IIT-JEE 2010 · Paper 2, Physics question 41

First pass through the lens. $u=-30$, $f=15$: the image would form 30 cm beyond the lens, i.e. 20 cm behind the mirror.

Mirror. That point is a virtual object 20 cm behind the plane mirror, so the reflected rays converge towards a point 20 cm in front of the mirror, which is 10 cm beyond the lens on the object side.

Second pass. The converging rays meet the lens first: a virtual object 10 cm away on the far side ($u=+10$ along the new direction):
$$\frac1v=\frac1f+\frac1u=\frac1{15}+\frac1{10}=\frac16\implies v=6\text{ cm}.$$
The final image is real, 6 cm from the lens, i.e. $10+6=16$ cm from the mirror.

(B)