Superposition
At the same place and time, add the wave quantities: y = y₁ + y₂.
Waves · superposition · phase
Interference
When two waves overlap, their displacements or fields superpose. Depending on their relative phase, the result can be reinforcement, partial cancellation or—when amplitudes match—complete cancellation at a point. A difference in the distances travelled creates a path difference, which often turns geometry into an observable pattern of bright and dark, loud and quiet, or large and small oscillations.
φ · phase differenceδ · path differenceλ · wavelength
The essential idea
Add the waves first—not their “effects” by guesswork
In a linear system where superposition applies, the displacement observed at one place and time is the algebraic sum of the displacements that each wave would produce separately. Crest plus crest gives a larger displacement. A crest meeting a trough reduces the resultant.
For sinusoidal waves of the same frequency, the phase difference φ tells you where one oscillation is in its cycle relative to the other. At φ = 0° they are in phase. At φ = 180° = π rad they are in opposition. Intermediate phase differences give intermediate reinforcement or cancellation.
Phase difference
It measures the offset within the cycle. One full cycle is 2π rad = 360°.
Path difference
If two coherent waves travel different distances, the extra path can create a phase shift.
The mathematical model
What does adding two sinusoids actually give?
At one point, write y₁ = A sin(ωt) and y₂ = rA sin(ωt + φ). The factor r is the amplitude of the second wave relative to the first. Because the two waves have the same frequency, their sum is another sinusoid of that frequency, with a new amplitude and phase.
Resultant amplitudeR = A√(1 + r² + 2r cos φ)
If r = 1R = 2A |cos(φ/2)|
At φ = 0°, the waves reinforce as strongly as possible and the amplitude is A(1 + r). At φ = 180°, the minimum amplitude is A|1 − r|. Complete cancellation is therefore possible only when the two amplitudes are equal.
Mini-lab
Change the phase and watch the sum update point by point
The three curves come directly from the equations above: y₁, y₂ and their sum y₁ + y₂. Change the phase difference and the second wave's relative amplitude. All three graphs use the same vertical scale, so amplitude comparisons remain meaningful.
Equivalent δ / λ*0
Resultant amplitude2.00 A
Effectconstructive maximum
Constructive maximumThe waves are in phase, so their amplitudes add as strongly as possible.
Partial reinforcement / cancellationAn intermediate phase gives a resultant between the two limiting cases.
MinimumAt 180° the waves are in opposition. Equal amplitudes can cancel completely.
How to read the graph: compare the two upper waves first, then look at their resultant below. Matching crests make the resultant larger; a crest facing a trough makes it smaller.
* The conversion φ = 2πδ/λ applies here when the phase difference comes only from a path difference between same-frequency waves, with no extra source or reflection phase shift.
From geometry to phase
Path difference becomes phase difference
Suppose two coherent waves start in phase and travel different distances before reaching a point M. If their path lengths are r₁ and r₂, the path difference is δ = r₂ − r₁. The extra distance becomes a phase offset.
φ = 2π δ / λ
An extra path of one wavelength corresponds to one complete phase cycle, φ = 2π, so the waves arrive in phase again. An extra half-wavelength gives φ = π, so they arrive in opposition.
Maximum reinforcement
δ = mλm is an integer. For sources initially in phase, the waves arrive in phase.
Maximum cancellation
δ = (m + 1/2)λWith equal amplitudes, the ideal minimum is complete cancellation.
Example 1 · predict before calculating
λ = 0.60 m and δ = 1.20 m: what should happen?
Before calculating: 1.20 m is exactly two wavelengths of 0.60 m. After two complete cycles, the waves should return to the same relative phase.
Calculation: δ/λ = 1.20 / 0.60 = 2, an integer, so δ = 2λ.
Conclusion: two sources that started in phase give a constructive maximum at that point.
Variation: if δ = 0.90 m, then δ/λ = 1.5 = 1 + 1/2. The waves arrive in opposition and give the strongest cancellation allowed by their amplitudes.
Amplitude ≠ intensity
Why can a bright fringe be four times as intense?
For two coherent light waves of intensities I₁ and I₂, the resulting intensity depends on their phase difference. The cosine term is the interference term.
I = I₁ + I₂ + 2√(I₁I₂) cos φ
If I₁ = I₂ = I₀, then I = 4I₀ cos²(φ/2). At the constructive maximum the field amplitude doubles and the intensity becomes 4I₀, because intensity is proportional to the square of field amplitude. At the ideal minimum, equal contributions can give zero intensity at that location.
Young's double-slit experiment
Turn a path difference into visible fringes
In Young's experiment, one wave illuminates two narrow slits. The slits act as two coherent secondary sources. Their waves overlap on a screen and form alternating bright and dark fringes.
For a distant screen and small angles, the geometry gives δ ≈ a sin θ ≈ ax/D, where a is the slit separation, D the slit-to-screen distance, and x the position of the observation point on the screen.
In the diagram, S is the original source illuminating both slits. M is one chosen point on the screen. At M, compare the two path lengths to decide whether the contributions reinforce or cancel.
Soriginal light source
adistance between the slits
Dslit-to-screen distance
M / xM is the observed point; x locates it on the screen
S
M
a
D
Distance between adjacent bright fringesi = λD / a
Important: this fringe-spacing formula uses the small-angle approximation and assumes D is large compared with the slit separation.
Example 2 · Young's slits
How far apart are neighbouring bright fringes?
Use light with λ = 600 nm, two slits separated by a = 0.30 mm, and a screen at D = 2.0 m.
Convert units first: λ = 6.00 × 10⁻⁷ m and a = 3.0 × 10⁻⁴ m.
i = λD/a = (6.00 × 10⁻⁷ × 2.0)/(3.0 × 10⁻⁴) = 4.0 × 10⁻³ m
So i = 4.0 mm. Adjacent bright maxima are about four millimetres apart.
Check the trend: increasing D spreads the fringes out. Decreasing a also spreads them out. The equation predicts both trends.
Practice
Your turn: two short problems
Constructive or destructive?
Two waves have λ = 4.0 cm and arrive with path difference δ = 10.0 cm. Identify the interference condition.
Show the solution
δ/λ = 10.0/4.0 = 2.5 = 2 + 1/2. The half-integer condition is satisfied, so the waves arrive in opposition for sources that started in phase. With equal amplitudes, the cancellation is complete.
Calculate the fringe spacing
λ = 500 nm, D = 1.50 m and a = 0.25 mm. Calculate i.
Show the solution
Answer: λ = 5.00 × 10⁻⁷ m and a = 2.5 × 10⁻⁴ m. Therefore i = λD/a = 3.0 × 10⁻³ m = 3.0 mm.
Why: convert both nanometres and millimetres to metres before substituting, so all quantities use consistent SI units.
Check: a millimetre-scale spacing is plausible for visible light, sub-millimetre slit separation and a metre-scale screen distance.
Keep these ideas separate
Superposition is the instantaneous addition of wave quantities. Interference is the pattern of reinforcement and cancellation produced by that addition. Coherence means the phase relationship is stable enough for the pattern to remain observable.
Go one step further
The familiar δ conditions need their assumptions
The relations δ = mλ and δ = (m + 1/2)λ follow directly when the two sources start in phase and no additional phase shift is introduced. A reflection or propagation through different media can add phase shifts, which must then be included separately.
If the frequencies differ, or if relative phase changes rapidly, the maxima and minima do not stay fixed. The simple fringe pattern moves or averages out and loses contrast—even though superposition still holds instant by instant.
Interference belongs to the broader study of waves. Continue with waves to review wavelength, frequency and propagation, or move to diffraction to see how finite apertures reshape wave patterns.