Virtual laboratory/Physics/Waves and acoustics/Wave interference — amplification and cancellation

Wave interference — amplification and cancellation

Change the path difference of two waves. When it is a multiple of λ, the waves meet in phase and reinforce; when it is an odd multiple of ½λ — in antiphase and cancel. Watch the resultant wave.

Wave interference — amplification and cancellation
Physics · Waves and acoustics · Wave interference (FIZ.FALE.PP.05)
Path difference Δ\Delta (in units of λ\lambda)0.00 λ
Waves meeting in antiphase (Δ=12λ\Delta = \tfrac{1}{2}\lambda):
path difference Δ
0.00 λ
phase difference φ
0°
amplitude A_res
2.00 A
effect
constr.
🎯 Challenge: constructive or destructive?
streak: 0

I will give you a path difference Δ — identify the interference result (without the slider). The waves have the same wavelength λ.

What just happened?

Two waves of the same wavelength are superposed. Whether they reinforce or cancel depends only on the path difference Δ that each has travelled to the meeting point.

Principle of superposition:

Ares=2A∣cos⁡φ2∣,φ=2π ΔλA_{res} = 2A\left|\cos\dfrac{\varphi}{2}\right|,\qquad \varphi = \dfrac{2\pi\,\Delta}{\lambda}

— the resultant amplitude depends on the phase difference φ, which follows directly from the path difference Δ. A — amplitude of a single wave, λ — wavelength.

When constructive, and when destructive?

When the path difference is a multiple of λ, the waves travel in phase and add up to a maximum. When it is an odd multiple of half a λ, they are in antiphase and cancel:

Δ=nλ⇒constructiveΔ=(n+12)λ⇒destructive\Delta = n\lambda \Rightarrow \text{constructive}\qquad \Delta = \left(n+\tfrac{1}{2}\right)\lambda \Rightarrow \text{destructive}

Where can you see it?

Interference fringes in Young's experiment, noise cancellation in ANC headphones, rainbow colours on an oil film — the same path difference decides everywhere.

Why you need this for the exam:

The conditions Δ=nλ\Delta = n\lambda (constructive) and Δ=(n+12)λ\Delta = (n+\tfrac{1}{2})\lambda (destructive) are the key to interference and diffraction problems. Trick: compute the path difference, divide by λ and check whether you get a whole number or a half-integer.

Remember: waves in phase (Δ = nλ) give amplification, antiphase (Δ = (n+½)λ) — cancellation. The path difference decides the result.

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Theory & formulas

Superposition of waves

Two coherent waves of the same frequency are superposed. The result depends on the path difference Δr\Delta r.

Amplification and cancellation

  • amplification (in phase): Δr=nλ\Delta r = n\lambda
  • cancellation (in antiphase): Δr=(2n+1)λ2\Delta r = (2n+1)\dfrac{\lambda}{2}

Condition for interference: the waves must be coherent (constant phase difference) and of the same frequency.

Notation: Δr\Delta r — path difference, λ\lambda — wavelength, n=0,1,2…n=0,1,2\ldots

Frequently asked questions

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When do waves cancel out?

When they meet in antiphase (path difference (n+½)λ) — the crest of one meets the trough of the other and the amplitudes cancel.

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When does amplification occur?

When the path difference is an integer multiple of λ (nλ) — the waves are in phase and the amplitudes add up.

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What is the condition for interference?

The waves must be coherent (constant phase difference) and of the same frequency.

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