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 the slit is

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

Answer 1

The correct option is D $2\pi$

When rays passes through edges of slits, the path difference is,

$\Delta x=a\sin \theta$

For first minima, $a\sin \theta =n\lambda =\lambda [\because n=1]$

$\Rightarrow \Delta x=\lambda$

The phase difference is,

$\Delta \varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)=\frac{2\pi }{\lambda }\lambda =2\pi$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.