Reflection and Refraction of Light, Schemes and Mind Maps of Law

Download Reflection and Refraction of Light and more Schemes and Mind Maps Law in PDF only on Docsity! 22 Refl ection and Refraction of Light Clicker Questions Question P1.01 Question The sun appears larger at sunset than it does when overhead because of: 1. Scattering of light. 2. Dispersion. 3. Diffraction. 4. Refraction. 5. An optical illusion. 6. None of the above. Commentary Purpose: To connect physics processes to your everyday experience. Discussion: A careful measurement of the sun’s angular size when overhead and setting would reveal that it is not in fact larger at sunset. It just looks larger: an optical illusion, due to the fact that your eye has objects on the horizon to compare it to when setting, but nothing to use as a size reference when it is overhead. The moon, also, looks larger on the horizon than overhead. Key Points: • How something appears can be strongly infl uenced by its surroundings. This is the basis for many optical illusions. • Not all of your everyday experiences can be explained by physics! Some must be explained by psychology or cognitive science. For Instructors Only This question follows a set on scattering and similar optical processes, such as Questions 60 and 61, well. Students may assume from context that the answer to this must be “scattering” or the like; this question helps keeps them thinking. 281 56157_22_ch22_p281-304.indd 281 3/19/08 7:54:43 PM 282 Chapter 22 Question P1.02 Description: Introducing refraction. Question Refraction occurs at the interface between two transparent media because: 1. The mass density of the material changes. 2. The frequency of the light changes. 3. The speed of light is different in the two media. 4. The direction of the light changes. 5. Some of the light is refl ected. 6. None of the above. Commentary Purpose: To probe your understanding of the cause of refraction. Discussion: Answer (4), the changing of the light’s direction, is what refraction means; it is not an explanation or cause. The light changes direction because the propagation speed of light in the two media is different (answer 3); as the wave front crosses the interface, the part already across the interface is moving at a different rate, “skewing” the front. The mass density of the materials may differ (answer 1), but this is not necessarily the case and does not directly cause the refraction. Some of the light may indeed refl ect (answer 5), but this is a separate phenomenon. Although the speed and wavelength of the light change, its frequency does not, so statement (2) is simply false. Key Points: • Refraction occurs when light crosses a boundary between two materials of different index of refraction: that is, materials in which light travels at different speeds. • The index of refraction indicates the speed of light in a medium, as compared to the speed of light in vacuum. • It is important to distinguish between a cause of a phenomenon, a description of it, phenomena associated with it, and phenomena similar to it. For Instructors Only Explaining how refraction arises from a change in the speed of light depends on the level of your class and the time you are willing to spend on it. A wavefront description based on Huygens’s construction works. So also can a “marching ants” picture. QUICK QUIZZES 1. (a). In part (a), you can see clear refl ections of the headlights and the lights on the top of the truck. The refl ection is specular. In part (b), although bright areas appear on the roadway in front of the headlights, the refl ection is not as clear, and no separate refl ection of the lights from the top of the truck is visible. The refl ection in part (b) is mostly diffuse. 2. Beams 2 and 4 are refl ected; beams 3 and 5 are refracted. 56157_22_ch22_p281-304.indd 282 3/19/08 7:54:44 PM Refl ection and Refraction of Light 285 8. The color traveling slowest is bent the most. Thus, X travels more slowly in the glass prism. 10. Total internal refl ection occurs only when light attempts to move from a medium of high index of refraction to a medium of lower index of refraction. Thus, light moving from air ( )n = 1 to water ( . )n = 1 33 cannot undergo total internal refl ection. 12. Objects beneath the surface of water appear to be raised toward the surface by refraction. Thus, the bottom of the oar appears to be closer to the surface than it really is, and the oar looks to be bent. PROBLEM SOLUTIONS 22.1 The total distance the light travels is ∆d D R R= − − ⎛ ⎝⎜ ⎞ ⎠⎟ = 2 2 3 center to center Earth Moon .84 10 6 38 10 1 76 10 7 52 108 6 6 8× − × − ×( ) = ×. . . m m Therefore, v = = × = ×∆ ∆ d t 7 52 10 3 00 10 8 8. . m 2 .51 s m s 22.2 (a) The energy of a photon is E hf hc= = λ, where Planck’s constant is h = × ⋅−6 63 10 34. J s and the speed of light in vacuum is c = ×3 00 108. m s. If λ = × −1 00 10 10. m, E = × ⋅( ) ×( ) × − − 6 63 10 3 00 10 10 34 8. . J s m s 1.00 m10 = × −1 99 10 15. J (b) E = ×( ) × ⎛ ⎝⎜ ⎞ ⎠⎟ =− −1 99 10 1 10 1 215. . J eV 1.602 J19 4 104× eV (c) and (d) For the x-rays to be more penetrating, the photons should be more energetic. Since the energy of a photon is directly proportional to the frequency and inversely proportional to the wavelength, the wavelength should decrease , which is the same as saying the frequency should increase . 22.3 (a) E hf= = × ⋅( ) ×( )−6 63 10 5 00 10 134 17. . J s Hz eV 1.60 × ⎛ ⎝⎜ ⎞ ⎠⎟ = ×−10 2 07 1019 3 J eV. (b) E hf hc= = = × ⋅( ) ×( ) × − λ 6 63 10 3 00 1034 8. . J s m s 3.00 10 1 10 6 63 10 19 2 9 nm nm m J− −⎛ ⎝⎜ ⎞ ⎠⎟ = ×. E = × × ⎛ ⎝⎜ ⎞ ⎠⎟ =− −6 63 10 1 10 4 1419 19. . J eV 1.60 J eV 56157_22_ch22_p281-304.indd 285 3/19/08 7:54:46 PM 286 Chapter 22 22.4 (a) λ0 8 73 00 10 10 5 50 10= = × × = × −c f . . m s 5.45 Hz m14 (b) From Table 22.1 the index of refraction for benzene is n = 1 501. . Thus, the wavelength in benzene is λ λ n n = = × = × − −0 7 75 50 10 3 67 10 . . m 1.501 m (c) E hf= = × ⋅( ) ×( )−6 63 10 5 45 10 1 1 60 34 14. . . J s Hz eV × ⎛ ⎝⎜ ⎞ ⎠⎟ =−10 2 2619 J eV. (d) The energy of the photon is proportional to the frequency, which does not change as the light goes from one medium to another. Thus, when the photon enters benzene, the energy does not change . 22.5 The speed of light in a medium with index of refraction n is v = c n, where c is its speed in vacuum. (a) For water, n = 1 333. , and v = × = ×3 00 10 1 333 2 25 10 8 8. . . m s m s (b) For crown glass, n = 1 52. , and v = × = ×3 00 10 1 52 1 97 10 8 8. . . m s m s (c) For diamond, n = 2 419. , and v = × = ×3 00 10 2 419 1 24 10 8 8. . . m s m s 22.6 (a) From λ f c= , the wavelength is given by λ = c f . The energy of a photon is E hf= , so the frequency may be expressed as f E h= , and the wavelength becomes λ = = =c f c E h hc E (b) Higher energy photons have shorter wavelengths. 22.7 From Snell’s law, n n2 2 1 1sin sinθ θ= . Thus, when θ1 45= ° and the fi rst medium is air (n1 1 00= . ), we have sin . sinθ2 2 1 00 45= ( ) ° n . (a) For quartz, n2 1 458= . , and θ2 1 1 00 45 1 458 29= ( ) °⎛ ⎝⎜ ⎞ ⎠⎟ = °−sin . sin . (b) For carbon disulfi de, n2 1 628= . , and θ2 1 1 00 45 1 628 26= ( ) °⎛ ⎝⎜ ⎞ ⎠⎟ = °−sin . sin . (c) For water, n2 1 333= . , and θ2 1 1 00 45 1 333 32= ( ) °⎛ ⎝⎜ ⎞ ⎠⎟ = °−sin . sin . 56157_22_ch22_p281-304.indd 286 3/19/08 7:54:46 PM Refl ection and Refraction of Light 287 22.8 (a) From geometry, 1 25 40 0. sin . m = d °, so d = 1 94. m (b) 50 0. ° above horizontal , or parallel to the incident ray 22.9 n n1 1 2 2sin sinθ θ= sin . sin .θ1 1 333 45 0= ° sin ( . )( . ) .θ1 1 333 0 707 0 943= = θ1 70 5= ° →. 19 5. ° above the horizontal 22.10 (a) n c= = × × = v 3 00 10 2 17 10 1 38 8 8 . . . m s m s (b) From Snell’s law, n n2 2 1 1sin sinθ θ= , θ θ 2 1 1 1 2 1 1 00 23 1 = ⎛ ⎝⎜ ⎞ ⎠⎟ = ( ) °− −sin sin sin . sin .n n 1 38 0 284 16 51 . sin . . ⎡ ⎣⎢ ⎤ ⎦⎥ = ( ) =− ° 22.11 (a) From Snell’s law, n n 2 1 1 2 1 00 30 0 19 24 1 5= = ( ) ° ° =sin sin . sin . sin . . θ θ 2 (b) λ λ 2 0 2 632 8 1 52 416= = = n . . nm nm (c) f c= = × × = ×−λ0 8 9 143 00 10 4 74 10 . . m s 632.8 10 m Hz in air and in syrup (d) v2 2 8 83 00 10 1 52 1 97 10= = × = ×c n . . . m s m s 56157_22_ch22_p281-304.indd 287 3/19/08 7:54:47 PM 290 Chapter 22 22.18 (a) From Snell’s law, the angle of refraction at the fi rst surface is θ θ 2 1 1 1 1 00 = ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ( )− −sin sin sin .n n air glass sin . . . 30 0 1 50 19 5 ° ° ⎡ ⎣⎢ ⎤ ⎦⎥ = (b) Since the upper and lower surfaces are parallel, the normal lines where the ray strikes these surfaces are parallel. Hence, the angle of incidence at the lower surface will be θ2 19 5= . ° . The angle of refraction at this surface is then θ θ 3 1 1 1 50 = ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ =− −sin sin sin .n n glass glass air ( ) °⎡ ⎣⎢ ⎤ ⎦⎥ = ° sin . . . 19 5 1 00 30 0 Thus, the light emerges traveling parallel to the incident beam. (c) Consider the sketch above and let h represent the distance from point a to c (that is, the hypotenuse of triangle abc). Then, h = = ° =2 00 2 00 19 5 2 12 2 . cos . cos . . cm cm cm θ Also, α θ θ= − = − =1 2 30 0 19 5 10 5. . .° ° °, so d h= = ( ) =sin . sin . .α 2 12 10 5 0 386 cm cm° (d) The speed of the light in the glass is v = = × = ×c nglass m s m s 3 00 10 1 50 2 00 10 8 8. . . (e) The time required for the light to travel through the glass is t h= = × ⎛ ⎝⎜ ⎞ ⎠⎟ = × v 2 12 10 1 10 1 06 1 . . cm 2.00 m s m cm8 2 0 10− s (f) Changing the angle of incidence will change the angle of refraction, and therefore the dis- tance h the light travels in the glass. Thus, the travel time will also change . 22.19 From Snell’s law, the angle of incidence at the air-oil interface is θ θ = ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ( ) − − sin sin sin . sin 1 1 1 48 n n oil oil air 20 0 1 00 30 4 . . . °⎡ ⎣⎢ ⎤ ⎦⎥ = ° and the angle of refraction as the light enters the water is ′ = ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ( )− −θ θ sin sin sin .1 1 1 48n n oil oil water sin . . . 20 0 1 333 22 3 °⎡ ⎣⎢ ⎤ ⎦⎥ = ° 56157_22_ch22_p281-304.indd 290 3/19/08 7:54:49 PM Refl ection and Refraction of Light 291 22.20 Since the light ray strikes the fi rst surface at normal inci- dence, it passes into the prism without deviation. Thus, the angle of incidence at the second surface (hypotenuse of the triangular prism) is θ1 45 0= °. , as shown in the sketch at the right. The angle of refraction is θ2 45 0 15 0 60 0= + =. . .° ° ° and Snell’s law gives the index of refraction of the prism material as n n 1 2 2 1 1 00 60 0 45 0 = = ( ) ( ) ( ) =sin sin . sin . sin . θ θ ° ° 1 22. 22.21 ∆t = ( ) −time to travel 6.20 m in ice time to travel 6.20 m in air( ) ∆t c = −6 20 6 20. . m m icev Since the speed of light in a medium of refractive index n is v = c n ∆t c c = ( ) −⎛ ⎝⎜ ⎞ ⎠⎟ = ( )( ) 6 20 1 6 20 0 309 3 . . . m 1.309 m . . . 00 10 6 39 10 6 398 9 × = × =− m s s ns 22.22 From Snell’s law, sin sin .θ = ⎛ ⎝⎜ ⎞ ⎠⎟ °n n medium liver 50 0 But, n n c c medium liver medium liver liver medium = =v v v v = 0 900. so, θ = ( ) °⎡⎣ ⎤⎦ = °−sin . sin . .1 0 900 50 0 43 6 From the law of refl ection, d = =12 0 6 00 . . cm 2 cm , and h d= = ( ) = tan cm tan 43.6 cm θ 6 00 6 30 . . ° 22.23 (a) Before the container is fi lled, the ray’s path is as shown in Figure (a) at the right. From this fi gure, observe that sinθ1 1 2 2 2 1 1 = = + = ( ) + d s d h d h d (a) d q1 q1 s1h After the container is fi lled, the ray’s path is shown in Figure (b). From this fi gure, we fi nd that sinθ2 2 2 2 2 2 2 2 1 4 1 = = + ( ) = ( ) + d s d h d h d From Snell’s law, n nair sin sinθ θ1 2= , or (b) q1 q 2 s2h d/2 1 00 1 4 1 2 2 . h d n h d( ) + = ( ) + and 4 1 2 2 2 2h d n h d n( ) + = ( ) + Simplifying, this gives 4 12 2 2−( )( ) = −n h d n or h d n n = − − 2 2 1 4 continued on next page 56157_22_ch22_p281-304.indd 291 3/19/08 7:54:50 PM 292 Chapter 22 (b) If d = 8 0. cm and n n= =water 1 333. , then h = ( ) ( ) − − ( ) =8 0 1 333 1 4 1 333 4 7 2 2. . . . cm cm 22.24 (a) A sketch illustrating the situation and the two triangles needed in the solution is given below: (b) The angle of incidence at the water surface is θ1 1 90 0 10 42 0= × ⎛ ⎝⎜ ⎞ ⎠⎟ =−tan . . m 1.00 m2 ° (c) Snell’s law gives the angle of refraction as θ θ 2 1 1 1 1 333 = ⎛ ⎝⎜ ⎞ ⎠⎟ = ( )− −sin sin sin . sn n water air in . . . 42 0 1 00 63 1 ° ° ⎛ ⎝⎜ ⎞ ⎠⎟ = (d) The refracted beam makes angle φ θ= − =90 0 26 92. .° ° with the horizontal. (e) Since tan ( .φ = ×h 2 10 102 m), the height of the target is h = ×( ) ( ) =2 10 10 26 9 1072. tan . m m° 22.25 As shown at the right, θ β θ1 2 180+ + = °. When β = 90°, this gives θ θ2 190= −° Then, from Snell’s law sin sin sin cos θ θ θ θ 1 2 1 190 = = −( ) = n n n n g g g air ° Thus, when β = °90 , sin cos tan θ θ θ1 1 1= = ng or θ1 1= ( )−tan ng 56157_22_ch22_p281-304.indd 292 3/19/08 7:54:51 PM Refl ection and Refraction of Light 295 22.34 As light goes from a medium having a refractive index n1 to a medium with refractive index n n2 1< , the critical angle is given the relation sinθc n n= 2 1. Table 22.1 gives the refractive index for various substances at λ0 589= nm. (a) For fused quartz surrounded by air, n n1 21 458 1 00= =. . and , giving θc = =−sin ( . . ) . .1 1 00 1 458 43 3° (b) In going from polystyrene ( . )n1 1 49= to air, θc = =−sin ( . . ) . .1 1 00 1 49 42 2° (c) From sodium chloride (n1 1 544= . ) to air, θc = ( ) = °−sin . . .1 1 00 1 544 40 4 . 22.35 When light is coming from a medium of refractive index n1 into water (n2 1 333= . ), the critical angle is given by θc n= −sin ( . )1 11 333 . (a) For fused quartz, n1 1 458= . , giving θc = ( ) =−sin . . .1 1 333 1 458 66 1° . (b) In going from polystyrene (n1 1 49= . ) to water, θc = ( ) =−sin . . .1 1 333 1 49 63 5° . (c) From sodium chloride (n1 1 544= . ) to water, θc = ( ) =−sin . . .1 1 333 1 544 59 7° . 22.36 Using Snell’s law, the index of refraction of the liquid is found to be n nair i r liquid = = ( ) °sin sin . sin . sin . θ θ 1 00 30 0 22 0 1 33 ° = . Thus, θc n n = ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞− −sin sin . . 1 1 1 00 1 33 air liquid ⎠⎟ = 48 5. ° 22.37 When light attempts to cross a boundary from one medium of refractive index n1 into a new medium of refractive index n n2 1< , total internal refl ection will occur if the angle of incidence exceeds the critical angle given by θc n n= ( )−sin 1 2 1 . (a) If n n n c1 21 53 1 00= = = = −. . , sin and then air θ 1 1 00 1 53 . . ⎛ ⎝⎜ ⎞ ⎠⎟ = 40.8° (b) If n n n c1 21 53 1 333= = = =. . , s and then water θ in . . − ⎛ ⎝⎜ ⎞ ⎠⎟ =1 1 333 1 53 60.6° 22.38 The critical angle for this material in air is θc n n = ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ − −sin sin . . 1 1 1 00 1 36 air pipe = °47 3. Thus, θ θr c= − =90 0 42 7. .° ° and from Snell’s law, θ θ i rn n = ⎛ ⎝⎜ ⎞ ⎠⎟ = ( )− −sin sin sin . sin1 1 1 36pipe air 42 7 1 00 67 2 . . . ° ° ⎛ ⎝⎜ ⎞ ⎠⎟ = 56157_22_ch22_p281-304.indd 295 3/19/08 8:32:08 PM 296 Chapter 22 22.39 The angle of incidence at each of the shorter faces of the prism is 45°, as shown in the fi gure at the right. For total internal refl ection to occur at these faces, it is necessary that the critical angle be less than 45°. With the prism surrounded by air, the critical angle is given by sin .θc n n n= =air prism prism1 00 , so it is necessary that sin . sinθc n = < °1 00 45 prism or nprism > ° = =1 00 45 1 00 2 2 2 . sin . 22.40 (a) The minimum angle of incidence for which total internal refl ection occurs is the critical angle. At the critical angle, the angle of refraction is 90°, as shown in the fi gure at the right. From Snell’s law, n ng i asin sinθ = 90°, the critical angle for the glass-air interface is found to be θ θi c a g n n = = ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝ − −sin sin sin . . 1 190 1 00 1 78 ° ⎜ ⎞ ⎠⎟ = 34 2. ° (b) When the slab of glass has a layer of water on top, we want the angle of incidence at the water-air interface to equal the critical angle for that combination of media. At this angle, Snell’s law gives n nw c asin sin .θ = =90 1 00° and sin .θc wn= 1 00 Now, considering the refraction at the glass-water interface, Snell’s law gives n ng i g csin sinθ θ= . Combining this with the result for sinθc from above, we fi nd the required angle of incidence in the glass to be θ θ i w c g w w g n n n n n = ⎛ ⎝⎜ ⎞ ⎠⎟ = ( )⎛ ⎝ − −sin sin sin .1 1 1 00 ⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ − −sin . sin . . 1 11 00 1 00 1 78ng = 34 2. ° (c) an d (d) Observe in the calculation of part (b) that all the physical properties of the intervening layer (water in this case) canceled, and the result of part (b) is identical to that of part (a). This will always be true when the upper and lower surfaces of the intervening layer are parallel to each other. Neither the thickness nor the index of refraction of the intervening layer affects the result. 56157_22_ch22_p281-304.indd 296 3/19/08 7:54:55 PM Refl ection and Refraction of Light 297 22.41 (a) Snell’s law can be written as sin sin θ θ 1 2 1 2 = v v . At the critical angle of incidence θ θ1 =( )c , the angle of refraction is 90° and Snell’s law becomes sinθc = v v 1 2 . At the concrete-air boundary, θc = ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ − −sin sin1 1 2 1 343v v m s 1 850 m s = 10 7. ° (b) Sound can be totally refl ected only if it is initially traveling in the slower medium. Hence, at the concrete-air boundary, the sound must be traveling in air . (c) Sound in air falling on the wall from most directions is 100% reflected , so the wall is a good mirror. 22.42 The sketch at the right shows a light ray entering at the painted corner of the cube and striking the center of one of the three unpainted faces of the cube. The angle of incidence at this face is the angle θ1 in the triangle shown. Note that one side of this triangle is half the diagonal of a face and is given by d 2 2 2 2 2 = + =


Also, the hypotenuse of this triangle is L d= + ⎛ ⎝⎜ ⎞ ⎠⎟ = + =



2 2 2 2 2 2 3 2 Thus, sinθ1 2 2 3 2 1 3 = = ( ) =d L

For total internal refl ection at this face, it is necessary that sin sinθ θ1 ≥ =c n n air cube or 1 3 1 00≥ . n giving n ≥ 3 22.43 If θc = °42 0. at the boundary between the prism glass and the surrounding medium, then sinθc n n = 2 1 gives n n m glass = °sin .42 0 From the geometry shown in the fi gure at the right, α β α= − = = − − =90 0 42 0 48 0 180 60 0 72 0. . . . .° ° ° ° ° °, and θ βr = − =90 0 18 0. .° °. Thus, applying Snell’s law at the fi rst surface gives θ θ θ 1 1 1= ⎛ ⎝⎜ ⎞ ⎠⎟ =− −sin sin sin sinn n n n r m r m glass glass ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ =−sin sin . sin . .1 18 0 42 0 27 5 ° ° ° 56157_22_ch22_p281-304.indd 297 3/19/08 7:54:56 PM 300 Chapter 22 22.51 In the fi gure at the right, observe that β θ= ° −90 1 and α θ= −90 1° . Thus, β α= . Similarly, on the right side of the prism, δ θ= ° −90 2 and γ θ= −90 2° , giving δ γ= . Next, observe that the angle between the refl ected rays is B = +( ) + +( )α β γ δ , so B = +( )2 α γ . Finally, observe that the left side of the prism is sloped at angle a from the vertical, and the right side is sloped at angle g . Thus, the angle between the two sides is A = +α γ , and we obtain the result B A= +( ) =2 2α γ . 22.52 (a) Observe in the sketch at the right that a ray originally traveling along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fi ber in the curve. Thus, if this ray is totally internally refl ected, all of the others are also totally refl ected. For this ray to be totally internally refl ected it is necessary that θ θ≥ c or sin sinθ θ≥ = =c n n n air pipe 1 But, sinθ = −R d R , so we must have R d R n − ≥ 1 which simplifi es to R nd n≥ −( )1 (b) As d R→ →0 0, . This is reasonable behavior. As n increases, R nd n d nmin = − = −1 1 1 decreases. This is reasonable behavior. As n → 1, Rmin increases. This is reasonable behavior. (c) R nd nmin . . = − = ( )( ) − = 1 1 40 100 1 40 1 350 m m µ µ 56157_22_ch22_p281-304.indd 300 3/19/08 7:55:00 PM Refl ection and Refraction of Light 301 22.53 Consider light which leaves the lower end of the wire and travels parallel to the wire while in the benzene. If the wire appears straight to an observer looking along the dry portion of the wire, this ray from the lower end of the wire must enter the observers eye as he sights along the wire. Thus, the ray must refract and travel parallel to the wire in air. The angle of refraction is then θ2 90 0 30 0 60 0= − =. . .° ° °. From Snell’s law, the angle of incidence was θ θ 1 1 2 1 1 00 = ⎛ ⎝⎜ ⎞ ⎠⎟ = ( ) − − sin sin sin . n n air benzene sin . . . 60 0 1 50 35 3 ° ° ⎛ ⎝⎜ ⎞ ⎠⎟ = and the wire is bent by angle θ θ= − = − =60 0 60 0 35 3 24 71. . . .° ° ° ° . 22.54 From the sketch at the right, observe that the angle of incidence at A is the same as the prism angle at point O. Given that θ = 60 0. °, application of Snell’s law at point A gives 1 50 1 00 60 0. sin . sin .β = ( ) ° or β = °35 3. From triangle AOB, we calculate the angle of incidence and refl ection, γ , at point B: θ β γ+ −( ) + −( ) =90 0 90 0 180. .° ° ° or γ θ β= − = − =60 0 35 3 24 7. . .° ° ° Now, we fi nd the angle of incidence at point C using triangle BCQ: 90 0 90 0 90 0 180. . .° ° ° °−( ) + −( ) + −( ) =γ δ θ or δ θ γ= − +( ) = − =90 0 90 0 84 7 5 26. . . .° ° ° ° Finally, application of Snell’s law at point C gives 1 00 1 50 5 26. sin . sin .( ) = ( ) ( )φ ° or φ = ( ) =−sin . sin . .1 1 50 5 26 7 91° ° 22.55 The path of a light ray during a refl ection and/or refraction process is always reversible. Thus, if the emerging ray is parallel to the incident ray, the path which the light follows through this cylinder must be symmetric about the center line as shown at the right. Thus, θ1 1 12 1 00 3= ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ =− −sin sin .d R m 2 .00 m 0 0. ° Triangle ABC is isosceles, so γ α= and β α γ α= − − = −180 180 2° ° . Also, β θ= −180 1° , which gives α θ= =1 2 15 0. °. Then, from applying Snell’s law at point A, n n cylinder air= = ( )sin sin . sin . sin θ α 1 1 00 30 0 15 ° . . 0 1 93 ° = 56157_22_ch22_p281-304.indd 301 3/19/08 7:55:01 PM 302 Chapter 22 22.56 The angle of refraction as the light enters the left end of the slab is θ θ 2 1 1 1 1 00 = ⎛ ⎝⎜ ⎞ ⎠⎟ = ( )− −sin sin sin . sinn n air slab 50 0 31 2 . . ° ° 1.48 ⎛ ⎝⎜ ⎞ ⎠⎟ = Observe from the fi gure that the fi rst refl ection occurs at x = d, the second refl ection is at x = 3d, the third is at x = 5d, and so forth. In general, the Nth refl ection occurs at x N d= −( )2 1 , where d = ( ) = ° = 0 310 0 310 2 31 2 0 256 2 . tan . tan . . cm 2 cm θ cm Therefore, the number of refl ections made before reaching the other end of the slab at x L= = 42 cm is found from L N d= −( )2 1 to be N L d = +⎛ ⎝⎜ ⎞ ⎠⎟ = +⎛ ⎝⎜ ⎞ ⎠⎟ =1 2 1 1 2 42 1 82 5 cm 0.256 cm . or 82 complete reflections 22.57 (a) If θ1 45 0= . °, application of Snell’s law at the point where the beam enters the plastic block gives 1 00 45 0. sin . sin( ) =° n φ [1] Application of Snell’s law at the point where the beam emerges from the plastic, with θ2 76 0= . °, gives n sin . sin90 1 00 76° °−( ) = ( )φ or 1 00 76. sin cos( ) =° n φ [2] Dividing Equation [1] by Equation [2], we obtain tan sin . sin .φ = ° ° =45 0 76 0 729 and φ = 36 1. ° Thus, from Equation [1], n = ° = ° ° =sin . sin sin . sin . . 45 0 45 0 36 1 1 20 φ (b) Observe from the fi gure above that sinφ = L d. Thus, the distance the light travels inside the plastic is d L= sinφ , and if L = =50 0 0 500. . cm m, the time required is ∆t d L c n nL c = = = = ( ) ×v sin sin . .φ φ 1 20 0 500 m 3.00 108 m s s ns( ) = × =− sin . . . 36 1 3 40 10 3 409 ° 56157_22_ch22_p281-304.indd 302 3/19/08 7:55:02 PM