Quick QuizzesAnswers to Even Numbered Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsChapter 22 Reflection and Refraction of Light Quick Quizzes 1. (a). In part (a), you can see clear reflections of the headlights and the lights on the top of the truck. The reflection is specular. In part (b), although bright areas appear on the roadway in front of the headlights, the reflection is not as clear, and no separate reflection of the lights from the top of the truck is visible. The reflection in part (b) is mostly diffuse. 2. Beams 2 and 4 are reflected; beams 3 and 5 are refracted. 3. (b). When light goes from one material into one having a higher index of refraction, it refracts toward the normal line of the boundary between the two materials. If, as the light travels through the new material, the index of refraction continues to increase, the light ray will refract more and more toward the normal line. 4. (c). Both the wave speed and the wavelength decrease as the index of refraction increases. The frequency is unchanged. 233234 CHAPTER 22 Answers to Even Numbered Conceptual Questions 2. Ceilings are generally painted a light color so they will reflect more light, making the room brighter. Textured materials are often used on the ceiling to diffuse the reflected light and reduce glare (specular reflections). 4. At the altitude of the plane the surface of Earth does not block off the lower half of the rainbow. Thus, the full circle can be seen. You can see such a rainbow by climbing on a stepladder above a garden sprinkler in the middle of a sunny day. 6. The spectrum of the light sent back to you from a drop at the top of the rainbow arrives such that the red light (deviated by an angle of 42°) strikes the eye while the violet light (deviated by 40°) passes over your head. Thus, the top of the rainbow looks red. At the bottom of the bow, violet light arrives at your eye and red light is deviated toward the ground. Thus, the bottom part of the bow appears violet. 8. A mirage occurs when light changes direction as it moves between batches of air having different indices of refraction. The different indices of refraction occur because the air has different densities at different temperatures. Two images are seen; One from a direct path from the object to you, and the second arriving by rays originally heading toward Earth but refracted to your eye. On a hot day, the Sun makes the surface of blacktop hot, so the air is hot directly above it, becoming cooler as one moves higher into the sky. The “water” we see far in front of us is an image of the blue sky. Adding to the effect is the fact that the image shimmers as the air changes in temperature, giving the appearance of moving water. 10. The upright image of the hill is formed by light that has followed a direct path from the hill to the eye of the observer. The second image is a result of refraction in the atmosphere. Some light is reflected from the hill toward the water. As this light passes through warmer layers of air directly above the water, it is refracted back up toward the eye of the observer, resulting in the observation of an inverted image of the hill directly below the upright image. 12. The color traveling slowest is bent the most. Thus, X travels more slowly in the glass prism. 14. Total internal reflection 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 reflection. 16. 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.Reflection and Refraction of Light 235 18. The cross section can be visualized by considering just the two rays of light on the edges of the beam. If the beam of light enters a new medium with a higher index of refraction, the rays bend toward the normal, and the cross section of the refracted beam will be larger than that of the incident beam as suggested by Fig. CQ22.18a. If the new index of refraction is lower, the rays bend away from the normal, and the cross section of the beam is reduced, as shown in Fig. CQ22.18b. Figure CQ22.18(b)21n2 < n1(a)21n2 > n1236 CHAPTER 22 Answers to Even Numbered Problems 2. 82.97 10 m s× 4. (a) 536 rev s (b) 31.07 10 rev s× 6. (a) 1.94 m (b) 50.0° above horizontal (parallel to incident ray) 8. 112.09 10 s−×10. (a) The longer the wavelength, the less it is deviated (or refracted) from the original path. (b) Using data from Figure 22.14, the angles of refraction are: (400 nm) 216.0θ= , (500 nm) °216.1θ=°, (650 nm) 216.3θ=° 12. (a) 327 nm (b) 287 nm 14. 67.4° 16. 53.4° 18. First surface: r30.0 , 19.5iθθ=19.5 , 30.0ir=° Second surface: °θθ=°=° 20. 101.06 10 s−×22. 107 m 24. 6.30 cm 26. 23.1° 28. 2.5 m 30. 0.40° 32. 4.6° 34. (a) 24.4° (b) 37.0° 36. 48.5° 38. 67.2° 40. 4.54 mReflection and Refraction of Light 237 44. (a) 1230.0 , 18.8θθ′=°=° (b) 1230.0 , 50.8θθ′=°= ° (c) See solution. (d) See solution. 46. (a) Any angle of incidence (b) 30.0° (c) not possible since 90≤° polystyrene carbon disulfidenn< 48. (a) 0.172 mm s (b) 0.345 mm s (c) and (d) Northward at 50.0° below horizontal. 50. 77.5° 52. (a) ()1Rndn≥− (b) yes; yes; yes (c) 350 mµ 54. 7.91° 56. 82 58. 62.2% of a circle 60. The graph is a straight line passing through the origin. From the slope of the graph, . 1.33watern =62. (a) ()12214ntd=+ (b) 2.10 cm (c) violet238 CHAPTER 22 Problem Solutions 22.1 The total distance the light travels is Therefore, () 86622 3.84 10 6.38 10 1.76 10 m 7.52 10 mcenter toEarth MooncenterdD R R∆= − −=×−×−× =×8887.52 10 m3.00 10 m s2.51 sdvt∆×== = ×∆ 22.2 If the wheel has 360 teeth, it turns through an angle of 1 in the time it takes the light to make its round trip. From the definition of angular velocity, we see that the time is 720 rev()51720 rev5.05 10 s27.5 rev stθω−== = × Hence, the speed of light is ()8527500 m22.97 10 m s5.05 10 sdt−== = ××c 22.3 The experiment is most convincing if the wheel turns fast enough to pass outgoing light through one notch and returning light through the next. Then,