2015 AIPMT-I Wave OpticsSingle slit diffraction Easy Question 2 of 4 in Single slit diffraction

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 the screen from the slit, the width of the central maxima will be

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

The answer is A.

Solution

First minima lie at $\sin\theta = \frac{\lambda}{a}$ on either side, at distance $\frac{D\lambda}{a}$ from the centre.
Width of the central maximum $= \frac{2D\lambda}{a}$

Concept Width of central maximum

Single slit diffraction in past papers

4 questions from this chapter have appeared across 3 exam years.