 # Physics exercises_solution: Chapter 33

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3 ## Physics exercises_solution: Chapter 33

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 33

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## Nội dung Text: Physics exercises_solution: Chapter 33

1. c 3 . 00  10 8 m s 33.1: a) v    2 . 04  10 8 m s. n 1 . 47 λ (6.50  10 7 m) b) λ  0   4.42  10 7 m. n 1.47 c 3.00  108 m s 33.2: a) λ vacuum    5.17  10 7 m. f 5.80  10 Hz 14 c 3.00  108 m s b) λ glass    3.40  10 7 m. fn (5.80  10 Hz)(1.52) 14 c 3.00  108 m/s 33.3: a) n   1.54. v 1.94  108 m/s 7 7 b) λ 0  nλ  (1.54) (3.55  10 m)  5.47  10 m. λ water nwater (4.38  10 7 m)(1.333) 33.4: λ water nwater  λ Benzene nBenzene  λ CS2   nBenzene 1.501 33.5 : a) Incident and reflected angles are always equal   r   a  47.5.     na    1.00   b)  b  2   b  2  arcsin n sin  a   2  arcsin 1.66 sin 42.5   66.0.    b    d 2.50 m 33.6: v   2.17  108 m s t 11.5  10 9 s c 3.00  10 8 m/s n   1.38 v 2.17  10 8 m/s 33.7: n a sin  a  nb sin  b  sin  a   sin 62.7  nb  na   sin    1.00 sin 48.1   1.194   b    n  c v so v  c n  (3.00  108 m/s) / 1.194  2.51  108 m s
2. 33.8 (a) Apply Snell’s law at both interfaces. 33.9: a) Let the light initially be in the material with refractive index na and let the third and final slab have refractive index nb Let the middle slab have refractive index n1 1st interface : na sin  a  n1 sin 1 2nd interface : n1 sin 1  nb sin  b Combining the two equations gives na sin  a  nb sin  b . b) For N slabs, where the first slab has refractive index na and the final slab has refractive index nb , na sin  a  n1 sin 1 , n1 sin 1  n2 sin  2 ,  , n N 2 sin  N 2  nb sin  b . This gives na sin  a  nb sin  b . The final direction of travel depends on the angle of incidence in the first slab and the indicies of the first and last slabs.  nair   1.00  33.10: a)  water  arcsin n  sin  air   arcsin   1.33 sin 35.0   25.5.   water  b) This calculation has no dependence on the glass because we can omit that step in the chain : nair sin θair  nglass sin θ glass  nwater sin  water . 33.11: As shown below, the angle between the beams and the prism is A/2 and the angle between the beams and the vertical is A, so the total angle between the two beams is 2A.
3. 33.12: Rotating a mirror by an angle while keeping the incoming beam constant leads to an increase in the incident angle  by θ. Therefore the angle between incoming and outgoing beams becomes 2θ  2 where an additional deflection of 2 arose from the mirror rotation.  na   1.70  33.13:  b  arcsin  n sin a   arcsin  1.58 sin 62.0   71.8.      b   na   1.33  33.14:  b  arcsin  n sin a   arcsin  1.52 sin 45.0   38.2. But this is the angle      b  from the normal to the surface, so the angle from the vertical is an additional 15 because of the tilt of the surface. Therefore the angle is 53.2. 33.15: a) Going from the liquid into air: nb 1.00  sin crit  na   1.48. na sin 42.5 n   1.48  So : θb  arcsin  a sin  a   arcsin  n  sin 35.0   58.1.  b   1.00  b) Going from air into the liquid: n   1.00   b  arcsin  a sin a   arcsin  n  sin 35.0   22.8.  b   1.48  33.16: If   critical angle, no light escapes, so for the largest circle, θ  θc nw sin θc  nair sin 90  (1.00) (1.00)  1.00 1 θc  sin 1 (1 / nw )  sin 1  48.6 1.333 tan  c  R / 10.0 m  R  (10.0 m) tan 48.6  11.3 m A  πR 2  π (13.3 m) 2  401 m 2
4. 33.17: For glass  water, θ crit  48.7 nb 1.333 na sin crit  nb sin 90, so na    1.77 sin  crit sin48.7 33.18: (a) Total internal reflection occurs at AC : n sin   (1.00) sin 90  1.00 (1.52) sin θ  1.00   41.1     90    90  41.1  48.9 If  is larger, is smaller and thus less than the critical angle, so this answer is the largest that  can be. (b) Same approach as in (a), except AC is now a glass-water boundary. n sin θ  nw sin 90  1.333 1.52 sin θ  1.333 θ  61.3   90  61.3  28.7 33.19: a) The slower the speed of the wave, the larger the index of refraction—so air has a larger index of refraction than water. n   v   344 m s  b)  crit  arcsin  b   arcsin  air   arcsin  n  v   1320 m s   15.1.   a  water    c) Air. For total internal reflection, the wave must go from higher to lower index of refraction—in this case, from air to water. n   1.00  33.20 :  crit  arcsin  b   arcsin  n    24.4.  a  2.42  nb 33.21 : a) tan  p   tan 54.5  1.40  nb  1.40. na n   1.00  b)  b  arcsin  a sin  a   arcsin  n  sin 54.5   35.6.  b   1.40 
5. 33.22 : From the picture on the next page,  r  37.0, and so : sin  a sin 53 nb  n a  1.33  1.77. sin  b sin 37 nb nb 1.00 33.23 : a) tan  p   na    1.65. na tan  p tan31.2 n   1.65  b)  b  arcsin  a sin  a   arcsin  n  sin 31.2   58.7.  b   1.00  n   1.66  33.24 : a ) In air  p  arctan  b   arctan  n    58.9.  a  1.00  n   1.66  b) In water  p  arctan  b   arctan  n    51.3.  a  1.33  1 33.25 : a) Through the first filter : I1  I 0 . 2 1 The second filter : I 2  I 0 cos 2 (41.0)  0.285 I 0 . 2 b) The light is linearly polarized. 33.26 : a) I  I max cos 2  I  I max cos 2 (22.5)  0.854 I max  b) I  I max cos 2  I  I max cos 2 (45.0)  0.500 I max  c) I  I max cos 2  I  I max cos 2 (67.5)  0.146 I max  33.27 : After the first filter the intensity is 1 I 0  10.0 W m 2 and the light is polarized 2 along the axis of the first filter. The intensity after the second filter is I  I 0 cos 2 , where I 0  10.0 W m 2 and ω  62.0  25.0  37.0. Thus, I  6.38 W m 2 .
6. 33.28: Let the intensity of the light that exits the first polarizer be I1, then, according to repeated application of Malus’ law, the intensity of light that exits the third polarizer is 75.0 W cm 2  I 1 cos 2 (23.0) cos 2 (62.0  23.0). 75.0 W cm 2 So we see that I1  , which is also the intensity incident cos 2 (23.0) cos 2 (62.0  23.0) on the third polarizer after the second polarizer is removed. Thus, the intensity that exits the third polarizer after the second polarizer is removed is 75.0 W cm 2 cos 2 (62.0)  32.3 W cm 2 . cos (23.0) cos (62.0  23.0) 2 2 1 1 33.29 : a) I1  I 0 , I 2  I 0 cos 2 (45.0)  0.250 I 0 , I 3  I 2 cos 2 (45.0)  0.125I 0 . 2 2 1 1 b) I1  I 0 , I 2  I 0 cos 2 (90.0)  0. 2 2 33.30: a) All the electric field is in the plane perpendicular to the propagation direction, and maximum intensity through the filters is at  to the filter orientation for the case of minimum intensity. Therefore rotating the second filter by 90 when the situation originally showed the maximum intensity means one ends with a dark cell. b) If filter P1 is rotated by 90, then the electric field oscillates in the direction pointing toward the P2 filter, and hence no intensity passes through the second filter: see a dark cell. c) Even if P2 is rotated back to its original position, the new plane of oscillation of the electric field, determined by the first filter, allows zero intensity to pass through the second filter. 33.31: Consider three mirrors, M1 in the (x,y)-plane, M2 in the (y,z)-plane, and M3 in the (x,z)-plane. A light ray bouncing from M1 changes the sign of the z-component of the velocity, bouncing from M2 changes the x-component, and from M3 changes the y- component. Thus the velocity, and hence also the path, of the light beam flips by 180 n  v   1480  33.32 : a)  b  arcsin  a sin  a   arcsin  b sin a   arcsin  n  v  sin 9.73   46.6.  b   a   344  v   344  b)  crit  arcsin  a v   arcsin     13.4.  b   1480  33.33: a) n1 sin 1  n2 sin  2 and n2 sin  2  n3 sin  3 , so n1 sin 1  n3 sin  3 sin  3  (n1 sin 1 ) / n3 b) n3 sin  3  n2 sin  2 and n2 sin  2  n1 sin 1 , so n1 sin 1  n3 sin  3 and the light makes the same angle with respect to the noral in the material with n1 as it did in part (a). c) For reflection,  r   a . These angles are still equal if  r becomes the incident angle; reflected rays are also reversible.
7. 33.34: It takes the light an additional 4.2 ns to travel 0.840 m after the glass slab is inserted into the beam. Thus, 0.840 m 0.840 m 0.840 m   (n  1)  4.2 ns. cn c c We can now solve for the index of refraction: (4.2  10 9 s) (3.00  10 8 m s) n  1  2.50. 0.840 m The wavelength inside of the glass is 490 nm λ   196 nm  200 nm. 2.50 n   1.00  33.35:  b  90  arcsin  a   90  arcsin  n    43.6.  b  1.38   n sin  b   1.38 sin( 43.6)  But na sin  a  nb sin  b   a  arcsin  b  n   arcsin     72.1.  a   1.00    33.36: na sin  a  nb sin  b  nb sin a   2           (1.00) sin a  sin 2  a   2 sin  a  cos  a   (1.80) sin  a   2  2  2  2    1.80   2 cos  a   (1.80)   a  2 arccos    51.7.  2  2  33.37: The velocity vector “maps out” the path of the light beam, so the geometry as shown below leads to:  va  v  va  vr and  a   r  arccos  y v   arccos  ry   va  vr , with the minus  v  y y  a  r   va   vr  sign chosen by inspection. Similarly,  arcsin  x   arcsin  x   v a x  v rx . v  v   a   r 
8. d air d glass (0.0180 m  0.00250 m) 0.00250 m 33.38: # λ  (# λ) air  # λ glass   n 7   λ λ 5.40  10 m 5.40  10 7 m (1.40)  3.52  10 4.  (0.00534 m) / 2   nb   1.0  1  0.00310 m   40.7  arcsin  n   arcsin  n   n  sin( 40.7 33.39: crit  arctan        a   Note: The radius is reduced by a factor of two since the beam must be incident at  crit , then reflect on the glass-air interface to create the ring.  1.5 m  33.40:  a  arctan  1.2 m   51    n   1.00    b  arcsin  a sin a   arcsin  n  sin 51   36.  b   1.33  So the distance along the bottom of the pool from directly below where the light enters to where it hits the bottom is: x  (4.0 m) tan  b  (4.0 m) tan 36  2.9 m.  xtotal  1.5 m  x  1.5 m  2.9 m  4.4 m.  8.0 cm   4.0 cm  33.41  a  arctan  16.0 cm   27 and  b  arctan 16.0 cm   14.         n sin  a   1.00 sin 27  So, na sin  a  nb sin  b  nb   a  sin     sin 14   1.8.    b   33.42: The beam of light will emerge at the same angle as it entered the fluid as seen by following what happens via Snell’s Law at each of the interfaces. That is, the emergent beam is at 42.5 from the normal.  n sin 90   1.000  33.43: a)  i  arcsin  a    arcsin     48.61.  nw   1.333  The ice does not come into the calculation since nair sin 90  nice sin  c  n w sin  i . b) Same as part (a).  n sin  b   1.33 sin 90  33.44: n a sin  a  nb sin  b  n a   b  sin      1.9.  a   sin 45 
9.  n sin  a  33.45: n a sin  a  nb sin  b   b  arcsin a  n    b   1.66 sin (25.0)   arcsin   44.6.  1.00  So the angle below the horizontal is  b  25.0  44.6  25.0  19.6, and thus the angle between the two emerging beams is 39.2.  n sin  b   1.62 sin 60  33.46: na sin  a  nb sin  b  na   b  sin     sin 90   1.40.    a    n sin  b   1.52 sin 57.2  33.47: n a sin  a  nb sin  b  n a   b  sin      1.28.  a   sin 90  33.48: a) For light in air incident on a parallel-faced plate, Snell’s Law yields: n sin  a  n sin  b  n sin  b  n sin  a  sin  a  sin  a   a   a .     b) Adding more plates just adds extra steps in the middle of the above equation that  always cancel out. The requirement of parallel faces ensures that the angle  n   n and the chain of equations can continue. c) The lateral displacement of the beam can be calculated using geometry: t  t sin( a   b )  d  L sin( a   b ) and L  d  . cos b  cos b   n sin  a   sin 66.0   d)  b  arcsin    arcsin    30.5  n   1.80  (2.40 cm) sin(66.0  30.5) d   1.62 cm. cos 30.5 33.49: a) For sunlight entering the earth’s atmosphere from the sun BELOW the horizon, we can calculate the angle  as follows: na sin  a  nb sin  b  (1.00) sin  a  n sin  b , where nb  n is the atmosphere’s index of refraction. But the geometry of the situation tells us: R nR  nR   R  sin  b   sin  a      a   b  arcsin   R  h   arcsin  R  h  .    Rh Rh      (1.0003)(6.4  10 6 m)   6.4  10 6 m  b)   arcsin    arcsin  6.4  10 6 m  2.0  10 4 m)   64.  10 6 m  2.0  10 4 m          0.22. This is about the same as the angular radius of the sun, 0.25.
10. 33.50: A quarter-wave plate shifts the phase of the light by   90 . Circularly polarized light is out of phase by 90 , so the use of a quarter-wave plate will bring it back into phase, resulting in linearly polarized light. 1 1 1 33.51: a) I  I 0 cos 2  cos 2 (90   )  I 0 (cos  sin  ) 2  I 0 sin 2 2 . 2 2 8 b) For maximum transmission, we need 2  90, so   45. 33.52: a) The distance traveled by the light ray is the sum of the two diagonal segments:   12  d  x 2  y12  (l  x) 2  y 2 .2  12 Then the time taken to travel that distance is just: d ( x 2  y12 )1 2  (l  x) 2  y 2  2 12 t  c c b) Taking the derivative with respect to x of the time and setting it to zero yields: dt 1 d  dx c dt  ( x 2  y12 )1 2  (l  x) 2  y 2  2 12  dt 1 dx c   x( x 2  y12 ) 1 2  (l  x) (l  x) 2  y 2  2 1 2 0 x (l  x)    sin  1  sin  2   1   2 . x  y1 2 2 (l  x) 2  y 2 2 33.53: a) The time taken to travel from point A to point B is just: d1 d 2 h2  x2 h 2  (l  x) 2 t   1  2 . v1 v 2 v1 v2 Taking the derivative with respect to x of the time and setting it to zero yields: d  h1  x h 2  (l  x) 2  2 2 dt x (l  x) 0   2   . dx dt  v1 v2  v1 h12  x 2 v2 h2  (l  x) 2 2   c c n1 x n2 (l  x) But v1  and v2     n1 sin 1  n2 sin  2 . n1 n2 h12  x 2 h2  (l  x) 2 2
11. 33.54: a) n decreases with increasing λ , so n is smaller for red than for blue. So beam a is the red one. b) The separation of the emerging beams is given by some elementary geometry. x x  x r  xv  d tan  r  d tan  v  d  , where x is the vertical beam tan  r  tan  v 1.00 mm separation as they emerge from the glass x   2.92 mm. From the ray sin 20 geometry, we also have  sin 70   sin 70   r  arcsin    35.7 and  v  arcsin    34.5, so :  1.61   1.66  x 2.92 mm d   9 cm. tan  r  tan  v tan 35.7  tan 34.5 A 33.55: a) na sin  a  nb sin  b  sin  a  nb sin . 2 A A  A  2 A But  a     sin     sin  n sin . 2 2  2 2 A A At each face of the prism the deviation is  , so 2    sin  n sin . 2 2  A b) From part (a),   2 arcsin  n sin   A  2  60.0     2 arcsin  (1.52) sin   60.0  38.9.  2  c) If two colors have different indices of refraction for the glass, then the deflection angles for them will differ:  60.0   red  2 arcsin  (1.61) sin   60.0  47.2  2   60.0   violet  2 arcsin  (1.66) sin   60.0  52.2    52.2  47.2  5.0.  2 
12. 33.56: Direction of ray A:  by law of reflection. Direction of ray B: At upper surface: n1 sin   n2 sin  The lower surface reflects at  . Ray B returns to upper surface at angle of incidence  : n2 sin   n1 sin  Thus n1 sin   n1 sin    Therefore rays A and B are parallel. 33.57: Both l-leucine and d-glutamic acid exhibit linear relationships between concentration and rotation angle. The dependence for l-leucine is: Rotation angle   (0.11100 ml g)C (g/100 ml), and for d - glutamic acid is : Rotation angle   (0.124100 ml g)C (g/100 ml). 33.58: a) A birefringent material has different speeds (or equivalently, wavelengths) in two different directions, so: λ λ D D 1 nD n D 1 λ0 λ1  0 and λ 2  0     1  2  D . n1 n2 λ1 λ 2 4 λ0 λ0 4 4(n1  n2 ) λ0 5.89  10 7 m b) D    6.14  10 7 m. 4(n1  n2 ) 4(1.875  1.635)
13. 33.59: a) The maximum intensity from the table is at   35, so the polarized component of the wave is in that direction (or else we would not have maximum intensity at that angle). 1 b) At   40 : I  24.8 W m  I 0  I p cos 2 (40  35) 2 2 2  24.8 W m  0.500 I 0  0.996 I p (1). 1 At   120 : I  5.2 W m  2 I 0  I p cos 2 (120  35) 2 2  5.2 W m  0.500 I 0  7.60  10 3 I p (2). Solving equations (1) and (2) we find: 2 2  19.6 W m  0.989 I p  I p  19.8 W m . Then if one subs this back into equation (1), we find: 2 5.049 = 0.500 I 0  I 0  10.1 W m . 33.60: a) To let the most light possible through N polarizers, with a total rotation of 90, we need as little shift from one polarizer to the next. That is, the angle between  successive polarizers should be constant and equal to . Then: 2N    4    2N    I1  I 0 cos 2  , I 2  I 0 cos   ,   I  I N  I 0 cos  .  2N   2N   2N  n  2  n b) If n  1, cos   1  n       1   2       2  2 2    (2 N )    2  cos 2N    1    1  1, for large N.  2N  2  2N  4N
14. 33.61: a) Multiplying Eq. (1) by sin  and Eq. (2) by sin  yields: x (1): sin   sin ωt cos α sin β  cos t sin  sin  a y (2): sin  sin t cos  sin   cos t sin  sin  a x sin   y sin  Subtracting yields:  sin t (cos  sin   cos  sin  ). a b) Multiplying Eq. (1) by cos  and Eq. (2) by cos  yields: x (1) : cos   sin t cos  cos   cos t sin  cos  a y (2) : cos   sin t cos  cos   cos t sin  cos  a x cos   y cos  Subtracting yields:   cos t (sin  cos   sin  cos  ). a (c) Squaring and adding the results of parts (a) and (b) yields: ( x sin   y sin  ) 2  ( x cos   y cos  ) 2  a 2 (sin  cos   sin  cos  ) 2 (d) Expanding the left-hand side, we have: x 2 (sin 2   cos 2  )  y 2 (sin 2   cos 2  )  2 xy (sin  sin   cos  cos  )  x 2  y 2  2 xy (sin  sin   cos  cos  )  x 2  y 2  2 xy cos(   ). The right-hand side can be rewritten: a 2 (sin  cos   sin  cos  ) 2  a 2 sin 2 (   ). Therefore: x 2  y 2  2 xy cos(   )  a 2 sin 2 (   ). Or: x 2  y 2  2 xy cos   a 2 sin 2  , where      . (e)   0 : x 2  y 2  2 xy  ( x  y ) 2  0  x  y, which is a straight diagonal line.  2 a2   : x  y  2 xy  , which is an ellipse. 2 4 2    : x 2  y 2  a 2 , which is a circle. 2 This pattern repeats for the remaining phase differences.
15. 33.62: a) By the symmetry of the triangles,  bA   aB , and  a   rB   aB   bA . C Therefore, sin  bC  n sin  a  n sin  bA  sin  aA   bC   aA . C b) The total angular deflection of the ray is:    aA   bA    2 aB   bC   a  2 aA  4 bA   . C 1  c) From Snell’s Law, sin  aA  n sin  bA   bA  arcsin  sin  aA  n  1     2 aA  4 bA    2 aA  4 arcsin  sin  aA    . n  d d  1  4  cos 1  d)  0  2  4 A  arcsin sin  aA    0  2      d aA d a  n  1  sin 2 1 n 2  n   sin 2  1   16 cos 2  1   41      4 cos 2  1  n 2  1  cos 2  1  n2     n2   1  3 cos 2  1  n 2  1  cos 2  1  (n 2  1). 3  1 2   1  e) For violet:  1  arccos  3 (n  1)   arccos  (1.342 2  1)   58.89   3        violet  139.2   violet  40.8.  1 2   1  For red:  1  arccos   3 (n  1)   arccos    3 (1.330 2  1)   59.58        red  137.5   red  42.5. Therefore the color that appears higher is red.
16. 33.63: a) For the secondary rainbow, we will follow similar steps to Pr. (34-51). The total angular deflection of the ray is:    aA   bA    2 bA    2 bA   aA   bA  2 aA  6 bA  2 , where we have used the fact from the previous problem that all the internal angles are equal and the two external equals are equal. Also using the Snell’s Law relationship, we have: 1   bA  arcsin  sin  aA  . n  1     2 aA  6 bA  2  2 aA  6 arcsin  sin  aA   2 . n  d d  1  6  cos 2  b)  0  2  6 A  arcsin  sin  aA    0  2    .  d aA d a  n  1  sin 2  2 n 2  n  1  n 2 (1  sin 2  2 n 2 )  (n 2  1  cos 2  2 )  9 cos 2  2  cos 2  2  (n 2  1). 8  1 2   1  c) For violet:  2  arccos      8 (n  1)   arccos  8 (1.342  1)   71.55 2       violet  233.2   violet  53.2.  1 2   1  For red:  2  arccos   8 (n  1)   arccos    8 (1.330 2  1)   71.94.        red  230.1   red  50.1. Therefore the color that appears higher is violet. 