1021_Physics ProblemsTechnical Physics

1021_Physics ProblemsTechnical Physics - 362 P36.33 Image...

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362 Image Formation P36.33 We are looking at an enlarged, upright, virtual image: M h h q p = == 2s o p q =− =− =+ 2 284 2 142 . . cm cm af 111 pq f += gives 1 1 1 .. cm cm + = f f = . c m . FIG. P36.33 P36.34 (a) : 11 30 0 1 12 5 p + = cm cm p = 882 c m M q p = 30 0 340 . . . a f , upright (b) See the figure to the right. FIG. P36.34(b) P36.35 : pq −− constant We may differentiate through with respect to p : = 0 22 dq dp dq dp q p M =− 2 2 2 . P36.36 The image is inverted: M h h q p = = = 180 75 0 . . m 0.024 0 m qp = 75 0 . . (b) pp = + 300 750 m p = 39 5 . mm (a) q = 296 m 1 00395 1 fpq =+= + m m f = 39 0 P36.37 (a) 1 20 0 32 0 cm cm
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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