A transparent solid cube of side 'a' has refractive index 3/2. A point source of light is embedded in it at its centre. Find the minimum area of the surface of the cube which must be painted black so that the source is not visible from outside.

Explanation

To solve the problem of determining the minimum area of the surface of a transparent solid cube that must be painted black so that a point source of light embedded at its center is not visible from outside, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- The cube has a side length of 'a'.
- The point source of light is located at the center of the cube.

2. **Critical Angle Calculation**:
- The refractive index of the cube (denser medium) is given as \( \mu_d = \frac{3}{2} \).
- The refractive index of air (rarer medium) is \( \mu_r = 1 \).
- The critical angle \( \theta_c \) can be calculated using Snell's law:
\[ \sin \theta_c = \frac{\mu_r}{\mu_d} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \]

3. **Finding the Radius of the Circle**:
- The radius \( R \) of the circular area from which the light source is visible can be determined using the geometry of the problem. The distance from the center of the cube to the face is \( \frac{a}{2} \).
- Using the relationship involving the critical angle:
\[ R = \left(\frac{a}{2}\right) \tan \theta_c \]
- To find \( \tan \theta_c \), we can use the identity:
\[ \tan \theta_c = \frac{\sin \theta_c}{\sqrt{1 - \sin^2 \theta_c}} = \frac{\frac{2}{3}}{\sqrt{1 - \left(\frac{2}{3}\right)^2}} = \frac{\frac{2}{3}}{\sqrt{\frac{5}{9}}} = \frac{2}{3} \cdot \frac{3}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \]
- Therefore, substituting back:
\[ R = \left(\frac{a}{2}\right) \cdot \frac{2\sqrt{5}}{5} = \frac{a\sqrt{5}}{5} \]

4. **Calculating the Area to be Painted**:
- Since the cube has 6 faces and the area of each circular region (where the source is visible) is given by:
\[ A_{circle} = \pi R^2 = \pi \left(\frac{a\sqrt{5}}{5}\right)^2 = \pi \cdot \frac{5a^2}{25} = \frac{\pi a^2}{5} \]
- The total area to be painted black on all 6 faces is:
\[ A_{total} = 6 \cdot A_{circle} = 6 \cdot \frac{\pi a^2}{5} = \frac{6\pi a^2}{5} \]

### Final Answer:
The minimum area of the surface of the cube that must be painted black so that the source is not visible from outside is:
\[ \frac{6\pi a^2}{5} \]