At the first minimum adjacent to the central maximum of a single-slit diffraction pattern the phase difference between the Huygens wavelet from the edge of the slit and the wavelet from the mid-point of the slit is

Options

A. 1. `(pi)/(8)` radian
B. 2. `(pi)/(4)` radian
C. 3. `(pi)/(2)` radian
D. 4. `pi` radian

Explanation

To solve the problem, we need to determine the phase difference between the Huygens wavelet from the edge of the slit and the wavelet from the midpoint of the slit at the first minimum adjacent to the central maximum in a single-slit diffraction pattern.

### Step-by-Step Solution:

1. **Understanding the Condition for Minima**:
In a single-slit diffraction pattern, the condition for the first minimum is given by the formula:
\[ d \sin \theta = n \lambda \]
where \( d \) is the width of the slit, \( \theta \) is the angle of diffraction, \( n \) is the order of the minimum (for the first minimum, \( n = 1 \)), and \( \lambda \) is the wavelength of the light used.

2. **Applying the Condition for the First Minimum**:
For the first minimum (where \( n = 1 \)):
\[ d \sin \theta = \lambda \]

3. **Calculating Path Difference**:
The path difference \( \Delta x \) between the wavelet from the midpoint of the slit and the wavelet from the edge of the slit at the angle \( \theta \) can be approximated as:
\[ \Delta x = \frac{d}{2} \sin \theta \]
Here, \( \frac{d}{2} \) is the distance from the midpoint to the edge of the slit.

4. **Substituting the Condition into Path Difference**:
From the first minimum condition:
\[ \Delta x = \frac{d}{2} \sin \theta = \frac{1}{2} \lambda \]

5. **Calculating Phase Difference**:
The phase difference \( \Delta \phi \) corresponding to a path difference \( \Delta x \) is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \Delta x = \frac{1}{2} \lambda \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \left(\frac{1}{2} \lambda\right) = \pi \]

6. **Conclusion**:
Therefore, the phase difference between the Huygens wavelet from the edge of the slit and the wavelet from the midpoint of the slit at the first minimum is:
\[ \Delta \phi = \pi \text{ radians} \]

### Final Answer:
The phase difference is \( \pi \) radians.