## College Physics (7th Edition)

$333.3X$
$fo=4.5mm=0.45cm$ $fe=3cm$ $L=18cm$ Thus, $m_{total}=-\frac{25(cm)L}{fo\times fe}=-\frac{25\times 18}{0.45\times3}=-333.3X$ Thus magnification of the viewed image is $333.3X$