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 solve the problem of finding the maximum intensity in an interference pattern between two coherent sources with given intensities \(I_1\) and \(I_2\) at phase differences of \(\frac{\pi}{3}\) and \(\frac{2\pi}{3}\), respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula for Intensity**:
The net intensity \(I\) at any point where there is a phase difference \(\phi\) between two coherent sources is given by:
\[ I = I_A + I_B + 2\sqrt{I_A I_B} \cos(\phi) \]
where \(I_A\) and \(I_B\) are the intensities of the two sources.

2. **Setting Up the Equations**:
For the first point where the phase difference is \(\frac{\pi}{3}\):
\[ I_1 = I_A + I_B + 2\sqrt{I_A I_B} \cos\left(\frac{\pi}{3}\right) \]
Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), we can substitute this value:
\[ I_1 = I_A + I_B + 2\sqrt{I_A I_B} \cdot \frac{1}{2} \]
Simplifying this gives:
\[ I_1 = I_A + I_B + \sqrt{I_A I_B} \quad \text{(Equation 1)} \]

For the second point where the phase difference is \(\frac{2\pi}{3}\):
\[ I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cos\left(\frac{2\pi}{3}\right) \]
Since \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\), we substitute:
\[ I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cdot \left(-\frac{1}{2}\right) \]
This simplifies to:
\[ I_2 = I_A + I_B - \sqrt{I_A I_B} \quad \text{(Equation 2)} \]

3. **Adding and Subtracting the Equations**:
Now, we can add Equation 1 and Equation 2:
\[ I_1 + I_2 = (I_A + I_B + \sqrt{I_A I_B}) + (I_A + I_B - \sqrt{I_A I_B}) \]
This simplifies to:
\[ I_1 + I_2 = 2(I_A + I_B) \]
Therefore:
\[ I_A + I_B = \frac{I_1 + I_2}{2} \quad \text{(Equation 3)} \]

Now, subtract Equation 2 from Equation 1:
\[ I_1 - I_2 = (I_A + I_B + \sqrt{I_A I_B}) - (I_A + I_B - \sqrt{I_A I_B}) \]
This simplifies to:
\[ I_1 - I_2 = 2\sqrt{I_A I_B} \]
Therefore:
\[ \sqrt{I_A I_B} = \frac{I_1 - I_2}{2} \quad \text{(Equation 4)} \]

4. **Finding the Maximum Intensity**:
The maximum intensity occurs when \(\cos(\phi) = 1\):
\[ I_{\text{max}} = I_A + I_B + 2\sqrt{I_A I_B} \]
Substituting Equation 3 and Equation 4 into this expression:
\[ I_{\text{max}} = \frac{I_1 + I_2}{2} + 2\left(\frac{I_1 - I_2}{2}\right) \]
Simplifying this gives:
\[ I_{\text{max}} = \frac{I_1 + I_2}{2} + (I_1 - I_2) = \frac{I_1 + I_2 + 2I_1 - 2I_2}{2} = \frac{3I_1 - I_2}{2} \]

### Final Answer:
Thus, the maximum intensity \(I_{\text{max}}\) is:
\[ I_{\text{max}} = \frac{3I_1 + I_2}{2} \]