p35_045 - f = (1 5 − 1 µ 1 − 40 cm − 1 40 cm ¶¸...

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45. (a) We use Eq. 35-10, with the conventions for signs discussed in § 35-5 and § 35-6. (b) For the bi-convex (or double convex) case, we have f = · ( n 1) µ 1 r 1 1 r 2 ¶¸ 1 = · (1 . 5 1) µ 1 40 cm 1 40 cm ¶¸ 1 =40cm . Since f> 0 the lens forms a real image of the Sun. (c) For the planar convex lens, we ±nd f = · (1 . 5 1) µ 1 1 40 cm ¶¸ 1 =80cm , and the image formed is real (since f> 0). (d) Now f = · (1 . 5 1) µ 1 40 cm 1 60 cm ¶¸ 1 = 240 cm , and the image formed is real (since f> 0). (e) For the bi-concave lens, the focal length is
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Unformatted text preview: f = · (1 . 5 − 1) µ 1 − 40 cm − 1 40 cm ¶¸ − 1 = − 40 cm , and the image formed is virtual (since f < 0). (f) In this case, f = · (1 . 5 − 1) µ 1 ∞ − 1 40 cm ¶¸ − 1 = − 80 cm , and the image formed is virtual (since f < 0). (g) Now f = · (1 . 5 − 1) µ 1 60 cm − 1 40 cm ¶¸ − 1 = − 240 cm , and the image formed is virtual (since f < 0)....
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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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