In a fraunhofer's diffraction by a slit, if slit width is a, wave length `lamda` focal length of lens is f, linear width of central maxima is-

Options

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

Answer

To find the linear width of the central maxima in Fraunhofer diffraction by a single slit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a single slit of width \( a \).
- The wavelength of light used is \( \lambda \).
- A lens with focal length \( f \) is placed in front of the slit to observe the diffraction pattern.

2. **Identifying the Minima**:
- The first minima in the diffraction pattern occurs at an angle \( \theta \) where the path difference is equal to the wavelength \( \lambda \).
- The condition for the first minima is given by:
\[ a \sin \theta = \lambda \]

3. **Small Angle Approximation**:
- For small angles, we can use the approximation \( \sin \theta \approx \tan \theta \approx \theta \) (in radians).
- Thus, we can rewrite the equation as:
\[ a \theta = \lambda \quad \Rightarrow \quad \theta = \frac{\lambda}{a} \]

4. **Calculating the Position of the Minima**:
- The distance from the slit to the first minima on either side can be calculated using the focal length \( f \) of the lens.
- The position of the first minima on the screen (distance \( l_1 \)) can be expressed as:
\[ l_1 = f \tan \theta \approx f \theta = f \left(\frac{\lambda}{a}\right) = \frac{f \lambda}{a} \]

5. **Finding the Linear Width of Central Maxima**:
- The linear width of the central maxima is defined as the distance between the first minima on both sides of the central peak.
- Therefore, the total linear width \( W \) is:
\[ W = 2l_1 = 2 \left(\frac{f \lambda}{a}\right) = \frac{2f \lambda}{a} \]

### Final Result:
The linear width of the central maxima is given by:
\[ W = \frac{2f \lambda}{a} \]