The maximum intensity of fringes in Young's experiment is I. If one of the slit is closed, then the intensity at that place becomes `I_o`. Which of the following relation is true?

Options

A. `I=I_(o)`
B. `I=2I_(o)`
C. `I=4I_(o)`
D. There is not relation bertween I and `I_(o)`

Answer

To solve the problem, we need to analyze the relationship between the maximum intensity of fringes in Young's double-slit experiment and the intensity when one of the slits is closed.

### Step-by-Step Solution:

1. **Understanding Maximum Intensity in Young's Experiment:**
- In Young's double-slit experiment, the maximum intensity of the interference pattern is given by:
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
- If both slits are identical, we can denote the intensity from each slit as \( I_1 = I_2 = I_d \). Thus, the maximum intensity becomes:
\[ I_{\text{max}} = 2I_d \]

2. **Intensity When One Slit is Closed:**
- When one of the slits is closed, the intensity at the point where the fringe pattern was observed becomes \( I_0 \). This intensity is equal to the intensity from the open slit:
\[ I_0 = I_d \]

3. **Relating Maximum Intensity to Closed Slit Intensity:**
- From the first step, we established that:
\[ I_{\text{max}} = 2I_d \]
- Since \( I_d = I_0 \), we can substitute:
\[ I_{\text{max}} = 2I_0 \]

4. **Finding the Relationship:**
- We know from the problem statement that the maximum intensity \( I \) is given by:
\[ I = 2I_0 \]
- Rearranging this gives us:
\[ I_0 = \frac{I}{2} \]

5. **Conclusion:**
- Therefore, the relationship that holds true is:
\[ I = 2I_0 \]

### Final Answer:
The correct relation is \( I = 2I_0 \).