In the diffractin pattern of a single slit aperture, the width of the central fringe compared to widths of te other fringes is-

Options

A. equal
B. less
C. little more
D. double

Answer

To solve the problem regarding the width of the central fringe compared to the widths of the other fringes in the diffraction pattern of a single slit aperture, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Diffraction Pattern**:
- In a single slit diffraction pattern, the central maximum is the brightest and widest fringe, flanked by minima on either side.

2. **Condition for Minima**:
- The condition for the minima in single slit diffraction is given by:
\[ a \sin \theta = n \lambda \]
where \( a \) is the width of the slit, \( \lambda \) is the wavelength of light, and \( n \) is the order of the minima (n = 1, 2, 3,...).

3. **Small Angle Approximation**:
- For small angles, we can use the approximation \( \sin \theta \approx \tan \theta \approx \frac{y}{D} \), where \( y \) is the distance from the central maximum to the minima on the screen, and \( D \) is the distance from the slit to the screen.

4. **Finding the Position of the First Minimum**:
- For the first minimum (n = 1), we can write:
\[ a \frac{y_1}{D} = \lambda \]
Rearranging gives:
\[ y_1 = \frac{\lambda D}{a} \]

5. **Width of the Central Maximum**:
- The width of the central maximum is the distance between the first minima on either side. Thus, the total width of the central maximum is:
\[ W_c = 2y_1 = 2 \left(\frac{\lambda D}{a}\right) = \frac{2\lambda D}{a} \]

6. **Width of Other Fringes**:
- The width of the other fringes (for example, the first order fringe) can be found similarly. The distance to the second minima (n = 2) would be:
\[ y_2 = \frac{2\lambda D}{a} \]
- The width of the first order fringe (between the first and second minima) is:
\[ W_o = y_2 - y_1 = \frac{2\lambda D}{a} - \frac{\lambda D}{a} = \frac{\lambda D}{a} \]

7. **Comparing Widths**:
- The width of the central maximum \( W_c = \frac{2\lambda D}{a} \) is twice the width of the other fringes \( W_o = \frac{\lambda D}{a} \).

8. **Conclusion**:
- Therefore, the width of the central fringe is **double** the width of the other fringes.

### Final Answer:
The width of the central fringe compared to the widths of the other fringes is **double**.