What is the path difference for destructive interference (if n is an integer)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( (2n+1)\displaystyle\frac{\lambda}{2} \)
Destructive interference occurs when the two waves arrive in opposite phase, so that a crest of one meets a trough of the other. This needs a path difference of an odd multiple of half a wavelength.
\( \displaystyle \Delta x = (2n+1)\frac{\lambda}{2}, \quad n = 0, 1, 2, \ldots \)
Why the other options are incorrect
Option (1) \( n\lambda \) — Incorrect. \( n\lambda \), a whole number of wavelengths, puts the waves in phase: it is the condition for constructive interference.
Option (2) \( (n+1)\displaystyle\frac{\lambda}{2} \) — Incorrect. \( (n+1)\lambda/2 \) gives both odd and even multiples of \( \lambda/2 \); for \( n = 1 \) it equals \( \lambda \), which is constructive.
Option (4) \( (n+1)\lambda \) — Incorrect. \( (n+1)\lambda \) is again a whole number of wavelengths, the condition for constructive interference.
NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “… On the other hand, if the point P is such that the path difference,
\( \displaystyle S_1P \sim S_2P = \left(n + \tfrac{1}{2}\right)\lambda \qquad (n = 0, 1, 2, 3, \ldots) \qquad (10.10) \)
we will have destructive interference and the resultant intensity will be zero.”