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 find the intensity at the point where two monochromatic light waves of amplitudes \(3A\) and \(2A\) are interfering with a phase difference of \(60^\circ\), we can follow these steps:

### Step 1: Understand the formula for intensity in terms of amplitude
The intensity \(I\) of a wave is proportional to the square of its amplitude. For two waves with amplitudes \(A_1\) and \(A_2\), the resultant intensity \(I\) due to interference can be expressed as:
\[ I \propto (A_1 + A_2)^2 \]

### Step 2: Calculate the resultant amplitude
Given:
- Amplitude of the first wave \(A_1 = 3A\)
- Amplitude of the second wave \(A_2 = 2A\)
- Phase difference \(\phi = 60^\circ\)

The resultant amplitude \(R\) can be calculated using the formula:
\[ R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi)} \]
Substituting the values:
\[ R = \sqrt{(3A)^2 + (2A)^2 + 2(3A)(2A) \cos(60^\circ)} \]

### Step 3: Substitute the values and calculate
Now, we substitute \(\cos(60^\circ) = \frac{1}{2}\):
\[ R = \sqrt{9A^2 + 4A^2 + 2(3A)(2A) \cdot \frac{1}{2}} \]
\[ R = \sqrt{9A^2 + 4A^2 + 6A^2} \]
\[ R = \sqrt{19A^2} = \sqrt{19}A \]

### Step 4: Calculate the intensity
The intensity \(I\) is now proportional to the square of the resultant amplitude:
\[ I \propto R^2 = (\sqrt{19}A)^2 = 19A^2 \]

### Conclusion
Thus, the intensity at that point will be proportional to:
\[ \text{Intensity} \propto 19A^2 \]