A monochromatic light of wavelength `6000 dot A` is incident on the single slit of width `0.01` mm. If the diffraction pattern is formed at the focus of the convex lens of focal length 20 cm, the linear width of the central maximum is :

Options

A. 60 mm
B. 24 mm
C. 120 mm
D. 12 mm

Answer

To solve the problem of finding the linear width of the central maximum in a single-slit diffraction pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Wavelength of light, \( \lambda = 6000 \, \text{Å} \)
- Width of the slit, \( d = 0.01 \, \text{mm} \)
- Focal length of the lens, \( f = 20 \, \text{cm} \)

2. **Convert Units**:
- Convert the wavelength from Angstroms to meters:
\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]
- Convert the slit width from mm to meters:
\[ d = 0.01 \, \text{mm} = 0.01 \times 10^{-3} \, \text{m} = 1 \times 10^{-5} \, \text{m} \]
- Convert the focal length from cm to meters:
\[ f = 20 \, \text{cm} = 20 \times 10^{-2} \, \text{m} = 0.2 \, \text{m} \]

3. **Use the Formula for Linear Width of Central Maximum**:
The formula for the linear width of the central maximum in a single-slit diffraction pattern is given by:
\[ W = \frac{2f\lambda}{d} \]

4. **Substitute the Values into the Formula**:
\[ W = \frac{2 \times 0.2 \, \text{m} \times 6 \times 10^{-7} \, \text{m}}{1 \times 10^{-5} \, \text{m}} \]

5. **Calculate the Width**:
- Calculate the numerator:
\[ 2 \times 0.2 \times 6 = 2.4 \]
- Now substitute back into the equation:
\[ W = \frac{2.4 \times 10^{-7}}{1 \times 10^{-5}} = 2.4 \times 10^{-2} \, \text{m} \]
- Convert to mm:
\[ W = 2.4 \times 10^{-2} \, \text{m} = 24 \, \text{mm} \]

6. **Final Answer**:
The linear width of the central maximum is \( \boxed{24 \, \text{mm}} \).