Answer
$333.3X$
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 $333.3X$
College Physics (7th Edition)
by Wilson, Jerry D.; Buffa, Anthony J.; Lou, Bo
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
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