The intensity of the central fringe obtained in the interference pattern due to two identical slit sources is I. When one of the slits is closed then the intensity at the same point is `I_0`. Then the correct relation between I and `I_0`is :

Options

A. `I=I_0`
B. `I=2I_0`
C. `I=4I_0`
D. `I=I_0//4`

Answer

To solve the problem, we need to analyze the intensity of light in the interference pattern created by two identical slits and how it changes when one of the slits is closed.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- In the double-slit experiment, when both slits are open, the intensity at the central maximum (where the path difference is zero) is denoted as \( I \).
- Since both slits are identical and contribute equally to the intensity, we can denote the intensity from each slit as \( I_0 \).

2. **Intensity with Both Slits Open**:
- The total intensity \( I \) at the central maximum when both slits are open can be expressed using the formula for interference:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cdot \cos(\delta) \]
- Here, \( I_1 = I_0 \) and \( I_2 = I_0 \) since both slits are identical, and at the central maximum, the phase difference \( \delta \) is zero (i.e., \( \cos(0) = 1 \)).
- Thus, substituting the values, we get:
\[ I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cdot 1 = 2I_0 + 2I_0 = 4I_0 \]

3. **Intensity with One Slit Closed**:
- When one of the slits is closed, the intensity at the same point (the central maximum) is simply the intensity from the open slit:
\[ I_0 = I_0 \]

4. **Establishing the Relationship**:
- From the above calculations, we have established that:
\[ I = 4I_0 \]
- This means that the intensity at the central maximum when both slits are open is four times the intensity from one slit when it is closed.

### Conclusion:
The correct relationship between \( I \) and \( I_0 \) is:
\[ I = 4I_0 \]