In the Young's double slit experiment using a monochromatic light of wavelength `lamda`, the path difference (in terms of an integer n) corresponding to any point having half the peak

Options

A. `(2n + 1) (lambda)/(2)`
B. `(2n + 1) (lambda)/(4)`
C. `(2n + 1) (lambda)/(8)`
D. `(2n + 1) (lambda)/(16)`

Explanation

To solve the problem of finding the path difference corresponding to a point having half the peak intensity in Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Intensity in Young's Double Slit Experiment**:
The intensity \( I \) at any point on the screen can be expressed as:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( I_0 \) is the peak intensity and \( \phi \) is the phase difference.

2. **Setting Up the Equation for Half Peak Intensity**:
Given that the intensity at the point is half the peak intensity, we have:
\[ I = \frac{I_0}{2} \]
Substituting this into the intensity formula gives:
\[ \frac{I_0}{2} = I_0 \cos^2\left(\frac{\phi}{2}\right) \]

3. **Simplifying the Equation**:
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{2} = \cos^2\left(\frac{\phi}{2}\right) \]

4. **Finding the Cosine Value**:
Taking the square root of both sides:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]
This value corresponds to specific angles, specifically:
\[ \frac{\phi}{2} = \frac{\pi}{4} + n\pi \quad \text{for integers } n \]
Thus:
\[ \phi = \frac{\pi}{2} + 2n\pi \]

5. **Relating Phase Difference to Path Difference**:
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the formula:
\[ \Delta x = \frac{\lambda}{2\pi} \phi \]
Substituting for \( \phi \):
\[ \Delta x = \frac{\lambda}{2\pi} \left(\frac{\pi}{2} + 2n\pi\right) \]

6. **Simplifying the Path Difference**:
This simplifies to:
\[ \Delta x = \frac{\lambda}{2\pi} \cdot \frac{\pi}{2} + \frac{\lambda}{2\pi} \cdot 2n\pi \]
\[ \Delta x = \frac{\lambda}{4} + n\lambda \]
Rearranging gives:
\[ \Delta x = \left(n + \frac{1}{4}\right)\lambda \]

7. **Final Expression**:
To express this in terms of an integer \( m = 2n + 1 \) (where \( n \) is an integer):
\[ \Delta x = \frac{(2n + 1)\lambda}{4} \]

### Conclusion:
The path difference corresponding to any point having half the peak intensity is given by:
\[ \Delta x = \frac{(2n + 1)\lambda}{4} \]