In a YDSE experiment, d = 1mm, `lambda`= 6000Å and D= 1m. The minimum distance between two points on screen having 75% intensity of the maximum intensity will be

Options

A. 0.50 mm
B. 0.40 mm
C. 0.30 mm
D. 0.20 mm

Explanation

To solve the problem, we need to find the minimum distance between two points on the screen that have 75% of the maximum intensity in a Young's Double Slit Experiment (YDSE). Here are the steps to derive the solution:

### Step-by-Step Solution:

1. **Understanding Maximum Intensity**:
The maximum intensity \( I_{\text{max}} \) in a YDSE setup is given by:
\[ I_{\text{max}} = 4I_0 \]
where \( I_0 \) is the intensity from each slit.

2. **Calculating 75% of Maximum Intensity**:
We need to find the intensity at points where the intensity is 75% of the maximum intensity:
\[ I = 0.75 I_{\text{max}} = 0.75 \times 4I_0 = 3I_0 \]

3. **Intensity Formula**:
The intensity at a point on the screen can be expressed as:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
Since \( I_1 = I_2 = I_0 \):
\[ I = 2I_0 + 2I_0 \cos \phi = 2I_0(1 + \cos \phi) \]

4. **Setting Up the Equation**:
Setting \( I = 3I_0 \):
\[ 2I_0(1 + \cos \phi) = 3I_0 \]
Dividing through by \( 2I_0 \):
\[ 1 + \cos \phi = \frac{3}{2} \]
Thus:
\[ \cos \phi = \frac{1}{2} \]

5. **Finding Phase Difference**:
The angle corresponding to \( \cos \phi = \frac{1}{2} \) is:
\[ \phi = \frac{\pi}{3} \text{ (or 60 degrees)} \]

6. **Relating Phase Difference to Path Difference**:
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Rearranging gives:
\[ \Delta x = \frac{\phi \lambda}{2\pi} \]

7. **Substituting Values**:
Given \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \):
\[ \Delta x = \frac{\frac{\pi}{3} \cdot 6000 \times 10^{-10}}{2\pi} = \frac{6000 \times 10^{-10}}{6} = 1000 \times 10^{-10} = 10^{-7} \, \text{m} \]

8. **Finding Minimum Distance on Screen**:
The distance \( y \) on the screen corresponding to this path difference is given by:
\[ y = \frac{\Delta x \cdot D}{d} \]
where \( D = 1 \, \text{m} \) and \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \):
\[ y = \frac{10^{-7} \cdot 1}{1 \times 10^{-3}} = 10^{-4} \, \text{m} = 0.1 \, \text{mm} \]

9. **Calculating Minimum Distance Between Two Points**:
Since we need the distance between two points having this intensity, we multiply by 2:
\[ \text{Minimum distance} = 2y = 2 \times 0.1 \, \text{mm} = 0.2 \, \text{mm} \]

### Final Answer:
The minimum distance between two points on the screen having 75% intensity of the maximum intensity is **0.2 mm**.