JEE Main 2025 Apr 7 Shift 2, Physics Q9: Magnification

A magnification of $+\frac{1}{4}$ (an erect diminished image) points to a convex mirror; the image forms behind the mirror while the object is in front, so the object-image separation equals $|u|+|v|$.

Using the magnification relation for a mirror in magnitude form, $\frac{|v|}{|u|}=\frac{1}{4}$, so $|u|=4|v|$. Combining with $|u|+|v|=40$: $5|v|=40\Rightarrow |v|=8$ cm and $|u|=32$ cm.

Applying the sign convention (object in front of mirror gives $u=-32$ cm; image behind gives $v=+8$ cm) to the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$: $\frac{1}{8}-\frac{1}{32}=\frac{4-1}{32}=\frac{3}{32}=\frac{1}{f}$.

$f=\frac{32}{3}\approx 10.7$ cm.