28_Ch 23 College Physics ProblemCH23 Mirrors and Lenses

# 28_Ch 23 College Physics ProblemCH23 Mirrors and Lenses -...

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294 CHAPTER 23 Thus, the final image is located 160 cm to the left of the lens The overall magnification is 12 123 qqq MMMM 3 3 p pp  ==   , or ( ) () 60.0 cm 400 cm 160 cm 0.800 100 cm 300 cm 160 cm M  =−   0 Since M < , the final image is inverted 23.52 (a) Using the sign convention from Table 23.2, the radii of curvature of the surfaces are . The lens maker’s equation then gives 15.0 cm and 10.0 cm RR =+ 11 1 . 5 0 1 15.0 cm 10.0 cm n fR R = or 12.0 cm f (b) If p → ∞ , then 12.0 cm qf ==− The thin lens equation gives, ( ) 12.0 cm 12.0 cm p pf q pf p −+ and the following results: (c) If 33 6 . 0 c + m pf , 9.00 cm q (d) If 12.0 cm + , 6.00 cm q (e) If 26 .
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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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