Answer
(a) $47cm $ to the right of the lens 2; $0.38$
(b) $3.0cm$ to the right of the lens 2; $-0.61$
(c) $15.8cm$ to the right of the lens 2; $-0.316$
Work Step by Step
(a) We know that
$\frac{1}{d_{i1}}=\frac{1}{f_1}-\frac{1}{d_{\circ 1}}$
$\frac{1}{d_{i1}}=\frac{1}{20cm}-\frac{1}{50cm}$
$\implies d_{i1}=33.333cm$
and $d_{\circ 2}=x-d_{i1}$
$\implies d_{\circ 2}=115cm-33.333cm=82cm$
Now $\frac{1}{d_{i2}}=\frac{1}{f_2}-\frac{1}{d_{\circ 2}}$
$\implies \frac{1}{d_{i2}}=\frac{1}{30cm}-\frac{1}{82cm}$
$\implies d_{i2}=47cm$ to the right of lens 2.
The magnification can be determined as
$m=(\frac{d_{i1}}{d_{\circ 1}})(\frac{d_{i2}}{d_{\circ 2}})$
We plug in the known values to obtain:
$m=(\frac{33.3cm}{50cm})(\frac{47cm}{82cm})$
$m=0.38$
(b) As $d_{\circ 2}=x-d_{i1}$
$d_{\circ 2}=30cm-33.33cm=-3.33cm$
and $\frac{1}{d_{i2}}=\frac{1}{f_2}-\frac{1}{d_{\circ 2}}$
$\implies \frac{1}{d_{i2}}=\frac{1}{30cm}-\frac{1}{-3.33cm}$
$\implies d_{i2}=3.0cm$ to the right of lens 2.
The magnification is given as
$m=(\frac{33.33cm}{50cm})(\frac{3cm}{-3.33cm})=-0.61$
(c) We know that
$d_{\circ 2}=x-d_{i1}$
$d_{\circ 2}=0cm-33.33cm=-33.3cm$
$\frac{1}{d_{i2}}=\frac{1}{f_2}-\frac{1}{d_{\circ 2}}$
$\frac{1}{d_{i2}}=\frac{1}{30cm}-\frac{1}{-33.3cm}=15.78cm$ to the right of lens 2.
The magnification is given as
$m=(\frac{33.3cm}{50cm})(\frac{15.8cm}{-33.3cm})$
$m=-0.316$
(d) We know that
$\frac{1}{f_{eff}}=\frac{1}{d_{\circ}}+\frac{1}{d_i}$
$\frac{1}{f_{eff}}=\frac{1}{50cm}+\frac{1}{15.8cm}$
$\implies f_{eff}=12cm$
Now the effective focal length using the formula is given as
$\frac{1}{f_{eff}}=\frac{1}{f_1}+\frac{1}{f_2}$
$\implies E_{eff}=12cm$
Thus $\frac{1}{f_{eff}}=\frac{1}{f_1}+\frac{1}{f_2}$ agrees.
