1036_Physics ProblemsTechnical Physics

# 1036_Physics ProblemsTechnical Physics - Chapter 36 P36.74...

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Chapter 36 377 P36.74 (a) Start with the second lens: This lens must form a virtual image located 19.0 cm to the left of it (i.e., q 2 19 0 =− . cm ). The required object distance for this lens is then p qf 2 22 19 0 20 0 19 0 20 0 380 39 0 = = −− = .. . cm cm cm cm cm a fa f . The image formed by the first lens serves as the object for the second lens. Therefore, the image distance for the first lens is qp 12 50 0 50 0 380 1570 =−= cm cm cm 39.0 cm 39.0 . The distance the original object must be located to the left of the first lens is then given by 111 1 10 0 39 0 157 39 0 118 pf q = = . cm cm cm 1 570 cm or p 1 13 3 == cm 118 cm (b) MMM q p q p F H G I K J F H G I K J = F H G I K J F H G I K J L N M M O Q P P L N M O Q P 1 1 2 2 118 19 0 39 0 380 590 cm 39.0 cm cm cm . af a f (c) Since M < 0 , the final image is inverted . P36.75 (a) P fpq ==+= + = 1 00224 1 44 6 . . m diopters bg (b) P + = 1 0330 1 303 . . m diopters P36.76 The object is located at the focal point of the upper mirror. Thus, the
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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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