Condition for destructive interference
Destructive interference occurs where the two waves arrive in antiphase, that is Δφ = (2n + 1)π. For sources emitting in phase this means a path difference of (n + ½)λ (n = 0, 1, 2, …), and the resultant amplitude is the difference of the two amplitudes, which is zero if they are equal.
Δφ = (2n + 1)π ⇔ path difference = (n + ½)λ (sources in phase)What this actually means
Half-integer multiples: 0.5λ, 1.5λ, 2.5λ and so on. A path difference of exactly 1λ or 2λ is constructive, never destructive.
The first minimum (n = 0) sits at 0.5λ, so a path difference of half a wavelength is the closest destructive point to the centre. There is no destructive point at zero path difference.
Say resultant amplitude is the difference of the amplitudes, not resultant is zero, unless you have been told the amplitudes are equal. Real double-slit and speaker setups rarely give perfect cancellation.
For antiphase sources everything flips: path difference nλ becomes destructive and (n + ½)λ becomes constructive. Check the wording of the stem before you commit.
Convert to wavelengths early. A path difference of 1.0 m with λ = 2.0 m is 0.5λ, which is destructive, and you can say so in one line without any further working.
Using nλ/2 for destructive interference, which wrongly includes the whole-number multiples that are actually maxima.
Prove it — watch it be true
- Drag the path-difference probe until the readout shows 0.5λ.
- Confirm the verdict reads DARK and the intensity profile shows a minimum there.
- Step out to 1.5λ and 2.5λ and confirm both are also minima, while 1.0λ and 2.0λ are maxima.