Light from a sodium lamp. `lambda`. = 600 nm, is diffracted by a slit of width d= 0.60 mm. The distance from the slit to the screen is D = 0.60 m. Then, the width of the central maximum is

Options

A. 2.4 mm
B. 1.2 mm
C. 0.6 mm
D. 4.8 mm

Answer

To find the width of the central maximum in the diffraction pattern produced by a single slit, we can use the formula for the angular width of the central maximum. The width of the central maximum (W) can be calculated using the formula:

\[ W = \frac{2 \lambda D}{d} \]

Where:
- \( \lambda \) is the wavelength of 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 of light, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Width of the slit, \( d = 0.60 \, \text{mm} = 0.60 \times 10^{-3} \, \text{m} \)
- Distance from the slit to the screen, \( D = 0.60 \, \text{m} \)

2. **Substitute the values into the formula:**
\[ W = \frac{2 \lambda D}{d} \]
\[ W = \frac{2 \times (600 \times 10^{-9}) \times (0.60)}{0.60 \times 10^{-3}} \]

3. **Simplify the expression:**
- The \( 0.60 \) in the numerator and denominator cancels out:
\[ W = \frac{2 \times (600 \times 10^{-9})}{0.60 \times 10^{-3}} \]
\[ W = \frac{1200 \times 10^{-9}}{0.60 \times 10^{-3}} \]

4. **Convert the denominator:**
- Convert \( 0.60 \, \text{mm} \) to meters:
\[ 0.60 \, \text{mm} = 0.60 \times 10^{-3} \, \text{m} \]

5. **Calculate the width:**
\[ W = \frac{1200 \times 10^{-9}}{0.60 \times 10^{-3}} = 1200 \times 10^{-9} \times \frac{10^{3}}{0.60} \]
\[ W = 1200 \times \frac{10^{-6}}{0.60} \]
\[ W = 1200 \times \frac{10^{-6}}{0.60} = 2000 \times 10^{-6} = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm} \]

6. **Final result:**
The width of the central maximum is \( 2 \, \text{mm} \).