College Physics (7th Edition)

a). $-340X$ b). $3900$%
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 = $\frac{m_{total}-m_{e}}{m_{e}}\times 100=\frac{340-8.69}{8.69}\times 100=3900$%