Question

Which of the following is the path difference of destructive interference?
(A) $\dfrac{{(2n + 1)\lambda }}{2}$
(B) $\dfrac{{(n + 1)\lambda }}{2}$
(C) $n\lambda$
(D) $(n + 1)\lambda$

Answer 1

Hint There are two types of interference constructive and destructive. Constructive interference when phase difference between two waves is an even multiple of $\pi ({180^0})$ and amplitude of wave at that point is sum of amplitude of both the waves. Destructive interference when phase difference between two waves is an odd multiple of $\pi ({180^0})$ and amplitude of wave at that point is the difference between amplitude of both the waves.

Complete step by step solution
Destructive interference when phase difference between two waves is an odd multiple of $\pi ({180^0})$ and amplitude of wave at that point is the difference between amplitude of both the waves.
We know that in terms of angle one cycle of wave is $2\pi$ and in terms of path-length one cycle of wave is equal to wave-length $\lambda$.
Hence, we can say that for destructive interference the path difference is an odd multiple of $\lambda /2$.
From the given options $\dfrac{{(n + 1)\lambda }}{2}$, $n\lambda$ and $(n + 1)\lambda$ may be odd or even multiple of $\lambda /2$ but $\dfrac{{(2n + 1)\lambda }}{2}$ is always odd multiple of $\lambda /2$.

Hence the correct answer is option A.

Note Interference occurs between waves from two different sources but the frequency of both waves is same and wave-length may be different. In constructive interference wave is amplified or amplitude of wave increases but in destructive interference amplitude is decreased. Wave-length and frequency of waves always remain the same during interference.