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?
Answer & explanation
Correct answer: option 1
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.