Question

In the figure shown, a point object $\text{O}$ is placed in air on the principal axis. The radius of curvature of the spherical surface is $60\text{cm}$. $\text{I}$ is the final image formed after all refractions and reflections.

• A. If $x=120\text{cm}$, then $\text{I}$ is formed on $\text{O}$ for any value of $y$
• B. If $x=120\text{cm}$, then $\text{I}$ is formed on $\text{O}$ only if $y=120\text{cm}$
• C. If $x=120\text{cm}$, then $\text{I}$ is formed on $\text{O}$ only if $y=110\text{cm}$
• D. If $x=120\text{cm}$, then $\text{I}$ is formed on $\text{O}$ only if $y=360\text{cm}$

Answer 1

The correct option is A If $x=120\text{cm}$, then $\text{I}$ is formed on $\text{O}$ for any value of $y$

Given,
refractive index of air, ${\mu }_a=1$,

refractive index of spherical surface material, ${\mu }_g=\frac{3}{2}$,

For the first refraction: On using mirror formula,

$\frac{{\mu }_g}{v}-\frac{{\mu }_a}{u}=\frac{{\mu }_g-{\mu }_a}{R}$

Taking case of option (a):

$\Rightarrow \frac{3/2}{v}-\frac{1}{(-120)}=\frac{3/2-1}{+60}$

$\Rightarrow \frac{3}{2v}=0$

$\Rightarrow v=\infty$

Thus, rays become parallel to principal axis after refraction. It will fall normally on the plane mirror and will retrace the path.

Hence, final image will form at the object $\text{O}$ itself and it does not depend on value of $y$.

Therefore, option (a) is correct.
$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: When a ray falls normally on a}\\ \text{mirror, it retraces its path.}\end{array}$