Angular width of central maximum in diffraction at a single slit is……………. .

Options

A. `lamda/a`
B. `(2lamda)/a`
C. `(2a)/lamda`
D. `(2a)/(3lamda)`

Answer

To find the angular width of the central maximum in diffraction at a single slit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Single Slit Diffraction**:
In a single slit diffraction experiment, light passing through a narrow slit produces a pattern of light and dark fringes on a screen. The central maximum is the brightest fringe, and it is flanked by dark fringes (minima).

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

3. **Small Angle Approximation**:
For small angles, we can use the approximation:
\[ \sin \theta \approx \theta \quad (\text{in radians}) \]
Thus, the condition for the first minima (n = 1) becomes:
\[ a \theta = \lambda \quad \Rightarrow \quad \theta = \frac{\lambda}{a} \]

4. **Identifying the Angular Width of Central Maximum**:
The angular width of the central maximum is defined as the angle between the first minima on either side of the central maximum. Therefore, the total angular width \( \theta_0 \) is:
\[ \theta_0 = 2\theta \]

5. **Substituting for \( \theta \)**:
Substituting the value of \( \theta \):
\[ \theta_0 = 2 \left(\frac{\lambda}{a}\right) = \frac{2\lambda}{a} \]

6. **Conclusion**:
Thus, the angular width of the central maximum in diffraction at a single slit is:
\[ \theta_0 = \frac{2\lambda}{a} \]

### Final Answer:
The angular width of the central maximum in diffraction at a single slit is \( \frac{2\lambda}{a} \).