A point source is placed at a depth h below the surface of water (refractive index = mu). (a) Show that light escapes through a circular area on the water surface with its centre directly above the point source. (b) Find the angle subtended by a radius of the area on the source.

Explanation

To solve the problem step by step, we will break it down into two parts as required by the question.

### Part (a): Show that light escapes through a circular area on the water surface with its center directly above the point source.

1. **Understanding the Setup**:
- We have a point source of light located at a depth \( h \) below the surface of the water. The refractive index of water is denoted as \( \mu \).
- The light rays emitted from the point source will travel upwards towards the water surface.

2. **Refraction of Light**:
- As light travels from a denser medium (water) to a rarer medium (air), it bends away from the normal.
- The critical angle \( \theta_C \) is the angle of incidence above which light cannot pass through the boundary and is totally internally reflected.

3. **Finding the Critical Angle**:
- According to Snell's Law:
\[ \mu \sin(\theta_C) = 1 \cdot \sin(90^\circ) \]
This simplifies to:
\[ \mu \sin(\theta_C) = 1 \]
Therefore, we find:
\[ \sin(\theta_C) = \frac{1}{\mu} \]

4. **Geometry of the Escape Circle**:
- When light reaches the critical angle, it will travel along the water surface at an angle of \( 90^\circ \) to the normal.
- The rays that escape will form a circular area on the water surface, with the center of this circle directly above the point source.

5. **Conclusion for Part (a)**:
- Thus, light escapes through a circular area on the water surface with its center directly above the point source.

### Part (b): Find the angle subtended by a radius of the area on the source.

1. **Setting Up the Triangle**:
- Consider the triangle formed by the point source, the edge of the circular area on the surface, and the point directly above the source on the surface.
- Let \( R \) be the radius of the circular area on the water surface.

2. **Using Trigonometry**:
- In the right triangle formed, we can relate the radius \( R \) and the depth \( h \) using the tangent of the critical angle \( \theta_C \):
\[ \tan(\theta_C) = \frac{R}{h} \]

3. **Finding the Angle**:
- From the earlier derived equation for the critical angle:
\[ \sin(\theta_C) = \frac{1}{\mu} \]
- We can also find \( \tan(\theta_C) \) using the identity:
\[ \tan(\theta_C) = \frac{\sin(\theta_C)}{\sqrt{1 - \sin^2(\theta_C)}} \]
- Substituting \( \sin(\theta_C) = \frac{1}{\mu} \):
\[ \tan(\theta_C) = \frac{\frac{1}{\mu}}{\sqrt{1 - \left(\frac{1}{\mu}\right)^2}} = \frac{1/\mu}{\sqrt{\frac{\mu^2 - 1}{\mu^2}}} = \frac{1}{\sqrt{\mu^2 - 1}} \]

4. **Final Expression**:
- Therefore, we can express \( R \) in terms of \( h \) and \( \mu \):
\[ R = h \tan(\theta_C) \]
- The angle subtended by the radius at the source is \( \theta_C \).

### Conclusion for Part (b):
- The angle subtended by a radius of the area on the source is given by \( \theta_C = \sin^{-1}\left(\frac{1}{\mu}\right) \).