
Given is a thin convex lens of glass (refractive index $\mu$) and each side having radius of curvature $R$.…

- $\mathrm{R} / \mu$
- $R /(2 \mu-3)$
- $\mu R$
- $\mathrm{R} /(2 \mu-1)$
Solution

$\begin{aligned}
-\frac{1}{f_{e q}} & =\frac{2}{f_l}-\frac{1}{f_m} \\ & =2(\mu-1) \frac{2}{R}+\frac{2}{R} \\ -\frac{1}{f_{e q}} & =\frac{2(2 \mu-1)}{R} \\ f_{e q} & =-\frac{R}{2(2 \mu-1)}
\end{aligned}$
For concave mirror, object should be at $2 f$ for the image to be at same point
$\text { Distance }=\frac{R}{(2 \mu-1)}$
Asked in: JEE Main 2025 (22 Jan Shift 1)