Physics Solution Manual for 1100 and 2101

In other words f 120 cm then according to 1 equation

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Unformatted text preview: ________________________________ CHAPTER 25 THE REFLECTION OF LIGHT: MIRRORS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (e) This is the definition of a wave front (see Section 25.1). 2. (b) Rays are radial lines pointing outward from the source and perpendicular to the wave fronts. They point in the direction of the velocity of the wave. 3. (c) When diffuse reflection occurs, the surface reflects different light rays in different directions. 4. (c) The ray of light strikes the mirror four units down from the top of the mirror with a 45° angle of incidence. The ray reflects from the mirror at an angle of 45° and passes through point C. 5. (a) The image is as far behind the mirror as the object is in front of the mirror. In addition, the image and the object lie on the horizontal line that is perpendicular to the mirror. 6. (d) The image of your friend is 2 m behind the mirror. The distance between you and the mirror is 5 m. Thus, the distance between you and your friend’s image is 7 m. 7. (b) Letters and words held up to a mirror are reversed left-to-right and right-to-left. 8. (d) As discussed in Section 25.4, rays that are parallel and near the principal axis of a concave mirror converge at the focal point after reflecting from the mirror. 9. (a) Parallel rays that are near the principal axis converge at the focal point after reflecting from a concave mirror. The radius of curvature is twice the focal length (see Equation 25.1), so R = 2f = 36 cm. 10. (d) This is how real and virtual images are defined. See Sections 25.3 and 25.5. 11. (a) Any ray that leaves the object and reflects from the mirror can be used in the method of ray tracing to locate the image. 12. (c) According to the discussion in Section 25.5, a concave mirror can produce an enlarged image, provided the object distance is less than the radius of curvature. A convex mirror cannot produce an enlarged image, regardless of where the object is located. 13. (b) A convex mirror always produces a virtual, upright image (see Section 25.5). Chapter 25 Answers to Focus on Concepts Questions 14. (e) A negative image distance means that the image is behind the mirror and, hence, is a virtual image. See the Reasoning Strategy at the end of Section 25.6. 15. (c) A convex lens always produces an upright image that is smaller than the object. 16. f = 4.0 cm 17. (b) The image distance is di = −mdo = −2(25 cm) = −50 cm (Equation 25.4). 18. f = 90.0 cm 1283 1284 THE REFLECTION OF LIGHT: MIRRORS CHAPTER 25 THE REFLECTION OF LIGHT: MIRRORS PROBLEMS M 1. 2. SSM WWW REASONING AND SOLUTION The drawing at the right shows a ray diagram in which the reflected rays have been projected behind the mirror. We can see by inspection of this drawing that, after the rays reflect from the plane mirror, the angle α between them is still 10° . object 10 α o reflected ray REASONING AND SOLUTION Referring to Figure 25.9b and Conceptual Example 2, we find the following locations for the three images: Image 1: Image 2: Image 3: 3. reflected ray x = − 2.0 m, y = + 1.0 m x = + 2.0 m, y = − 1.0 m x = + 2.0 m, y = + 1.0 m REASONING The drawing shows Bird the image of the bird in the plane mirror, as seen by the camera. Note that the image is as far behind the mirror as the bird is in front of it. We can find the distance d between the camera and the image by noting that this distance is the hypotenuse 4.3 m of a right triangle. The base of the triangle has a length of 3.7 m + 2.1 m and the height of the triangle is 4.3 m. 2.1 m Image Hedge 3.7 m 4.3 m 2.1 m Chapter 25 Problems 1285 SOLUTION The distance d from the camera to the image of the bird can be obtained by using the Pythagorean theorem: 2 2 d = ( 3.7 m + 2.1 m ) + ( 4.3 m ) = 7.2 m 4. REASONING According to the discussion about relative velocity in Section 3.4, the velocity vIY of your image relative to you is the vector sum of the velocity vIM of your image relative to the mirror and the velocity vMY of the mirror relative to you: vIY = vIM + vMY. As you walk perpendicularly toward the stationary mirror, you perceive the mirror moving toward you in the opposite direction, so that vMY = –vYM. The velocities vYM and vIM have the same magnitude. This is because the image in a plane mirror is always just as far behind the mirror as the object is in front of it. For instance, if you move 1 meter perpendicularly toward the mirror in 1 second, the magnitude of your velocity relative to the mirror is 1 m/s. But your image also moves 1 meter toward the mirror in the same time interval, so that the magnitude of its velocity relative to the mirror is also 1 m/s. The two velocities, however, have opposite directions. SOLUTION According to the discussion in the REASONING, vIY = vIM + vMY = vIM – vYM (1) Remembering that the magnitudes of both velocities vYM and vIM are the same and that the direction in which you walk is positive, we have vYM = +0.90 m/s and vIM = –0.90 m/s The velocity vIM is negative, because its direction is opposite to the direction in which you walk. Substituting...
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