 # Physics exercises_solution: Chapter 35

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7 ## Physics exercises_solution: Chapter 35

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 35

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## Nội dung Text: Physics exercises_solution: Chapter 35

1. 35.1: Measuring with a ruler from both S1 and S 2 to there different points in the antinodal line labeled m = 3, we find that the difference in path length is three times the wavelength of the wave, as measured from one crest to the next on the diagram. 35.2: a) At S1 , r2  r1  4λ, and this path difference stays the same all along the y - axis, so m  4. At S 2 , r2  r1  4λ, and the path difference below this point, along the negative y -axis, stays the same, so m  4. b) c) The maximum and minimum m-values are determined by the largest integer less d than or equal to . λ 1 d) If d  7 λ  7  m  7, so there will be a total of 15 antinodes between the 2 sources. (Another antinode cannot be squeezed in until the separation becomes six times the wavelength.) 35.3: a) For constructive interference the path diference is mλ, n  0,  1,  2, . . . The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. Thus only the path difference of zero is possible. This occurs midway between the two sources, 2.50 m from A. b) For destructive interference the path difference is (m  1 )λ, m  0,  1,  2, . . . 2 A path difference of  λ/ 2  3.00 m is possible but a path difference as large as 3λ / 2  9.00 m is not possible. For a point a distance x from A and 5.00  x from B the path difference is x  (5.00 m  x). x  (5.00 m  x)  3.00 m gives x  4.00 m x  (5.00 m  x)  3.00 m gives x  1.00 m
2. 35.4: a) The path difference is 120 m, so for destructive interference: λ  120 m  λ  240 m. 2 b) The longest wavelength for constructive interference is λ  120 m. 35.5: For constructive interference, we need r2  r1  mλ  (9.00 m  x)  x  mλ mλ mc m(3.00  108 m s)  x  4.5 m   4.5m   4.5m   4.5 m  m(1.25 m). 2 2f 2(120  10 6 Hz)  x  0.75 m, 2.00 m, 3.25 m, 4.50 m, 5.75 m, 7.00 m, 8.25 m. For m  3, 2, 1, 0,  1,  2,  3. (Don’t confuse this m with the unit meters, also represented by an “m”). 35.6: a) The brightest wavelengths are when constructive interference occurs: d 2040 nm 2040 nm d  mλ  λ   λ 3   680 nm, λ 4   510 nm and m 3 4 2040 nm λs   408 nm. 5 b) The path-length difference is the same, so the wavelengths are the same as part (a). 35.7: Destructive interference occurs for: d 2040 nm 2040 nm λ  λ3   583 nm and λ 4   453 nm. m 1 2 3.5 4.5 35.8: a) For the number of antinodes we have: mλ mc m(3.00  108 m s) sin      0.2317 m, so, setting   90, d df (12.0 m)(1.079  108 Hz) the maximum integer value is four. The angles are  13.4,  27.6,  44.0, and  67.9 for m  0,  1,  2,  3,  4. (m  1 2)λ b) The nodes are given by sin    0.2317 (m  1 2). So the angles are d  6.65,  20.3,  35.4, 54.2 for m  0,  1,  2,  3. Rλ dy (4.60  10 4 m)(2.82  10 3 m) 35.9: y  λ   5.90  10 7 m. d R 2.20 m 35.10: For bright fringes: Rmλ (1.20 m)(20)(5.02  10 7 m) d   1.14  10 3 m  1.14 mm. ym 0.0106 m
3. Rmλ Rλ(3  2) (0.750 m) (5.00  10 7 m) 35.11: Recall y m   y 23  y3  y 2   d d 4.50  10 4 m  y 23  8.33  10 4 m  0.833 mm. 35.12: The width of a bright fringe can be defined to be the distance between its two adjacent destructive minima. Assuming the small angle formula for destructive interference (m  1 )λ ym  R 2 , d the distance between any two successive minima is λ (400  10 9 m) y n1  y n  R  (4.00 m)  8.00 mm. d (0.200  10 3 m) Thus, the answer to both part (a) and part (b) is that the width is 8.00 mm. 35.13: Use the information given about the bright fringe to find the distance d between Rλ Rλ (3.00 m)(600  10 9 m) the two slits: y1  1 (Eq.35.6), so d  1   3.72  10 4 m. d y1 4.84  10 3 m (R is much greater than d, so Eq.35.6 is valid.) The dark fringes are located by d sin   (m  1 )λ, m  0,  1,  2, . . . The first order dark 2 fringe is located by sin   λ 2 2d , where λ 2 is the wavelength we are seeking. λ R y  R tan   R sin   2 2d Rλ1 Rλ 2 We want λ2 such that y  y1 . This gives  and λ 2  2λ1  1200 nm. d 2d 35.14: Using Eq.35.6 for small angles, mλ ym  R , d we see that the distance between corresponding bright fringes is Rm (5.00 m)(1) y  λ  3 (660  470)  (10 9 m)  3.17 mm. d (0.300  10 m)
4. 35.15: We need to find the positions of the first and second dark lines:  λ   5.50  10 7 m  1  arcsin   arcsin  2(1.80  10 6 m   8.79   2d     y1  R tan 1  (0.350 m) tan(8.79)  0.0541 m.  3λ   3(5.50  10 7 m)  Also  2  arcsin    arcsin  2(1.80  10 6 m)   27.3   2d     y 2 R tan θ2  (0.350 m) tan(27.30)  0.1805 m. The fringe separation is then y  y 2  y1  0.1805 m  0.0541 m  0.1264 m. 35.16: (a) Dark fringe implies destructive interference. 1 d sin   λ 2 λ 624  10 9 m d   1.64  10 6 m 2 sin  2 sin 11.0 (b) Bright fringes: d sin θmax  mmax λ 6 10 m The largest that θ can be is 90, so mmax  d / λ  1.64109 m  2.6 Since m is an integer, its 624 maximum value is 2. There are 5 bright fringes, the central spot and 2 on each side of it. Dark fringes: d sin θ  m  1 λ. This equation has solutions for θ  11.0;  34.9; and 2  72.6. Therefore, there are 6 dark fringes. 35.17: Bright fringes for wavelength λ are located by d sin   mλ. First-order (m  1) is closest to the central bright line, so sin   λ / d . λ  400 nm gives sin  (400  10 9 m)/(0.100  10 3 m) and   0.229 λ  700 nm gives sin  (700  10 9 m)/(0.100  10 3 m) and   0.401 The angular width of the visible spectrum is thus 0.401  0.229  0.172. Rλ Rλ (1.80 m) (4.50  10 7 m) 35.18: y d   3  1.93  10 4 m  0.193 m. d y 4.20  10 m 35.19: The phase difference  is given by   (2d / λ) sin  (Eq.35.13)   [2 (0.340  10 3 m) (500  10 9 m)] sin 23.0  1670 rad
5. 35.20:  Path difference  2 λ  524cm  486 cm    2     119 radians   2 cm  35.21: a) Eq.(35.10): I  I 0 cos 2 ( 2)  I 0 (cos 30.0) 2  0.750 I 0 b) 60.0  ( / 3) rad Eq. (35.11) :   (2 / λ )(r2  r1 ), so (r2  r1 )  ( / 2 )λ  ( / 3) / 2 λ  λ / 6  80 nm 35.22: a) The source separation is 9.00 m, and the wavelength of the wave is c 3.00  108 m/s λ   20.0 m. So there is only one antinode between the sources f 1.50  107 Hz (m  0), and it is a perpendicular bisector of the line connecting the sources.    d    (9.00 m)  b) I  I 0 cos 2    I 0 cos 2  sin    I 0 cos 2   (20.0 m) sin    2  λ     I 0 cos 2 ((1.41) sin  ) So, forθ  30, I  0.580 I 0 ;θ  45 , I  0.295 I 0 ; θ  60, I  0.117 I 0 ;θ  90, I  0.026 I 0 . 35.23: a) The distance from the central maximum to the first minimum is half the distance to the first maximum, so: Rλ (0.700 m) (6.60  10 7 m) y  4  8.88  10 4 m. 2d 2(2.60  10 m) b) The intensity is half that of the maximum intensity when you are halfway to the first minimum, which is 4.44  10 4 m. Remember, all angles are small.
6. c 3.00  108 m/s 35.24: a) λ    2.50 m, and we have: f 1.20  108 Hz 2 2  (r1  r2 )  (1.8 m)  4.52 rad. λ 2.50 m    4.52 rad  b) I  I 0 cos 2    I 0 cos 2    0.404 I 0 . 2  2  Rλ (0.900 m) (5.50  10 7 m) 35.25: a) To the first maximum: y1   4  3.81  10 3 m. d 1.30  10 m So the distance to the first minimum is one half this, 1.91 mm. b) The first maximum and minimum are where the waves have phase differences of zero and pi, respectively. Halfway between these points, the phase difference between the  waves is . So : 2     I I  I 0 cos 2    I 0 cos 2    0  2.00  10 6 W/m 2 . 2 4 2  d  35.26: From Eq. (35.14), I  I 0 cos 2  sin  . So the intensity goes to zero when the  λ   d cosine’s argument becomes an odd integer of . That is : sin   (m  1 / 2)  2 λ d sin   λ(m  1 / 2), which is Eq. (35.5). 35.27: By placing the paper between the pieces of glass, the space forms a cavity whose height varies along the length. If twice the height at any given point is one wavelength (recall it has to make a return trip), constructive interference occurs. The distance between the maxima (i.e., the # of meters per fringe) will be λl λ  λ   5.46  10 7 m  x      arctan   arctan  2((1 / 1500) m)   4.095  10 rad  0.0235  4 2h 2 tan   2x    35.28: The distance between maxima is λl (6.56  10 7 m) (9.00 cm) x    0.0369 cm. 2h 2(8.00  10 5 m) 1 So the number of fringes per centimeter is  27.1 fringes/cm. x
7. 35.29: Both parts of the light undergo half-cycle phase shifts when they reflect, so for λ λ 6.50  10 7 m destructive interference t   0   1.14  10 7 m  114 nm. 4 4n 4(1.42) 35.30: There is a half-cycle phase shift at both interfaces, so for destructive interference: λ λ 480 nm t  0   80.5 nm. 4 4n 4(1.49) 35.31: Destructive interference for λ1  800 nm incident light. Let n be the refractive index of the oil. There is a λ / 2 phase shift for the reflection at the air-oil interface but no phase shift for the reflection at the oil-water interface. Therefore, there is a net λ / 2 phase difference due to the reflections, and the condition for destructive interference is 2t  m(λ / n). Smallest nonzero thickness means m  1, so 2tn  λ1. The condition for constructive interference with incident wavelength λ is 2t  (m  1 )(λ / n) and 2tn  (m  1 )λ. 2 2 But 2tn  λ1 , so λ  λ1 /( m  1 ), where λ1  800 nm. 2 for m  0, λ  1600 nm for m  1, λ  533 nm for m  2, λ  320 nm, and so on The visible wavelength for which there is constructive interference is 533 nm. 35.32: a) The number of wavelengths is given by the total extra distance traveled, divided by the wavelength, so the number is x 2tn 2(8.76  10 6 m) (1.35)    36.5. λ λ0 6.48  10 7 m b) The phase difference for the two parts of the light is zero because the path difference is a half-integer multiple of the wavelength and the top surface reflection has a half-cycle phase shift, while the bottom surface does not. 35.33: Both rays, the one reflected from the pit and the one reflected from the flat region between the pits, undergo the same phase change due to reflection. The condition for destructive interference is 2t  (m  1 )(λ / n), where n is the refractive index of the 2 plastic substrate. The minimum thickness is for m  0, and equals t  λ /( 4n)  790 nm/[(4)(1.8)]  110 nm  0.11m. 35.34: A half-cycle phase change occurs, so for destructive interference λ λ 480nm t  0   180 nm. 2 2n 2(1.33)
8. 35.35: a) To have a strong reflection, constructive interference is desired. One part of the light undergoes a half-cycle phase shift, so:  1 λ 2dn 2(290 nm) (1.33) 771 nm 2d   m    λ    . For an integer  2 n  1  1  1 m   m   m    2  2  2 value of zero, the wavelength is not visible (infrared) but for m  1 , the wavelength is 514 nm, which is green. b) When the wall thickness is 340 nm, the first visible constructive interference 904 nm occurs again for m  1 and yields λ  = 603 nm, which is orange.  1 m    2 35.36: a) Since there is a half-cycle phase shift at just one of the interfaces, the minimum thickness for constructive interference is: λ λ 550 nm t  0   74.3 nm. 4 4n 4(1.85) b) The next smallest thickness for constructive interference is with another half 3λ 3λ 0 3550 nm  wavelength thickness added: t     223 nm. 4 4n 4(1.85) mλ 1800(6.33  10 7 m) 35.37: x   5.70  10 4 m  0.570 mm. 2 2 mλ1 818(6.06  10 7 m) 35.38: a) For Jan, the total shift was x1    2.48  10 4 m. 2 2 mλ 2 818(5.02  10 7 m) For Linda, the total shift was x2    2.05  10 4 m. 2 2 b) The net displacement of the mirror is the difference of the above values: x  x1  x2  0.248 mm  0.205 mm  0.043 mm. 35.39: Immersion in water just changes the wavelength of the light from Exercise Rλ y vacuum 0.833 mm 35.11, so: y     0.626 mm, using the solution from Exercise dn n 1.33 35.11.
9. 35.40: Destructive interference occurs 1.7 m from the centerline. r1  (12.0 m) 2  (6.2 m) 2 =13.51 m r2  (12.0 m) 2  (2.8 m) 2 =12.32 m For destructive interference, r1  r2  λ / 2  1.19 m and λ  2.4 m. The wavelength we have calculated is the distance between the wave crests. Note: The distance of the person from the gaps is not large compared to the separation of the gaps, so the path length is not accurately given by d sin  . 35.41: a) Hearing minimum intensity sound means that the path lengths from the individual speakers to you differ by a half-cycle, and are hence out of phase by 180 at that position. b) By moving the speakers toward you by 0.398 m, a maximum is heard, which means that you moved the speakers one-half wavelength from the min and the signals are back in phase. Therefore the wavelength of the signals is 0.796 m, and the frequency is v 340 m/s f   =427 Hz. λ 0.796 m c) To reach the next maximum, one must move an additional distance of one wavelength, a distance of 0.796 m.
10.  1 35.42: To find destructive interference, d  r2  r1  (200 m) 2  x 2  x   m  λ  2 2  1   1  (200 m)  x  x   m  λ   2 x  m  λ 2 2 2  2   2 20,000 m 2 1  1 x   m  λ.  1 2 2  m  λ  2 c 3.00  108 m s The wavelength is calculated by λ    51.7 m. f 5.80  10 6 Hz  m  0 : x  761 m, and m  1 : x  219 m, and m  2 : x  90.1 m, and m  3; x  20.0 m. 35.43: At points on the same side of the centerline as point A, the path from B is longer than the path from A, and the path difference d sin θ puts speaker A ahead of speaker B in phase. Constructive interference occurs when  1 d sin   λ 6   m  λ, m  0, 1, 2,   2  2  2 sin    m   λ d    m   0.2381, m  0, 1, 2, ...  3  3 m  0, 9.13; m  1, 23.4; m  2, 39.4; m  3, 60.8; m  4, no solution At points on the other side of the centerline, the path from A is longer than the path from B , and the path difference d sin θ puts speaker A behind speaker B in phase. Constructive interference occurs when  1 d sin   λ 6   m  λ, m  0, 1, 2, ...  2  2  1 sin    m  λ d    m  0.2381, m  0, 1, 2,...  3  3 m  0, 4.55; m  1, 18.5; m  2, 33.7; m  3, 52.5; m  4, no solution
11. 35.44: First find out what fraction the 0.159 ms time lag is of the period. 0.159  10 3 s t   (0.159  10 3 s) f  (0.159  10 3 s) (1570 Hz) T t  0.250, so the speakers are 1 4 period out of phase. Let A be ahead of B in phase. 330 m s λv f   0.210 m 1570 Hz On A' s side of centerline : Since A is ahead by 1 4 period, the path difference must retard B’s phase enough so the waves are in phase. 3 7 d sin   λ, λ,... 4 4 3  0.210 m  sin 1     1  21.9 4  0.422 m    7  0.210 m  sin  2      2  60.6 4  0.422 m    On B' s side of centerline : The path difference must now retard A’s sound by 4 λ, 4 λ,  1 5 1 5  d sin   λ, λ,  gives  7.2,  38.5 4 4 35.45: a) If the two sources are out of phase by one half-cycle, we must add an extra half a wavelength to the path difference equations Eq. (35.1) and Eq. (35.2). This exactly changes one for the other, for m  m  1 and m  1  m, since m 2 2 in any integer. b) If one source leads the other by a phase angle  , the fraction of a cycle difference is  . Thus the path length difference for the two sources must be adjusted for both 2 destructive and constructive interference, by this amount. So for constructive inference: r1  r2  (m   2 )λ, and for destructive interference, r1  r2  (m  1 2   2 )λ.
12. 35.46: a) The electric field is the sum of the two wave functions, and can be written: E p (t)  E2 (t )  E1 (t )  E cos(t )  E cos(t   )  E p (t )  2 Ecos( / 2) cos(ωt   / 2). b) E p (t )  A cos(t   / 2), so comparing with part (a), we see that the amplitude of the wave (which is always positive) must be A  2 E | cos( / 2) | .  c) To have an interference maximum,  2m . So, for example, using m  1, the 2  relative phases are E 2 :   0; E1 :     4 ; E p :    2 , and all waves are in 2 phase.   1 d) To have an interference minimum,    m  . So, for example using 2  2 m  0, relative phases are E 2 :   0; E1 :      ; E p :    2   2, and the resulting wave is out of phase by a quarter of a cycle from both of the original waves. e) The instantaneous magnitude of the Poynting vector is: | S |  0 cE p (t )   0 c(4 E 2 cos 2 ( 2) cos 2 (t   2)). 2 1 For a time average, cos 2 (t   2)  , so | S av | 2 0 cE 2 cos 2 ( 2). 2 35.47: a) r  mλ r1  x 2  ( y  d ) 2 . r2  x 2  ( y  d ) 2 . So r  x 2  ( y  d ) 2  x 2  ( y  d ) 2  mλ. b) The definition of hyperbola is the locus of points such that the difference between P to S 2 and P to S1 is a constant. So, for a given m and λ we get a hyperbola. Or, in the case of all m for a given λ , a family of hyperbola. c) x 2  ( y  d ) 2  x 2  ( y  d ) 2  (m  1 )λ. 2
13. 35.48: a) E p  E12  E 2  2 E1 E 2 cos(   ) 2 2  E 2  4 E 2  4 E 2 cos   5E 2  4 E 2 cos  . 1  5   4   I  0 cE p 2   0 c  E 2    E 2  cos  . 2  2   2   9   0  I 0   0 cE 2 . 2 5 4  So I  I 0   cos  . 9 9  1 b) I min  I 0 which occurs when   n (n odd). 9 35.49: For this film on this glass, there is a net λ 2 phase change due to reflection and the condition for destructive interference is 2t  m(λ n), where n  1.750. Smallest nonzero thickness is given by t  λ 2n . At 20.0C, t 0  (582.4 nm) [(2) (1.750)]  166.4 nm. At 170C, t 0  (588.5 nm) [(2) (1.750)]  168.1 nm. t  t 0 (1  T ) so   (t  t 0 ) (t 0 T )  (1.7 nm) [(166.4 nm) (150C)]  6.8  10 5 (C ) 1
14. 35.50: For constructive interference: d sin   mλ1  d sin   3(700 nm)  2100 nm.  1 d sin  2100 nm For destructive interference: d sin    m   λ 2  λ 2   .  2 m 1 2 m 1 2 So the possible wavelengths are λ 2  600 nm, for m  3, and λ 2  467 nm, for m  4. Both d and  drop out of the calculation since their combination is just the path difference, which is the same for both types of light. 35.51: First we need to find the angles at which the intensity drops by one-half from the value of the m th bright fringe.  d  I d d m  I  I 0 cos 2  sin    0  sin    (m  1 2) .  λ  2 λ λ 2 λ 3λ λ  m  0 :  m  ; m  1:  m     m  4d 4d 2d  so there is no dependence on the m - value of the fringe. 35.52: There is just one half-cycle phase change upon reflection, so for constructive interference: 2t  (m1  1 )λ1  (m2  1 )λ 2 . But the two different wavelengths differ by 2 2 just one m - value, m2  m1  1.  1  1 λ  λ2 λ  λ2   m1   λ1   m1   λ 2  m1 (λ 2  λ1 )  1  m1  1  2  2 2 2( λ2  λ1 ) 477.0 nm  540.6 nm  m1   8. 2(540.6 nm  477.0 nm)  1 λ 17(477.0 nm)  2t   8   1  t   1334 nm.  2 n 4(1.52)
15. 35.53: a) There is a half-cycle phase change at the glass, so for constructive interference: 2  x  1 2 d  x  2 h     x   m  λ 2 2  2  1  x 2  4h 2  x   m  λ.  2 Similarly for destructive interference: x 2  4h 2  x  mλ. b) The longest wavelength for constructive interference is when m  0 : x 2  4h 2  x (14 cm) 2  4(24 cm)2  14 cm λ   72 cm. m 12 12 35.54: a) At the water (or cytoplasm) to guanine interface, is a half-cycle phase shift for the reflected light, but there is not one at the guanine to cytoplasm interface. Therefore there will always be one half-cycle phase difference between two neighboring reflected beams. For the guanine layers: 1 λ 2t g n g 2(74 nm) (1.80) 266 nm 2t g  (m  ) λ    λ  533 nm(m  0). 2 ng (m  2 ) 1 (m  1 ) 2 (m  1 ) 2 For the cytoplasm layers:  1 λ 2t c nc 2(100 nm) (1.333) 267 nm 2t c   m    λ     λ  533 nm(m  0).  2  nc (m  2 )1 (m  1 ) 2 (m  1 ) 2 b) By having many layers the reflection is strengthened, because at each interface some more of the transmitted light gets reflected back, increasing the total percentage reflected. c) At different angles, the path length in the layers change (always to a larger value than the normal incidence case). If the path length changes, then so do the wavelengths that will interfere constructively upon reflection.
16. 35.55: a) Intensified reflected light means we have constructive interference. There is one half-cycle phase shift, so:  1 λ 2tn 2(485 nm) (1.53) 1484 nm 2t   m    λ    .  2 n (m  2 ) 1 (m  1 ) 2 (m  1 ) 2  λ  593 nm(m  2), and λ  424 nm (m  3). b) Intensified transmitted light means we have destructive interference at the upper surface. There is still a one half-cycle phase shift, so: mλ 2tn 2(485 nm) (1.53) 1484 nm 2t  λ   . n m m m  λ  495 nm (m  3) is the only wavelength of visible light that is intensified. We could also think of this as the result of internal reflections interfering with the outgoing ray without any extra phase shifts. 35.56: a) There is one half-cycle phase shift, so for constructive interference:  1 λ 2tn 2(380 nm) (1.45) 1102 nm 2t   m   0  λ    .  2 n (m  2 ) 1 (m  1 ) 2 (m  1 ) 2 Therefore, we have constructive interference at λ  441 nm (m  2), which corresponds to blue-violet. b) Beneath the water, looking for maximum intensity means that the reflected part of the wave at the wavelength must be weak, or have interfered destructively. So: mλ 0 2tn 2(380 nm) (1.45) 1102 nm 2t   λ0    . n m m m Therefore, the strongest transmitted wavelength (as measured in air) is λ  551 nm (m  2), which corresponds to green. 35.57: For maximum intensity, with a half-cycle phase shift,  1 (2m  1)λ 2t   m  λ and t  R  R 2  r 2   R  R2  r 2  2 4 2 (2m  1)λ  (2m  1)λ  (2m  1)λR  R r  R 2 2  R2  r 2  R2     4  4  2 2 (2m  1)λR  (2m  1)λ  (2m  1)λR r   r , for R  λ. 2  4  2 The second bright ring is when m  1 : (2(1)  1) (5.80  10 7 m) (0.952 m) r  9.10  10 4 m  0.910 mm. 2 So the diameter of the third bright ring is 1.82 mm.
17. 35.58: As found in Problem (35.51), the radius of the mth bright ring is in general: (2m  1)λR r , 2 for R  λ. Introducing a liquid between the lens and the plate just changes the λ wavelength from λ  . n So: (2m  1)λR r 0.850 mm r ( n)     0.737 mm. 2n n 1.33 35.59: a) Adding glass over the top slit increases the effective path length from that slit to the screen. The interference pattern will therefore change, with the central maximum shifting downwards. 2d b) Normally the phase shift is   sin  , but now there is an added shift from the λ glass, so the total phase shift is now 2πd  2Ln 2L  2d 2L(n  1) 2  sin      sin    (d sin   L(n  1)). λ  λ λ  λ λ λ    So the intensity becomes I  I 0 cos 2  I 0 cos 2  (d sin   L(n  1)) . 2 λ   c) The maxima occur at (d sin   L(n  1))  m  d sin   mλ  L(n  1) λ 35.60: The passage of fringes indicates an effective change in path length, since the wavelength of the light is getting shorter as more gas enters the tube. 2L 2L 2L mλ m    (n  1)  (n  1)  . λn λ λ 2L So here: 48(5.46  10 7 m) (n  1)   2.62  10 4. 2(0.0500 m)
18. 35.61: There are two effects to be considered: first, the expansion of the rod, and second, the change in the rod’s refractive index. The extra length of rod replaces a little of the air so that the change in the number of wavelengths due to this is given by: 2nglass L 2nair L 2(nglass  1) L0T N1    λ0 λ0 λ0 2(1.48  1) (0.030 m) (5.00  10 6 C) (5.00 C)  N1   1.22. 5.89  10 7 m The change in the number of wavelengths due to the change in refractive index of the rod is: 2nglass L0 2(2.50  10 5 C) (5.00 C min ) (1.00 min) (0.0300 m) N 2    12.73. λ0 5.89  10 7 m So the total change in the number of wavelengths as the rod expands is N  12.73  1.22  14.0 fringes/minute. 35.62: a) Since we can approximate the angles of incidence on the prism as being small, Snell’s Law tells us that an incident angle of θ on the flat side of the prism enters the prism at an angle of θ n , where n is the index of refraction of the prism. Similarly on leaving the prism, the in-going angle is θ n  A from the normal, and the outgoing, relative to the prism, is n(θ n  A). So the beam leaving the prism is at an angle of θ   n(θ n  A)  A from the optical axis. So θ  θ   (n  1) A. At the plane of the source S 0 , we can calculate the height of one image above the source: d  tan(   )a  (   )a  (n  1) Aa  d  2aA(n  1). 2 b) To find the spacing of fringes on a screen, we use: Rλ Rλ (2.00 m  0.200 m) (5.00  10 7 m) y     1.57  10 3 m. d 2aA(n  1) 2(0.200 m) (3.50  10 3 rad) (1.50  1.00) 