Question

For a parallel beam of monochromatic light of wavelength $\lambda $, diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of he screen from the slit, the width of the central maxima will be

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

Answer 1

The Correct Option is C

The problem involves the concept of single-slit diffraction, where a parallel beam of monochromatic light passes through a slit, causing diffraction. The central region of diffraction patterns on a screen is the central maxima, and we are asked to find its width.

Diffraction from a single slit is governed by the formula for the angular position $\theta$ of the minima, which is given by:

a \sin \theta = n \lambda

where a is the slit width, \lambda is the wavelength of light, and n is the order of the minimum.

For the first order minima on either side of the central maxima, n = \pm 1. Thus, the equation for the first order minima is:

a \sin \theta = \lambda

For small angles, \sin \theta \approx \theta (in radians), so:

\theta = \frac{\lambda}{a}

The angular width of the central maxima consists of two angles: one for the minimum on the left and one on the right, giving the total angular spread as 2 \theta:

2 \theta = 2 \frac{\lambda}{a}

The linear width of the central maxima on the screen is then given by:

w = 2 \theta D = 2 \frac{\lambda}{a} \cdot D

Simplifying, we get the width of the central maxima as:

w = \frac{2 D \lambda}{a}

Therefore, the width of the central maxima is \frac{2 D \lambda}{a}.

This matches option \frac{2 D \lambda}{a}, confirming that it is the correct answer.