Wave Interference Calculator
Computes the resultant amplitude, intensity ratio, or phase fraction when two coherent waves overlap. Use it for optics experiments, acoustics design, or double-slit homework problems.
About this calculator
When two waves with amplitudes A₁ and A₂, the same wavelength λ, a path difference Δ, and a phase shift φ overlap, the resultant amplitude is A = √(A₁² + A₂² + 2A₁A₂ cos(2πΔ/λ + φ)). This follows directly from the phasor addition of two sinusoids. Pure constructive interference occurs when 2πΔ/λ + φ = 0, 2π, 4π… giving A = A₁ + A₂. Pure destructive interference occurs when the argument equals π, 3π… giving A = |A₁ − A₂|. The intensity ratio between maximum and minimum intensities is (A₁ + A₂)² / (A₁ − A₂)², a key metric for fringe visibility in optics. The phase fraction output, (Δ mod λ)/λ, tells you where in the cycle the path difference falls — 0 means perfectly in phase, 0.5 means half a wavelength out of phase.
How to use
Two sound waves both have amplitude 3 m, wavelength 2 m, a path difference of 0.5 m, and zero phase shift. Select 'Resultant Amplitude'. A = √(3² + 3² + 2·3·3·cos(2π·0.5/2 + 0)) = √(9 + 9 + 18·cos(π/2)) = √(18 + 18·0) = √18 ≈ 4.24 m. Now try path difference = 0 (fully constructive): A = √(9 + 9 + 18·cos(0)) = √36 = 6 m, equal to A₁ + A₂ as expected. A path difference of 1 m (half wavelength) gives A = √(18 + 18·cos(π)) = √0 = 0 — total destructive interference.
Frequently asked questions
What is the condition for constructive vs destructive interference?
Constructive interference occurs when the path difference is an integer multiple of the wavelength (Δ = 0, λ, 2λ…), making the cosine term equal to +1 and the amplitudes add fully. Destructive interference occurs when the path difference is a half-integer multiple (Δ = λ/2, 3λ/2…), making the cosine equal to −1 so amplitudes partially or fully cancel. If a phase shift φ is present, it shifts these conditions by φ/(2π) wavelengths. Equal amplitudes produce complete cancellation in the destructive case.
How does the intensity ratio formula measure fringe visibility in optics?
Fringe visibility (or contrast) quantifies how bright the bright fringes are compared to the dark fringes. The ratio (A₁ + A₂)² / (A₁ − A₂)² compares maximum intensity (fully constructive) to minimum intensity (fully destructive). When both amplitudes are equal the denominator is zero, meaning the dark fringes have zero intensity — perfect visibility. As the amplitudes become more unequal, the dark fringes brighten and the fringes become harder to distinguish, lowering the signal quality in instruments like interferometers.
Why does path difference cause a phase shift between two waves?
A wave travels one full cycle — 2π radians — for every wavelength λ it covers. If one wave travels an extra distance Δ compared to the other, it accumulates an additional phase of 2πΔ/λ radians by the time both waves meet. This extra phase shifts the cosine in the interference formula, directly controlling whether the waves reinforce or cancel. Engineers exploit this in noise-cancelling headphones (destructive interference of sound) and anti-reflection coatings (destructive interference of reflected light).