Question

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

• A. If $d_1=120\text{cm}$, then the ${}^{\prime }{I}_f^{\prime }$ is formed on ${}^{\prime }{O}^{\prime }$ for any value of $d_2$.
• B. If $d_1=240\text{cm}$, then the ${}^{\prime }{I}_f^{\prime }$ is formed on ${}^{\prime }{O}^{\prime }$ only if $d_2=360\text{cm}$.
• C. If $d_1=240\text{cm}$, then the ${}^{\prime }{I}_f^{\prime }$ is formed on ${}^{\prime }{O}^{\prime }$ for all values of $d_2$.
• D. If $d_1=240\text{cm}$, then the ${}^{\prime }{I}_f^{\prime }$ cannot be formed on ${}^{\prime }{O}^{\prime }$.

Answer 1

Answer: (A), (B)

If $d_1=240\text{cm}$, then the ${}^{\prime }{I}_f^{\prime }$ is formed on ${}^{\prime }{O}^{\prime }$ only if $d_2=360\text{cm}$.

For $d_1=120\text{cm}$ ,

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

$\Rightarrow v=\infty$

so, the ray is incident normally on the mirror.

So for any value of $d_2$, ray retraces its path. So $I_f$ is at $O$.

For $d_1=240cm$

$\frac{3/2}{v}-\frac{1}{(-240)}=\frac{3/2-1}{60}$

$\Rightarrow v=360\text{cm}$

If first image is formed at mirror ray retraced its path to form image at $O$.