2015 AIPMT-II Wave OpticsSingle slit diffraction Medium Question 3 of 4 in Single slit diffraction

At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the mid point of the slit is:

A $\frac{\pi}{8}$ radian
B $\frac{\pi}{4}$ radian
C $\frac{\pi}{2}$ radian
D $\pi$ radian

The answer is D.

Solution

For the first minimum, $a\sin\theta = \lambda$.
Path difference between the edge and the mid-point wavelets: $\Delta x = \frac{a}{2}\sin\theta = \frac{\lambda}{2}$
$\Delta\phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{2} = \pi$ radian

Concept Phase difference at first minimum

Single slit diffraction in past papers

4 questions from this chapter have appeared across 3 exam years.