In Young’s double-slit experiment, the distance between the two identical slits is `6.1` times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is

Options

A. 3
B. 6
C. 24
D. 12

Answer

To solve the problem, we need to determine the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern in Young's double-slit experiment, given that the distance between the two slits is 6.1 times larger than the slit width.

### Step-by-Step Solution:

1. **Identify Variables**:
- Let \( w \) be the slit width.
- The distance between the two slits is \( d = 6.1w \).

2. **Determine the Angular Width of the Central Maximum**:
- In a single-slit diffraction pattern, the angular position of the first minima on either side of the central maximum is given by:
\[ w \sin \theta = \lambda \quad \text{(for first minima)} \]
- For small angles, \( \sin \theta \approx \theta \), so:
\[ \theta \approx \frac{\lambda}{w} \]
- The total angular width of the central maximum is from \(-\theta\) to \(+\theta\), which gives:
\[ \text{Total angular width} = 2\theta = 2 \frac{\lambda}{w} \]

3. **Determine the Angular Width of Each Fringe in the Double-Slit Experiment**:
- The angular width of each fringe in the double-slit experiment is given by:
\[ \text{Angular width of one fringe} = \frac{\lambda}{d} \]
- Substituting \( d = 6.1w \):
\[ \text{Angular width of one fringe} = \frac{\lambda}{6.1w} \]

4. **Calculate the Number of Maxima within the Central Maximum**:
- The number of maxima that fit within the angular width of the central maximum can be calculated by dividing the total angular width of the central maximum by the angular width of one fringe:
\[ \text{Number of maxima} = \frac{\text{Total angular width}}{\text{Angular width of one fringe}} = \frac{2 \frac{\lambda}{w}}{\frac{\lambda}{6.1w}} \]
- Simplifying this expression:
\[ \text{Number of maxima} = \frac{2 \frac{\lambda}{w}}{\frac{\lambda}{6.1w}} = 2 \times 6.1 = 12.2 \]

5. **Round to the Nearest Integer**:
- Since the number of maxima must be a whole number, we round \( 12.2 \) to the nearest integer, which is \( 12 \).

### Final Answer:
The number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is **12**.