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 step by step, we can follow these logical steps:

### Step 1: Understand the Initial Condition
In Young's Double Slit Experiment (YDSE), we have two identical slits. The intensity of the central bright fringe (where the two waves constructively interfere) is given as \( I_0 \).

### Step 2: Relate Individual Intensities to Total Intensity
Since the slits are identical, we can denote the intensity from each slit as \( I_1 \) and \( I_2 \). Because they are identical, we have:
\[ I_1 = I_2 = I \]

### Step 3: Use the Formula for Maximum Intensity
The maximum intensity \( I_{max} \) at the central fringe due to interference from both slits is given by the formula:
\[ I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \]
Substituting \( I_1 \) and \( I_2 \) with \( I \):
\[ I_{max} = (\sqrt{I} + \sqrt{I})^2 = (2\sqrt{I})^2 = 4I \]

### Step 4: Set Up the Equation
Since we know that the maximum intensity at the central fringe is \( I_0 \), we can equate:
\[ 4I = I_0 \]

### Step 5: Solve for Individual Intensity
From the equation \( 4I = I_0 \), we can solve for \( I \):
\[ I = \frac{I_0}{4} \]

### Step 6: Consider the Case When One Slit is Covered
When one of the slits is covered, only one slit contributes to the intensity at the central point. Therefore, the intensity at the same point when one slit is covered will be equal to the intensity of the remaining open slit:
\[ I_{new} = I \]

### Step 7: Substitute the Value of \( I \)
Substituting the value of \( I \) we found:
\[ I_{new} = \frac{I_0}{4} \]

### Conclusion
Thus, the intensity at the central point when one of the slits is covered is:
\[ I_{new} = \frac{I_0}{4} \]