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SM_PDF_chapter26

# SM_PDF_chapter26 - Image Formation by Mirrors and Lenses...

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713 Image Formation by Mirrors and Lenses CHAPTER OUTLINE 26.1 Images Formed by Flat Mirrors 26.2 Images Formed by Spherical Mirrors 26.3 Images Formed by Refraction 26.4 Thin Lenses 26.5 Context Connection Medical Fiberscopes ANSWERS TO QUESTIONS Q26.1 With a concave spherical mirror, for objects beyond the focal length the image will be real and inverted. For objects inside the focal length, the image will be virtual, upright, and magnified. Try a shaving or makeup mirror as an example. Q26.2 With a convex spherical mirror, all images of real objects are upright, virtual, and smaller than the object. As seen in Question 26.1, you only get a change of orientation when you pass the focal point—but the focal point of a convex mirror is on the non-reflecting side! Q26.3 The mirror equation and the magnification equation apply to plane mirrors. A curved mirror is made flat by increasing its radius of curvature without bound, so that its focal length goes to infinity. From 1 1 1 0 p q f + = = we have 1 1 p q = − ; therefore, p q = − . The virtual image is as far behind the mirror as the object is in front. The magnification is M q p p p = − = = 1. The image is right side up and actual size. Q26.4 Stones at the bottom of a clear stream always appear closer to the surface because light is refracted away from the normal at the surface. Example 26.5 in the textbook shows that its apparent depth is three quarters of its actual depth. Q26.5 For definiteness, we consider real objects ( p > 0). (a) For M q p = − to be negative, q must be positive. This will happen in 1 1 1 q f p = if p f > , if the object is farther than the focal point. (b) For M q p = − to be positive, q must be negative. From 1 1 1 q f p = we need p f < . continued on next page

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714 Image Formation by Mirrors and Lenses (c) For a real image, q must be positive. As in part (a), it is sufficient for p to be larger than f . (d) For q < 0 we need p f < . (e) For M > 1, we consider separately M < − 1 and M > 1. If M q p = − < − 1, we need q p > 1 or q p > or 1 1 q p < . From 1 1 1 p q f + = , 1 1 1 p p f + > or 2 1 p f > or p f 2 < or p f < 2 . Now if > q p 1 or > q p or q p < − we may require q < 0, since then 1 1 1 p f q = with 1 0 f > gives 1 1 p q > − as required or > p q . For q < 0 in 1 1 1 q f p = we need p f < . Thus the overall condition for an enlarged image is simply p f < 2 . (f) For M < 1, we have the reverse of part (e), requiring p f > 2 . Q26.6 Using the same analysis as in Question 26.5 except f < 0. (a) Never. (b) Always. (c) Never, for light rays passing through the lens will always diverge. (d) Always. (e) Never. (f) Always.
Chapter 26 715 Q26.7 We assume the lens has a refractive index higher than its surroundings. For the biconvex lens in Figure 26.20(a), R 1 0 > and R 2 0 < . Then all terms in n R R F H G I K J 1 1 1 1 2 a f are positive and f > 0. For the other two lenses in part (a) of the figure, R 1 and R 2 are both positive but R 1 is less than R 2 . Then 1 1 1 2 R R > and the focal length is again positive.

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