Question

In a Fraunhofer's diffraction by a slit, if slit width is a, wavelength $\lambda$, focal length of lens is f, linear width of central maxima is :

• A. $\frac{f\lambda}{a}$
• B. $\frac{fa}{\lambda}$
• C. $\frac{2f\lambda}{a}$
• D. $\frac{f\lambda}{2a}$

Answer 1

Answer: (C)

(3) $\frac{2f\lambda}{a}$

Answer 2

The linear width of the central maximum in a Fraunhofer diffraction pattern by a single slit is determined by the angular spread of the central maximum and the focal length of the lens used to observe the pattern.

1. Condition for the first minimum: In single-slit diffraction, the condition for the first minimum (on either side of the central maximum) is given by: $a \sin \theta = \lambda$ where $a$ is the slit width, $\lambda$ is the wavelength of light, and $\theta$ is the angle of diffraction.

2. Small angle approximation: For Fraunhofer diffraction, the angles are typically small. Therefore, we can use the approximation $\sin \theta \approx \theta$ (when $\theta$ is in radians). So, $a \theta = \lambda$ This gives the angular position of the first minimum from the center: $\theta = \frac{\lambda}{a}$

3. Angular width of the central maximum: The central maximum extends from the first minimum on one side to the first minimum on the other side. Therefore, the total angular width of the central maximum is $2\theta$. Angular width $= 2\theta = \frac{2\lambda}{a}$

4. Linear width of the central maximum: When the diffraction pattern is observed on a screen placed at the focal plane of a lens with focal length $f$, the linear width ($W$) of the central maximum is given by: $W = f \times (\text{angular width})$ $W = f \times \frac{2\lambda}{a}$ $W = \frac{2f\lambda}{a}$

Comparing this with the given options, option (3) matches our derived formula.

The final answer is $\boxed{\text{(3)}}$

Explanation of the solution:

The angular position of the first minimum in single-slit diffraction is $\theta = \frac{\lambda}{a}$. The central maximum spans from $-\theta$ to $+\theta$, so its angular width is $2\theta = \frac{2\lambda}{a}$. The linear width on a screen at focal length $f$ is $W = f \times (2\theta) = \frac{2f\lambda}{a}$.