A point source S is placed at a distance of 12 cm from a converging lens of focal length 10 cm on its principal axis. Where should a diverging mirror of focal length 14 cm be placed from the lens so that a real image is formed on the source itself?

32 cm

The correct answer is Option (1) → 32 cm

Given:

Converging lens focal length = $10\text{ cm}$

Point source distance = $12\text{ cm}$ (to the left of the lens)

Use sign convention: $u = -12$

Lens formula:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{v}-\frac{1}{-12}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{12}$

$=\frac{6-5}{60}=\frac{1}{60}$

So image forms at:

$v=60\text{ cm}$ (to the right of the lens)

Now a diverging mirror has focal length $f_m=-14\text{ cm}$. To return rays to the source as a real image, the first image must lie at the centre of curvature of the mirror.

Centre of curvature = $2f_m = 2(-14) = -28\text{ cm}$

This means the mirror must be placed so that the distance from the lens to the mirror is:

$x = 60 + (-28)$

$x = 32\text{ cm}$

The correct mirror position is $32\text{ cm}$ to the right of the lens.