Q 12-09-076JEE MainJEE Main 2025 (28 Jan, Shift 2)Easy
A concave mirror produces an image of an object such that the distance between the object and the image is $20\ \text{cm}$. If the magnification of the image is $-3$, then the magnitude of the radius of curvature of the mirror is
Answer: (C) $15\ \text{cm}$
$m = -\dfrac{v}{u} = -3 \Rightarrow v = 3u$: both in front of the mirror, the image three times as far.
$|v| - |u| = 20\ \text{cm} \Rightarrow 2|u| = 20 \Rightarrow u = -10\ \text{cm}$, $v = -30\ \text{cm}$.
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = -\frac{1}{30} - \frac{1}{10} = -\frac{4}{30} \Rightarrow f = -7.5\ \text{cm}$$
$|R| = 2|f| = 15\ \text{cm}$
Solution by Sreeraj P, M.Sc Physics