Question

A container contains water up to a height of 20 cm and there is a point source at the centre of the bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m above the water surface.

1. Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm.
2. Find the maximum value of r for which the shadow of the ring is formed on the ceiling. Refractive index of water = `4/3`.

Answer 1

Given: Height (h) of the water in the container = 20 cm

The ceiling of the room is 2.0 m above the water surface.

Radius of the rubber ring = r

Refractive index of water = 4/3

From the figure, we can infer:

\[\sin i = \frac{15}{25}\]

(a) Radius of the shadow:

Using Snell’s law, we get:

\[\frac{\sin i}{\sin r} = \frac{1}{\mu} = \frac{3}{4}\]

\[ \Rightarrow \sin i = \frac{3}{5}\]

From the figure, we have:

\[\tan r = \frac{x}{2}\]

So, 

\[\sin r = \frac{\tan r}{\sqrt{1 + \tan^2 r}}\]

\[ = \frac{\frac{x}{2}}{\sqrt{1 + \frac{x^2}{4}}}\]

\[ \Rightarrow \frac{x}{\sqrt{4 + x^2}} = \frac{4}{5}\]

⇒ 25x² = 16(4 + x²)

⇒ 9x² = 64 

\[ \Rightarrow x = \frac{8}{3} m\]

Total radius of the shadow = \[\frac{8}{3} + 0 . 15 = 2 . 81 m\]
(b) Condition for the maximum value of r:

Angle of incidence should be equal to the critical angle, i.e., \[i = \theta_c\]

Let us take R as the maximum radius.
Now,

\[\sin   \theta_c  = \frac{\sin  \theta_c}{\sin  r}\]

\[ = \frac{R}{\sqrt{R^2 + 20}} = \frac{3}{4}  (\sin  r   = 1)\]

⇒ 16R² = 9R² + 9 × 400

⇒ 7R² = 9R² + 9 × 400

⇒ R = 22.67 cm