p35_031 - | m 1 | = i 1 p 1 = D − x 1 x 1 Now x 1 = 1 2 D...

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31. (a) If the object distance is x , then the image distance is D x and the thin lens equation becomes 1 x + 1 D x = 1 f . We multiply each term in the equation by fx ( D x )andobta in x 2 Dx + Df =0. Solv ingfor x , we Fnd that the two object distances for which images are formed on the screen are x 1 = D p D ( D 4 f ) 2 and x 2 = D + p D ( D 4 f ) 2 . The distance between the two object positions is d = x 2 x 1 = p D ( D 4 f ) . (b) The ratio of the image sizes is the same as the ratio of the lateral magniFcations. If the object is at p = x
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Unformatted text preview: | m 1 | = i 1 p 1 = D − x 1 x 1 . Now x 1 = 1 2 ( D − d ), where d = p D ( D − f ), so | m 1 | = D − ( D − d ) / 2 ( D − d ) / 2 = D + d D − d . Similarly, when the object is at x 2 , the magnitude of the lateral magniFcation is | m 2 | = I 2 p 2 = D − x 2 x 2 = D − ( D + d ) / 2 ( D + d ) / 2 = D − d D + d . The ratio of the magniFcations is m 2 m 1 = ( D − d ) / ( D + d ) ( D + d ) / ( D − d ) = µ D − d D + d ¶ 2 ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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