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 a single-slit diffraction pattern, we can use the formula:

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

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

### Step-by-step solution:

1. **Identify the given values**:
- Wavelength (\(\lambda\)) = 600 nm = \(600 \times 10^{-9}\) m
- Distance from the slit to the screen (\(D\)) = 0.60 m
- Width of the slit (\(d\)) = 0.60 mm = \(0.60 \times 10^{-3}\) m

2. **Substitute the values into the formula**:
\[ \text{Width of central maximum} = \frac{2 \times (600 \times 10^{-9}) \times 0.60}{0.60 \times 10^{-3}} \]

3. **Calculate the numerator**:
\[ 2 \times (600 \times 10^{-9}) \times 0.60 = 720 \times 10^{-9} \text{ m} \]

4. **Calculate the denominator**:
\[ 0.60 \times 10^{-3} = 0.00060 \text{ m} \]

5. **Divide the numerator by the denominator**:
\[ \text{Width of central maximum} = \frac{720 \times 10^{-9}}{0.00060} \]

6. **Perform the division**:
\[ \text{Width of central maximum} = \frac{720 \times 10^{-9}}{6 \times 10^{-4}} = 1.2 \times 10^{-3} \text{ m} \]

7. **Convert to mm**:
\[ 1.2 \times 10^{-3} \text{ m} = 1.2 \text{ mm} \]

### Final Answer:
The width of the central maximum is **1.2 mm**.