Phys 181b Problem Set 10 Solution

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

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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 11 2 2 1 21 2 (/ ) (/) / ip ip i ip 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).
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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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Phys 181b Problem Set 10 Solution - 4. When S is barely...

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