1031_Physics ProblemsTechnical Physics

1031_Physics ProblemsTechnical Physics - 372 P36.62 Image...

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372 Image Formation P36.62 111 1 10 0 1 12 5 qf p =−= .. cm cm so q 1 50 0 = . cm (to left of mirror). This serves as an object for the lens (a virtual object), so 1 16 7 1 25 0 222 p cm cm af and q 2 50 3 =− . cm , meaning 50.3 cm to the right of the lens. Thus, the final image is located 25 3 . cm to right of mirror . M q p M q p MMM 1 1 1 2 2 2 12 50 0 400 50 3 25 0 201 805 == . . . . . . cm 12.5 cm cm cm a f a f Thus, the final image is virtual, upright , 8.05 times the size of object, and 25.3 cm to right of the mirror. P36.63 We first find the focal length of the mirror. 1 10 0 1 800 9 40 0 fpq =+= + = . cm cm cm and f = 444 c m . Hence, if p = 20 0 cm, 1 1 20 0 15 56 88 8 qfp = . . cm cm cm . Thus, q = 571 c m , real. *P36.64 A telescope with an eyepiece decreases the diameter of a beam of parallel rays. When light is sent through the same device in the opposite direction, the beam expands. Send
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