⊕ Constructive and Destructive Interference
Adding two waves of the same frequency with a phase shift: reinforcement, cancellation, a phasor diagram and a calculator for the amplitude of the sum.
Phase difference Δφ0°
Amplitude of wave 21,00
Frequency4,5
Animation speed×1,0
Examples:
Phasor diagram: A₁ + A₂ = sum
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Phase difference Δφ
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Path difference Δ = (Δφ / 360°) · λ
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Amplitude of the sum (A₁ = 1)
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Intensity I / (I₁ + I₂)
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Phase of the sum relative to wave 1
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Type of interference
Sum: A₁ sin(ωt) + A₂ sin(ωt + Δφ) = A sin(ωt + θ)
Amplitude: A = √(A₁² + A₂² + 2 A₁ A₂ cos Δφ) Intensity: I ∝ A²
Maximum: Δφ = 2π m (Δ = m λ) Minimum: Δφ = (2m + 1) π (Δ = (m + ½) λ)
Amplitude: A = √(A₁² + A₂² + 2 A₁ A₂ cos Δφ) Intensity: I ∝ A²
Maximum: Δφ = 2π m (Δ = m λ) Minimum: Δφ = (2m + 1) π (Δ = (m + ½) λ)
- The top wave (blue) and the middle wave (purple) have the same frequency, but the second is shifted in phase by Δφ. The bottom green wave is their sum. At Δφ = 0° the amplitudes add, at 180° they subtract, and with equal amplitudes the sum vanishes completely.
- If the amplitudes differ, there is no complete cancellation: the minimum equals |A₁ − A₂|. Try antiphase with the amplitude of wave 2 set to 0.5.
- The phasor diagram shows the same thing differently: each wave is a rotating vector, and the sum of the waves is the sum of the vectors. The length of the green vector equals the amplitude of the sum.
- Interference neither creates nor destroys energy, it only redistributes it in space: at maxima the intensity is greater than the sum I₁ + I₂, at minima it is smaller. Averaged over all Δφ, it equals I₁ + I₂.
Enter the amplitudes of two waves of the same frequency and the phase difference. You will get the amplitude and phase of the sum, the intensity relative to the sum of the intensities, and the range of possible amplitudes.
Amplitude of the first wave A₁
Amplitude of the second wave A₂
Phase difference Δφ, °
Wavelength λ, nm (optional)
Examples:
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Amplitude of the sum A
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Intensity I / (I₁ + I₂)
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Phase of the sum θ, °
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Possible amplitudes of the sum (min … max)
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Path difference Δ, nm
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Type of interference
A = √(A₁² + A₂² + 2 A₁ A₂ cos Δφ), tg θ = A₂ sin Δφ / (A₁ + A₂ cos Δφ)
Δ = (Δφ / 360°) · λ, Amin = |A₁ − A₂|, Amax = A₁ + A₂
Δ = (Δφ / 360°) · λ, Amin = |A₁ − A₂|, Amax = A₁ + A₂
Amplitude and intensity of the sum for two identical waves (A₁ = A₂ = 1)
| Δφ | Path difference Δ / λ | Amplitude of the sum | Intensity I / I₁ | Type |
|---|---|---|---|---|
| 0° | 0,00 λ | 2,00 | 4,00 | constructive |
| 45° | 0,12 λ | 1,85 | 3,41 | intermediate |
| 90° | 0,25 λ | 1,41 | 2,00 | intermediate |
| 120° | 0,33 λ | 1,00 | 1,00 | intermediate |
| 135° | 0,38 λ | 0,77 | 0,59 | intermediate |
| 180° | 0,50 λ | 0,00 | 0,00 | destructive |
| 270° | 0,75 λ | 1,41 | 2,00 | intermediate |
| 360° | 1,00 λ | 2,00 | 4,00 | constructive |
Constructive and Destructive Interference Online: Simulation and Calculator
A free interactive tool that shows how two waves of the same frequency add up with a phase shift. Change the phase difference, the amplitude of the second wave and the frequency, and watch reinforcement (constructive interference) or cancellation (destructive interference) appear. The phasor diagram shows the same thing with rotating vectors, and the calculator works out the amplitude, phase and intensity of the sum and the corresponding path difference.
Key concepts
Constructive interference
Waves in phase: crests coincide with crests. The amplitude of the sum equals A₁ + A₂, and the intensity is at its maximum.
Δφ = 2π m, Δ = m λ
Destructive interference
Waves in antiphase: a crest meets a trough. The amplitude of the sum is |A₁ − A₂|, and for equal amplitudes the wave vanishes.
Δφ = (2m + 1) π, Δ = (m + ½) λ
Path difference
The difference between the distances the waves have traveled to the meeting point. It determines the phase difference if the sources are coherent.
Δφ = 2π Δ / λ
Where it is used
- Noise-cancelling headphones: sound in antiphase cancels the noise
- Interferometers, including gravitational-wave detectors
- Colors of thin films and anti-reflective coatings on optics
- Antennas and radio emitters: shaping the radiation pattern
- Young’s experiment and diffraction gratings
- Room acoustics: «dead zones» and sound reinforcement
How to use
- Move the phase difference slider from 0° to 360° and watch how the green sum changes
- Set 180° and equal amplitudes: the sum vanishes
- Reduce the amplitude of wave 2: there is no complete cancellation any more, and the minimum equals the difference of the amplitudes
- Look at the phasor diagram: the length of the green vector is the amplitude of the sum
- In the «Calculator» tab, enter the amplitudes and the phase shift, and get the path difference from the wavelength
Frequently asked questions
What are constructive and destructive interference?
They are the two extreme cases of adding coherent waves. In constructive interference the waves are in phase (Δφ = 0, 2π, …) and the amplitude of the sum equals the sum of the amplitudes. In destructive interference the waves are in antiphase (Δφ = π, 3π, …) and the amplitude of the sum equals the absolute value of the difference of the amplitudes, and for equal waves it is zero.
How is the path difference related to the phase difference?
Δφ = 2π Δ / λ, where Δ is the difference of the distances from the sources to the point. A maximum occurs at Δ = mλ, a minimum at Δ = (m + ½)λ, where m is an integer. So a shift by half a wavelength turns reinforcement into cancellation.
Where does the energy go in destructive interference?
It does not disappear, it is redistributed: where the waves cancel each other, the intensity is lower than the sum, and in the neighboring maxima it is higher. The intensity averaged over space equals I₁ + I₂, so the law of conservation of energy holds.
Why is there no complete cancellation when the amplitudes differ?
The sum of two sinusoids in antiphase has the amplitude |A₁ − A₂|. The waves cancel each other completely only when A₁ = A₂. For the narrower wave with amplitude 0.5, an amplitude of 0.5 remains in antiphase.
Can you see interference from two ordinary light bulbs?
Usually not: their light is not coherent, the phase difference changes randomly and very quickly, so the pattern is smeared out. Interference is observed with laser light or when a wave from one source is split into two, as in Young’s experiment or in thin films.