31_Ch 23 College Physics ProblemCH23 Mirrors and Lenses

31_Ch 23 College Physics ProblemCH23 Mirrors and Lenses -...

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Mirrors and Lenses 297 Combining fractions gives () 11 1 1.50 m 1.50 m 1.50 m 0.900 m 0.600 m 1 p pp p = −+ ( or )( ) ( ) 1 0.900 m 0.600 m 1.50 m 1 p p +− = (a) Simplifying yields ( ) ( ) 2 m 0.600 m 0.900 m 1.50 m 0.540 p p = or 2 1 0.540 m 1.80 m p == 0.300 m and 21 0.900 m 1.20 m =+ = (b) Then, q and the thin lens equation gives 1.50 m 1.20 m p =− = ( ) 0.300 m 1.20 m 0.240 m 1.50 m pq f = + (c) The second image distance is 0.600 m 0.300 m qp = −= + and the magnification for this configuration is 2 2 2 0.300 m 1.20 m q M p = − 0.250 Thus, the second image is real, inverted, and diminished 23.57 From the thin lens equation, the image distance for the first lens is
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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