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. (a) Find the radius of the shadow of the ring formed on the ceiling if *r* = 15 cm. (b) Find the maximum value of *r* for which the shadow of the ring is formed on the ceiling. Refractive index of water = 4/3.

#### Solution

Given,

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

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:

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

Using Snell's law, we get:

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

\[ \Rightarrow \sin i = \frac{4}{5}\]

From the figure, we have:

\[So, \]

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

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

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

\[\Rightarrow 25 x^2 = 16(4 + x^2 )\]

\[ \Rightarrow 9 x^2 = 64\]

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

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,

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

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

\[ \Rightarrow 16 R^2 = 9 R^2 + 9 \times 400\]

\[ \Rightarrow 7 R^2 = 9 R^2 + 9 \times 400\]

\[ \Rightarrow R = 22 . 67 cm\]