Q. Consider Fraunhofer diffraction pattern obtained with a single slit at normal incidence. At the angular position of first diffraction minimum, the phase difference between the wavelets from the opposite edges of the slit is
A) $\frac{\pi}{4}$
B) $\frac{\pi}{2}$
C) $\pi$
D) $2\pi$
Explanation
(d) The angular position of first diffraction minimum is
$\sin\theta = \frac{\lambda}{a}$
The phase difference
$\phi = \frac{2\pi}{\lambda} \times \Delta x$
$\phi = \frac{2\pi}{\lambda} \times \lambda$
$\phi = 2\pi$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.