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Interference questions in sound and light all reduce to one comparison: how far out of step are the two waves when they meet? Sometimes the question gives you a path difference, sometimes a phase difference, sometimes a time delay. This post shows how to convert between them, how to add a phase the sources already have, and where students go wrong.
Where is this in NCERT and the syllabus?
The 2026-27 reprint of the NCERT Class 12 chapter Wave Optics uses the conversion in Section 10.4: "a path difference of 2λ corresponds to a phase difference of 4π", and a path difference of 2.5λ to 5π. The Syllabus for JEE (Main) 2026 lists the principle of superposition of waves in Unit 10 and Young's double-slit experiment in Unit 16, and the NEET (UG) 2026 syllabus lists both too. Check your own year's syllabus.
Why does one wavelength of path equal 2π of phase?
A wave y = a sin(kx − ωt) has k = 2π/λ. Move a distance Δx along the wave and the phase changes by kΔx. So
Δφ = (2π/λ) Δx, and Δx = (λ/2π) Δφ
In time, the same idea gives Δφ = (2π/T) Δt: a delay of one period is a phase difference of 2π.
The conditions follow:
Constructive: Δφ = 0, 2π, 4π, …, so Δx = 0, λ, 2λ, …
Destructive: Δφ = π, 3π, 5π, …, so Δx = λ/2, 3λ/2, …
These path rules hold only when the sources start in phase. That caveat is where most errors come from.
Why do students find this hard?
Wosilait, Heron, Shaffer and McDermott (1999) found that university students who had studied physics at the introductory level and beyond often did not have a functional understanding of the wave model for light. In practice, students memorise "path difference λ means bright" without seeing it as a statement about phase, so they get stuck when the sources are out of step or a sheet of glass is added.
Worked example 1: two loudspeakers
Two loudspeakers driven in phase emit a 686 Hz tone. Take the speed of sound as 343 m/s. A listener is 3.00 m from one speaker and 3.75 m from the other.
λ = v/f = 343 / 686 = 0.500 m.
Δx = 3.75 − 3.00 = 0.75 m = 1.5λ.
Δφ = 2π × 1.5 = 3π, an odd multiple of π, so the listener is at a point of destructive interference.
Now swap the two wires on one speaker, so its cone moves out while the other moves in. That adds a source phase difference of π. The total becomes 3π + π = 4π, and the same spot becomes loud.
Worked example 2: sources out of step by π/2
Two coherent sources emit waves of wavelength 0.40 m. Source 1 leads source 2 by π/2. Let Δx = r₂ − r₁, the extra distance from source 2.
Total phase difference: Δφ = π/2 + (2π/0.40) Δx.
Constructive: set Δφ = 2nπ. Then Δx = nλ − λ/4 = −0.10 m, 0.30 m, 0.70 m, …
Destructive: set Δφ = (2n + 1)π. Then Δx = nλ + λ/4 = 0.10 m, 0.50 m, …
The point with Δx = 0, equidistant from both sources, is neither bright nor dark. It gets the in-between intensity, 2I₀ for equal sources, since I = 4I₀ cos²(Δφ/2) and cos²(π/4) = ½.
Worked example 3: points on one wave, and a time delay
A wave has λ = 60 cm and period 0.020 s.
Two points 15 cm apart along the wave: Δφ = (2π/60) × 15 = π/2.
One point observed 0.0025 s apart: Δφ = (2π/0.020) × 0.0025 = π/4.
Worked example 4: a glass sheet in light
In a double-slit set-up with λ = 500 nm, a sheet of μ = 1.6 and thickness 2.5 μm covers one slit. Light through the sheet travels an optical path μt instead of t, an extra (μ − 1)t = 0.6 × 2.5 μm = 1.5 μm = 3λ. That is a phase difference of 6π added to every point. The central bright fringe moves to where the third bright fringe used to be.
Common mistakes and how to avoid them
Using the path rule when sources are out of phase. Add the source phase first.
Forgetting the medium. Inside glass the wavelength is λ/μ, so use optical path μ × distance.
Mixing degrees and radians. Keep phase in radians: π, not 180, inside 2π/λ.
Treating λ/2 as "one wavelength out". λ/2 is a phase of π, the dark condition.
Once you have the total phase, the brightness follows from I = 4I₀ cos²(Δφ/2), worked in our post on interference intensity. For phase along a single wave, see the guide to reading the travelling wave equation. If a phase difference is hard to picture, the Fermi Simulation Lab has interactive simulations for ideas easier to see than read.
A practice routine
Convert ten path differences, given as fractions of λ, into phases, and back.
Solve two loudspeaker problems, one with in-phase and one with out-of-phase sources.
Solve one thin-sheet problem and say which way the fringes move.
For each answer, write the total phase before deciding bright or dark.
Try it now: two in-phase sources of λ = 0.80 m, a point 0.20 m further from one. Find Δφ and the intensity as a fraction of 4I₀.
