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 35 - Optical Instruments - Exercises and Problems - Page 1017: 17

Answer

$9.2\,mm$

Work Step by Step

Lateral magnification $m=-\frac{s'}{s}$ where $s'$ is the image distance and $s$ is the object distance. Using the formula $\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$, we get $\frac{1}{9.0\,mm}=\frac{1}{s}+\frac{1}{40s}=\frac{41}{40s}$ $\implies s=\frac{41}{40}(9.0\,mm)=9.2\,mm$
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