A thin biconvex lens is prepared from the glass ($\mu = 1.5$) both curved surfaces of which have equal radii of $20$ cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of ______ cm.
Answer: (A) $10$
Focal length of the lens: $\dfrac{1}{f_L} = (1.5 - 1)\left(\dfrac{1}{20} + \dfrac{1}{20}\right) = \dfrac{1}{20}$, so $f_L = 20$ cm.
The silvered surface is a concave mirror of radius $20$ cm, so $f_m = 10$ cm.
Light passes through the lens, reflects, and passes through the lens again, so the system acts as a mirror of power
$$P = \frac{2}{f_L} + \frac{1}{f_m} = \frac{2}{20} + \frac{1}{10} = \frac{1}{5}\ \text{cm}^{-1}$$
That is a concave mirror of focal length $F = 5$ cm. Image and object coincide when the object is at its centre of curvature: $2F = 10$ cm.
Solution by Sreeraj P, M.Sc Physics
