The angular width of the central maximum of the diffraction patternn in a single slit (of width a) experiment, with `lamda` as the wavelenth of light, is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a single slit of width \( a \) through which light of wavelength \( \lambda \) passes, creating a diffraction pattern on a screen.

2. **Identifying the Minima**:
- In a single slit diffraction pattern, the first minima occur at angles where the path difference between light from the edges of the slit is equal to one wavelength. The condition for minima is given by:
\[ a \sin \theta = n \lambda \]
- Here, \( n \) is the order of the minima (for the first minima, \( n = 1 \)).

3. **Finding the Angle for the First Minimum**:
- For the first minimum (\( n = 1 \)):
\[ a \sin \theta = \lambda \]
- Rearranging gives:
\[ \sin \theta = \frac{\lambda}{a} \]

4. **Using Small Angle Approximation**:
- For small angles, \( \sin \theta \approx \tan \theta \approx \theta \) (in radians). Thus, we can approximate:
\[ \theta \approx \frac{\lambda}{a} \]

5. **Calculating the Angular Width of the 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 \( W \) is:
\[ W = 2\theta \]
- Substituting the value of \( \theta \):
\[ W = 2 \left(\frac{\lambda}{a}\right) = \frac{2\lambda}{a} \]

6. **Conclusion**:
- The angular width of the central maximum in the diffraction pattern is:
\[ W = \frac{2\lambda}{a} \]

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

---