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. (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.

Answer 1

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

(a)

From the figure, we can infer:
$\sin i=\frac{15}{25}$

Using Snell's law, we get:

$$\begin{gathered} \frac{\sin i}{\sin r}=\frac{1}{\mathrm{\mu }}=\frac{3}{4} \\ \Rightarrow \sin i=\frac{4}{5} \end{gathered}$$

From the figure, we have:

$$\begin{gathered} \tan r=\frac{x}{2} \\ \mathrm{So}, \\ \sin r=\frac{\tan r}{\sqrt{1+{\tan }^{2}r}} \\ =\frac{{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{1+{\displaystyle \frac{x^2}{4}}}} \\ \Rightarrow \frac{{\displaystyle x}}{\sqrt{4+{\displaystyle x^2}}}=\frac{4}{5} \end{gathered}$$
$$\begin{gathered} \Rightarrow 25x^2=16(4+x^2) \\ \Rightarrow 9x^2=64 \\ \Rightarrow x=\frac{8}{3}\mathrm{m} \end{gathered}$$
Total radius of the shadow = $\frac{8}{3}+0.15=2.81\mathrm{m}$

(b)
Condition for the maximum value of r:
Angle of incidence should be equal to the critical angle, i.e., $i={\theta }_{\mathrm{c}}$.
Let us take R as the maximum radius.
Now,
$$\begin{gathered} \sin {\theta }_{\mathrm{c}}=\frac{\sin {\theta }_{\mathrm{c}}}{\sin r} \\ =\frac{R}{\sqrt{R^2+20}}=\frac{3}{4}(\sin r\mathit{}=1) \\ \Rightarrow 16R^2=9R^2+9\times 400 \\ \Rightarrow 7R^2=9{\mathrm{R}}^2+9\times 400 \\ \Rightarrow R=22.67\mathrm{cm} \end{gathered}$$