Question

Radius of curvature of both surfaces of lens is $60\text{cm}$. Find the focal length of the silvered lens.

Answer 1

From diagram,
${\mu }_1=1;{\mu }_2=1.5$
$R_1=R_2=R=60\text{cm}$

Now applying lens maker's formula for convex lens

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{f_{lens}}=(\frac{1.5}{1}-1)(\frac{1}{60}-\frac{1}{-60})$

$\Rightarrow f_{lens}=60\text{cm}$

For mirror,

$\frac{1}{f_m}=-\frac{2}{R}$

$\Rightarrow f_m=-30\text{cm}$

So, effective power of given combination is given by

$-P_{eff}=P_l+P_m+P_l$

$-\frac{1}{f_{eff}}=-\frac{1}{f_m}+\frac{2}{f_l}$

$\frac{1}{f_{eff}}=-\frac{1}{30}-\frac{2}{60}$

$\Rightarrow f_{eff}=-15\text{cm}$

Negative sign implies combination behaves like a concave mirror.

Hence, option $\left(b\right)$ is correct.