Chapter 25 - Vision and Optical Instruments - Learning Path Questions and Exercises - Exercises - Page 872: 35

$25X$

$di=0.5cm$ $P=250D$ $M=-100X$ $fo=\frac{1}{P}=\frac{100}{250}=0.4cm$ $\frac{1}{do}+\frac{1}{di}=\frac{1}{fo}$ or, $\frac{1}{di}+\frac{1}{0.5}=\frac{1}{0.4}$ $di=2cm$ Lateral magnification of objective = $Mo=-\frac{di}{do}=-\frac{2}{0.5}=-4X$ Also, $M=Mo\times m_{e}$ Thus, $m_{e}=\frac{M}{Mo}=\frac{-100}{-4}=25X$ The magnifying power of the eyepiece is $25X$.