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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 ( ) ( )( ) 1 1 1 1.50 m 1.50 m 1.50 m 0.900 m 0.600 m 1 p p p p = + ( or )( ) ( ) 1 1 1 0.900 m 0.600 m 1.50 m 1 p p p p + = (a) Simplifying yields ( ) ( ) 2 1 1 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 2 1 0.900 m 1.20 m p p = + = (b) Then, q and the thin lens equation gives 1 1 1.50 m 1.20 m p = = ( )( ) 1 1 1 1 0.300 m 1.20 m 0.240 m 1.50 m p q f p q = = = + (c) The second image distance is 2 1 0.600 m 0.300 m q p = = + 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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