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.