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**: In Fraunhofer diffraction, we have a slit of width \( a \) and a wavelength \( \lambda \). A lens of focal length \( f \) is used to observe the diffraction pattern on a screen placed at the focal plane of the lens.

2. **Condition for Minima**: The condition for the minima in the diffraction pattern is given by:
\[ a \sin \theta = n \lambda \]
where \( n \) is the order of the minima (for the first minima, \( n = 1 \)).

3. **Small Angle Approximation**: For small angles, we can use the approximation \( \sin \theta \approx \tan \theta \approx \frac{y}{f} \), where \( y \) is the distance from the center of the central maxima to the position of the minima on the screen.

4. **Position of the First Minima**: For the first minima (\( n = 1 \)):
\[ a \frac{y_1}{f} = \lambda \]
Rearranging gives:
\[ y_1 = \frac{\lambda f}{a} \]

5. **Width of the Central Maxima**: The central maxima extends from the first minima on one side to the first minima on the other side. Therefore, the total width of the central maxima is:
\[ \text{Width} = 2y_1 = 2 \left(\frac{\lambda f}{a}\right) = \frac{2\lambda f}{a} \]

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