Question

A double convex lens has two surface of equal radii ${}^{\prime }{R}^{\prime }$ and refractive index $n=1.5$. Find the focal length ${}^{\prime }{f}^{\prime }$

Answer 1

Given

Radius of curvature $=R$

focal length $=f$

Refractive index, $\mu =1.5$

From lens maker's formula

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

$\frac{1}{f}=(1.5-1)[\frac{1}{R}+\frac{1}{R}]$

$=0.5\times \frac{2}{R}$

$\frac{1}{f}=\frac{1}{R}$

$f=R$