Monochromatic light of wavelength 580 nm is incident on a slit of width 0.30 mm. The screen is 2m from the slit . The width of the central maximum is

Options

A. `3.35 times 10^-3 m`
B. `2.25 times 10^-3 m`
C. `6.20 times 10^-3 m`
D. `7.7 times 10^-3 m`

Answer

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

\[ \text{Width of central maximum} = \frac{2 \lambda D}{d} \]

Where:
- \(\lambda\) is the wavelength of the light,
- \(D\) is the distance from the slit to the screen,
- \(d\) is the width of the slit.

### Step-by-step Solution:

1. **Identify the given values**:
- Wavelength, \(\lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m}\)
- Distance from the slit to the screen, \(D = 2 \text{ m}\)
- Width of the slit, \(d = 0.30 \text{ mm} = 0.30 \times 10^{-3} \text{ m}\)

2. **Substitute the values into the formula**:
\[ \text{Width of central maximum} = \frac{2 \times (580 \times 10^{-9} \text{ m}) \times (2 \text{ m})}{0.30 \times 10^{-3} \text{ m}} \]

3. **Calculate the numerator**:
\[ 2 \times (580 \times 10^{-9}) \times 2 = 2320 \times 10^{-9} \text{ m} \]

4. **Calculate the denominator**:
\[ 0.30 \times 10^{-3} = 0.00030 \text{ m} \]

5. **Now, divide the numerator by the denominator**:
\[ \text{Width of central maximum} = \frac{2320 \times 10^{-9}}{0.00030} = \frac{2320 \times 10^{-9}}{3 \times 10^{-4}} = \frac{2320}{3} \times 10^{-5} \text{ m} \]

6. **Perform the division**:
\[ \frac{2320}{3} \approx 773.33 \text{ (approximately)} \]
So,
\[ \text{Width of central maximum} \approx 7.73 \times 10^{-3} \text{ m} \text{ or } 7.73 \text{ mm} \]

7. **Final result**:
The width of the central maximum is approximately \(7.73 \text{ mm}\).