Two coherent sources of intensity ratio `alpha` interfere in interference pattern `(I_(max)-I_(min))/(I_(max)+I_(min))` is equal to

Options

A. `(2alpha)/(1+alpha)`
B. `(2sqrt(alpha))/(1+alpha)`
C. `(2alpha)/(1+sqrt(alpha)`
D. `(1+alpha)/(2alpha)`

Explanation

To solve the problem, we need to find the expression for the ratio \((I_{\text{max}} - I_{\text{min}}) / (I_{\text{max}} + I_{\text{min}})\) given two coherent sources with intensity ratio \(\alpha\).

### Step-by-Step Solution:

1. **Understanding Intensities**:
- Let the intensities of the two coherent sources be \(I_1\) and \(I_2\).
- Given the intensity ratio \(\alpha = \frac{I_1}{I_2}\).

2. **Relating Intensity to Amplitude**:
- The intensity \(I\) is proportional to the square of the amplitude \(A\):
\[ I \propto A^2 \]
- Therefore, we can express the amplitudes as:
\[ A_1 = \sqrt{I_1}, \quad A_2 = \sqrt{I_2} \]

3. **Finding Maximum and Minimum Intensities**:
- The maximum intensity \(I_{\text{max}}\) for two coherent sources is given by:
\[ I_{\text{max}} = (A_1 + A_2)^2 \]
- The minimum intensity \(I_{\text{min}}\) is given by:
\[ I_{\text{min}} = (A_1 - A_2)^2 \]

4. **Calculating \(I_{\text{max}} - I_{\text{min}}\)**:
- We can expand both expressions:
\[ I_{\text{max}} = A_1^2 + 2A_1A_2 + A_2^2 \]
\[ I_{\text{min}} = A_1^2 - 2A_1A_2 + A_2^2 \]
- Therefore, the difference is:
\[ I_{\text{max}} - I_{\text{min}} = (A_1^2 + 2A_1A_2 + A_2^2) - (A_1^2 - 2A_1A_2 + A_2^2) = 4A_1A_2 \]

5. **Calculating \(I_{\text{max}} + I_{\text{min}}\)**:
- The sum is:
\[ I_{\text{max}} + I_{\text{min}} = (A_1^2 + 2A_1A_2 + A_2^2) + (A_1^2 - 2A_1A_2 + A_2^2) = 2A_1^2 + 2A_2^2 = 2(A_1^2 + A_2^2) \]

6. **Forming the Ratio**:
- Now we can form the desired ratio:
\[ \frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}} = \frac{4A_1A_2}{2(A_1^2 + A_2^2)} = \frac{2A_1A_2}{A_1^2 + A_2^2} \]

7. **Substituting Amplitude Ratios**:
- From the intensity ratio \(\alpha\), we know:
\[ \frac{A_1^2}{A_2^2} = \alpha \implies \frac{A_1}{A_2} = \sqrt{\alpha} \quad \text{or} \quad \frac{A_2}{A_1} = \frac{1}{\sqrt{\alpha}} \]
- Let \(A_1 = k\) and \(A_2 = k/\sqrt{\alpha}\) (where \(k\) is a constant amplitude).

8. **Calculating \(A_1^2 + A_2^2\)**:
- We find:
\[ A_1^2 + A_2^2 = k^2 + \frac{k^2}{\alpha} = k^2\left(1 + \frac{1}{\alpha}\right) = k^2\frac{\alpha + 1}{\alpha} \]

9. **Substituting Back into the Ratio**:
- Now substitute back into the ratio:
\[ \frac{2A_1A_2}{A_1^2 + A_2^2} = \frac{2k \cdot \frac{k}{\sqrt{\alpha}}}{k^2\frac{\alpha + 1}{\alpha}} = \frac{2k^2/\sqrt{\alpha}}{k^2\frac{\alpha + 1}{\alpha}} = \frac{2\sqrt{\alpha}}{\alpha + 1} \]

10. **Final Result**:
- Thus, the final expression for the ratio is:
\[ \frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}} = \frac{2\sqrt{\alpha}}{\alpha + 1} \]