Answer
(a) $0.21$
(b) $-32$
Work Step by Step
(a) We know that
$d_i=(\frac{1}{f_{objective}}-\frac{1}{d_{\circ}})^{-1}$
We plug in the known values to obtain:
$d_i=(\frac{1}{75}-\frac{1}{122})^{-1}$
$d_i=0.19mm$
(b) As $L=d_i+f_{eye\space}=0.19+0.020=0.21m$
Now we can find the magnification as
$M_{total}=\frac{d_iN}{f_{0bjective}\space f_{eye\space piece}}$
We plug in the known values to obtain:
$M_{total}=\frac{(19)(25)}{(7.5)(2.0)}=-32$
