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 find the angular width of the central maximum between two minima in a single-slit diffraction pattern, we can follow these steps:

### 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, \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \)

2. **Understanding the Angular Width:**
- The angular width \( \theta \) of the central maximum can be approximated using the formula:
\[ \theta \approx \frac{2\lambda}{d} \]
- This formula is valid for small angles where \( \sin \theta \approx \theta \).

3. **Substituting the Values:**
- Plugging in the values of \( \lambda \) and \( d \):
\[ \theta \approx \frac{2 \times 6328 \times 10^{-10} \, \text{m}}{0.1 \times 10^{-3} \, \text{m}} \]

4. **Calculating \( \theta \):**
- Calculate the numerator:
\[ 2 \times 6328 \times 10^{-10} = 12656 \times 10^{-10} \, \text{m} \]
- Now calculate \( \theta \):
\[ \theta \approx \frac{12656 \times 10^{-10}}{0.1 \times 10^{-3}} = \frac{12656 \times 10^{-10}}{1 \times 10^{-4}} = 126.56 \times 10^{-6} \, \text{radians} \]

5. **Convert Radians to Degrees:**
- To convert radians to degrees, use the conversion factor \( \frac{180}{\pi} \):
\[ \theta \text{ (in degrees)} = 126.56 \times 10^{-6} \times \frac{180}{\pi} \]
- Calculating this gives:
\[ \theta \approx 126.56 \times 10^{-6} \times 57.2958 \approx 0.0073 \, \text{degrees} \]

6. **Final Result:**
- The angular width of the central maximum is approximately \( 0.0073 \, \text{degrees} \).