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.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( a \) be the width of each slit.
- Let \( d \) be the distance between the two slits.
- According to the problem, \( d = 6.1a \).

2. **Width of the Central Maximum**:
- The width of the central maximum of a single-slit diffraction pattern is given by the formula:
\[ \text{Width of central maximum} = \frac{2\lambda D}{a} \]
where \( \lambda \) is the wavelength of light and \( D \) is the distance from the slits to the screen.

3. **Width of the Central Maximum in Double-Slit Experiment**:
- The width of the central maximum in the double-slit experiment can also be expressed in terms of the distance between the slits:
\[ \text{Width of central maximum} = \frac{2\lambda D}{d} = \frac{2\lambda D}{6.1a} \]

4. **Setting the Widths Equal**:
- To find the number of maxima (\( n \)) within the central maximum of the single-slit diffraction pattern, we equate the widths:
\[ n \cdot \frac{2\lambda D}{6.1a} = \frac{2\lambda D}{a} \]

5. **Simplifying the Equation**:
- We can cancel \( 2\lambda D \) from both sides:
\[ n \cdot \frac{1}{6.1} = 1 \]
- Thus, we find:
\[ n = 6.1 \]

6. **Rounding Down**:
- Since \( n \) must be an integer (as it represents the number of maxima), we round down \( 6.1 \) to get:
\[ n = 6 \]

### Conclusion:
The number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is **6**.