In single slit diffraction experiment, the width of the central maximum inversely proportional to

Options

A. Distance between source and screen
B. Slit width
C. Wavelength of light used
D. Both (1) & (3)

Answer

To solve the question regarding the relationship between the width of the central maximum in a single slit diffraction experiment and the slit width, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
In a single slit diffraction experiment, light passes through a slit of width \( a \) and creates a diffraction pattern on a screen placed at a distance \( D \).

2. **Identifying the Central Maximum**:
The central maximum is the brightest part of the diffraction pattern, located in the center. The width of the central maximum is determined by the positions of the first-order minima on either side of the central maximum.

3. **Finding the Position of Minima**:
The condition for the first-order minima in single slit diffraction is given by:
\[ a \sin \theta = m \lambda \]
where \( m \) is the order of the minima (for first-order, \( m = 1 \)), \( \lambda \) is the wavelength of the light, and \( \theta \) is the angle from the central maximum to the first-order minima.

4. **Calculating Angular Width**:
For small angles (where \( \theta \) is small), we can use the approximation \( \sin \theta \approx \tan \theta \approx \theta \) (in radians). Thus, the position of the first-order minima can be approximated as:
\[ \theta \approx \frac{\lambda}{a} \]
Therefore, the angular width \( 2\theta \) between the first-order minima is:
\[ 2\theta \approx \frac{2\lambda}{a} \]

5. **Finding Linear Width**:
The linear width \( W \) of the central maximum on the screen can be calculated using the formula:
\[ W = 2D \tan \theta \approx 2D \theta \]
Substituting the expression for \( \theta \):
\[ W \approx 2D \left(\frac{\lambda}{a}\right) = \frac{2D\lambda}{a} \]

6. **Concluding the Relationship**:
From the equation \( W \approx \frac{2D\lambda}{a} \), we can see that the width of the central maximum \( W \) is inversely proportional to the slit width \( a \). Therefore, as the slit width \( a \) increases, the width of the central maximum \( W \) decreases.

### Final Answer:
The width of the central maximum in a single slit diffraction experiment is inversely proportional to the width of the slit \( a \).

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