Two waves of same frequency but of amplitude a and 2a respectively superimpose over each other. The intensity at a point where the phase difference is `(3pi)/(2)`, will be proportional to :

Options

A. `9a^(2)`
B. `3a^(2)`
C. `a^(2)`
D. `5a^(2)`

Explanation

To solve the problem, we need to determine the intensity of the resultant wave formed by the superposition of two waves with different amplitudes and a given phase difference.

### Step-by-Step Solution:

1. **Identify the Amplitudes**:
- Let the amplitude of the first wave be \( A_1 = a \).
- Let the amplitude of the second wave be \( A_2 = 2a \).

2. **Calculate the Intensities of Individual Waves**:
- The intensity \( I_1 \) of the first wave is proportional to the square of its amplitude:
\[ I_1 \propto A_1^2 = a^2 \]
- The intensity \( I_2 \) of the second wave is:
\[ I_2 \propto A_2^2 = (2a)^2 = 4a^2 \]

3. **Determine the Resultant Intensity**:
- The resultant intensity \( I \) when two waves interfere can be given by the formula:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- Here, \( \phi \) is the phase difference between the two waves, which is given as \( \frac{3\pi}{2} \).

4. **Substituting Values**:
- Substitute \( I_1 \) and \( I_2 \):
\[ I = a^2 + 4a^2 + 2\sqrt{a^2 \cdot 4a^2} \cos\left(\frac{3\pi}{2}\right) \]
- Simplifying the terms:
\[ I = 5a^2 + 2\sqrt{4a^4} \cos\left(\frac{3\pi}{2}\right) \]
- Since \( \cos\left(\frac{3\pi}{2}\right) = 0 \):
\[ I = 5a^2 + 2 \cdot 2a^2 \cdot 0 = 5a^2 \]

5. **Final Result**:
- The intensity at the point where the phase difference is \( \frac{3\pi}{2} \) is proportional to \( 5a^2 \).

### Conclusion:
The intensity at that point is proportional to \( 5a^2 \).