Light of wavelength `6328 Å` is incident normally on a slit having a width of `0.2 mm`. The distance of the screen from the slit is `0.9 m`. The angular width of the central maximum is

Options

A. 2.4 mm
B. 1.2 mm
C. 0.36 mm
D. 4.8 mm

Explanation

To find the angular width of the central maximum in a single-slit diffraction pattern, we can use the formula for the angular width of the central maximum, which is given by:

\[ \theta = \frac{2\lambda}{a} \]

where:
- \(\theta\) is the angular width of the central maximum,
- \(\lambda\) is the wavelength of the light,
- \(a\) is the width of the slit.

### Step-by-step Solution:

1. **Identify the given values:**
- Wavelength, \(\lambda = 6328 \, \text{Å} = 6328 \times 10^{-10} \, \text{m}\)
- Width of the slit, \(a = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m}\)

2. **Substitute the values into the formula:**
\[ \theta = \frac{2 \times 6328 \times 10^{-10}}{0.2 \times 10^{-3}} \]

3. **Calculate the numerator:**
\[ 2 \times 6328 \times 10^{-10} = 12656 \times 10^{-10} \, \text{m} \]

4. **Calculate the denominator:**
\[ 0.2 \times 10^{-3} = 2 \times 10^{-4} \, \text{m} \]

5. **Now divide the numerator by the denominator:**
\[ \theta = \frac{12656 \times 10^{-10}}{2 \times 10^{-4}} = \frac{12656}{2} \times 10^{-6} = 6328 \times 10^{-6} \, \text{radians} \]

6. **Convert the angle from radians to degrees:**
\[ \theta \text{ (in degrees)} = 6328 \times 10^{-6} \times \frac{180}{\pi} \]
\[ \theta \approx 6328 \times 10^{-6} \times 57.2958 \approx 0.363 \, \text{degrees} \]

7. **Final result:**
The angular width of the central maximum is approximately \(0.363 \, \text{degrees}\).