Question

$O$ is a point object kept on the principal axis of a concave mirror $M$ of radius of curvature $20cm$. $P$ is a prism of angle ${1.8}^{\circ }$. Light falling on the prism (at small angle of incidence) gets refracted through the prism and then falls on the mirror. Refractive index of the prism is $3/2$. Find the distance between the images formed by the concave mirror due to this light.

Answer 1

For thins prism $\delta =(\mu -1)\alpha$
$=(1.5-1){1.8}^{\circ }=\left(0.5\right)1.8\times \frac{\pi }{180}rad=\frac{\pi }{200}rad$
$\therefore DistanceO{O}_1=10\times \frac{\pi }{200}=\frac{\pi }{20}cm$
Similarly, distance $O{O}_2=\frac{\pi }{20}cm$
$\therefore O_1{O}_2=\frac{\pi }{10}cm$
For the mirror, $O_1$ and $O_2$ are two point objects, at a distance of $30cm$.
Now, applying mirror formula
$\frac{1}{v}+\frac{1}{-30}=\frac{1}{-10}\Rightarrow v=-15cm$
Latent magnification is
$-\frac{v}{u}=-\frac{(-15)}{-30}=-\frac{1}{2}$
$(-ve$ sign indicates inverted image)
Therefore, if $O_1^{\prime }$ and $O_2^{\prime }$ are the images formed, then distance between them,
$O_1^{\prime }{O}_2^{\prime }=\frac{1}{2}{O}_1{O}_2=\frac{1}{2}\cdot \frac{\pi }{10}=\frac{\pi }{20}cm$.