Question

An equiconvex lens of refractive index $\mu$ and radius of curvature $R$ has its one surface silvered. A point source $O$ is placed before the silvered lens so that its image is coincident with it, the distance of the object from the lens is:

• A. $\frac{R}{(\mu -1)}$
• B. $\frac{2R}{(\mu -1)}$
• C. $\frac{R}{(2\mu -1)}$
• D. $\frac{2R}{(2\mu -1)}$

Answer 1

The correct option is C $\frac{R}{(2\mu -1)}$

The given lens can be considered as a system of a convex lens and a concave mirror. We must find the equivalent focal length of the system, say $f_{eq}$.

To find the equivalent length of the system, we shall consider the power of the system. If the power of the lens and the mirror can be defined as $P_l$ and $P_m$ respectively, then the equivalent power of the lens-mirror system can be given as-

$P_{eq}=P_l+P_m+P_l$ Since the light is incident on the lens, then mirror and then the lens again

Therefore,

$\frac{1}{f_{eq}}=\frac{2(\mu -1)}{R}+\frac{2}{R}+\frac{2(\mu -1)}{R}$

$\frac{1}{f_{eq}}=\frac{4(\mu -1)+2}{R}$

$\frac{1}{f_{eq}}=\frac{2\times (2\mu -1)}{R}$

Now, the whole system is acting like a spherical mirror, so we can apply the spherical mirror formula-

$\frac{1}{f_{eq}}=\frac{1}{u}+\frac{1}{v}$

Given that, the object is placed at a point where the object and image coincide. We assume that the object distance is $x$.

Therefore,

$\frac{1}{f_{eq}}=\frac{1}{x}+\frac{1}{x}$

$\Rightarrow \frac{2\times (2\mu -1)}{R}=\frac{2}{x}$

$\therefore x=\frac{R}{2\mu -1}$

So, option (C) is correct.