Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 34 - Ray Optics - Exercises and Problems - Page 991: 32

Answer

$s' = -15~cm$ $h' = 1.5~cm$

Work Step by Step

We can find the image distance: $\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$ $\frac{1}{s'} = \frac{1}{f}-\frac{1}{s}$ $\frac{1}{s'} = \frac{1}{30~cm}-\frac{1}{10~cm}$ $\frac{1}{s'} = -\frac{1}{15~cm}$ $s' = -15~cm$ We can find the image height: $h' = \vert M \vert ~h$ $h' = \vert (\frac{s'}{s}) \vert ~(h)$ $h' = \vert (\frac{-15~cm}{10~cm})\vert ~(1.0~cm)$ $h' = 1.5~cm$
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