Visible light of wavelength 500 nm falls normally on a single slit and produces a diffraction pattern. It is found that the diffraction pattern is on a screen 1 m away from slit. If the first minimum is produced at a distance of 2.5 mm from the centre of screen, then the width of the slit is

Options

A. `0.1 mm`
B. `0.4 mm`
C. `0.3 mm`
D. `0.2 mm`

Explanation

To solve the problem, we will use the formula for the position of the first minimum in a single-slit diffraction pattern. The formula is given by:

\[ y = \frac{D \cdot \lambda}{d} \]

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

### Step-by-step Solution:

1. **Identify the given values:**
- Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- Distance from the slit to the screen, \( D = 1 \, \text{m} \)
- Distance to the first minimum, \( y = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \)

2. **Rearrange the formula to solve for \( d \):**
\[ d = \frac{D \cdot \lambda}{y} \]

3. **Substitute the known values into the equation:**
\[ d = \frac{1 \, \text{m} \cdot 500 \times 10^{-9} \, \text{m}}{2.5 \times 10^{-3} \, \text{m}} \]

4. **Calculate \( d \):**
- First, calculate the numerator:
\[ 1 \cdot 500 \times 10^{-9} = 500 \times 10^{-9} \, \text{m} \]
- Now divide by \( 2.5 \times 10^{-3} \):
\[ d = \frac{500 \times 10^{-9}}{2.5 \times 10^{-3}} = \frac{500}{2.5} \times 10^{-6} = 200 \times 10^{-6} \, \text{m} \]
- Convert to mm:
\[ d = 0.2 \, \text{mm} \]

5. **Final Result:**
The width of the slit \( d \) is \( 0.2 \, \text{mm} \).

### Conclusion:
The correct answer is \( d = 0.2 \, \text{mm} \).