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 regarding the path difference in the Young's double slit experiment corresponding to any point having half the peak intensity, 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 in terms of the maximum intensity \( I_0 \) and the phase difference \( \phi \):
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]

2. **Setting Up the Condition for Half Peak Intensity**:
We need to find the condition when the intensity is half of the peak intensity:
\[ I = \frac{I_0}{2} \]
Substituting this into the intensity equation 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, we have:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]

5. **Determining the Phase Difference**:
The angle whose cosine is \( \frac{1}{\sqrt{2}} \) is:
\[ \frac{\phi}{2} = \frac{\pi}{4} \quad \Rightarrow \quad \phi = \frac{\pi}{2} \]

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

7. **Solving for Path Difference**:
Rearranging gives:
\[ \Delta x = \frac{\lambda}{4} \]

8. **Considering the Integer n**:
The general expression for the path difference at points of half peak intensity can be expressed as:
\[ \Delta x = \frac{\lambda}{4} (2n + 1) \]
where \( n \) is an integer.

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