JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Geometrical Optics

First let the lens form an image of the object placed $1\text{m}$ in front of it. With $f=0.5\text{m}$ and $u=-1\text{m}$, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{0.5}-1=1$, so $v=1\text{m}$: the image lies $1\text{m}$ behind the lens, which is $1\text{m}$ in front of the plane mirror standing $2\text{m}$ behind the lens. The mirror reflects it to a point $1\text{m}$ behind the mirror, and this acts as an object for the lens a second time, now $3\text{m}$ away on the mirror side. Applying the lens formula again, $\frac{1}{v}=\frac{1}{0.5}-\frac{1}{3}=\frac{5}{3}$, so $v=0.6\text{m}$ on the object side of the lens, a real image. Its distance from the mirror is $2+0.6=2.6\text{m}$, so the final image is real and $2.6\text{m}$ from the mirror.