A thin plano-convex lens acts like a concave mirror of focal length 0.2 m when silvered at its plane surface. Refractive index of the material of lens is 1.5. Radius of curvature of convex surface of the lens is :

Given,

Focal length, $F=0.2m=20cm$

Refractive index, $\mu =1.5$

We know

$\frac{1}{f}=\frac{2}{f_l}+\frac{1}{f_m}$

But, $f_m=\infty$

$\frac{1}{f}=\frac{2}{f_l}$

$f_l=2\times 20=40cm$

From lens maker's formula

$\frac{1}{f}=(\mu -1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

$\frac{1}{40}=(1.5-1)\left[\frac{1}{R_1}-\frac{1}{\infty }\right]$

$\Rightarrow$ $\frac{1}{40}=\frac{0.5}{R_1}$

$R_1=0.5\times 40=20cm$

$R_1=0.2m$

Radius of curvature of convex surface $=0.2m$