Chapter 25 - Vision and Optical Instruments - Learning Path Questions and Exercises - Exercises - Page 872: 32

$16cm$

Power of objective $Po=+100D$ Power of eye piece $Pe=+50D$ Total desired magnification $=-200X$ Respective focal lengths $fe=\frac{1}{P}=\frac{1}{50}=2cm$ $fo=\frac{1}{P}=\frac{1}{100}=1cm$ $m_{total}=-\frac{25\times(cm)L}{fofe}$ or, $-200=-\frac{25\times L}{1\times 2}$ or, $L=16cm$