The intensity ratio of two coherent sources of light is p. They are interfering in some region and produce interference patten. Then the fringe visibility is

Options

A. `(1+P)/(2sqrtP)`
B. `(2sqrtP)/(1+P)`
C. `(P)/(1+P)`
D. `(2P)/(1+P)`

Answer

To solve the problem of finding the fringe visibility when the intensity ratio of two coherent sources of light is given as \( p \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Fringe Visibility**:
The fringe visibility \( V \) is defined as:
\[ V = \frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}} \]

2. **Intensity Ratio**:
We are given that the intensity ratio of the two coherent sources is:
\[ \frac{I_1}{I_2} = p \]
This implies:
\[ I_1 = p \cdot I_2 \]

3. **Finding \( I_{\text{max}} \) and \( I_{\text{min}} \)**:
- The maximum intensity \( I_{\text{max}} \) is given by:
\[ I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2 = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
- The minimum intensity \( I_{\text{min}} \) is given by:
\[ I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2 = I_1 + I_2 - 2\sqrt{I_1 I_2} \]

4. **Substituting the Values**:
- Substitute \( I_1 \) and \( I_2 \) into the formulas for \( I_{\text{max}} \) and \( I_{\text{min}} \):
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
\[ I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2} \]

5. **Calculating \( I_{\text{max}} - I_{\text{min}} \)**:
- The difference \( I_{\text{max}} - I_{\text{min}} \) is:
\[ I_{\text{max}} - I_{\text{min}} = (I_1 + I_2 + 2\sqrt{I_1 I_2}) - (I_1 + I_2 - 2\sqrt{I_1 I_2}) = 4\sqrt{I_1 I_2} \]

6. **Calculating \( I_{\text{max}} + I_{\text{min}} \)**:
- The sum \( I_{\text{max}} + I_{\text{min}} \) is:
\[ I_{\text{max}} + I_{\text{min}} = (I_1 + I_2 + 2\sqrt{I_1 I_2}) + (I_1 + I_2 - 2\sqrt{I_1 I_2}) = 2(I_1 + I_2) \]

7. **Substituting into the Visibility Formula**:
- Now substituting these into the visibility formula:
\[ V = \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2} \]

8. **Expressing in Terms of \( p \)**:
- Since \( I_1 = p \cdot I_2 \), we can express \( I_1 + I_2 \) and \( \sqrt{I_1 I_2} \):
\[ I_1 + I_2 = pI_2 + I_2 = (p + 1)I_2 \]
\[ \sqrt{I_1 I_2} = \sqrt{pI_2 \cdot I_2} = \sqrt{p} \cdot I_2 \]

9. **Final Visibility Expression**:
- Substituting these into the visibility expression:
\[ V = \frac{2\sqrt{p} \cdot I_2}{(p + 1)I_2} = \frac{2\sqrt{p}}{p + 1} \]

### Conclusion:
Thus, the fringe visibility \( V \) is given by:
\[ V = \frac{2\sqrt{p}}{p + 1} \]