Question

The focal length of an equiconvex lens in air is equal to either of its radii of curvature. the refractive index of a material of the lens is:

• A. 1.2
• B. 1.25
• C. 1.5
• D. 1.8

Answer 1

The correct option is C 1.5

Given Focal length, f=R

Being an equiconvex lens, let $R_1=R,R_2=-R$

According to lens makers formula :

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

Substituting the values we get:

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

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

$\frac{1}{2}+1=\mu$

$\Rightarrow \mu =\frac{3}{2}=1.5$

The refractive index of the material of lens is 1.5