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: 31


a). $-340X$ b). $3900$%

Work Step by Step

a). $fo=0.5cm$ $fe=3.25m$ $L=22cm$ Also, $m_{total}=-\frac{25(cm)L}{fofe}=-\frac{25\times22}{0.5\times3.25}=-340X$ b). To compare total magnification with that of eyepiece alone as magnifying glass in percentage :- $m_{e}=1+\frac{25}{fe}=1+\frac{25}{3.25}=8.69X$ Hence, the required percentage = $\frac{m_{total}-m_{e}}{m_{e}}\times 100=\frac{340-8.69}{8.69}\times 100=3900$%
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