2 L v1 T c=n2 L = 2L jn n j : c=n1 2 1 a = (8:50 10 6 m) (1:60 1:50) = 1:70 : 2 500 10 9 m b = (8:50 10 6 m) (1:72 1:62) = 1:70 : 2 500 10 9 m c = (3:25 10 6 m) (1:79 1:59) = 1:30 : 2 500 10 9 m 988 CHAPTER 36 INTERFERENCE (d) Since a = b the brightness must be the same for (a) and (b). As for (c) since c =2 and a =2 each diers from an integer by 0:30, the brightness in case (c) is also the same as that in (a) and (b). (a) Use the formula obtained in 13P. Let = 2L jn1 n2 j = (2n + 1) which yields (n = 0; 1; 2; ) ; 14P Lmin = Ljn=0 = 2jn n j = 2j1:620 nm:65j = 1550 nm = 1:55 m : 45 1 1 2 (b) For the next greater one n = 1, so L1 = 2jn 3 n j = 3(1:55m) = 4:65 m : 1 2 15P (a) The wavelength 0 in vacuum is related with the wavelength in a medium by 0 = =n, where n is the index of refraction of the medium. Thus the phase dierence in terms of the wavelength is given by L1 L2 + L1 L2 0 =n2 0 0 =n1 = 600:1 nm [(1:40)(4:00 m) + (4:00 m 3:50 m) (1:60)(3:50 m)] 0 = 0:800 : (b) Intermediate, closer to constructive interference, since the phase dierence in terms of the wavelength is 0.800, which is closer to 1 (fully constructive case) than to 0.5 (fully destructive case). Use Eq. 36-14 with m = 3: (a) 1 m = sin = sin d 16E 1 (b) = (0:216 rad)(180 = rad) = 12:4 : 2(550 10 9 m) = 0:216 rad : 7:70 10 6 m CHAPTER 36 INTERFERENCE 989 For the rst dark fringe 1 = , and for the second one 2 = 3, etc. For the mth one m = (2m + 1). In Fig. 36-8(a) sin y=D so the fringe separation (say, between two adjacent bright fringes) is m = D m = D ; y = (D sin ) = D sin = D d d d Where we used Eq. 36-14. To keep y / D=d a constant we need to double D if d is doubled. 19E 18E 17E The condition for a maximum in the two-slit interference pattern is d sin = m, where d is the slit separation, is the wavelength, m is an integer, and is the angle made by the interfering rays with the forward direction. If is small, sin may be approximated by in radians. Then d = m and the angular separation of adjacent maxima, one associated with the integer m and the other associated with the integer m + 1, is given by = =d. The separation on a screen a distance D away is given by y = D = D=d. Thus 10 9 m)(5:40 y = (500 1:20 10 3 m m) = 2:25 10 3 m = 2:25 mm : In the case of a distant screen the angle is close to zero so sin . Thus from Eq. 36-14 sin = m = m = ; d d d or d = = 589 10 9 m=0:018 rad = 3:3 10 5 m = 33 m: 21E 20E The angular positions of the maxima of a two-slit interference pattern are given by d sin = m, where d is the slit separation, is the wavelength, and m is an integer. If is small, sin may be approximated by in radians. Then d = m. The angular separation of two adjacent maxima is = =d. Let 0 be the wavelength for which the angular separation is 10:0% greater. Then 1:10=d = 0 =d or 0 = 1:10 = 1:10(589 nm) = 648 nm. 990 CHAPTER 36 INTERFERENCE (b) Since y1 = D sin 1 = (50:0 cm) sin(0:010 rad) = 5:0 mm, the separation is y = y1 y0 = y1 0 = 5:0 mm: 0 For the fth maximum y5 = D sin 5 = D(4=d), and for the seventh minimum y7 = 0 D sin 7 = D[(7 + 1=2)=d]. So 0 y5 = D (7 + 1=2) D 4 = 3D y = y7 d d 2d 9 m)(20 10 2 m) = 1:6 10 3 m = 1:6 mm : = 3(546 1010 10 3 m) 2(0: 24E 23E (a) For the maximum adjacent to the central one m = 1, so 1 m 1 (1)() = 0:010 rad : 1 = sin d m=1 = sin 100 22E From the formula = =d obtained in 20E we have, in our case, 0 0 = = v=f = (c=n)=f = = = 0:20 = 0:15 : d d d nd n 1:33 Interference maxima occur at angles such that d sin = m, where d is the separation of the sources, is the wavelength, and m is an integer. Since d = 2:0 m and = 0:50 m, this means that sin = 0:25m. You want all values of m (positive and negative) for which j0:25mj 1. These are 4, 3, 2, 1, 0, +1, +2, +3, and +4. For each of these except 4 and +4 there are 2 dierent values for . A single value of ( 90 ) is associated with m = 4 and a single value ( 90 ) is associated with m = +4. There are 16 dierent angles in all and therefore 16 maxima. Initially, source A leads source B by 90 , which is equivalent to 1/4 wavelength. However, source A also lags behind source B since rA is longer than rB by 100 m, which is 100 m=400 m = 1=4 wavelength. So the net phase dierence between A and B at the detector is zero. The maxima of a two-slit interference pattern are at angles given by d sin = m, where d is the slit separation, is the wavelength, and m is an integer. If is small, sin may 27P 26E 25P CHAPTER 36 INTERFERENCE 991 be replaced by in radians. Then d = m. The angular separation of two maxima associated with dierent wavelengths but the same value of m is = (m=d)(2 1 ) and the separation on a screen a distance D away is y = D tan D = mD (2 1 ) d 0 = 5:03(1:10m) m (600 10 9 m 480 10 9 m) = 7:2 10 5 m : 3 The small angle approximation tan was made. must be in radians. For the rst maximum m = 0 and for the tenth one m = 9. So the separation is y = (D=d)m = 9D=d. Solve for : 28P dy = (0:15 10 3 m)(18 10 3 m) = 6:0 10 7 m = 600 nm : = 9D 9(50 10 2 m) 29P Let the distance in question be x. Then where m = 0; 1; 2; : Solve for x: jAB j = jB A j = 2 ( d2 + x2 xm ) = (2m + 1) ; m p 2 + xm = (2md+ 1) (2m 4 1) : To obtain the largest value of xm put m = 0: x0 = d 30P 2 = (3:00)2 4 = 8:75 : 4 (a) Use y = D=d (see 18E) to solve for d: m)(632 d = D = 2(20:0 10:0 cm:8 nm) = 0:253 mm : y (b) In this case the interference pattern will be shifted. For example, since at the location of the original central maximum the phase dierence is now = (kL) = kL = (2=)(2:50) = 5:0; there will now be a minimum instead of a maximum. 992 CHAPTER 36 INTERFERENCE Let the m = 10 bright fringe on the screen be a distnace y from the central maximum. Then from Fig. 31-8(a) 31P r1 r2 = (y + d=2)2 + D2 p p (y d=2)2 + D2 = 10 ; from which we may solve for y. To the order of (d=D)2 we nd where y0 = 10D=d. Let y y0 to nd the percent error to be 2 y0 (y0 + d2 =4) = 1 10 2 + 1 d 2 2y D2 2 D 8 D 0 2 2 1 5:89 1 mm = 2 2000 m + 8 2:0mm = 0:03% : m 40 y(y2 + d2 =4) ; y = y0 + 2D2 For constant phase dierence 32P = k(r1 r2 ) = 2 (r1 r2 ) ; so r1 r2 = =2 = const.. The curve described by this equation is a hyperbola. Consider the two waves, one from each slit, that produce the seventh bright fringe in the absence of the mica. They are in phase at the slits and travel dierent distances to the seventh bright fringe, where they are out of phase by 2m = 14. Now a piece of mica with thickness x is placed in front of one of the slits and the waves are no longer in phase at the slits. In fact, their phases at the slits dier by 2x 2x = 2x (n 1) ; 33P m where m is the wavelength in the mica and n is the index of refraction of the mica. The relationship m = =n was used to substitute for m . Since the waves are now in phase at the screen 2x (n 1) = 14 or 9 x = n7 1 = 7(550:58 10 1 m) = 6:64 10 6 m = 6:64 m : 1 CHAPTER 36 INTERFERENCE 993 Use the result of 33P. Now 34P jj = j1 so t t 2 j = 2 (n1 1) 2 (n2 1) = 2m ; t = jn mn j = j5(480 nm)j = 8:0 103 nm = 8:0 m : 1:7 1:4 1 2 See the phasor diagram to the right. We have y = A cos(!t + ), where 35E y = 17 and A = A2 + A2 = (A2 cos 30 + A1 )2 + (A2 sin 30 )2 x y p = (8:0 cos 30 + 10)2 + (8:0 sin 30 )2 1 q p A2 A A1 x A :0 30 = tan Ay = tan 1 8:08cossin + 10 = 13 : 30 x Thus y = y1 + y2 = A cos(!t + ) = 17 cos(!t + 13 ). 36E The phasor diagram is shown to the right. Here E1 = 1:00, E2 = 2:00, and = 60 . The resultant amplitude Em is given by the trigonometric law of cosines: 2 2 2 Em = E1 + E2 2E1 E2 cos(180 ) ; Em E2 so Em = (1:00)2 + (2:00)2 2(1:00)(2:00) cos 120 = 2:65 : p E1 The angular separation between adjacent maxima satis es (d sin ) (d) = d = (m) = m = ; 37E 994 CHAPTER 36 INTERFERENCE or The intensity pattern is shown below. 1 nm = 0600 mm = 0:0010 rad : d :60 I() /4I0 0.5 0 0 0.001 0.002 0.003 0.004 (rad) The phasor is shown to the right. For the resultant phasor 38E y A1 A2 Ax = A2 cos 30 + A1 + A3 cos 45 = 15 cos 30 + 10 + 5 cos 45 = 26:5 A3 A x and Ay = A2 sin 30 A3 sin 45 = 15 sin 30 5 sin 45 = 3:96 : Thus and = tan 1 Ay = tan Ax q 1 3:96 = 8:5 ; 26:5 y = A sin(!t + ) = A2 + A2 sin(!t + ) x y p = (26:5)2 + (3:96)2 sin(!t + 8:5 ) = 27 sin(!t + 8:5 ) : CHAPTER 36 INTERFERENCE 995 39P (a) To get to the detector the wave from A travels a distance x and the wave from B travels p a distance d2 + x2 . The dierence in phase of the two waves is = 2 p d2 + x2 x ; where is the wavelength. For a maximum in intensity this must be a multiple of 2. Solve p d2 + x2 x = m p for x. Here m is an integer. Write the equation as d2 + x2 = x + m, then square both sides to obtain d2 + x2 = x2 + m2 2 + 2mx. The solution is m x = d 2m : The largest value of m that produces a positive value for x is m = 3. This corresponds to the maximum that is nearest A, at 2 00 2 x = (4:00 m) 9(1:m) m) = 1:17 m : (2)(3)(1:00 2 2 2 For the next maximum m = 2 and x = 3:00 m. For the third maximum m = 1 and x = 7:50 m. (b) Minima in intensity occur where the phase dierence is rad; the intensity at a minimum, however, is not zero because the amplitudes of the waves are dierent. Although the amplitudes are the same at the sources, the waves travel dierent distances to get to the points of minimum intensity and each amplitude decreases in inverse proportion to the distance traveled. 40P According to Eqs. 36-21 and 36-22, the intensity is given by I = 4I0 where cos2 ; 2 Here d is the slit separation and is the wavelength. The intensity at the center of the interference pattern is 4I0 , so you want the value of for which I = 2I0 . First solve 2I0 = 4I0 cos2 ( 1 ) for : 2 1 1 p = rad : = 2 cos 2 2 = 2d sin : 996 CHAPTER 36 INTERFERENCE = 2d sin 2 for . Since is small, sin , provided is measured in radians. Then = 2d 2 and Now solve Another point of half intensity, at = =4d, is symmetrically placed relative to the central point of the pattern, so the width of the pattern at half intensity is 2(=4d) = =2d. 41P = 4 : d Take the electric eld of one wave, at the screen, to be E1 = E0 sin(!t) and the electric eld of the other to be 2E0 E E2 = 2E0 sin(!t + ) ; where the phase dierence is given by = 2d sin : E0 t Here d is the center-to-center slit separation and is the wavelength. The resultant wave can be written E = E1 + E2 = E sin(!t + ), where is a phase constant. The phasor diagram is shown above. The resultant amplitude E is given by the trigonometric law of cosines: 2 2 2 E 2 = E0 + (2E0 )2 4E0 cos(180 ) = E0 (5 + 4 cos ) : The intensity is given by I = I0 (5 + 4 cos ), where I0 is the intensity that would be produced by the rst wave if the second were not present. Since cos = 2 cos2 (=2) 1, this may also be written I = I0 1 + 8 cos2 (=2) . If you choose to express I in terms of Im , the intensity of the central maximum, note that at the central maximum = 0, where Im = I0 [(1 + 8 cos2 (0)] = 9I0 , or I0 = Im =9. Thus I () = 1 Im 1 + 8 cos2 9 2 : CHAPTER 36 INTERFERENCE 997 (a) Let = k(2L) = (2=)(2L) = and solve for L: 42E L = = 6204nm = 155 nm : 4 (b) Suppose that the silver has to be moved by L. Then the additional phase dierence between the two light waves as a result of the motion of the silver is 0 = (2=)(2L). Let 0 = 2 to obtain L = = 6202nm = 310 nm : 2 Now = + k(2L) = + (2=)(2L) = 2m, so 43E L = (2m 1) ; 4 44E m = 1; 2; 3 The wave re ected from the front surface suers a phase change of rad since it is incident in air on a medium of higher index of refraction. The phase of the wave re ected from the back surface does not change on re ection since the medium beyond the soap lm is air and has a lower index of refraction than the lm. If L is the thickness of the lm this wave travels a distance 2L further than the wave re ected from the front surface. The phase dierence of the two waves is 2L(2=f )+ , where f is the wavelength in the lm. If is the wavelength in vacuum and n is the index of refraction of the soap lm, then f = =n and the phase dierence is 2nL 2 + = 2(1:33)(1:21 10 6 m) 585 210 9 m + = 12 rad : Since the phase dierence is an even multiple of the interference is completely constructive. For constructive interference use Eq. 36-34: 2n2 L = (m + 1=2). For the two smallest valuses of L let m = 0 and 1: 624 nm L0 = =2 = 4(1:33) = 117 nm = 0:117 m ; 2n2 1 L1 = (1 + n=2) = 23n = 3L0 = 3(0:1173m) = 0:352 m : 2 2 2 45E 998 CHAPTER 36 INTERFERENCE Since the lens has a greater index of refraction than the lm there is a -phase shift upon re ecting from the lens- lm boundary, which cancels with the -phase shift due to the re ection from the lm-air boundary. Therefore there is no net -phase shift, and the condition for destructive interference is 2n2 L = (n + 1=2). For the smallest value of L let m = 0: 680 nm Lmin = 4 = 4(1:30) = 131 nm = 0:131 m : n2 Use the formula obtained in 46E: 47E 46E Lmin = 4 = 4(125) = 0:200 : n2 : Use Eq. 36-34 for constructive interference: 2n2 L = (m + 1=2), or 50)(410 1230 nm nL = m2+21=2 = 2(1:m + 1=2nm) = m + 1=2 ; where m = 0; 1; 2; : The only value of m which, when substituted into the equation above, would yield a wavelength which falls within the visible light range, is m = 1. So = 12301nm = 492 nm : 1 + =2 49E 48E Light re ected from the front surface of the coating suers a phase change of rad while light re ected from the back surface does not change phase. If L is the thickness of the coating, light re ected from the back surface travels a distance 2L further than light re ected from the front surface. The dierence in phase of the two waves is 2L(2=c ) , where c is the wavelength in the coating. If is the wavelength in vacuum, then c = =n, where n is the index of refraction of the coating. Thus the phase dierence is 2nL(2=) . For fully constructive interference this should be a multiple of 2. Solve 2 = 2m 2nL for L. Here m is an integer. The solution is L = (2m4+ 1) : n CHAPTER 36 INTERFERENCE 999 To nd the smallest coating thickness, take m = 0. Then 9 L = 4 = 5604(210 m = 7:00 10 8 m : n :00) 50E For complete destructive interference, you want the waves re ected from the front and back of the coating to dier in phase by an odd multiple of rad. Each wave is incident on a medium of higher index of refraction from a medium of lower index, so both suer phase changes of rad on re ection. If L is the thickness of the coating, the wave re ected from the back surface travels a distance 2L further than the wave re ected from the front. The phase dierence is 2L(2=c ), where c is the wavelength in the coating. If n is the index of refraction of the coating, c = =n, where is the wavelength in vacuum, and the phase dierence is 2nL(2=). Solve 2 = (2m + 1) 2nL for L. Here m is an integer. The result is To nd the least thickness for which destructive interference occurs, take m = 0. Then 9 L = 4 = 6004(110 m = 1:2 10 7 m : n :25) Let the thickness of the structure at a certain section be t = L. The condition for constructive interference is 1 t = L = (m +n =2) ; 2 2 51P L = (2m4+ 1) : n which gives where m = 0; 1; 2; : You can check that no integer values of m would produce any of the given values of (1; 2; 1=2; 3 and 1=10). So none of the sections will provide the right thickness for constructive interference. If the expression m = 2Ln2 gives the condition for constructive interferene then there should be no net -phase shift due to the two re ections from the lm-media boundary. 52P + = ( 2)(600 = (m 2n1L2) = 2(1m + 1=00 10nm) = 2m + 1 ; 3 nm) :50)(4: 40 2 1000 CHAPTER 36 INTERFERENCE This requires that either n2 > n1 and n2 > n3 , or n2 < n1 and n2 < n3 . Here n1 ; n3 are the indices of refraction for the upper and lower media, respectively. You can easily check that this condition is satis ed in cases (a) and (c). (a) In this case there are -phase shifts for both of the waves re ected, so there is no net -phase shift, thus m = 2Ln2 . Solve for : 53P = 2Ln2 = 2(460 nm)(1:20) = 1104 nm : m m m In the visible light range the only possible value for m is 2, so = 1104 nm=2 = 552 nm: (b) Now there is a -phase shift upon re ection o the kerosen-air boundary but no -phase shift o the water-kerosen boundary. So a net -phase shift is present, and for constructive interference Ln 1104 nm = m2+ 12=2 = 2(460 nm)(1:20) = m + 1=2 : m + 1=2 In the visible light range m can only be 2, in which case = 1104 nm=(1 + 1=2) = 442 nm: There is no net -phase shift in this case so for constructive interference = 2Ln2 =m, i.e., 8 > > < 54P Ln 1 = 700 nm = 2m 2 1 2Ln2 : > > = 500 nm = : 2 m1 + 1 Solve for L: (700 nm)(500 L = 2n (1 2 ) = 2(1:30)(700 nm nm)nm) = 673 nm : 500 2 1 2 55P For the maximum at 1 = 600 nm 2 1 = 600 nm = m2Ln1=2 ; + 1 and for the minimum at 2 = 450 nm Ln 2 2 = 450 nm = 2m 2 = mLn21 : 2 1+ CHAPTER 36 INTERFERENCE 1001 Solve for L: (600 nm)(450 L = 4n (1 2 ) = 4(1:30)(600 nm nm)nm) = 338 nm : 450 2 1 2 (a) In this case there is no -phase shift (see 52P) and the condition for constructive interference is m = 2Ln2 . Solve for L: (525 nm) m L = 2n = m2(1:55) = (169 nm)m : 2 For the minimum value of L, let m = 1 to obtain Lmin = 169 nm: (b) The light of wavelength (other than the 525 nm- one) that would also be preferentially transmitted satis es m0 = 2n2 L, or = 2n2 L = 2(1:55)(169 nm) = 525 nm : 56P visible spectrum will be preferentially transmitted. They are, in fact, re ected. (c) For a sharp reduction of transmission let 525 = m2n2 L=2 = m0 +nm2 ; 0+1 1= where m0 = 0; 1; 2; 3; . In the visible light range m0 = 1 and = 350 nm. This corresponds to the blue-violet light. 57P m0 Here m0 = 2; 3; 4; . . . (note that m0 = 1 corresponds to the = 525 nm light so it should not be included here). Since the minimum value of m0 is 2, you can easily verify that no m0 will give a value of which falls into the visible light range. So no other parts of the m0 m0 Light re ected from the upper oil surface (in contact with air) changes phase by rad. Light re ected from the lower surface (in contact with glass) changes phase by rad if the index of refraction of the oil is less than that of the glass and does not change phase if the index of refraction of the oil is greater than that of the glass. First suppose the index of refraction of the oil is greater than the index of refraction of the glass. The condition for fully destructive interference is 2no d = m, where d is the thickness of the oil lm, no is the index of refraction of the oil, is the wavelength in vacuum, and m is an integer. For the shorter wavelength 2no d = m1 1 and for the longer 2no d = m2 2 . Since 1 is less than 2 , m1 is greater than m2 and since fully destructive interference does not occur for any wavelengths between, m1 = m2 + 1. Solve (m2 + 1)1 = m2 2 for m2 . The result is 500 m2 = 1 = 700 nm nm nm = 2:50 : 500 2 1 Since m2 must be an integer the oil cannot have an index of refraction that is greater than that of the glass. 1002 CHAPTER 36 INTERFERENCE Now suppose the index of refraction of the oil is less than that of the glass. The condition for fully destructive interference is then 2no d = (2m + 1). For the shorter wavelength 2mo d = (2m1 + 1)1 and for the longer 2no d = (2m2 + 1)2 . Again m1 = m2 + 1, so (2m2 + 3)1 = (2m2 + 1)2 . This means the value of m2 is 3(500 nm) 700 nm 3 m2 = 2(1 2 ) = 2(700 nm 500 nm) = 2:00 : 2 1 This is an integer. Thus the index of refraction of the oil is less than that of the glass. In this case there is a -phase shift (see 52P). So for destructive interference at 1 = 600 nm we have m1 1 = 2n2 L, and for constructimve interference at 2 = 700 nm (m2 +1=2)2 = 2n2 L. Thus (m2 + 1=2)2 = m1 1 , or 58P m2 + 1=2 = 1 = 700 nm = 7 : m1 2 600 nm 6 This gives m1 = m2 = 3. Thus nm) L = m1 1 = 3(70025) = 840 nm = 0:840 m : 2n2 2(1: 59P In this case the path traveled by ray no.2 is longer than that of ray no.1 by 2L cos r , in stead of 2L. Here sin i = sin r = n2 , or r = sin 1 (sin i =n2 ). So we replace 2L by 2L cos r in Eqs. 36-34 and 36-35 to obtain 1 2n2 L cos r = m + 2 for the maxima and for the minima. 60P (m = 0; 1; 2; ) (m = 0; 1; 2; ) 2n2 L cos r = m You can check easily that the condition for no net -phase shift in the transmitted light is n2 > n1 , n3 or n2 < n1 , n3 ; and that in the re ected light is n1 < n2 < n3 or n3 < n2 < n1 . (a) There is no net -phase shift for transmission in this case so for maximum transmission m = 2n2 L, or Lmin = =2n2 . (b) There is no net -phase shift for re ection so for minimum re ection (m +1=2) = 2n2 L, or Lmin = =4n2 . CHAPTER 36 INTERFERENCE 1003 (c) Now m = 2n2 L for maximum re ection, so Lmin = =2n2 . Use Eq. 36-21: I = 4I0 cos2 (=2) = Imax cos2 (=2), where = k(2n2 L)+ = (2=)(2n2 L) +. At = 450 nm 61P I = cos2 = cos2 2n2 L + Imax 2 2 = cos2 2(1:38)(99:6 nm) + = 0:883 ; 450 nm 2 and at = 650 nm max I = cos2 2(1:38)(99:6 nm) + = 0:942 : I 650 nm 2 62P Consider the interference of waves re ected from the top and bottom surfaces of the air lm. The wave re ected from the upper surface does not change phase on re ection but the wave re ected from the bottom surface changes phase by rad. At a place where the thickness of the air lm is L the condition for fully constructive interference is 2L = (m + 1 ), where 2 (= 683 nm) is the wavelength and m is an integer. The largest value of m for which L is less than 0:048 m is 140. Note that for m = 140, L= and for m = 141, (m + 1 ) (140:5)(683 10 9 m) 2 = = 4:798 10 5 m 2 2 9 L = (141:5)(683 10 m) = 4:83 10 5 m = 48:1 m : 2 At the thin end of the air lm there is a bright fringe associated with m = 0. There are therefore 141 bright fringes in all. (a) The left end is dark, becasue one of the two re ected beams of light undergoes a -phase shift when it re ects from the upper surface of the bottom glass plate, while the other does not suer such a phase shift since it is re ected from the lower surface of the upper plate. (b) The blue end, because the wavelengths there are shorter, which correspond to lower values of the thickness of the air gap between the two glass plates that satisfy the condition for fully destrcutive interference to occur. 63P 1004 CHAPTER 36 INTERFERENCE (a) Every time one more destructive (constructive) fringe appears the increase in thickness of the air gap is =2. Now that there are 6 more destructive fringes in addition to the one at point A, the thickness at B is tB = 6(=2) = 3(600 nm) = 1:80 m: (b) We must now replace by 0 = =nw . Since tB is unchanged tB = N (0 =2) = N (=2nw ), or N = 2tB nw = 2(3)nw = 6nw = 6(1:33) = 8 : Assume the wedge-shaped lm is in air, so the wave re ected from one surface undergoes a phase change of rad while the wave re ected from the other surface does not. At a place where the lm thickness is L the condition for fully constructive interference is 2nL = (m + 1 ), where n is the index of refraction of the lm, is the wavelength in 2 vacuum, and m is an integer. The ends of the lm are bright. Suppose the end where the lm is narrow has thickness L1 and the bright fringe there corresponds to m = m1 . Suppose the end where the lm is thick has thickness L2 and the bright fringe there corresponds to m = m2 . Since there are 10 bright fringes m2 = m1 + 9. Subtract 2nL1 = (m1 + 1 ) from 2 2nL2 = (m1 + 9 + 1 ) to obtain 2n L = 9, where L = L2 L1 is the change in the 2 lm thickness over its length. Thus 9 = 9(630 10 9 m) = 1:89 10 6 m : L = 2n 2(1:50) Use the result of 64P, part (a). The dierence in thickness is given by m = 480 nm (16 6) = 2400 nm = 2:4 m : t = 2 2 In a medium with index of refraction n the wavelength 0 of an electromagnetic wave is given by 0 = =n, where is the corresponding wavelength in vacuum. This means that x0 = x=n. The total length of the wedge x then satis es x = N x = N 0 x0 = N 0 x=n, where N = 4000 and N 0 = 4001. Solve for n: 0 4001 n = N = 4000 = 1:00025 : N Consider the interference pattern formed by waves re ected from the upper and lower surfaces of the air wedge. The wave re ected from the lower surface undergoes a rad 68P 67P 66P 65P 64P CHAPTER 36 INTERFERENCE 1005 phase change while the wave re ected from the upper surface does not. At a place where the thickness of the wedge is d the condition for a maximum in intensity is 2d = (m + 1 ), 2 where is the wavelength in air and p is an integer. Thus d = (2m + 1)=4. As the m geometry of Fig. 40{31 shows, d = R R2 r2 , where R is the radius of curvature of the p lens and r is the radius of a Newton's ring. Thus (2m + 1)=4 = R R2 r2 . Solve for r. First rearrange the terms so the equation becomes p + R2 r2 = R (2m 4 1) : Now square both sides and solve for r2 . When you take the square root you should get 2 2 r = (2m + 1)R (2m + 1) : 2 16 r If R is much larger than a wavelength the rst term dominates the second and r = (2m + 1)R : 2 r (a) Solve m from the last formula obtained in 68P: 3 2 r2 1 m = R 2 = (5:(20 10 m=2) m) 1 = 33 : 0 m)(589 10 9 2 69P (b) Counting the largest one, the total number of bright rings is 33 + 1 = 34. Replace by 0 = =nw : r2 m0 = R0 1 = nw r2 2 R 1 = (1:33)(20 10 3 m=2)2 2 (5:0 m)(589 10 9 m) 1 = 45 : 2 So the number of bright rings is 45 + 1 = 46. Solve m From the formula r = (2m + 1)R=2 obtained in 68P to nd m = r2 =R 1=2. Now, when m is changed to m + 20, r becomes r0 , so m + 20 = r02 =R 1=2. Taking the dierence between the two equations above, we eliminate m and nd 02 2 2 (0 162 cm)2 F = r 20r = (0:368 cm) 10 :7 cm) = 1:00 m : 20(546 70P p 1006 CHAPTER 36 INTERFERENCE From the result of 68P 71P dr r = rm+1 rm dm m m=1 s r r d (2m + 1)R = 1 R 1 R : = dm 2 2 m + 1=2 2 m 72P S 1 R a 2 2 x B a a I D A Here L1 = jSRj = D2 + (a x)2 and L2 = jSB j + jBRj. To simplify the calculation, introduce the image of point S in the water, point I . Obviously jSB j = jIB j, and I , B and R are located on the same straight line. Thus L2 = jIB j + jBRj = jIRj, where jIRj may be calculated by noting that in the right-angled triangle p p jIRj = jIAj2 + jRAj2 = D2 + (a + x)2 : For D a x we may use the approximation 2 p 2 + (a x)2 D 1 + 1 a x D ; 2 D so the condition for maximum reception reduces to 1 a + x 2 D 1 + 1 a x 2 = 2ax = m + 1 ; L2 L1 D 1 + 2 D 2 D D 2 Refer to the ray diagram shown above. Two paths lead from the source (S ) to the receiver (R): path 1, the direct one; and path 2, the one through a re ection from the water surface. Such a re ection causes a phase change of , so the condition for maximum reception is given by 1 ; L2 L1 = m + 2 (m = 0; 1; 2; . . . ) p CHAPTER 36 INTERFERENCE 1007 or 1 x = m + 2 D : 2a 73E A shift of one fringe corresponds to a change in the optical path length of one wavelength. When the mirror moves a distance d the path length changes by 2d since the light traverses the mirror arm twice. Let N be the number of fringes shifted. Then 2d = N and 3 = 2d = 2(0:23379210 m) = 5:88 10 7 m = 588 nm : N According to Eq. 36-41, the number of fringes shifted (N ) due to the insertion of the lm of thickness L is N = (2L=)(n 1). So 74P :0) L = 2(N1) = (589 :nm)(71) = 5:2 m : n 2(1 40 75P where L is the length of the tube. The factor 2 arises because the light traverses the tube twice, once on the way to a mirror and once after re ection from the mirror. Each shift by 1 fringe corresponds to a change in phase of 2 rad so if the interference pattern shifts by N fringes as the tube is evacuated, 4(n 1)L = 2N and Thus n = 1:00030. Let 1 be the phase dierence of the waves in the two arms when the tube has air in it and let 2 be the phase dierence when the tube is evacuated. These are dierent because the wavelength in air is dierent from the wavelength in vacuum. If is the wavelength in vacuum then the wavelength in air is =n, where n is the index of refraction of air. This means 2n 2 = 4(n 1)L ; 1 2 = 2L 10 m) n 1 = N = 60(500 10 2 m) = 3:0 10 4 : 2L 2(5:0 9 1008 CHAPTER 36 INTERFERENCE Denote the two wavelengths as and 0 , respectively. Then from = 2kd2 = 2(2=)d2 0 = 4d2 4d2 = 2 ; or 76P 0 1 1 1 1 d2 = 2 0 1 1 1 1 = 2 589:10 nm 589:59 nm = 3:54 105 nm = 354 m : 77P Let the position of the mirror measured from the point at which d1 = d2 be x. The phase dierence between the two light paths is = 2kx = 2(2=)d2 = 4x=. Then from Eq. 36-21 2 = I 2 2x : I (d2 ) = Imax cos 2 max cos (a) In a reference frame xed on Earth the ether travels leftward with speed v. Thus the speed of the light beam in this reference frame is c v as the beam travels rightward from M to M1 and c + v as it travels back from M1 to M . The total time for the round trip is therefore given by t1 = c d1 v + c d1 v = c22cd1 2 : + d 78P (b) In a reference frame xed on te ether the mirrors travels rightward with speed v, while the speed of the light beam remains c. Thus in this reference frame the total distance the beam has to travel is given by d2 0 = 2 d2 + v t2 2 2 [see Fig. 36-54(h) { (j)]. Thus s 2 2 t2 = c = 2 d2 + v t2 2 c d2 0 s 2 ; which we solve for t2 : t2 = p 2d2 2 : c2 v CHAPTER 36 INTERFERENCE 1009 (c) Use the binomial expansion (1 + x)n = 1 + nx + 1 + nx In our case let x = v=c 1, then 2 1 L1 = c2c dv2 = 2d1 1 2 2 (jxj 1) : v c 1 2d1 1 + v c 2 ; and L2 = p 22cd2 c L = L1 Thus if d1 = d2 = d then v2 = 2d2 1 2 v c 1=2 1 2 2d2 1 + 2 v : c L2 2d 1 + v c 2 2 1 2 2d 1 + 2 v = dv2 : c c (d) In terms of the wavelength, the phase dierence is given by L = dv2 : c2 (e) We now must reverse the indices 1 and 2 so the new phase dierence is L = dv2 : c2 The shift in phase dierence between these two cases is shift = L L = 2dv2 : c2 (f) Assume that v is about the same as the orbital speed of the Earth. So v 29:8 km/s (see Appendix C). Thus 3 2 dv2 shift = 2c2 = (5002(10 m)(29:8 10 m/s) 2 = 0:40 : 10 9 m)(3:00 108 m/s) ...
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