Question

A bi convex lens has focal length 2/3 times radius of curvature of either surface calculate refractive index of the lens material

Answer 1

According to lens makers formula:

Answer 2

use formula,

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

a/c to question,

lens is bi convex lens of focal length 2/3 times of radius of curvature of either surface.

Let us consider R is the radius of curvature of bi convex lens.

so, $R_1=R,R_2=-R$

$f=\frac{2}{3}R$

now, $\frac{1}{\frac{2}{3}R}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)$

$\frac{3}{2R}=(\mu-1)\frac{2}{R}$

$\frac{3}{4}=(\mu-1)$

$\mu=\frac{7}{4}$

hence, refractive index of medium is 7/4