JEE Main 2025 Apr 4 Shift 1, Physics Q23: Thin Lens and Mirrors

Magnification $m=-\frac{v}{u}=-\frac{1}{3}$, giving $v=\frac{u}{3}$.

The image is real and inverted (since $m$ is negative and less than 1 in magnitude — a concave mirror with object beyond the centre of curvature). Taking the standard sign convention with the object on the left ($u<0$, $v<0$ for real image): the magnitudes satisfy $|v|=|u|/3$ and the distance between them is $|u|-|v|=30$, so $|u|-|u|/3=30$, giving $|u|=45$ cm and $|v|=15$ cm.

Mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=-\frac{1}{15}-\frac{1}{45}=-\frac{3+1}{45}=-\frac{4}{45}$, so $f=-\frac{45}{4}$ cm.

Magnitude $|f|=\frac{45}{4}$ cm, hence $x=45$.