Light of wavelength 5000Å is incident normally on a slit. The first minimumof the diffraction pattern is formed at adistance of 5 mm from centra maximum. The screen is situated at a distance of 2 m from the slit. The slit width is

Options

A. `0.2 mm`
B. `0.8` mm
C. `0.4 mm`
D. `2.0 mm`

Explanation

To solve the problem, we need to use the formula for the position of the minima in a single-slit diffraction pattern. The position of the first minimum (y) is given by the formula:

\[ y = \frac{\lambda L}{a} \]

where:
- \( y \) is the distance from the central maximum to the first minimum,
- \( \lambda \) is the wavelength of the light,
- \( L \) is the distance from the slit to the screen, and
- \( a \) is the width of the slit.

### Given Data:
- Wavelength \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)
- Distance to the screen \( L = 2 \, \text{m} \)
- Distance to the first minimum \( y = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)

### Step-by-Step Solution:

1. **Write down the formula for the first minimum:**
\[ y = \frac{\lambda L}{a} \]

2. **Rearrange the formula to solve for the slit width \( a \):**
\[ a = \frac{\lambda L}{y} \]

3. **Substitute the known values into the equation:**
\[ a = \frac{(5 \times 10^{-7} \, \text{m}) \cdot (2 \, \text{m})}{5 \times 10^{-3} \, \text{m}} \]

4. **Calculate the numerator:**
\[ 5 \times 10^{-7} \, \text{m} \cdot 2 \, \text{m} = 10 \times 10^{-7} \, \text{m} = 1 \times 10^{-6} \, \text{m} \]

5. **Now divide by \( y \):**
\[ a = \frac{1 \times 10^{-6} \, \text{m}}{5 \times 10^{-3} \, \text{m}} = \frac{1}{5} \times 10^{-3} \, \text{m} = 0.2 \times 10^{-3} \, \text{m} = 0.2 \, \text{mm} \]

6. **Final answer:**
\[ a = 0.2 \, \text{mm} \]

### Conclusion:
The slit width is \( 0.2 \, \text{mm} \).