Q 12-09-171JEE MainJEE Main 2021 (27 Aug, Shift 1)Medium
An object is placed beyond the centre of curvature $C$ of the given concave mirror. If the distance of the object is $d_1$ from $C$ and the distance of the image formed is $d_2$ from $C$, the radius of curvature of this mirror is:
Answer: (B) $\frac{2d_1d_2}{d_1-d_2}$
The object is beyond $C$, so the real image lies between $F$ and $C$. Using sign convention:
$u = -(R + d_1)$, $v = -(R - d_2)$, $f = -\dfrac{R}{2}$.
$$\frac{1}{R - d_2} + \frac{1}{R + d_1} = \frac{2}{R}$$
$$R(2R + d_1 - d_2) = 2(R - d_2)(R + d_1) \Rightarrow R(d_1 - d_2) = 2d_1d_2$$
$$R = \frac{2d_1d_2}{d_1 - d_2}$$
Solution by Sreeraj P, M.Sc Physics