p35_025 - x , we set dx/dp = 0 and solve for p . Since dx...

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25. For an object in front of a thin lens, the object distance p and the image distance i are related by (1 /p )+(1 /i )=(1 /f ), where f is the focal length of the lens. For the situation described by the problem, all quantities are positive, so the distance x between the object and image is x = p + i . We substitute i = x p into the thin lens equation and solve for x : x = p 2 p f . To ±nd the minimum value of
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Unformatted text preview: x , we set dx/dp = 0 and solve for p . Since dx dp = p ( p − 2 f ) ( p − f ) 2 , the result is p = 2 f . The minimum distance is x min = p 2 p − f = (2 f ) 2 2 f − f = 4 f . This is a minimum, rather than a maximum, since the image distance i becomes large without bound as the object approaches the focal point....
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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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