Question

A hemispherical portion of the surface of a solid glass sphere $(\mu =1.5)$ 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.

• A. Image is formed at a distance of $\frac{r}{2}$ from the centre towards the reflecting surface.
• B. Image is formed at a distance $\frac{r}{4}$ from the centre away from the reflecting surface.
• C. Image is formed on the reflecting surface of the sphere.
• D. Image is formed at a distance $\frac{r}{3}$ from the refracting surface towards the centre.

Answer 1

The correct option is C Image is formed on the reflecting surface of the sphere.


As shown in the figure, $\text{OQ}$ $=3r$,
$\text{OP}$ $=r$

So, $\text{PQ}$ $=2r$

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

$\Rightarrow \frac{1.5}{v}-\frac{1}{-2r}=\frac{0.5}{r}$

$\Rightarrow \frac{1.5}{v}=\frac{0.5}{r}-\frac{1}{2r}$

$\Rightarrow v=\infty$

For the reflection in concave mirror,

$u=\infty$

So, $v=$ Focal length of mirror $=$ $r/2$

For the refraction of the reflected image,

Here, $u=\frac{-3r}{2}$

$\frac{1}{v}-\frac{1.5}{-3r/2}=\frac{-0.5}{r}$

[Here ${\mu }_1=1.5,{\mu }_2=1,R=-r$]

$\Rightarrow v=-2r$

Negative sign indicates that image should be formed on the reflecting surface of the sphere.