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.
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As shown in the figure, OQ = 3r, OP = r
So, PQ = 2r
For refraction at APB
We know, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{1.5}{\text{v}}-\Big(\frac{1}{-2\text{r}}\Big)=\frac{0.5}{\text{r}}=\frac{1}{2\text{r}}$ [because u = -2r]
$\Rightarrow\text{v}=\infty$
For the reflection in concave mirror
$\text{u}=\infty$
So, v = focal length of mirror $=\frac{\text{r}}{2}$
For the refraction of APB of the reflected image.
Here, $\text{u}=\frac{-3\text{r}}{2}$
$\frac{1}{\text{v}}-\frac{1.5}{\frac{-3\text{r}}{2}}=\frac{-0.5}{-\text{r}}$ $\big[\text{Here}, \ \mu_1=1.5 \ \text{and} \ \text{R}=-\text{r}\big]$
$\Rightarrow\text{v}=-2\text{r}$
As, negative sign indicates images are formed inside APB. So, image should be at C. So, the final image is formed on the reflecting surface of the sphere.
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