Q 12-09-239JEE MainJEE Main 2025 (7 Apr, Shift 2)Medium
A mirror is used to produce an image with magnification of $\dfrac14$. If the distance between the object and its image is $40\ \text{cm}$, then the focal length of the mirror is:
Answer: (C) $10.7\ \text{cm}$
An erect, diminished image ($m = +\tfrac14$) is formed by a convex mirror, with the image behind the mirror: $|v| = \dfrac{|u|}{4}$ and $|u| + |v| = 40$.
So $|u| = 32\ \text{cm}$ and $v = +8\ \text{cm}$, $u = -32\ \text{cm}$:
$$\frac1f = \frac1v + \frac1u = \frac18 - \frac1{32} = \frac{3}{32} \Rightarrow f = \frac{32}{3} \approx 10.7\ \text{cm}$$
(Reading $m$ as $-\tfrac14$ with a concave mirror gives the same magnitude: $|u| = 53.3$, $|v| = 13.3$, $|f| = 10.7\ \text{cm}$.)
Solution by Sreeraj P, M.Sc Physics