University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 34 - Geometric Optics - Problems - Exercises - Page 1152: 34.15

Answer

(a) $s'=10$ $cm$ and $y'=-2.20$ $mm$ (b)$s'=-4.29$ $cm$ and $y'=0.943$ $mm$

Work Step by Step

(a) Given: $R=12.0$ $cm$ $y=3.30$ $mm$ $s=15.0$ $cm$ $f=\frac{R}{2}=6.0$ $cm$ Using the formula: $\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$ Rearranging gives: $s'=\frac{fs}{s-f}=\frac{6.0\times{15.0}}{15.0-6.0}=10$ $cm$ $y'=-\frac{s'}{s}\times{y}=-\frac{10}{15}\times{3.30}$ $mm$ $=-2.2$ $mm$ (b) Given: $R=-12.0$ $cm$ $y=3.30$ $mm$ $s=15.0$ $cm$ $f=\frac{R}{2}=-6.0$ $cm$ Using the formula: $\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$ Rearranging gives: $s'=\frac{fs}{s-f}=\frac{(-6.0)\times{15.0}}{15.0-(-6.0)}=-4.29$ $cm$ $y'=-\frac{s'}{s}\times{y}=-\frac{(-4.29)}{15}\times{3.30}$ $mm$ $=0.943$ $mm$
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