If two light waves reaching at a point produce destructive interference, then condition of phase difference is

If two light waves reaching at a point produce destructive interference, then condition of phase difference is
  1. $0,2 \pi, 4 \pi, 6 \pi \ldots \ldots$
  2. $\frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4} \ldots \ldots$
  3. $\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2} \ldots \ldots$
  4. $\pi, 3 \pi, 5 \pi \ldots \ldots$

Solution

For destructive interference, the condition on the phase difference $\Delta \phi$ between the two light waves is: $\Delta \phi=(2 m+1) \pi \quad \text { where } m=0,1,2,3, \ldots$ Explanation: Destructive interference occurs when the two waves are out of phase by an odd multiple of $\pi$ radians (or $180^{\circ}$ ). This phase difference results in the crest of one wave aligning with the trough of the other, causing cancellation. *

Asked in: MHT CET 2020 (20 Oct Shift 2)

Practice more Wave Optics questions on Aicharya