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 in a Young's Double Slit Experiment (YDSE) where the intensity is 75% of the maximum intensity.

### Step-by-Step Solution:

1. **Understand the relationship between intensity and phase difference**:
The intensity \( I \) at any point on the screen in a YDSE can be expressed as:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference.

2. **Set up the equation for 75% intensity**:
Given that the intensity is 75% of the maximum intensity, we can write:
\[ \frac{3}{4} I_{\text{max}} = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
Dividing both sides by \( I_{\text{max}} \) gives:
\[ \cos^2\left(\frac{\phi}{2}\right) = \frac{3}{4} \]

3. **Solve for phase difference**:
Taking the square root of both sides, we have:
\[ \cos\left(\frac{\phi}{2}\right) = \pm \frac{\sqrt{3}}{2} \]
The angles corresponding to this cosine value are:
\[ \frac{\phi}{2} = 30^\circ \quad \text{or} \quad \frac{\phi}{2} = 150^\circ \]
Therefore, the phase difference \( \phi \) can be:
\[ \phi = 60^\circ \quad \text{or} \quad \phi = 300^\circ \]

4. **Convert phase difference to radians**:
Converting degrees to radians:
\[ \phi = \frac{\pi}{3} \quad \text{or} \quad \phi = \frac{5\pi}{3} \]

5. **Relate phase difference to path difference**:
The path difference \( \Delta x \) corresponding to the phase difference \( \phi \) is given by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Rearranging gives:
\[ \Delta x = \frac{\phi \lambda}{2\pi} \]

6. **Calculate the distance on the screen**:
The distance \( y \) on the screen can also be related to the path difference:
\[ y = \frac{D \Delta x}{d} \]
where \( D \) is the distance from the slits to the screen, \( d \) is the distance between the slits, and \( \lambda \) is the wavelength.

7. **Substituting values**:
Given:
- \( d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- \( \lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} \)
- \( D = 1 \text{ m} \)

For \( \phi = \frac{\pi}{3} \):
\[ \Delta x = \frac{\frac{\pi}{3} \cdot 6000 \times 10^{-10}}{2\pi} = \frac{6000 \times 10^{-10}}{6} = 1000 \times 10^{-10} \text{ m} \]

Now substituting \( \Delta x \) into the equation for \( y \):
\[ y = \frac{1 \cdot 1000 \times 10^{-10}}{1 \times 10^{-3}} = 1000 \times 10^{-7} \text{ m} = 1 \times 10^{-4} \text{ m} \]

8. **Finding the minimum distance between two points**:
The minimum distance between two points on the screen corresponding to the two phase differences is:
\[ \Delta y = 2y = 2 \times 1 \times 10^{-4} \text{ m} = 2 \times 10^{-4} \text{ m} = 0.2 \text{ mm} \]

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