A point source is placed at a depth `h` below the surface of water (refractive index `= mu`). The medium above the surface area of water is air from water.

Explanation

To solve the problem of finding the surface area illuminated by a point source placed at a depth \( h \) below the surface of water (with refractive index \( \mu \)), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a point source of light located at a depth \( h \) in water, which has a refractive index \( \mu \).
- The medium above the water is air, which has a refractive index of \( 1 \).

2. **Applying Snell's Law**:
- When light travels from a denser medium (water) to a rarer medium (air), Snell's law states:
\[ \mu \sin I = n \sin R \]
- Here, \( n = 1 \) for air, and when light reaches the critical angle, \( R = 90^\circ \), so \( \sin R = 1 \).

3. **Finding the Critical Angle**:
- At the critical angle \( \theta_c \):
\[ \mu \sin I = 1 \cdot 1 \]
- Therefore,
\[ \sin I = \frac{1}{\mu} \]

4. **Using Geometry to Relate Angles and Distances**:
- In the triangle formed by the point source and the radius \( r \) at the water's surface, we have:
\[ \sin I = \frac{r}{\sqrt{r^2 + h^2}} \]
- Setting the two expressions for \( \sin I \) equal gives us:
\[ \frac{r}{\sqrt{r^2 + h^2}} = \frac{1}{\mu} \]

5. **Cross-Multiplying and Squaring**:
- Cross-multiplying yields:
\[ \mu r = \sqrt{r^2 + h^2} \]
- Squaring both sides results in:
\[ \mu^2 r^2 = r^2 + h^2 \]
- Rearranging gives:
\[ h^2 = \mu^2 r^2 - r^2 = r^2 (\mu^2 - 1) \]

6. **Solving for \( r^2 \)**:
- Thus, we can express \( r^2 \) as:
\[ r^2 = \frac{h^2}{\mu^2 - 1} \]

7. **Calculating the Surface Area**:
- The surface area \( A \) illuminated by the point source is given by:
\[ A = \pi r^2 \]
- Substituting \( r^2 \) into the equation gives:
\[ A = \pi \left(\frac{h^2}{\mu^2 - 1}\right) \]

8. **Final Result**:
- Therefore, the surface area illuminated by the point source is:
\[ A = \frac{\pi h^2}{\mu^2 - 1} \]