Physics exercises_solution: Chapter 36

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Physics exercises_solution: Chapter 36

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 36

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Nội dung Text: Physics exercises_solution: Chapter 36

  1. xλ y a (1.35  10 3 m) (7.50  10 4 m) 36.1: y1  λ 1   5.06  10 7 m. a x 2.00 m xλ xλ (0.600 m) (5.46  10 7 m) 36.2: y1  a  3  3.21  10 5 m. a y1 10.2  10 m 36.3: The angle to the first dark fringe is simply: λ  633  10 9 m  θ  arctan   = arctan   0.24  10 3 m   0.15.  a   2 xλ 2(3.50 m) (6.33  10 7 m) 36.4: D  2 y1    5.91  10 3 m. a 7.50  10 4 m λ  9.00 cm  36.5: The angle to the first minimum is  = arcsin   = arcsin   12.00 cm   48.6.  a   So the distance from the central maximum to the first minimum is just y1  x tan   (40.0 cm) tan (48.6)  45.4 cm. 36.6: a) According to Eq. 36.2 mλ mλ λ sin (θ )   sin (90.0)  1   . a a a Thus, a  λ  580 nm  5.80  10 4 mm. b) According to Eq. 36.7 I  sin a (sin  ) λ   sin  (sin  )  2 2    4   0.128. I 0  a (sin  ) λ    (sin  )  4 36.7: The diffraction minima are located by sin   m λ a , m  1,  2, . . . λ  v f  (344 m s ) (1250 Hz)  0.2752 m; a  1.00 m m  1,    16.0; m  2, θ  33.4; m  3, θ  55.6; no solution for larger m
  2. 36.8: a) E  Emax sin(kx  ω t ) 2 2 2 k λ  1  5.24  10 7 m λ k 1.20  10 m 7 c 3.0  108 m s fλc f    5.73  1014 Hz λ 5.24  10 7 m b) a sin   λ λ 5.24  10 7 m a   1.09  10 6 m sin  sin 28.6 c) a sin   mλ (m  1,2,3,. . . ) 7 sin  2  2 D   2 1.09 106 m λ 5.24 10 m  2  74  36.9: sin θ  λ a locates the first minimum y  x tan θ , tan θ  y x  (36.5 cm) (40.0 cm) and   42.38 a  λ sin θ  (620  10 9 m) (sin 42.38)  0.920 μm xλ xλ (2.50 m) (5.00  10 7 m) 36.10: a) y1  a  3  4.17  10 4 m. a y1 3.00  10 m xλ (2.50 m)(5.00  10 5 m) b) a   3  4.17  10 2 m  4.2 cm. y1 3.00  10 m xλ (2.50 m)(5.00  10 10 m) c) a   3  4.17  10 7 m. y1 3.00  10 m xλ (3.00 m) (6.33  10 7 m) 36.11: a) y1    5.43  10 3 m. So the width of the a 3.50  10 4 m brightest fringe is twice this distance to the first minimum, 0.0109 m. 2 xλ 2(3.00 m)(6.33  10 7 m) b) The next dark fringe is at y 2    0.0109 m . a 3.50  10 4 m So the width of the first bright fringe on the side of the central maximum is the distance from y 2 to y1 , which is 5.43  10 3 m .
  3. 2a 2a y 2π (4.50  10 4 m) 36.12: β  sin     7 y  (1520 m 1 ) y. λ λ x (6.20  10 m)(3.00 m) β (1520 m 1 ) (1.00  10 3 m) a) y  1.00  10 3 m :   0.760. 2 2 2 2  sin ( β 2)   sin (0.760)   I  I0  β 2   I 0  0.760   0.822 I 0      β (1520 m 1 )(3.00  10 3 m) b) y  3.00  10 3 m :   2.28. 2 2 2 2  sin ( β 2)   sin (2.28)   I  I0  β 2   I 0  2.28   0.111I 0 .      β (1520 m ) (5.00  10 3 m) 1 c) y  5.00  10 3 m :   3.80. 2 2 2 2  sin ( β 2)   sin (3.80)   I  I0   β 2   I0     0.0259 I 0 .    3.80  xλ (3.00 m) (5.40  10 7 m) 36.13: a) y1    6.75  10 3 m. a 2.40  10 4 m 2a 2πa y1 2 2πa xλ b) β  sin       π. λ λ x λ 2ax 2 2  sin ( β 2)  6 2  sin (π 2)  6 2  I  I0   β 2   (6.00  10 W m )  π 2   2.43  10 W m .        2a 36.14: a)   0 :   sin 0 o  0. λ 2πa 2πa 2λ b) At the second minimum from the center β  sin θ    4π . λ λ a 2πa 2π (1.50  10 4 m) c) β  sin θ  sin 7.0  191 rad. λ 6.00  10 7 m 2a 2a 2 (3.20  10 4 m) 36.15:   sin   λ  sin   sin 0.24  5.36  10 6 m. λ   2
  4. 36.16: The total intensity is given by drawing an arc of a circle that has length E 0 and finding the length of the cord which connects the starting and ending points of the curve. So graphically we can find the electric field at a point by examining the geometry as shown below for three cases. 2a 2a λ Ep 2 a)   sin      . From the diagram,   E0  E p  E 0 . λ λ 2a 2  2 2 4I So the intensity is just: I    I 0  20 .    This agrees with Eq. (36.5). 2a 2a λ b)   sin     2 . From the diagram, it is clear that the total amplitude λ λ a is zero, as is the intensity. This also agrees with Eq. (36.5). 2a 2a 3λ Ep 2 c)   sin     3 . From the diagram, 3  E0  E p  E0 . So λ λ 2a 2 3 the intensity is just: 2  2  4 I    I0  I0.  3  9 2 This agrees with Eq. (36.5).
  5. 2a 2a 2 (1.05  10 4 m) 36.17: a)   sin   λ  sin   sin 3.25  λ  56.0 rad 6.68  10 7 m. 2 2  sin(  2)   sin (56. 0 2)  b) I  I 0    2    I0   56. 0 2   (9.36  10 ) I 0 .  5     36.18: a) Ignoring diffraction, the first five maxima will occur as given by:  mλ   mλ  d sin   mλ    arcsin    arcsin  , for m  1, 2, 3, 4, 5.  d   4a  2a 2a mλ m 2d 2d mλ b)   sin     , and   sin     2m. λ λ d 2 λ λ d So including diffraction, the intensity: 2 2 2   sin(  2)  2 2m  sin ( m / 4)   sin (m / 4)  I  I 0 cos  2   2   I 0 cos 2  m / 4   I 0  m / 4  .  2      So for 2 2  sin ( 4)   sin ( 2)  m  1 : I1     4  I 0  0.811I 0 ; m  2 : I 2    2  I 0  0.405I 0 ;        2 2  sin (3 4)   sin ( )  m  3 : I3    3 4  I 0  0.0901I 0 ; m  4 : I 4     I 0  0      2  sin (5  4)  m  5 : I5    5  4  I 0  0.0324I 0 .    d 36.19: a) If  3, then there are five fringes: m  0,  1,  2. a b) The m  6 interference fringe coincides with the second diffraction minimum, so there are two fringes (m   4, m   5) within the first diffraction maximum on one side of the central maximum. 36.20: By examining the diagram, we see that every fourth slit cancels each other.
  6. 36.21: a) If the slits are very narrow, then the first maximum is at d sin 1  1. λ λ  5.80  10 7 m   1  arcsin    arcsin   5.30  10  4 m   0.0627.  d    Also, the second maximum is at d sin  2  2 λ  2λ   2(5.80  10 7 m)    2  arcsin    arcsin   5.30  10 4 m    0.125.   d    2   sin(  2)   b) I  I 0 cos2    2  but cos 2  1, since we are at the 2 slit maximum. So  2  2 2  sin(a sin θ1 λ)   sin(a d )  I1  I 0   a sin θ λ   I 0  a d      1    2  sin( (3.20  10 4 m) (5.30  10 4 m))   I1  I 0    (3.20  10 4 m) (5.30  10 4 m)   0.249 I 0 .    And 2 2  sin(a sin  2 λ )   sin(2 a d )  I2  I0   a sin  λ   I 0  2a d      2    2  sin (2 (3.20  10 4 m) (5.30  10 4 m))   I2  I0   2 (3.20  10 4 m) (5.30  10 4 m)   0.0256 I 0 .    2   sin(  2)  36.22: We will use I  I 0 cos  2  , and must calculate the phases  and  . 2  2    a 2d Using  2  sin  , and   sin  , we have: λ λ a)   1.25  10 rad :  2  0.177(10.1), and   1.01(57.9)  I  0.757 I 0 . 4 b)   2.50  10 4 rad :  2  0.355(20.3), and   2.03(116.3)  I  0.268 I 0 . c)   3.00  10 4 rad :  2  0.426(24.1), and   2.43(139.2)  I  (0.114) I 0 .
  7. 36.23: With four slits there must be four vectors in each phasor diagram, with the orientation of each successive one determined by the relative phase shifts. So: We see that destructive interference occurs from adjacent slits in case (ii) and from alternate slits in cases (i) and (iii). md λ 36.24: Diffraction dark fringes occur for sin   , and interference maxima occur a mi λ for sin   . Setting them equal to each other yields a missing bright spot whenever d the destructive interference matches the bright spots. That is: mi λ md λ d   mi  md  3md . That is, the missing parts of the pattern occur for d a a mi  3, 6, 9. . .  3m, for m  integers.
  8. 36.25: a) Interference maxima: Diffraction minima: d sin θi  mλ and a sin θd  nλ. If the mth interference maximum corresponds to the nth diffraction minimum then θi  θ d . d m or  a n so n 1 a  d  (0.840 mm)  0.280 mm. m 3 m b) The diffraction minima will squelch the interference maxima for all  3 up to n the highest seen order. For λ  630 nm, the largest value of m will be when θ  90. d 8.40  10 4 m mmax    1333. λ 6.30  10 7 m a 8.40  10 4 m nmax    444. λ 3(6.30  10 7 m) So after m  3, m  6, 9,. . ., 1332 for n  1, 2, 3,. . ., 444 will also be missing. c) By changing λ we only change the highest order seen. d 8.40  10 4 m mmax    2000. λ 4.20  10 7 m a 2000 nmax    666. λ 3 So m  3, 6, 9,. . ., 1999 for n  1, 2, 3,. . ., 666.
  9. 36.26: The third bright band is missing because the first order single slit minimum occurs at the same angle as the third order double slit maximum. tan θ  3 cm 90 cm θ  1.91 Single-slit dark spot: a sin θ  λ λ 500 nm a   1.50  10 4 nm(width) sin θ sin 1.91 Double-slit bright fringe: d sin   3λ 3λ 3(500 nm) d   4.50  10 4 nm(separation) sin θ sin 1.91 36.27: a) Find d : d sin θ  mλ   78.4for m  3 and λ  681 nm, so d  mλ sin   2.086  10 4 cm The number of slits per cm is 1 d  4790 slits cm b) 1st order: m  1, so sin   λ d  (681  10 9 m) (2.086  10 6 m) and   19.1 2nd order: m  2, so sin  2λ / d and θ  40.8 c) For m  4, sin   4 / d is greater than 1.00, so there is no 4th-order bright band. 36.28: First-order: d sin θ1  λ Fourth-order: d sin θ1  4λ dsinθ4 4λ  dsinθ1 λ sin θ4  4 sinθ1  4 sin 8.94 θ 4  38.4
  10. 36.29: a) d   900  cm  1.111  10 5 m 1 For λ  700 nm, λ d  6.3  10 4. The first-order lines are located at sin   λ d ; sin θ is small enough for sin θ  θ to be an excellent approximation. b) y  x λ d , where x  2.50 m. The distance on the screen between 1st order bright bands for two different wavelengths is y  x (y ) d , so λ  d (y ) x  (1.111  10 5 m) (3.00  10 3 m) (2.50 m)  13.3 nm 36.30: a) λ λ 6.5645  10 7 m R  Nm  N    1820 slits. λ mλ 2(6.5645  10 7 m  6.5627  10 7 m)  mλ  b) d  (500 slits mm) 1  (500,000 slits m) 1    sin 1    d  1  sin 1 ((2)(6.5645  10 7 m)  500,000)  41.0297  2  sin 1 ((2)(6.5627  10 7 m)  500,000)  41.0160  θ  0.0137.  mλ   m(6.328  10 7 m)  36.31:   arcsin    arcsin   1.60  10 6 m   arcsin ((0.396)m)   d     m  1 : 1  23.3; m  2 :   52.3. All other m-values lead to angles greater than 90 o . 1 36.32: 5000 slits cm  d  1  2.00  10 6 m. 5.00  10 m 5 d sin  (2.00  10 6 m) sin 13.5 a) d sin   mλ  λ    4.67  10 7 m. m 1  mλ   2(4.67  10 7 m)  b) m  2 :   arcsin    arcsin   2.00  10  6 m   27.8 . o  d     1 36.33: 350 slits mm  d   2.86  10 6 m. Then : 3.50  10 5 m 1  mλ   m(5.20  10 7 m)  d sin   mλ    arcsin    arcsin   2.86  10 6 m   arcsin ((0.182)m)   d     m  1 :   10.5; m  2 :   21.3; m  3 :   33.1.
  11. 1 36.34: 350 slits mm  d  1  2.86  10 6 m., and d sin  mλ. 3.50  10 m 5 λ  4.00  10 7 m   m  1 :  400  arcsin    arcsin   2.86  10 6 m   8.05 .   d    λ  7.00  10 m  7  700  arcsin    arcsin   2.86  10 6 m   14.18 .    d    1  14.18  8.05  6.13.  3λ   3(4.00  10 7 m)   m  3 :  400  arcsin    arcsin   2.86  10 6 m   24.8.  d     3λ   3(7.00  10 7 m)   700  arcsin    arcsin   2.86  10 6 m   47.3.  d     1  47.3  24.8  22.5. 1 36.35: 4000 slits cm  d  1  2.50  10 6 m. So for the  - hydrogen 4.00  10 m 5 line, we have:  mλ   m(6.56  10 7 m)    arcsin    arcsin   2.50  10 6 m   arcsin ((0.262)m) .   d     m  1 : θ1  15.2; m  2 : θ  31.6. And for the  - hydrogen line, the angle is given by:  mλ   m(4.86  10 7 m)    arcsin    arcsin   2.50  10 6 m   arcsin ((0.194)m).   d     m  1 : 1  11.2; m  2 :   22.9; so, a) 1  4.00, b)  2  8.77 .  λ 587.8002 nm 587.8002 36.36:  Nm  N    λ mλ (587.9782 nm  587.8002 nm) 0.178  N  3302 slits. N 3302 slits   2752 . 1.20 cm 1.20 cm cm 36.37: For x-ray diffraction, mλ 2(8.50  10 11 m) 2d sin   mλ  d  d   2.32  10 10 m. 2 sin  2 sin 21.5
  12. 36.38: For the first order maximum in Bragg reflection: 2d sin  2(4.40  10 10 m) sin 39.4  2d sin   mλ  λ    5.59  10 10 m. m 1 36.39: The best resolution is 0.3 arcseconds, which is about 8.33  10 5 . 1.22λ 1.22(5.5  10 7 m) a) D    0.46 m  0.5 m. sin 1 sin(8.33  10 5) b) The Keck telescopes are able to gather more light than the Hale telescope, and hence they can detect fainter objects. However, their larger size does not allow them to have greater resolutionatmospheric conditions limit the resolution. 1.22λ 1.22(5.5  10 7 m) 36.40: D    2.31  10 3 m  2.3 mm. sin 1 sin(1 / 60)  λ 1.22 λ h 1.2  10 6 m 36.41: sin 1  1.22 D  1.22 λ  1.22(0.036 m) D sin 1 W 2.8  10 4 m  D  1.88 m. λ D sin 1 D1 (8.00  10 6 m)(1.00  10 8 ) 36.42: sin 1  1.22 λ   D 1.22 1.22 1.22  λ  0.0656 m  6.56 cm. λ 6.20  10 7 m 36.43: sin 1  1.22  1.22  0.102. The screen is 4.5 m away, so the D 7.4  10 6 m diameter of the Airy ring is given by trigonometry: D  2 y  2 x tan θ  2 x sin θ  2(4.5 m)(0.102)  91.8 cm. 36.44: The image is 25.0 cm from the lens, and from the diagram and Rayleigh’s criteria, the diameter of the circles is twice the “height” as given by: 2 s 2 fy 2(0.180 m)(8.00  10 3 m) D  2 | y  | y   1.15  10 4 m  0.115 mm. s s 25.0 m λ 1.22 λ R 36.45: sin 1  1.22 D  1.22λ D sin 1 W 75.93  1011 m  1.22(5.0  10 m)  1.45 m. 2.50  10 5 m
  13. λ y yD (4.00  10 3 m)(0.0720 m) 36.46: sin 1  1.22  s   429 m. D s 1.22λ 1.22(5.50  10 7 m) 36.47: Let y be the separation between the two points being resolved and let s be their λ y distance from the telescope. Then the limit of resolution corresponds to 1.22  D s a) Let the two points being resolved be the opposite edges of the crater, so y is the diameter of the crater. For the moon, s  3.8  10 8 m. y  1.22λ s D Hubble: D = 2.4 m and λ  400 nm gives the maximum resolution, so y = 77 m Arecibo: D = 305 m and λ  0.75 m; y  1.1  10 6 m yD b) s  1.22λ Let y  0.30 (the size of a license plate) s  (0.30 m) (2.4 m) [(1.22)(400  10 9 m)]  1500 km 36.48: Smallest resolving angle is for short-wavelength light(400 nm) λ 400  10 9 m   1.22  (1.22)  9.61  10 8 rad D 5.08 m 10,000 mi  R 10,000 mi 16,000 km R   1.7  1011 km  8 9.6  10 rad This is less than a light year, so there are no stars this close. 36.49: Let y be the separation between the two points being resolved and let s be their distance from the telescope. The limit of resolution corresponds to 1.22 λ D  y s s  4.28 ly  4.05  1016 m Assume visible light, with λ  400 m y  1.22 λs D  1.22(400  10 9 m) (4.05  1016 m (10.0 m)  2.0  109 m The diameter of Jupiter is 1.38  10 8 m, so the resolution is insufficient, by about one order of magnitude.
  14. 36.50: a) For dark spots, a sin θ  mλ, so sin θ  1 a. Heating the sheet causes the slit width to increase due to thermal expansion, so sin  and hence  will decrease. Therefore the bright region gets narrower. b) At the lower temperature: λ 5 cm a1  where tan1   1  0.35809 sin 1 800 cm 500 nm a1   80,002 nm sin 0.35809 At the higher temperature: 5 cm  0.001 cm tan  2    2  0.35802 800 cm λ 500 nm a2   a1   80,018 nm sin 1 sin 0.35802 Thermal expansion: a  a1 T a 80, 018 nm  80, 002 nm   a1T (80, 002 nm)(80C)  2.5  10 6 C 1
  15. 1 sin( a sin θ λ) sin x 36.51: a) I  I 0 2     0.7071. Solving for x through 2  a sin θ λ x trial and error, and remembering to use radians throughout, one finds x  1.39 rad and   2 x  2.78 rad. Also,         2  , and 2a λ λ  2.78 rad  λ  sin   sin     2 rad   0.442 a .  λ 2a a   a i)  2  sin    0.221     0.223 rad    0.446 rad  25.3. λ a ii)  5  sin    0.0885     0.0886 rad    0.177 rad  10.1. λ a iii)  10  sin    0.0442     0.0443 rad    0.885 rad  5.07 . λ λ b) For the first minimum, sin  0  . a a 1 i)  2   0  arcsin   0.524 rad  2 0  1.05 rad  60.2. λ  2 a 1 ii)  5   0  arcsin   0.201 rad  2 0  0.402 rad  23.0 . λ 5 a 1 iii)  10   0  arcsin   0.100 rad  2 0  0.200 rad  11.5. λ  10  Both methods show the central width getting smaller as the slit width a is increased. 36.52: If the apparatus of Exercise 36.4 is placed in water, then all that changes is the λ 2 xλ 2 xλ D 5.91  10 3 m wavelength λ  λ  . So : D  2 y1       n a an n 1.33 4.44  10 3 m  4.44 mm. 36.53: sin θ  λ a locates the first dark band λ λ liquid sin  air  air ; sin  liquid  a a  sin  liquid  λ liquid  λ air   sin    0.4836   air  λ  λ air n (Eq.33.5), so n  λ air λ liquid  1 0.4836  2.07
  16. 36.54: For bright spots, N sin   λ 1 Red: N sin θR  700 nm 1 Violet: N sin θV  400 nm 1 sin  R 7  sin  V 4  R   V  15   R  V  15 sin(V  15) 7  sin  V 4 sin θV cos 15  cos θ V sin 15 Expand: 7 4 sin θV cos15  cot θ V sin 15  7 4 tan θV  0.330  θV  18.3 θR  θV  15  18.3  15  33.3 Line density: 1 N sin  R  700nm sin θR sin 33.3 N  9  7.84  105 lines m 700 nm 700  10 m  7840 lines cm The spectrum begins at 18 .3o and ends at 33.3o xλ (1.20 m)(5.40  10 7 m) 36.55: a) y1   4  1.80  10 3 m. a 3.60  10 m sin(a sin  λ ) 1 a sin θ b)    1.39 a sin  λ 2 λ (1.39)(5.40  10 7 m)  sin    6.64  10 4  (3.60  10 m) 4  y  x tan   x sin   (1.20 m)(6.64  10 4 )  7.97  10 4 m  0.797 mm.
  17. 2  sin γ  36.56: a) I  I 0   γ  . The maximum intensity occurs when the derivative of the    intensity function with respect to  is zero. 2 dI d  sin γ   sin γ   cos γ sin γ   I0   γ   2 γ   γ  γ 2   0     dγ dγ      cos γ sin γ   2  γ cos γ  sin γ γ γ  γ  tan γ. b) The graph above is a plot of f(  ) =   tan  . So when it equals zero, one has an intensity maximum. Getting estimates from the graph, and then using trial and error to narrow in on the value, we find that the three smallest  -values are  = 4.49 rad 7.73 rad, and 10.9 rad.
  18. 2d 2dθ λ 36.57: The phase shift for adjacent slits is   sin θ    . λ λ 2d So, with the principal maxima at phase shift values of   2πm, and (N  1) minima between the maxima, the phase shift between the minima adjacent to the maximum, and 2 the maximum itself, must be  . N 2 Therefore total phase shifts of these minima are 2m  . N Hence the angle at which they are found, and the angular width, will be: λ  2  mλ λ 2λ    2m     θ  . 2d  N  d dN dN
  19. 36.58: a) E p  E p x  E p y . So, from the diagram at right, 2 2 2 we have: 2 Ep 2  (1  cos   cos 2 ) 2  (sin   sin 2 ) 2 E 0  (2 cos 2   cos  ) 2  (sin   2 sin  cos  ) 2  (cos 2   sin 2  )(1  2 cos  ) 2 2 Ep   (1  2 cos  ) 2  E p  E0 (1  2 cos  ). E02 2 2d   2d sin θ   b)   sin θ  I p  I 0 1  2 cos    . This is graphed below:  λ   λ  c) (i) At θ  0, I p  I 0 (1  2 cos(0 0 )) 2  9 I 0 . 2d sin θ (ii) The principal maximum is when I 0  2m  d sin θ  mλ λ  2d sin   2d sin θ 2m (iii) & (iv) The minima occur at 2 cos    1    λ  λ 3 mλ  d sin θ  , with m not divisible by 3. Thus there are two minima between every 3 principal maximum.  2d sin θ  I (v) The secondary maxima occur when cos    1  I p  I 0  max .  λ  9 2d sin θ mλ Also  m  d sin θ  . λ 2 All of these findings agree with the N - slit statements in Section 35.5. d) Below are phasor diagrams for specific phase shifts.
  20. 36.59: a) For eight slits, the phasor diagrams must have eight vectors: 3 5 7 b) For   ,  , and   , totally destructive interference occurs between 4 4 4 3 slits four apart. For   , totally destructive interference occurs with every second slit. 2
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