Question

All lights are switched off, except for a bright point-light source kept at the bottom of a swimming pool filled with clear water of refractive index $4/3$. As a result, only a circular patch of $6\, m$ diameter of the water surface is visible to spectators standing around the swimming pool. Which of the following gives the nearest value of the depth of the pool?

• A. 1.6 m
• B. 2.0 m
• C. 2.6 m
• D. 3.0 m

Answer 1

The Correct Option is C

Given situation is shown in the figure.
$\mu_W = \frac{4}{3} ; r = \frac{D}{2} = \frac{6}{2} = 3 \, m$
$h = ?$
This situation is seen if there is phenomena of total internal reflection of light.
$\therefore \:\:\: \sin \, C = \frac{1}{ \mu_W} = \frac{3}{4}$
Again from figure,
$\sin C = \frac{r}{\sqrt{r^{2}+h^{2}}} ; \frac{3}{4} =\frac{3}{\sqrt{3^{2} +h^{2}}}$
or $16 = 9 + h^2 ; h^2 = 7 , h = \sqrt{7} \, m = 2.6 \, m$

Answer 2

This problem involves the concept of refraction of light at the interface of two media with different refractive indices. The given scenario involves a point-light source placed at the bottom of a swimming pool filled with the clear water of refractive index \(\frac{4}{3}\), and we are asked to find the depth of the pool based on the observation that only a circular patch of 6m diameter of the water surface is visible to spectators standing around the pool.

The key idea here is that when light travels from one medium to another with a different refractive index, it bends at the interface. The amount of bending depends on the angle of incidence, the refractive indices of the two media, and the depth of the interface. In this case, the light from the point source bends as it travels from water to air at the surface of the pool, and only the light rays that are bent at a particular angle (which corresponds to the circular patch of 6m diameter) reach the eyes of the spectators.

Now, to find the depth of the pool, we can use the following steps:

1. The diameter of the circular patch of the visible water surface is 6m. This means the radius is 3m.
2. The light from the point source travels from the bottom of the pool to the surface and then to the eyes of the spectators. As it travels from water to air, it bends away from the normal to the interface.
3. Let h be the depth of the pool that we want to find. Using Snell's law of refraction, we can relate the angle of incidence of the light at the bottom of the pool to the angle of refraction at the water-air interface. The angle of incidence is given by \(sinθ = \frac{h}{3}\), where θ is the angle that the incident ray makes with the normal at the bottom of the pool. The angle of refraction is given by \(sinφ = \frac{4}{3}sinθ\), where φ is the angle that the refracted ray makes with the normal at the water-air interface.
4. To find the angle θ, we can use the fact that the diameter of the circular patch is 6m, which means the angle subtended by the patch at the bottom of the pool is \(2θ = 2\text{ arc tan}\frac{3}{h}\).
5. Now, using the relation \(sinφ = \frac{4}{3}sinθ\), we can find sinφ in terms of h. Solving for sinφ = 1 (since the light rays must be bent to the critical angle to reach the eyes of the spectators), we get \(h ≈ 2.7m\).

Therefore, the depth of the pool is approximately 2.7m, which is the nearest value given in the options.