Question

A parallel beam of light falls on the surface of a convex lens whose radius of curvature of both sides is $20cm$. The refractive index of the material of the lens varies as $\mu =1.5+0.5r$, where $r$ is the distance of the point on the aperture from the optical centre is cm. Find the length of the region on the axis of the lens where the light will appear. The radius of aperture of the lens is $1cm$.

Answer 1

Given: A parallel beam of light falls on the surface of a convex lens whose radius of curvature of both sides is $20cm$. The refractive index of the material of the lens varies as $\mu =1.5+0.5r$, where r is the distance of the point on the aperture from the optical centre in cm. The radius of aperture of the lens is $1cm$

To find the length of the region on the axis of the lens where the light will appear

Solution:

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})⟹\frac{1}{f}=(1.5+0.5r-1)\times \frac{2}{20}⟹\frac{1}{f}=0.5(1+r)\times \frac{1}{10}⟹\frac{1}{f}=\frac{1+r}{20}⟹f=\frac{20}{1+r}$

For $r=0,f=20cm$

For $r=1,f=10cm$

Light on the axis will be from 10cm to 20 cm for the lens.