Answer
$+77.0D$
Work Step by Step
$fo=8mm=0.8cm$,
$L=15cm$,
$m_{total}=-360X$
$m_{total}=-\frac{25(cm)L}{f_{o}f_{e}}$
$-360=\frac{-25\times15}{0.8\times f_{e}}$
Thus, $f_{e}=1.3cm=0.013m$
Power=$\frac{1}{f}=\frac{1}{0.013}=+77.0D$
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: 29
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