Two monochromatic light waves of amplitude 3A and 2A interfering at a point have a phase difference of `60^(@)`. The intensity at that point will be proportional to:

Options

A. `5A^(2)`
B. `13A^(2)`
C. `1A^(2)`
D. `19A^(2)`

Answer

To solve the problem of finding the intensity of two interfering monochromatic light waves with given amplitudes and a phase difference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Amplitude of the first wave, \( A_1 = 3A \)
- Amplitude of the second wave, \( A_2 = 2A \)
- Phase difference, \( \Delta \phi = 60^\circ \)

2. **Convert Phase Difference to Radians** (if necessary):
- Since we are using cosine in our calculations, we can keep the phase difference in degrees. However, it is useful to know that \( \cos(60^\circ) = \frac{1}{2} \).

3. **Use the Formula for Resultant Amplitude**:
The resultant amplitude \( A_R \) of two waves can be calculated using the formula:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\Delta \phi)} \]

4. **Substitute the Values**:
- Calculate \( A_1^2 \):
\[ A_1^2 = (3A)^2 = 9A^2 \]
- Calculate \( A_2^2 \):
\[ A_2^2 = (2A)^2 = 4A^2 \]
- Calculate \( 2A_1 A_2 \cos(\Delta \phi) \):
\[ 2A_1 A_2 \cos(60^\circ) = 2 \cdot (3A) \cdot (2A) \cdot \frac{1}{2} = 6A^2 \]

5. **Combine the Results**:
- Now substitute these values into the resultant amplitude formula:
\[ A_R = \sqrt{9A^2 + 4A^2 + 6A^2} = \sqrt{19A^2} \]
- Thus, \( A_R = \sqrt{19}A \).

6. **Calculate the Intensity**:
- The intensity \( I \) is proportional to the square of the amplitude:
\[ I \propto A_R^2 = (\sqrt{19}A)^2 = 19A^2 \]

7. **Conclusion**:
- Therefore, the intensity at that point will be proportional to \( 19A^2 \).

### Final Answer:
The intensity at that point will be proportional to \( 19A^2 \).