Consider Fraunhoffer diffraction pattern obtained with a single slit illuminated at normal incidence. AT the angular position of the first diffraction minimum the phase difference (in radian) between the wavelengths from the opposite edges of the slit is

Options

A. `pi"/"4`
B. `pi"/"2`
C. `pi`
D. `2pi`

Answer

To find the phase difference between the wavelengths from the opposite edges of a single slit at the angular position of the first diffraction minimum, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a single slit of width \( a \) illuminated by a monochromatic light source at normal incidence.
- The diffraction pattern is observed on a screen at a distance.

2. **Condition for First Minimum**:
- The condition for the first diffraction minimum in a single slit diffraction pattern occurs at an angle \( \theta \) given by:
\[ a \sin \theta = \lambda \]
- Here, \( \lambda \) is the wavelength of the light used.

3. **Path Difference**:
- The path difference \( \Delta x \) between the light waves coming from the edges of the slit at the angle \( \theta \) is equal to \( \lambda \) at the first minimum:
\[ \Delta x = \lambda \]

4. **Phase Difference Calculation**:
- The phase difference \( \phi \) corresponding to a path difference \( \Delta x \) is given by the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
- Substituting \( \Delta x = \lambda \) into the equation:
\[ \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]

5. **Conclusion**:
- Therefore, the phase difference between the wavelengths from the opposite edges of the slit at the angular position of the first diffraction minimum is:
\[ \phi = 2\pi \text{ radians} \]

### Final Answer:
The phase difference at the angular position of the first diffraction minimum is \( 2\pi \) radians.

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