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1015_Physics ProblemsTechnical Physics

# 1015_Physics ProblemsTechnical Physics - 356 P36.15 Image...

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356 Image Formation *P36.15 M q p = − q Mp p q f R R R = − = − = − + = = + = = = − 0 013 30 0 39 1 1 1 2 1 30 1 0 39 2 2 2 53 0 790 1 . . . . . cm cm cm cm m cm a f FIG. P36.15 The cornea is convex, with radius of curvature 0 790 . cm . *P36.16 With M h h q p = = + = + = − 4 00 0 400 . . cm 10.0 cm q p = − 0 400 . the image must be virtual. (a) It is a convex mirror that produces a diminished upright virtual image. (b) We must have p q p q + = = 42 0 . cm p q = + 42 0 . cm p p = 42 0 0 400 . . cm p = = 42 0 30 0 . . cm 1.40 cm The mirror is at the 30.0 cm mark . (c) 1 1 1 1 30 1 0 4 30 1 0 050 0 p q f f + = = + = = − cm cm cm . . a f f = − 20 0 . cm The ray diagram looks like Figure 36.15(c) in the text.
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