Question 382: Ray Optics

Question

An optical arrangement consists of two concave mirrors $M_1$ and $M_2$, and a convex lens $L$ with a common principal axis, as shown in the figure. The focal length of $\mathrm{L}$ is $10 \mathrm{~cm}$. The radii of curvature of $M_1$ and $M_2$ are $20 \mathrm{~cm}$ and $24 \mathrm{~cm}$, respectively. The distance between $L$ and $M_2$ is $20 \mathrm{~cm}$. A point object $S$ is placed at the mid-point between $\mathrm{L}$ and $M_2$ on the axis. When the distance between $\mathrm{L}$ and $\mathrm{M}_1$ is $n / 7 \mathrm{~cm}$, one of the images coincides with $\mathrm{S}$. The value of $n$ is _______.

Accepted Answer

Case-1 : if $\mathrm{I}^{	ext {st }}$ refraction on lens :
Given that the object is positioned at the focal point, the resulting light rays will transition to a parallel alignment.
Upon the first reflection at Mirror $ M_1 $:
As the incoming light rays are parallel, the resultant image will be formed at the focal point.
$\mathrm{II}^{	ext {nd }}$ refraction from $\mathrm{L}$ :-
$$
\begin{aligned}
& \mathrm{u}=-(\mathrm{d}-10) \\
& \mathrm{f}=10 \mathrm{~cm} \\
& \frac{1}{\mathrm{v}}-\frac{1}{\mu}=\frac{1}{\mathrm{f}} \\
& \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{~d}-10}=\frac{1}{10} \\
& \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{(\mathrm{~d}-10)} .......(i)
\end{aligned}
$$
This $v$ will be object for $M_2$, and image should be at $10 \mathrm{~cm}$
$$
\begin{aligned}
& \frac{1}{\mu}+\frac{1}{v_1}=\frac{1}{f} \\
& -\frac{1}{(20-v)}-\frac{1}{10}=-\frac{1}{12} \\
& \frac{1}{12}-\frac{1}{10}=\frac{1}{20-v} \\
& -\frac{2}{120}=\frac{1}{20-v} \\
& 20-v=-60 \\
& v=80 \mathrm{~cm}
\end{aligned}
$$
From equation (i), we get
$$
\begin{aligned}
& \frac{1}{80}=\frac{1}{10}-\frac{1}{d-10} \\
& \frac{1}{d-10}=\frac{1}{10}-\frac{1}{80} \\
& \frac{1}{d-10}=\frac{80-10}{800}=\frac{70}{800} \\
& d-10=\frac{80}{7} \Rightarrow d=10+\frac{80}{7}=\frac{150}{7} \\
& n=150
\end{aligned}
$$
Case-2 : If $1^{	ext {st }}$ reflection on mirror $m_2$ :
For $\mathrm{m}_2$
$$
\begin{aligned}
& \frac{1}{\mathrm{~V}_1}+\frac{1}{-10}=\frac{1}{-12} \\
& \mathrm{~V}_1=60 \mathrm{~cm}
\end{aligned}
$$
Then refraction on lens $\mathrm{L}$
$$
\begin{aligned}
& \mathrm{u}_2=80 \mathrm{~cm} \\
& \frac{1}{\mathrm{~V}_2}-\frac{1}{-60}=\frac{1}{10} \\
& \mathrm{~V}_2=\frac{80}{7}
\end{aligned}
$$
Then reflection on $\mathrm{m}_2$
Either $V_2$ is at centre (normal incidence)
$$
\begin{aligned}
& \mathrm{d}-\frac{80}{7}=20 \\
& \mathrm{~d}=\frac{220}{7} \\
& \frac{\mathrm{n}}{7}=\frac{220}{7} \\
& \mathrm{n}=220
\end{aligned}
$$
$V_2$ is at pole of $m_2$
$$
\begin{aligned}
& \mathrm{d}-\frac{80}{7}=0 \\
& \mathrm{~d}=\frac{80}{7} \\
& \frac{\mathrm{n}}{7}=\frac{80}{7} \\
& \mathrm{n}=80
\end{aligned}
$$

Ray Optics

Integer / Numeric Answer Type