Physics · Classical Mechanics · Oscillations & Waves
Wave Interference Calculator
Calculates the resultant amplitude and intensity of two interfering waves based on their individual amplitudes, wavelength, and path difference.
Calculator
Formula
A_R is the resultant amplitude; A_1 and A_2 are the amplitudes of the two individual waves; \delta is the phase difference in radians; \Delta x is the path difference between the two waves; \lambda is the wavelength. Intensity scales as I \propto A^2, so I_R = I_{\max} \cos^2(\delta/2) when A_1 = A_2, and more generally I_R \propto A_R^2.
Source: Hecht, E. (2017). Optics (5th ed.). Pearson. §9.3 — Superposition Principle and Wave Interference.
How it works
Wave interference arises from the superposition principle: when two or more waves overlap in the same medium, the net displacement at every point is the algebraic sum of the individual displacements. For two coherent sinusoidal waves of the same frequency, the outcome depends entirely on their phase relationship. When the waves are perfectly in phase (phase difference δ = 0, 2π, 4π, …), their crests and troughs align and the amplitudes add constructively. When they are perfectly out of phase (δ = π, 3π, …), crests meet troughs and the amplitudes subtract destructively. All intermediate cases produce partial interference.
The resultant amplitude is given by the vector addition formula: A_R = √(A₁² + A₂² + 2A₁A₂ cos δ), where the phase difference δ = 2π Δx / λ converts path difference Δx and wavelength λ into the angular mismatch between the two waves. Intensity is proportional to amplitude squared, so the relative intensity I_R / I_max = A_R² / (A₁ + A₂)², which equals 1 for full constructive interference and 0 for full destructive interference when the amplitudes are equal. This calculator outputs both the resultant amplitude and the relative intensity so users can quickly characterise the interference condition.
Practical applications span a wide range of fields. In physical optics, thin-film interference and anti-reflection coatings exploit path differences of a fraction of a wavelength to eliminate unwanted reflections. In acoustics, engineers designing concert halls and noise-cancellation headphones must account for interference between direct and reflected sound paths. In radio engineering, antenna arrays are spaced and phased to steer and shape radiation patterns through constructive interference in desired directions and destructive interference elsewhere. In quantum mechanics, the same mathematics describes probability-amplitude interference in electron diffraction and neutron scattering experiments.
Worked example
Suppose two coherent sound waves of the same frequency travel from two loudspeakers to a listener. Speaker 1 produces a wave with amplitude A₁ = 2.0 m (displacement units) and Speaker 2 produces A₂ = 1.5 m. The sound wavelength is λ = 0.40 m and the path difference to the listener is Δx = 0.10 m.
Step 1 — Compute the phase difference:
δ = 2π × Δx / λ = 2π × 0.10 / 0.40 = 2π × 0.25 = π/2 ≈ 1.5708 rad
A phase difference of π/2 (90°) means the waves are a quarter-wavelength out of step — neither fully constructive nor fully destructive.
Step 2 — Compute the resultant amplitude:
A_R = √(A₁² + A₂² + 2 A₁ A₂ cos δ)
= √(4.00 + 2.25 + 2 × 2.0 × 1.5 × cos(π/2))
= √(4.00 + 2.25 + 6.00 × 0)
= √6.25 = 2.50 m
Step 3 — Compute relative intensity:
I_max = (A₁ + A₂)² = (2.0 + 1.5)² = 12.25
I_R / I_max = A_R² / I_max = 6.25 / 12.25 ≈ 0.5102
The listener hears about 51% of the maximum possible intensity — a partial interference condition midway between the extremes. If the path difference were zero (full constructive interference), the intensity ratio would be 1.0; if Δx = 0.20 m (half-wavelength, δ = π), the resultant amplitude would drop to |2.0 − 1.5| = 0.5 m and the ratio to 0.0204 — nearly silent.
Limitations & notes
This calculator assumes the two waves are perfectly coherent (same frequency, stable phase relationship) and have the same polarisation direction. Real sources — including thermal light and most acoustic sources — are only partially coherent, so the observed fringe visibility will be lower than predicted. The formula also treats waves as one-dimensional plane waves; in three dimensions, amplitude falls with distance (1/r for spherical waves), which is not accounted for here. Path difference is assumed to be the only source of phase offset; any intrinsic initial phase difference between sources must be added to δ manually. Finally, the formula is linear and does not account for nonlinear medium effects that arise at very large amplitudes, nor for the vector nature of electromagnetic waves when polarisations differ.
Frequently asked questions
What is the condition for constructive interference?
Constructive interference occurs when the path difference Δx is an integer multiple of the wavelength: Δx = nλ (n = 0, 1, 2, …). This gives a phase difference δ = 2πn, so cos δ = 1 and the resultant amplitude reaches its maximum value A₁ + A₂.
What is the condition for destructive interference?
Destructive interference occurs when the path difference is a half-integer multiple of the wavelength: Δx = (n + ½)λ. This gives δ = (2n+1)π, so cos δ = −1 and the resultant amplitude is |A₁ − A₂|. If the two amplitudes are equal, the resultant is zero and the waves completely cancel.
Why does intensity depend on amplitude squared?
For mechanical waves, intensity is the power transported per unit area, which is proportional to the square of the displacement amplitude (I ∝ A²ω²ρv). For electromagnetic waves, intensity is the time-averaged Poynting vector magnitude, which is proportional to the square of the electric field amplitude. In both cases the A² relationship is fundamental to wave energy transport.
How does path difference relate to phase difference?
One full wavelength of path difference corresponds to one full cycle (2π radians) of phase difference. Therefore δ = (2π/λ) × Δx. For example, a path difference of half a wavelength (Δx = λ/2) gives δ = π radians (180°), placing the waves completely out of phase.
Can this calculator be used for light as well as sound?
Yes. The superposition formula is universal for linear waves regardless of the physical medium. For visible light, typical wavelengths are in the range 400–700 nm; for audible sound in air at 20 °C, wavelengths range from about 17 mm (20 kHz) to 17 m (20 Hz). Simply enter the appropriate wavelength and path difference in consistent units and the calculator gives the correct result.
Last updated: 2025-01-15 · Formula verified against primary sources.