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 873: 45


a) 60cm and 0.80cm, 40cm and 0.90cm, b) $-75X$ and $-44X$.

Work Step by Step

a). We know that $m=-\frac{fo}{fe}$ m is maximum when $fo$ is maximum and $fe$ is minimum. So $fo=60cm$ and $fe=0.80cm$ will give maximum magnification. m is minimum when $fo$ is minimum and $fe$ is maximum. So $fo=40cm$ and $fe=0.90cm$ can be used for minimum magnification. b). for maximum magnification $m=-\frac{fo}{fe}=-\frac{60}{0.8}=-75X$ for minimum magnification $m=-\frac{fo}{fe}=-\frac{40}{0.9}=-44X$
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