Chapter 26: Geometrical OpticsAnswers to Even-Numbered Conceptual Questions2. Three images are formed of object B. One extends from (–3 m, 1 m) to (–3 m, 2 m) to (– 4 m, 2 m). Another image forms an “L” from (3 m, –1 m) to (3 m, –2 m) to (4 m, –2 m). Finally, the third image extends from (–3 m, –1 m) to (–3 m, –2 m) to (– 4 m, –2 m).4. The concave side of the dish collects the parallel rays coming from a geosynchronous satellite and focuses them at the focal point of the dish. The convex side of the dish would send the parallel rays outward on divergent paths. The situation is analogous to that of light in an optical telescope.6. No. Light bends toward the normal when it enters a medium in which its speed of propagation is less than itwas in the first medium – as when light passes from air to water. On the other hand, light bends away from the normal if it enters a medium in which its speed is increased – as when light passes from water to air.8. You are actually seeing light from the sky, which has been bent upward by refraction in the low-density air near the hot ground. See figure 26-23 for a case where one would see a tree reflected in the "pool of water."10. The Sun is already well below the horizon when you see it setting. The reason is that as the Sun’s light entersthe atmosphere from the vacuum of space, it is bent toward the normal; that is, toward the surface of Earth. Therefore, the light from the Sun can still reach us even when a straight line from our eyes to the Sun would go below the horizon.12. The oil used in the bottle to the left has an index of refraction that is equal to the index of refraction of the glass in the eye dropper. Therefore, light is undeflected when it passes from the oil to the glass or from the glass to the oil. Because light propagates the same as if the eye dropper were not present, the dropper is invisible.14. Note that the word “SECRET,” which is in red, is inverted. On the other hand, the word “CODE,” which is inblue, appears not to be inverted. One might think that the different index of refraction for blue light versus red light is responsible for this behavior, but recall that we said the word “CODE” appears not to be inverted. In fact, it is inverted, just like the word “SECRET,” but because all of its letters have a vertical symmetry, it looks the same when inverted.Solutions to Problems and Conceptual Exercises1. Picture the Problem: The image shows a laser beam with an angle of 28between the incident and reflected rays. The image also shows the path of the laser beam when the mirror is rotated by 5.Strategy: Use the law of reflection to set the angle between the incident and reflected rays equal to twice the incident angle, and then solve for the incident angle. After the mirror is rotated, set the new angle between the incident and reflected waves equal to twice the new incident angle, which is 5 greater than the initial incident angle.Solution: 1. Calculate the initial incident angle: θi+θr=2θi=28°θi=14°Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.26 – 1Chapter 26: Geometrical Optics James S. Walker, Physics, 4th Edition2. Calculate the final angle between incident and reflected beams: ′θi+′θr=2′θi=2 θi+5°( )=2 14°+5°( )= 38°Insight: Because the incident and reflected angles are equal, increasing the incident angle by an angle will increase the angle between incident and reflected rays by 2.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.26 – 2Chapter 26: Geometrical Optics James S. Walker, Physics, 4th Edition2. Picture the Problem: The image shows two mirrors oriented at 120 with respect to each other. A light ray strikes the first mirror with an incident angle of 55. The reflected light then reflects off the second mirror..Strategy: Set the reflected angle from the first mirror equal to the incident angle. The two mirrors and the ray form a triangle. The sum of the interior angles of a triangle is 180. Two of the interior angles are the complementary angles 90°−θ( )for the incident and reflected rays. Use this relation to calculate the angle of incidence for mirror 2, and set angle of reflection for mirror 2 equal to the angle of incidence for mirror 2.Solution: 1. Write the angle of reflection for mirror 1: θ1r=θ1i=55°2. Use the interior angles of the triangle to calculate the incident angle of the second mirror: 90°−θ1r( )+120°+ 90°−θ2i( )=180°θ2i=120°−θ1r=120°−55°=65°3. Set the reflected angle equal to the incident angle: θ2r=θ2i= 65°Insight: For any incident angle, the sum of the incident angle and the reflected angle will equal the angle between the two mirrors. 3. Picture the Problem: The image shows a light beam reflecting off ofa mirror. The mirror is then rotated by an angle , causing the path ofthe reflected ray to change by the angle r.Strategy: Use the diagram together with the Law of Reflection to find the angle of each reflected ray with respect to the incident ray. Subtract the two angles to obtain the angle between the reflected beams. Solution: 1. Write an expression for θi+θr: θi+θr=θi+θi=2θi2. Repeat for ′θi+′θr, recalling that ′θr=′θi: ′θi+′θr= θi+θ( )+′θr= θi+θ( )+ θi+θ( )=2θi+2θ3. Calculate the angle between reflected rays: Δθr= 2θi+2θ( )−2θi= 2θInsight: The change in reflected angle is dependent only on the rotation angle, not in the initial angle of incidence.4. Picture the Problem: The image shows the reflection of sunlight off of a small mirror located 1.40 m above the floor at two different times.Strategy: Use the inverse tangent function to calculate the angle of reflection from the height of the mirror and the distance of the image on thefloor for each of the times. Set the angles of reflection equal to the angles of incidence and subtract the result to calculate the change in the Sun’s angle of elevation.Solution: 1. (a) Because the distance from the wall increased, the
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