Question

A convex lens of radii of curvature 20 cm and 30 cm respectively. It is silvered at the surface which has smaller radius of curvature. Then it will behave as $({\mu }_g=1.5)$

Answer 1

Focal length for lens, $\frac{1}{f_L}=({\mu }_g-1)[\frac{1}{R_1}-\frac{1}{R_2}]$
Here, ${\mu }_g=1.5,R_1=20cm,R_2=-30cm$
$\therefore \frac{1}{f_L}=(1.5-1)[\frac{1}{20}-\frac{1}{-30}]=0.5\times \left[\frac{5}{60}\right]$
or $f_L=\frac{120}{5}=24cm$
Focal length for mirror, $f_m=\frac{R}{2}=\frac{-20}{2}=-10cm$
$\therefore$ Equivalent focal length
$f_{eq}=\frac{-2}{f_L}+\frac{1}{f_m}=\frac{-2}{24}+\frac{1}{-10}=\frac{-11}{60}\Rightarrow f_{eq}=\frac{-60}{11}cm$
Hence, this system behaves like a concave mirror of local length $\frac{-60}{11}$ cm.