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
The light can come back to the source only if it retraces its path. A ray retraces its path after reflection only when it strikes the mirror normally, that is, when it is directed towards the mirror's centre of curvature. So the point to which the lens makes the rays converge must coincide with the centre of curvature of the convex mirror.
First find where the lens forms the image, using the Cartesian sign convention (distances measured from the lens, positive in the direction of the light).
\( u = -12\,\mathrm{cm} \)
\( f = +10\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \;\Rightarrow\; \frac{1}{v} = \frac{1}{10} - \frac{1}{12} = \frac{1}{60} \)
\( \displaystyle v = 60\,\mathrm{cm} \)
The radius of curvature of the diverging (convex) mirror is
\( \displaystyle R = 2f = 2 \times 14 = 28\,\mathrm{cm} \)
The centre of curvature of a convex mirror lies 28 cm behind it, and this point must be 60 cm from the lens. So the distance between the lens and the mirror is
\( \displaystyle d = 60 - 28 = 32\,\mathrm{cm} \)