College Physics (7th Edition)

a) 60cm and 0.80cm, 40cm and 0.90cm, b) $-75X$ and $-44X$.
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 $m=-\frac{fo}{fe}=-\frac{40}{0.9}=-44X$