In a YDSE with identical slits, the intensity of the central bright fringe is `I_(0)`. If one of the slits is covered, the intensity at the same point is

Options

A. `2 I_(0)`
B. `I_(0)`
C. `I_(0) //2`
D. `I_(0) //4`

Explanation

To solve the problem, we need to analyze the Young's Double Slit Experiment (YDSE) and the effect of covering one of the slits on the intensity of the central bright fringe.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- In a YDSE with two identical slits, the intensity of the central bright fringe (when both slits are open) is given as \( I_0 \).

2. **Intensity Contribution from Each Slit**:
- Let the intensity due to each slit be \( I_1 \). Since the slits are identical, the intensity from both slits contributes to the total intensity at the central maximum.

3. **Formula for Total Intensity**:
- The total intensity \( I \) at the central maximum when both slits are open can be expressed as:
\[ I = I_1 + I_1 + 2\sqrt{I_1 \cdot I_1} \cos \phi \]
- For the central bright fringe, the phase difference \( \phi \) is zero, hence \( \cos \phi = 1 \).

4. **Calculating Total Intensity with Two Slits**:
- Substituting \( \cos \phi = 1 \) into the intensity formula, we get:
\[ I = I_1 + I_1 + 2\sqrt{I_1 \cdot I_1} = 2I_1 + 2I_1 = 4I_1 \]
- Therefore, we have:
\[ I_0 = 4I_1 \]

5. **Intensity with One Slit Covered**:
- When one of the slits is covered, only one slit contributes to the intensity at the central maximum. Thus, the intensity at the same point becomes:
\[ I' = I_1 \]

6. **Relating \( I' \) to \( I_0 \)**:
- Since we know \( I_0 = 4I_1 \), we can express \( I_1 \) in terms of \( I_0 \):
\[ I_1 = \frac{I_0}{4} \]
- Therefore, the intensity with one slit covered is:
\[ I' = I_1 = \frac{I_0}{4} \]

7. **Final Answer**:
- The intensity at the same point when one of the slits is covered is:
\[ I' = \frac{I_0}{4} \]

### Conclusion:
The intensity at the central bright fringe when one of the slits is covered is \( \frac{I_0}{4} \).