Question

A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is

• A. zero
• B. $ \pi $
• C. $ \frac{\pi}{2} $
• D. $ 2\,\pi $

Answer 1

The Correct Option is D

The phase difference $(\phi)$ between the wavelets from the top edge and the bottom edge of the slit is $\phi=\frac{2 \pi}{\lambda}(d \sin \,\theta)$, where $d$ is the slit width.
The first minima of the diffraction pattern occurs at
$\sin \,\theta=\frac{\lambda}{d}$, so $\phi=\frac{2 \pi}{\lambda}\left(d \times \frac{\lambda}{d}\right)=2 \pi$