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,
\[\sin \theta_c = \frac{\sin \theta_c}{\sin r}\]
\[ = \frac{R}{\sqrt{R^2 + 20}} = \frac{3}{4} (\sin r = 1)\]
\[ \Rightarrow 16 R^2 = 9 R^2 + 9 \times 400\]
\[ \Rightarrow 7 R^2 = 9 R^2 + 9 \times 400\]
\[ \Rightarrow R = 22 . 67 cm\]
