Consider an interference pattern between two coherent sources. If `I_1 and I_2` be intensities at points where the phase difference are `pi/3 and (2pi)/3` and respectively , then the intensity at maxima is

Options

A. `(I_2-3I_1)/2`
B. `(I_1-3I_2)/2`
C. `(3I_2-I_1)/2`
D. `(3I_1-I_2)/2`

Answer

To find the intensity at maxima for the given interference pattern between two coherent sources, we can follow these steps:

### Step 1: Write the formula for intensity in interference
The intensity \( I \) at a point due to two coherent sources can be expressed as:
\[ I = I_A + I_B + 2\sqrt{I_A I_B} \cos \phi \]
where \( I_A \) and \( I_B \) are the intensities from the two sources, and \( \phi \) is the phase difference.

### Step 2: Calculate intensity \( I_1 \) for phase difference \( \frac{\pi}{3} \)
For the first case where the phase difference \( \phi = \frac{\pi}{3} \):
\[ I_1 = I_A + I_B + 2\sqrt{I_A I_B} \cos\left(\frac{\pi}{3}\right) \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ I_1 = I_A + I_B + 2\sqrt{I_A I_B} \cdot \frac{1}{2} \]
\[ I_1 = I_A + I_B + \sqrt{I_A I_B} \]

### Step 3: Calculate intensity \( I_2 \) for phase difference \( \frac{2\pi}{3} \)
For the second case where the phase difference \( \phi = \frac{2\pi}{3} \):
\[ I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cos\left(\frac{2\pi}{3}\right) \]
Since \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \):
\[ I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cdot \left(-\frac{1}{2}\right) \]
\[ I_2 = I_A + I_B - \sqrt{I_A I_B} \]

### Step 4: Add and subtract the equations for \( I_1 \) and \( I_2 \)
Now, we can add the two equations:
\[ I_1 + I_2 = (I_A + I_B + \sqrt{I_A I_B}) + (I_A + I_B - \sqrt{I_A I_B}) \]
\[ I_1 + I_2 = 2I_A + 2I_B \]
\[ I_A + I_B = \frac{I_1 + I_2}{2} \]

Now, subtract the two equations:
\[ I_1 - I_2 = (I_A + I_B + \sqrt{I_A I_B}) - (I_A + I_B - \sqrt{I_A I_B}) \]
\[ I_1 - I_2 = 2\sqrt{I_A I_B} \]

### Step 5: Express \( I_A \) and \( I_B \) in terms of \( I_1 \) and \( I_2 \)
From the equations derived:
\[ I_A + I_B = \frac{I_1 + I_2}{2} \]
\[ 2\sqrt{I_A I_B} = I_1 - I_2 \]
Thus, we can express \( \sqrt{I_A I_B} \):
\[ \sqrt{I_A I_B} = \frac{I_1 - I_2}{2} \]

### Step 6: Substitute back to find maximum intensity
The maximum intensity \( I_{max} \) can be expressed as:
\[ I_{max} = I_A + I_B + 2\sqrt{I_A I_B} \]
Substituting the values we derived:
\[ I_{max} = \frac{I_1 + I_2}{2} + 2 \cdot \frac{I_1 - I_2}{2} \]
\[ I_{max} = \frac{I_1 + I_2 + I_1 - I_2}{2} \]
\[ I_{max} = \frac{2I_1}{2} = I_1 \]

### Final Result
The intensity at maxima is:
\[ I_{max} = I_1 + I_2 \]