Question

A plano convex lens of refractive index $1.5$ and radius of curvature $30\text{cm}$ is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens, an object needs to be placed in order to have a same size image of the object?

Answer 1

The effective focal length of silvered lens is given by
$$\frac{-1}{f}=-\frac{1}{f_m}+\frac{2}{f_L}$$
Here, focal length of plano convex lens,
$f_L=\frac{R}{\mu -1}$

Substituting the given data we get,

$f_L=\frac{30}{1.5-1}=60\text{cm}$

Since, the curved surface is silvered and also focal length of concave mirror,
$f_m=\frac{R}{2}$

$f_m=\frac{-30}{2}=-15\text{cm}$

$\therefore -\frac{1}{f}=-\frac{1}{-15}+\frac{2}{60}=\frac{1}{10}\text{cm}$

$\therefore f=-10\text{cm}$
The equivalent lens behave as a concave mirror.
For the image to be of same size of object, it must be placed at the centre of curvature of the concave mirror, $u=|2f|=20\text{cm}$

Hence, option (c) is the correct answer.

Why the question? Note:- The image will be of same size as of object when the object is placed at distance $2F$.