For a parallel beam of monochromatic light of wavelength `'lambda'` diffraction is produced by a single slit whose width `'a'` is of the order of the wavelength of the light. If `'D'` is the distance of the screen from the slit, the width of the central maxima will be

Options

A. `(2Dlambda)/a`
B. `(Dlambda)/a`
C. `(Da)/lambda`
D. `(2Da)/lambda`

Explanation

To find the width of the central maxima in a single-slit diffraction pattern produced by a monochromatic light beam, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a single slit of width \( a \) and a parallel beam of monochromatic light of wavelength \( \lambda \). The screen is placed at a distance \( D \) from the slit.

2. **Identifying the Angle for First Minimum**:
The first minimum in the diffraction pattern occurs at an angle \( \theta \) given by the formula:
\[ a \sin \theta = m \lambda \]
where \( m = 1 \) for the first minimum. Thus, we can write:
\[ a \sin \theta_1 = \lambda \]
For small angles, \( \sin \theta \approx \tan \theta \approx \theta \) (in radians), so we can approximate:
\[ \theta_1 \approx \frac{\lambda}{a} \]

3. **Calculating the Position of the First Minimum**:
The position \( y_1 \) of the first minimum on the screen can be related to the angle \( \theta_1 \) and the distance \( D \) from the slit to the screen:
\[ y_1 = D \tan \theta_1 \approx D \theta_1 \approx D \frac{\lambda}{a} \]

4. **Finding the Width of the Central Maxima**:
The width of the central maxima \( \Delta y \) is defined as the distance between the first minima on either side of the central peak. Since the first minima are located at \( y_1 \) and \( -y_1 \), the width of the central maxima is:
\[ \Delta y = 2y_1 = 2D \frac{\lambda}{a} \]

5. **Final Result**:
Therefore, the width of the central maxima is given by:
\[ \Delta y = \frac{2\lambda D}{a} \]

### Conclusion:
The width of the central maxima in the diffraction pattern produced by a single slit is:
\[ \Delta y = \frac{2\lambda D}{a} \]