College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 25 - Vision and Optical Instruments - Learning Path Questions and Exercises - Exercises - Page 872: 30



Work Step by Step

$fo=4.5mm=0.45cm$ $fe=3cm$ $L=18cm$ Thus, $m_{total}=-\frac{25(cm)L}{fo\times fe}=-\frac{25\times 18}{0.45\times3}=-333.3X$ Thus magnification of the viewed image is $333.3X$
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