Read the passage and answer the question.
A wave is represented by a variable \(y\). It gives the displacement of the particles of the medium in which the wave travels, at any instant of time. Consider two waves \(y_1\) and \(y_2\) that superpose, and let \(y\) be the resultant of the two waves. Here, the resultant means that \(y\) is the vector sum of the displacements due to each wave.
For constructive interference, the phase difference between the two waves should be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2n\pi\), \(n = 0, 1, 2, \ldots\)
For two waves of equal amplitude, the resultant intensity at a point is \( I=4I_0\cos^2(\phi/2) \), where \( \phi \) is the phase difference. The intensity is largest when \( \cos(\phi/2)=\pm1 \). The two displacements are then in phase and add up fully.
Option (1) \(\pi/2\) — Incorrect. With \( \phi=\pi/2 \), \( \cos^2(\pi/4)=1/2 \) and the intensity is only \( 2I_0 \). This is neither a maximum nor a minimum.
Option (2) \(n\pi/2\), \(n = 0, 1, 2, \ldots\) — Incorrect. This set also has \( \pi/2 \), \( \pi \), \( 3\pi/2 \) and so on. Only the values with \( n=0,4,8,\ldots \) give a maximum, and \( \pi \) gives zero intensity.
Option (3) \(\pi\) — Incorrect. A phase difference of \( \pi \) puts a crest of one wave on a trough of the other. This is destructive interference.
Option (4) \(2n\pi\), \(n = 0, 1, 2, \ldots\) — Correct. For \( \phi=0,\ 2\pi,\ 4\pi,\ldots \), \( \cos^2(\phi/2)=1 \). The waves meet in phase and the intensity is maximum.
NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “If φ = 0, ± 2π, ± 4π,… … we will have constructive interference leading to maximum intensity.”