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 edges of the slit is.

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $2\pi$

Answer 1

The correct option is D $2\pi$

The condition for the minima is,

$\Delta x=a\sin \theta =n\lambda$

For the first diffraction to be minima, $n=1$

$\therefore$ path difference, $\Delta x=\lambda$

Now, using $\varphi =\frac{2\pi }{\lambda }\Delta x$

$\varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$

$\therefore$ Correct option is $\left(D\right)$

$\begin{array}{c}\text{Why this question?}\\ \text{Asked in IIT-JEE 1995 and 1998}.\end{array}$