Question

A concave mirror is placed on a horizontally table with its axis directed vertically upward. Let $O$ be the pole of the mirror and $C$ its centre of curvature. A point object is placed at $C$. It has a real image, also located at $C$. If the mirror is now filled with water, the image will be

• A. Real and will remain at $C$
• B. Real and located at a point between $C$ and $\infty$
• C. Real and located at a point between $C$ and $O$
• D. Real and located at a point between $C$ and $F$

Answer 1

The correct option is D Real and located at a point between $C$ and $F$

When there is no water in the mirror, the rays of light are incident normally on the mirror and retrace their path. So, we get an image coincide with the object as shown in the figure.

When the mirror is filled with water, then the equivalent focal length $F$ is given by

$\frac{1}{F}=\frac{1}{f_{waterlens}}+\frac{1}{f_{concavemirror}}+\frac{1}{f_{waterlens}}$

or $\frac{1}{F}=2\times \frac{1}{f_{waterlens}}+\frac{1}{f_{concavemirror}}$

or $\frac{1}{F}=2(\mu -1)\left(\frac{1}{R}\right)+\frac{1}{R}$

or $\frac{1}{F}=\frac{2(\mu -1)}{R}+\frac{2}{R}$

or $\frac{1}{F}=\frac{2\mu }{R}$ or $F=\frac{R}{2\mu }$

Clearly, focal length of the new optical system in less than the original. So, the object is effectively at a distance greater than twice the focal length. So, the real image will be formed between $F$ and $2F$.