A point like object is placed at a distance of $1$ m in front of a convex lens of focal length $0.5$ m. A plane mirror is placed at a distance of $2$ m behind the lens. The position and nature of the image formed by the system is:
Answer: (A) $2.6$ m from the mirror, real
**Lens (first pass):** $u = -1$ m, $f = 0.5$ m:
$$\frac1v = \frac1f + \frac1u = 2 - 1 = 1 \Rightarrow v = 1\ \text{m}$$
The image is $1$ m behind the lens, i.e. $1$ m in front of the mirror.
**Mirror:** the plane mirror forms its image $1$ m behind the mirror, i.e. $3$ m from the lens.
**Lens (second pass):** the reflected light travels back through the lens, with the object $3$ m away:
$$\frac1v = \frac{1}{0.5} - \frac13 = \frac53 \Rightarrow v = 0.6\ \text{m}$$
This image is on the side of the original object, so it is $2 + 0.6 = 2.6$ m from the mirror.
The returning rays actually converge at this point, so the image is real. (The official answer key lists '2.6 m from the mirror, virtual'.)
Solution by Sreeraj P, M.Sc Physics