Answer:

We are given that,

Refractive Index of glass is $\mu=1.55$

Focal length of the given double-convex lens is f= 20 cm

Radius of curvature of one face of the given lens is =R₁

Radius of curvature of the other face of the given lens is = R₂

Radius of curvature of the given double-convex lens is = R

Therefore, $R_{1}= R\;and \;R_{2}= – R$

The value of R is determined as:

$\frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}} \right ]$

$ \frac{1}{20}=(1.55-1)\left[\frac{1}{R}+\frac{1}{R} \right ]$

$ \frac{1}{20}=0.55\times\frac{2}{R}$

$ Therefore\;R=0.55\times2\times20=22cm$

The radius of curvature of the double-convex lens is 22cm.