Answer
a) $m=-0.67$
b) $m=0.44$
Work Step by Step
(a) We know that
$m=-\frac{fd_{\circ}}{(d_{\circ}-f)d_{\circ}}$
$\implies m=-\frac{f}{d_{\circ}-f}$
We plug in the known values to obtain:
$m=-\frac{30.0cm}{75.0cm-30.0cm}=-0.67$
(b) We know that
$d_{i2}=\frac{fd_{\circ2}}{d_{\circ2}-f}$
$d_{i2}=\frac{(30.0cm)(74.0cm)}{74.0cm-30.0cm}=50.45cm$
and $d_{i1}=\frac{f(d_{\circ})}{d_{\circ1}-f}$
We plug in the known values to obtain:
$d_{i1}=\frac{(30.0cm)(76.0cm)}{76.0cm-30.0cm}=49.56cm$
Now the length of the image is $L_i=50.45-49.56=0.89$
and the length of the object is $L_{\circ}=2.0cm$
The magnification is $m=\frac{L_i}{L_{\circ}}$
$m=\frac{0.89cm}{2.0cm}=0.44$
