Question

Consider Fraunhofer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum, the phase difference (in radius) between the wavelets from the opposite edge of the slit is:

• A. $\pi$
• B. $2\pi$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B. $2\pi$.

Path difference when the wavelets are diffracted from the opposite edge of the slit = $△x=\lambda$.

We know that,

Phase difference = $\varphi =\frac{2\pi }{\lambda }\times △x$

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

$\Rightarrow \varphi =2\pi$.