The phase difference between two interfering waves producing 1st dark fringe in Young's double slit experiment is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \pi \)
A dark fringe forms where the two waves arrive out of phase (destructive interference), that is, where the phase difference is an odd multiple of π.
\( \Delta\phi = (2n - 1)\pi, \quad n = 1, 2, 3, \ldots \)
For the first dark fringe, n = 1:
\( \Delta\phi = \pi \quad (\text{path difference } \lambda/2) \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{\pi}{2} \) — Incorrect. A phase difference of \( \pi/2 \) gives half the maximum intensity, not a dark fringe.
Option (3) \( 3\pi \) — Incorrect. A phase difference of \( 3\pi \) gives the second dark fringe.
Option (4) \( 5\pi \) — Incorrect. A phase difference of \( 5\pi \) gives the third dark fringe.
NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “… Now, for any other arbitrary point G (Fig. 10.10) let the phase difference between the two displacements be \( \phi \).”
Coherent and Incoherent Addition of Waves – “… On the other hand, if \( \phi = \pm\pi, \pm 3\pi, \pm 5\pi \ldots \) [which corresponds to the condition given by Eq. (10.10)] we will have destructive interference leading to zero intensity.”