Question

A concave mirror of radius of curvature $60\text{cm}$ is placed at the bottom of a tank containing water upto a height of $20\text{cm}$. The mirror focus upwards with its axis vertical , light falls normally on the surface of water and the image of the sun is formed. If $\mu =\frac{4}{3}$, then with the observer in air, the distance of the image from the surface of water is

• A. $30\text{cm}$
• B. $20\text{cm}$
• C. $7.5\text{cm}$ below
• D. $7.5\text{cm}$ above
  

Answer 1

The correct option is D $7.5\text{cm}$ above


Given, Radius of curvature of the concave mirror is $R=60\text{cm}$

For spherical mirrors , $f=\frac{R}{2}=\frac{60}{2}=30\text{cm}$

As we know, incident rays parallel to the principal axis of spherical mirrors , after reflection converges at its focus. Hence, image of the sun will be formed at the focus of the concave mirror.

Hence , image is formed at $d=30-20=10\text{cm}$ above the water surface

Due to the presence of water, apparent shift happens in the image position.

$\text{Refractive index}\left(\mu \right)=\frac{\text{Real depth}}{\text{apparent depth}}$

$\text{Apparent depth}=\frac{\text{Real depth}}{\mu }=\frac{10}{\mu }=\frac{10}{\frac{4}{3}}$

$\therefore \text{Image of the sun will appear at}7.5\text{cm}\text{above the water surface.}$

Hence, option (d) is the correct answer.

Why this question? To understand the working of combination of refraction & spherical mirrors.