Answer
Please see the work below.
Work Step by Step
(a) We know that
$\frac{1}{d_{i1}}=\frac{1}{f_1}-\frac{1}{d_{\circ 1}}$
We plug in the known values to obtain:
$\frac{1}{d_{i1}}=\frac{1}{20cm}-\frac{1}{50cm}=-14.3cm$
and $d_{\circ 2}=x-d_{i1}=115cm-(-14.3cm)=129.3cm$
Now $\frac{1}{d_{i2}}=\frac{1}{f_2}-\frac{1}{d_{\circ 2}}$
$\implies \frac{1}{d_{i2}}=\frac{1}{30cm}-\frac{1}{129.3cm}$
$\implies d_{i2}=39.1cm$ to the right of lens2
The lateral magnification is given as
$m=(\frac{-14.286cm}{50cm})(\frac{39.06cm}{129.3cm})=-0.0863$
(b) We know that
$\frac{1}{d_{i2}}=\frac{1}{f_2}-\frac{1}{d_{\circ 2}}$
$\frac{1}{d_{i2}}=\frac{1}{30cm}-\frac{1}{-44.28cm}$
$\implies d_{i2}=93cm$ to the right of lens 2
and $m=(\frac{-14.286cm}{50cm})(\frac{39cm}{44.29cm})$
$m=-0.6$
(c) We know that
$\frac{1}{d_{i2}}=\frac{1}{30cm}-\frac{1}{14.286cm}$
$\implies d_{i2}=27.3cm$ to the left of lens 2
and $m=(\frac{-14.286cm}{50cm})(\frac{-27.27cm}{14.286cm})$
$m=0.545$
(d) We know that
$\frac{1}{f_{eff}}=\frac{1}{f_1}+\frac{1}{f_2}$
$\frac {1}{f_{eff}}=\frac{1}{30cm}+\frac{1}{-20cm}$
$\implies f_{eff}=-60cm$
Thus, $\frac{1}{f_{eff}}=\frac{1}{f_1}+\frac{1}{f_2}$ agrees with part (c).
