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Chapter 22 Reflection and Refraction of Light

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Chapter 22 Reflection and Refraction of Light Chapter 22 Reflection and Refraction of Light Problem Solutions 22.1 The total distance the light travels is d2 Dcenter to R Earth R Moon center 2 3.84 108 6.38 10 6 1.76 10 6 m 7.52 10 8 m d 7.52 108 m Therefore, v 3.00 108 m s t 2.51 s 22.2 (a) The energy of a photon is sinc nair n prism 1.00 n prism , where Planck’ s constant is 1.00 8 sinc sin 45 and the speed of light in vacuum is c 3.00 10 m s . If nprism 1.00 1010 m , 6.63 1034 J s 3.00 10 8 m s E 1.99 1015 J 1.00 10-10 m 1 eV (b) E 1.99 1015 J 1.24 10 4 eV 1.602 10-19 J (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 . 1 eV 22.3 (a) E hf 6.63 1034 J s 5.00 10 17 Hz 2.07 10 3 eV 1.60 1019 J 355 356 CHAPTER 22 34 8 hc 6.63 10 J s 3.00 10 m s 1 nm (b) E hf 6.63 1019 J 3.00 1029 nm 10 m 1 eV E 6.63 1019 J 4.14 eV 1.60 1019 J c 3.00 108 m s 22.4 (a) 5.50 107 m 0 f 5.45 1014 Hz (b) From Table 22.1 the index of refraction for benzene is n 1.501. Thus, the wavelength in benzene is 5.50 107 m 0 3.67 107 m n n 1.501 1 eV (c) E hf 6.63 1034 J s 5.45 10 14 Hz 2.26 eV 1.60 1019 J (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. 3.00 108 m s (a) For water, n 1.333, and v 2.25 108 m s 1.333 3.00 108 m s (b) For crown glass, n 1.52, and v 1.97 108 m s 1.52 3.00 108 m s (c) For diamond, n 2.419, and v 1.24 108 m s 2.419 22.6 (a) From fc, the wavelength is given by cf. The energy of a photon is E hf , so the frequency may be expressed as f E h , and the wavelength becomes c c hc f E h E (b) Higher energy photons have shorter wavelengths. Reflection and Refraction of Light 357 22.7 From Snell’ s law, nn2sin 2 1 sin 1 . Thus, when 1 45 and the first medium is air ( ), we have . n1 1.00 sin2 1.00 sin45 n2 1.00 sin45 (a) For quartz, n 1.458 , and sin1 29 2 2 1.458 1.00 sin45 (b) For carbon disulfide, n 1.628, and sin1 26 2 2 1.628 1.00 sin45 (c) For water, n 1.333, and sin1 32 2 2 1.333 22.8 (a) From geometry, 1.25 md sin40.0 , so d 1.94 m (b) 50.0 above horizontal , or parallel to the incident ray 22.9 nn1sin 1 2 sin 2 sin1 1.333sin 45.0 sin1 (1.333)(0.707) 0.943 1 70.5 19.5 above the horizontal c 3.00 108 m s 22.10 (a) n 1.38 v 2.17 108 m s 358 CHAPTER 22 (b) From Snell’ s law, nn2sin 2 1 sin 1 , 1n11sin 11.00 sin23.1 1 2 sin sin sin 0.284 16.5° n2 1.38 n11sin 1.00 sin30.0 22.11 (a) From Snell’ s law, n2 1.52 sin2 sin19.24 0 632.8 nm (b) 2 416 nm n2 1.52 8 c 3.00 10 m s 14 (c) f 9 4.74 10 Hz in air and in syrup 0 632.8 10 m 8 c 3.00 10 m s 8 (d) v2 1.97 10 m s n2 1.52 22.12 (a) When light refracts from air n1 1.00 into the Crown glass, Snell’ s law gives the angle of refraction as 1 2sin sin25.0 n Crown glass For first quadrant angles, the sine of the angle increases as the angle increases. Thus, from the above equation, note that 2 will increase when the index of refraction of the Crown glass decreases. From Figure 22.14, this means that the longer wavelengths have the largest angles of refraction, and deviate the least from the original path. Figure 22.14 Reflection and Refraction of Light 359 (b) From Figure 22.14, observe that the index of refraction of Crown glass for the given wavelengths is: 400 nm: nCrown glass 1.53; 500 nm: nCrown glass 1.52; and 650 nm: nCrown glass 1.51 1 Thus, Snell’ s law gives: 400 nm: 2 sin sin25.0 1.53 16.0 1 500 nm: 2 sin sin25.0 1.52 16.1 1 650 nm: 2 sin sin25.0 1.51 16.3 22.13 From Snell’ s law, 2 26.5 and from the law of reflection, sin1 0.499 Hence, the angle between the reflected and refracted rays is 1 30.0 22.14 Using a protractor to measure the angle of incidence and the angle of refraction in Active Figure 22.6b gives 1255 and 33 . Then, from Snell’ s law, the index of refraction for the Lucite is 1 60.0 8 c 3.00 10 m s 8 (a) v2 2.0 10 m s n2 1.5 (b) 90.02 30.0 7 0 6.328 10 m 7 (c) 2 4.2 10 m n2 1.5 22.15 The index of refraction of zircon is 3 30.0 c 3.00 108 m s (a) v 1.56 108 m s n 1.923 (b) The wavelength in the zircon is 11nglasssin 3 1.66 sin30.0 4 sin sin 38.5 n2 1.333 360 CHAPTER 22 (c) The frequency is c 1 22.16 The angle of incidence is 2.00 m tan1 26.6 1 4.00 m Therefore, Snell’ s law gives nn2 glasssin 1 1.66 sin60.0 1.44 and the angle the refracted ray makes with the surface is 90.02 90.0 36.6 53.4 22.17 The incident light reaches the left-hand mirror at distance 122 above its bottom edge. The reflected light first reaches the right-hand mirror at height d 2 0.087 5 m 0.175 m It bounces between the mirrors with distance d between points of contact a given mirror. Since the full 1.00 length of the right-hand mirror is available for reflections, the number of reflections from this mirror will be 1.00 m N 5.71 5 full reflections right 0.175 m Since the first reflection from the left-hand mirror occurs at a height of d 2 0.087 5 m , the total number of that can occur from this mirror is 1.00 m 0.087 5 m N 1 6.21 6 full reflections left 0.175 m 22.18 (a) From Snell’ s law, the angle of refraction at the first surface is 90 1 Reflection and Refraction of Light 361 (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 2 19.5 . The angle of refraction at this surface is then 11nglasssin glass 1.50 sin19.5 3 sin sin 30.0 nair 1.00 Thus, the light emerges traveling parallel to the incident beam. (c) Consider the sketch at the right and let h represent the distance from point a to c (that is, the hypotenuse of triangle abc). Then, 2.00 cm 2.00 cm h 2.12 cm cos2 cos19.5 Also, 1230.0 19.5 10.5 , so dhsin 2.12 cm sin10.5 0.386 cm (d) The speed of the light in the glass is 8 c 3.00 10 m s 8 v 2.00 10 m s nglass 1.50 (e) The time required for the light to travel through the glass is h 2.12 cm 1 m t 1.06 1010 s v 2.00 1082 m s 10 cm (f) Changing the angle of incidence will change the angle of refraction, and therefore the distance h the light travels in the glass. Thus, the travel time will also change . 362 CHAPTER 22 22.19 From Snell’ s law, the angle of incidence at the air-oil interface is 1 noilsin oil sin nair 1.48 sin 20.0 sin1 30.4 1.00 and the angle of refraction as the light enters the water is n sin 1.48 sin20.0 sin11oil oil sin 22.3 nwater 1.333 22.20 Since the light ray strikes the first surface at normal incidence, it passes into the prism without deviation.
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