Question

A hemispherical portion of the surface of a solid glass sphere (p rt: 1.5) of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.

Answer 1

For refracting surface, ${\mu }_1=1$, u = -2r

${\mu }_2=\frac{3}{2}$

v=?, R=r

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\frac{3}{2r}+\frac{1}{2r}=\frac{0.5}{r}$

$\frac{3}{2r}=\frac{1}{2r}\Rightarrow v=\infty$

Now for reflecting surface, $u=-\infty$

$f=\frac{r}{2},v=?$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}+\frac{1}{\infty }=\frac{2}{r}$

$\Rightarrow v=\frac{r}{2}$

from reflecting surface
For reflecting surface,

$u=-(2r-\frac{r}{2})=\frac{-3r}{2}$

$\mu =?$ ${\mu }_1=\frac{3}{2}$

${\mu }_2=1$

so, $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{r}$

$\frac{1}{v}+\frac{3\times 2}{2\times 3r}=\frac{-0.5}{-r}$

$\frac{1}{v}=\frac{1}{2r}-\frac{1}{r}=\frac{-1}{2r}$

So final image will be at reflecting surface of sphere.