Question

Define the term 'critical angle'. A small bulb is placed at the bottom of a tank containing a liquid of refractive index $\mu$ at a depth $H$. It is observed that the light emerges from a circular area of radius $r$ of the surface. Obtain the expression for $r$ in terms of $H$ and $\mu$.

Hint:

At the edge of the lit circle the ray meets the surface at the critical angle, so r = H tan C with sin C = 1/mu.

Answer 1

Step 1: Define the critical angle:
The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is \(90^{\circ}\). For a liquid-air surface, it is written \(C\), and \(\sin C = \dfrac{1}{\mu}\).

Step 2: Understand the situation:
Let the bulb \(B\) sit at the bottom of the tank, a depth \(H\) below the surface. Draw the vertical line from the bulb to the surface, meeting it at point \(O\). Rays reaching the surface at an angle of incidence less than \(C\) leave the liquid. Rays at an angle greater than \(C\) suffer total internal reflection.

Step 3: Find the edge of the lit circle:
The edge of the lit circle is at distance \(r\) from \(O\). A ray from the bulb to this edge makes the angle \(C\) with the vertical. So in the right-angled triangle with sides \(H\) (vertical) and \(r\) (horizontal):
\[ \tan C = \dfrac{r}{H} \quad \Rightarrow \quad r = H \tan C \]

Step 4: Express tan C using the refractive index:
\[ \sin C = \dfrac{1}{\mu}, \qquad \cos C = \sqrt{1 - \dfrac{1}{\mu^2}} = \dfrac{\sqrt{\mu^2 - 1}}{\mu} \]
\[ \tan C = \dfrac{\sin C}{\cos C} = \dfrac{1}{\sqrt{\mu^2 - 1}} \]

Final Answer:
The radius of the circle of light on the surface is: \[ \boxed{r = \dfrac{H}{\sqrt{\mu^2 - 1}}} \]