Answer
(a) $7.85mm $
(b) $10cm $
(c) $5.1cm $
Work Step by Step
(a) We know that
$\frac{1}{d_{\circ}}=\frac{1}{f_{obj}}-\frac{1}{L+f_{obj}}$
This can be rearranged as:
$ d_{\circ}=\frac{(f_{obj})(L+f_{obj})}{L}$
We plug in the known values to obtain:
$ d_{\circ}=\frac{(7.50mm)(160mm+7.50mm)}{160mm}$
$\implies d_{\circ}=7.85mm $
(b) We know that
$ f_e=\frac{-(L+f_{obj})N}{f_{obj}M_{total}}$
We plug in the known values to obtain:
$ f_e=-\frac{(0.160m+0.0075m)(0.25m)}{(0.00750m)(-55)}$
$ f_e=0.1015m $
$\implies f_e=10cm $
(c) We know that
$ f_e=-\frac{(L+f_{obj})N}{f_{obj}M_{total}}$
We plug in the known values to obtain:
$ f_e=\frac{(0.160m+0.0075m)(0.25m)}{(0.0075m)(-110)}$
$ f_e=0.050m $
$ f_e=5.1cm $
