The light of wavelength 6328 `overset(o)(A)` is incident on a slit of width 0.1 mm perpendicularly situated at a distance of 9 m and the central maxima between two minima, the angular width is approximately:

Options

A. `0.36^@`
B. `0.18^@`
C. `0.72^@`
D. `0.08^@`

Answer

To solve the problem, we need to find the angular width of the central maximum between two minima in a single-slit diffraction pattern. The steps are as follows:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength of light, \( \lambda = 6328 \, \text{Å} = 6328 \times 10^{-10} \, \text{m} \)
- Width of the slit, \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \)
- Distance to the screen (not directly needed for angular width calculation), \( L = 9 \, \text{m} \)

2. **Use the Formula for Angular Width:**
The angular width of the central maximum is given by the formula:
\[ \sin \theta = \frac{\lambda}{d} \]
For small angles, we can approximate \( \sin \theta \approx \theta \) (in radians).

3. **Calculate \( \theta \):**
Substitute the values into the formula:
\[ \theta = \frac{\lambda}{d} = \frac{6328 \times 10^{-10}}{0.1 \times 10^{-3}} \]
Simplifying this gives:
\[ \theta = \frac{6328 \times 10^{-10}}{0.1 \times 10^{-3}} = 6.328 \times 10^{-3} \, \text{radians} \]

4. **Convert Radians to Degrees:**
To convert radians to degrees, use the conversion factor \( \frac{180}{\pi} \):
\[ \theta_{\text{degrees}} = 6.328 \times 10^{-3} \times \frac{180}{\pi} \approx 0.363 \, \text{degrees} \]

5. **Calculate Total Angular Width:**
The total angular width of the central maximum is twice the angle \( \theta \):
\[ \text{Total Angular Width} = 2 \times \theta_{\text{degrees}} = 2 \times 0.363 \approx 0.726 \, \text{degrees} \]

6. **Final Result:**
Rounding to two decimal places, the angular width is approximately:
\[ \text{Angular Width} \approx 0.72 \, \text{degrees} \]

### Conclusion:
The angular width of the central maximum between the two minima is approximately \( 0.72 \, \text{degrees} \).