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Phys 181b Problem Set 10 Solution

# Phys 181b Problem Set 10 Solution - 4 When S is barely able...

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4. When S is barely able to see B the light rays from B must reflect to S off the edge of the mirror. The angle of reflection in this case is 45°, since a line drawn from S to the mirror’s edge makes a 45° angle relative to the wall. By the law of reflection, we find x d x d / tan . . 2 45 2 30 15 = ° ¡ = = = m 2 m.

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80. (a) The image from lens 1 (which has f 1 = + 15 cm) is at i 1 = –30 cm (by Eq. 34-9). This serves as an “object” for lens 2 (which has f 2 = + 8 cm) with p 2 = d i 1 = 40 cm. Then Eq. 34-9 (applied to lens 2) yields i 2 = + 10 cm. (b) Eq. 34-11 yields M = m 1 m 2 1 1 2 2 1 2 1 2 ( / )( / ) / i p i p i i p p = − = = –0.75. (c) The fact that the (final) image distance is a positive value means the image is real (R). (d) The fact that the magnification is a negative value means the image is inverted (I). (e) The image it is on the side opposite from the object (relative to lens 2).
91. (a) When the eye is relaxed, its lens focuses far-away objects on the retina, a distance i behind the lens. We set p = in the thin lens equation to obtain 1/ i = 1/ f , where f is the focal length of the relaxed effective lens. Thus,

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