In case of diffraction, wavelength X is used to illuminate the slit of width d and the pattern is viewed on screen placed at a distance D meter away, the width of central maxima is :

Options

A. `(lambdaD)/d`
B. `(lambdaD)/(2d)`
C. 0
D. `(2lambdaD)/d`

Answer

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

### Step-by-Step Solution:

1. **Understand the Condition for Minima**:
The condition for the formation of minima in a single-slit diffraction pattern is given by the equation:
\[ d \sin \theta = m \lambda \]
where \(d\) is the width of the slit, \(\lambda\) is the wavelength of the light used, \(m\) is the order of the minima (with \(m = 1\) for the first minima), and \(\theta\) is the angle of diffraction.

2. **Apply the First Minima Condition**:
For the first minima, we set \(m = 1\):
\[ d \sin \theta = \lambda \]
Rearranging gives:
\[ \sin \theta = \frac{\lambda}{d} \]

3. **Use the Small Angle Approximation**:
For small angles, we can use the approximation \(\sin \theta \approx \theta\) (in radians):
\[ \theta \approx \frac{\lambda}{d} \]

4. **Calculate the Width of Central Maxima**:
The width of the central maxima is defined as the distance between the first minima on either side of the central peak. The distance between the two first minima is \(2y\), where \(y\) is the distance from the center to the first minima. The relationship between \(y\), \(\theta\), and the distance \(D\) from the slit to the screen is given by:
\[ y = D \tan \theta \approx D \theta \]
Thus, substituting \(\theta\):
\[ y \approx D \left(\frac{\lambda}{d}\right) \]
Therefore, the total width of the central maxima (distance between the two first minima) is:
\[ \text{Width} = 2y = 2D \left(\frac{\lambda}{d}\right) = \frac{2\lambda D}{d} \]

### Final Answer:
The width of the central maxima is:
\[ \text{Width of Central Maxima} = \frac{2\lambda D}{d} \]