JEE Main 2025 Apr 4 Shift 2, Physics Q19: Reflection at Plane and Spherical Surfaces

For the convex mirror with $u=-30$ cm and $f=+30$ cm:

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$\therefore \frac{1}{v}=\frac{1}{30}-\frac{1}{-30}=\frac{1}{30}+\frac{1}{30}=\frac{1}{15}$, so $v=+15$ cm (image behind the convex mirror).

So the image formed by the convex mirror is 15 cm behind it, i.e. $30+15=45$ cm from the object.

A plane mirror produces an image at equal distance on its opposite side. For its image to coincide with the convex mirror's image at 45 cm from the object, the plane mirror must sit midway between the object and that image point $\to$ at $22.5$ cm from the object.

Distance between the two mirrors $=30-22.5=7.5$ cm.