A screen is at a distance of 2m from a narrow slit illuminated with light of 600nm. The first minimum lies 5mm on either side of the central maximum. The width of slit is

Options

A. 0.024 mm
B. 0.24mm
C. 2.4 mm
D. 24mm

Answer

To solve the problem, we need to find the width of the slit using the given information about the distance to the screen, the wavelength of light, and the position of the first minimum.

### Step-by-step Solution:

1. **Identify the Given Values**:
- Distance from the slit to the screen (D) = 2 m = 2000 mm (since we will convert everything to mm for consistency).
- Wavelength of light (λ) = 600 nm = 0.6 mm (conversion: 1 nm = 10^-6 mm).
- Distance from the central maximum to the first minimum (β) = 5 mm.

2. **Understand the Formula**:
The position of the first minimum in a single-slit diffraction pattern is given by the formula:
\[ \beta = \frac{n \cdot \lambda \cdot D}{d} \]
where:
- \(n\) = order of the minimum (for the first minimum, \(n = 1\)),
- \(λ\) = wavelength of the light,
- \(D\) = distance from the slit to the screen,
- \(d\) = width of the slit.

3. **Rearranging the Formula**:
We need to rearrange the formula to solve for the width of the slit (d):
\[ d = \frac{n \cdot \lambda \cdot D}{\beta} \]

4. **Substituting the Known Values**:
Substitute \(n = 1\), \(λ = 0.6 \, \text{mm}\), \(D = 2000 \, \text{mm}\), and \(β = 5 \, \text{mm}\) into the equation:
\[ d = \frac{1 \cdot 0.6 \cdot 2000}{5} \]

5. **Calculating the Width of the Slit**:
\[ d = \frac{1200}{5} = 240 \, \text{mm} \]

6. **Converting to Appropriate Units**:
Since the options are in mm, we can express the answer in mm:
\[ d = 0.24 \, \text{mm} \]

7. **Final Answer**:
The width of the slit is \(0.24 \, \text{mm}\).

### Conclusion:
The correct answer is option (b) \(0.24 \, \text{mm}\).