Tài liệu Physics exercises_solution: Chapter 33 doc

33.1: a) .sm1004.2 47 . 1 sm1000.3 8 8    n c v b) .m1042.4 47 . 1 )m1050.6( λ λ 7 7 0      n 33.2: a) .m1017.5 Hz105.80 sm1000.3 λ 7 14 8 vacuum      f c b) .m1040.3 Hz)(1.52)1080.5( sm1000.3 λ 7 14 8 glass      fn c 33.3: a) .54.1 m/s101.94 m/s1000.3 8 8     v c n b) .m1047.5)m1055.3()54.1(λλ 77 0   n 33.4: 1.501 m)(1.333)1038.4( λ λλλ 7 Benzene waterwater BenzeneBenzenewaterwater 2    n n nn CS .5.47equalalwaysareanglesreflectedandIncident a)       ar   :33.5 b) .0.665.42sin 66.1 00.1 arcsin 2 sinarcsin 22                         a b a bb n n 33.6: sm1017.2 s105.11 m50.2 8 9     t d v 38.1 m/s102.17 m/s1000.3 8 8     v c n 33.7: bbaa nn   sinsin  sm1051.2194.1/)m/s1000.3(so 194.1 1.48sin 7.62sin 00.1 sin sin 88                     ncvvcn nn b a ab   33.8 (a) Apply Snell’s law at both interfaces. 33.9: a) Let the light initially be in the material with refractive index n a and let the third and final slab have refractive index n b Let the middle slab have refractive index n 1 11 sinsin:interface1st  nn aa  bb nn  sinsin:interface2nd 11  .sinsingivesequations two theCombining bbaa nn   b) For N slabs, where the first slab has refractive index n a and the final slab has .sinsin,,sinsin,sinsin,indexrefractive 22221111 bbNNaab nnnnnnn     ofangleon thedepends travelofdirection finalThe.sinsingivesThis bbaa nn   incidence in the first slab and the indicies of the first and last slabs. 33.10: a) .5.250.35sin 33.1 00.1 arcsinsinarcsin air water air water                   n n b) This calculation has no dependence on the glass because we can omit that step in the .sinsinsin:chain waterwateglassglassairair  r nθnθn  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. 33.12: Rotating a mirror by an angle  while keeping the incoming beam constant leads rotation.mirror thefromarose2ofdeflectionadditionalan where22becomesbeamsoutgoingand incomingbetween angle theTherefore.byangleincident in theincreasean to   θ θ 33.13: .71.862.0sin 1.58 1.70 arcsinsinarcsin                  a b a b n n  33.14: 38.2.45.0sin 1.52 1.33 arcsinsinarcsin                  a b a b n n  But this is the angle .53.2isangle theThereforesurface. theof tilt the ofbecause15additionalan is vertical thefromangle thesosurface, the tonormal thefrom   33.15: a) Going from the liquid into air: 1.48. 42.5sin 1.00 sin crit    a a b n n n  .58.135.0sin 1.00 1.48 arcsinsinarcsin:So b                  a b a n n θ  b) Going from air into the liquid: .22.835.0sin 1.48 1.00 arcsinsinarcsin                  a b a b n n  33.16: c θθ   circle,largest for theso escapes, light no angle, critical If  222 c 1 w 1 c aircw m401m)(13.3 m11.3648tan)m0.10(m0.10/tan 48.6 1.333 1 sin)/1(sin 00.1)00.1()00.1(90sinsin      ππRA .RR n θ nθn  33.17:    7.48water,glassFor crit θ 1.77 sin48.7 1.333 sin so,90sinsin crit crit      b aba n nnn 33.18: (a) 00.190sin)00.1(sin:ACat occursreflectioninternalTotal     n   41.1 1.00sin(1.52)  θ           48.941.19090    theisanswer thissoangle,critical than theless thusandsmaller islarger,isIf   largest that  can be. (b) Same approach as in (a), except AC is now a glass-water boundary.      61.3 1.333sin1.52 1.33390sinsin w θ θ nθn       28.761.390  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. .15.1 sm1320 sm344 arcsinarcsinarcsinb) water air crit                             v v n n a b  c) Air. For total internal reflection, the wave must go from higher to lower index of refraction—in this case, from air to water. .24.4 2.42 1.00 arcsinarcsin crit                  a b n n  :33.20 .40.140.154.5tantana)  b a b p n n n  :33.21 .35.654.5sin 1.40 1.00 arcsinsinarcsinb)                  a b a b n n  :soand,0.37page,next on thepicture theFrom   r  :33.22 1.77. 37sin 53sin 1.33 sin sin     b a ab nn   1.65. tan31.2 1.00 tan tana)    p b a a b p n n n n   :33.23 .58.731.2sin 1.00 1.65 arcsinsinarcsinb)                  a b a b n n  .58.9 1.00 1.66 arctanarctanair In )a                  a b p n n  :33.24 .51.3 1.33 1.66 arctanarctanIn water b)                  a b p n n  . 2 1 :filterfirst eThrough th a) 01 II :33.25 .285.0)0.41(cos 2 1 :filtersecondThe 0 2 02 III  b) The light is linearly polarized.   max 2 max 2 max 0.854)(22.5coscos a) IIIII  :33.26   max 2 max 2 max 500.0)(45.0coscos b) IIIII    max 2 max 2 max 146.0)5.67(coscos c) IIIII  polarizedislight theandmW10.0isintensity filter thefirst After the 2 0 2 1 I:33.27 where,cosisfilter secondafter theintensity Thefilter.first theofaxis thealong 2 0  II  .37.025.062.0andmW10.0 2 0  ωI .mW6.38Thus, 2 I 33.28: Let the intensity of the light that exits the first polarizer be I 1 , then, according to repeated application of Malus’ law, the intensity of light that exits the third polarizer is ).0.230.62(cos)0.23(coscmW0.75 22 1 2  I incident intensity thealsois which , )0.230.62(cos)0.23(cos cmW0.75 that see weSo 22 2 1   I 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 .cmW32.3 )23.0(62.0cos)(23.0cos )(62.0coscmW75.0 2 22 22    .125.0)(45.0cos,250.0)0.45(cos 2 1 , 2 1 a) 0 2 230 2 0201 IIIIIIII :33.29 0.)(90.0cos 2 1 , 2 1 b) 2 0201  IIII 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 P 1 is rotated by 90 , then the electric field oscillates in the direction pointing toward the P 2 filter, and hence no intensity passes through the second filter: see a dark cell. c) Even if P 2 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, M 1 in the (x,y)-plane, M 2 in the (y,z)-plane, and M 3 in the (x,z)-plane. A light ray bouncing from M 1 changes the sign of the z-component of the velocity, bouncing from M 2 changes the x-component, and from M 3 changes the y- component. Thus the velocity, and hence also the path, of the light beam flips by 180 .46.69.73sin 344 1480 arcsinsinarcsinsinarcsin a)                           a a b a b a b v v n n  :33.32 b) .13.4 1480 344 arcsinarcsin crit                  b a v v  33.33: a) 331133222211 sinsinso,sinsinandsinsin       nnnnnn    withmaterialin thenoral therespect to with anglesame themakeslight theandsin sinso,sinsinandsinsin b) /)sin(sin 33 11112222333113         n nnnnnnn     1 n as it did in part (a). c) For reflection, . ar   These angles are still equal if r  becomes the incident angle; reflected rays are also reversible. 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, ns.4.2 m0.840 )1( m0.840m0.840  c n cnc We can now solve for the index of refraction: 2.50.1 m0.840 )sm10(3.00s)10(4.2 89     n The wavelength inside of the glass is nm.200nm196 2.50 nm490 λ   33.35: .6.43 38.1 00.1 arcsin90arcsin90                  b a b n n  But .1.72 00.1 )6.43sin(38.1 arcsin sin arcsinsinsin                   a bb abbaa n n nn   33.36:        2 nnn a bbbaa   sinsinsin .51.7 2 1.80 arccos2)80.1( 2 cos2 2 sin(1.80) 2 cos 2 sin2 2 2sinsin(1.00)                                           a a aaaa a     33.37: The velocity vector “maps out” the path of the light beam, so the geometry as shown below leads to: ,arccosarccosand yy yy ra r r a a rara vv v v v v vv                     with the minus sign chosen by inspection. Similarly, .arcsinarcsin xx xx ra r r a a vv v v v v                    33.38:          m105.40 m0.00250 m105.40 m)0.00250m(0.0180 λλ λ #)λ(#λ# 77 glass air glass air n d d .103.52(1.40) 4  33.39: 7 .40sin( 10.1 arcsinarcsin7.40 m00310.0 2/)m00534.0( arctan crit                           n nn n a b  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. 33.40:           51 m2.1 m5.1 arctan a  .3651sin 1.33 1.00 arcsinsinarcsin                  a b a b n n  So the distance along the bottom of the pool from directly below where the light enters to where it hits the bottom is: .m2.936tanm)0.4(tanm)0.4(  b x  .m4.4m2.9m5.1m5.1 total  xx 33.41 .14 cm16.0 cm4.0 arctanand27 cm16.0 cm8.0 arctan                    ba  So, .8.1 14sin 27sin00.1 sin sin sinsin                    b aa bbbaa n nnn    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  5.42 from the normal. 33.43: a) .61.48 333.1 000.1 arcsin 90sin arcsin                   w a i n n  The ice does not come into the calculation since .sinsin90sin iceair iwc nnn   b) Same as part (a). 33.44: .9.1 45sin 90sin33.1 sin sin sinsin                    a bb abbaa n nnn    33.45:          b aa bbbaa n n nn   sin arcsinsinsin .6.44 00.1 )0.25(sin66.1 arcsin          So the angle below the horizontal is ,6.190.256.440.25  b  and thus the angle between the two emerging beams is .2.39  33.46: .40.1 90sin 60sin62.1 sin sin sinsin                    a bb abbaa n nnn    33.47: .28.1 90sin 2.57sin52.1 sin sin sinsin                    a bb abbaa n nnn    33.48: a) For light in air incident on a parallel-faced plate, Snell’s Law yields: .sinsinsinsinsinsin aaaaabba nnnn             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 nn     and the chain of equations can continue. c) The lateral displacement of the beam can be calculated using geometry: . cos )sin( cos and)sin( b ba b ba t d t LLd             d)                   5.30 80.1 0.66sin arcsin sin arcsin n n a b   .cm62.1 5 . 30 cos )5.300.66sin()cm40.2(       d 33.49: a) For sunlight entering the earth’s atmosphere from the sun BELOW the horizon, we can calculate the angle  as follows: nnnnn bbabbaa  where,sinsin)00.1(sinsin  is the atmosphere’s index of refraction. But the geometry of the situation tells us:                         hR R hR nR hR nR hR R baab arcsinarcsinsinsin  . b)                        m102.0m1064. m106.4 arcsin m)102.0m106.4 m)10(6.4(1.0003) arcsin 46 6 46 6  .22.0    This is about the same as the angular radius of the sun, .25.0  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. 33.51: a) .2sin 8 1 )sin(cos 2 1 )90(coscos 2 1 2 0 2 0 22 0  IIII  b) For maximum transmission, we need .45so,902       33.52: a) The distance traveled by the light ray is the sum of the two diagonal segments:     .)( 21 2 2 2 21 2 1 2 yxlyxd  Then the time taken to travel that distance is just:   c yxlyx c d t 21 2 2 2212 1 2 )()(   b) Taking the derivative with respect to x of the time and setting it to zero yields:         .sinsin )( )( 0)()()( 1 )()( 1 2121 2 2 22 1 2 21 2 2 2212 1 2 21 2 2 2212 1 2            yxl xl yx x yxlxlyxx cdx dt yxlyx dt d cdx dt 33.53: a) The time taken to travel from point A to point B is just: . )( 2 22 2 1 22 1 2 2 1 1 v xlh v xh v d v d t     Taking the derivative with respect to x of the time and setting it to zero yields: .sinsin )( )( andBut . )( )( )( 0 2211 22 2 2 22 1 1 2 2 1 1 22 22 22 11 2 22 2 1 22 1  nn xlh xln xh xn n c v n c v xlhv xl xhv x v xlh v xh dt d dx dt                        [...]... (1.66) sin   60.0  52.2    52.2  47.2  5.0 2   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... 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 λ0 D D 1 nD n D 1 λ1  0 and λ 2  0     1  2  D 4(n1  n2 ) n1 n2 λ1 λ 2 4 λ0 λ0 4 λ0 5.89  10 7 m b) D    6.14  10 7 m 4(n1  n2 ) 4(1.875  1.635) 33. 59: a) The maximum intensity from the table is at  ... violet:  1  arccos   3   3        violet  139.2   violet  40.8  1 2   1  (n  1)   arccos  For red:  1  arccos  (1 .330 2  1)   59.58  3   3        red  137.5   red  42.5 Therefore the color that appears higher is red 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... cos 2  2  cos 2  2  (n 2  1) 8  1 2   1  2    c) For violet:  2  arccos   8 (n  1)   arccos  8 (1.342  1)   71.55       violet  233. 2   violet  53.2  1 2   1  (n  1)   arccos  For red:  2  arccos  (1 .330 2  1)   71.94  8   8        red  230.1   red  50.1 Therefore the color that appears higher is violet ... 120 : I  5.2 W m  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...  I 0 cos    2N   2N   2N  n  2  n      1   2     b) If n  1, cos   1    2 2   n 2  cos 2N (2 N )    2     1, for large N    1    1 2  2N  4N  2N  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   sin...  y, which is a straight diagonal line  2 a2 2   : x  y  2 xy  , which is an ellipse 4 2    : x 2  y 2  a 2 , which is a circle 2 This pattern repeats for the remaining phase differences C 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 b) The total angular deflection of the ray.. .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... 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  2.92 mm x  d  9 cm tan  r  tan  v tan 35.7  tan 34.5 33. 55: a) na sin  a  nb sin  b  sin  a  nb sin But  a  A 2 A A  2 A A     sin     sin  n sin 2 2 2 2  At each face of the prism the deviation is  , so 2    sin A A  n sin . now a glass-water boundary.      61.3 1 .333 sin1.52 1 .333 90sinsin w θ θ nθn       28.761.390  33. 19: a) The slower the speed of the wave, the.                           a a b a b a b v v n n  :33. 32 b) .13.4 1480 344 arcsinarcsin crit                  b a v v  33. 33: a) 331 1332 22211 sinsinso,sinsinandsinsin       nnnnnn   