= Interference
{disambiguate=wave propagation}
{wiki}
= Wave interference
{synonym}
= Interference
{synonym}
<Wave interference> is the reinforcement or cancellation that occurs when overlapping <wave amplitudes> add by the <superposition principle>. For two <monochromatic waves> of the same <frequency>, mutual <optical coherence> means that their relative <wave phase> is stable during the observation. Equal <wave phases> give <constructive interference>; a half-cycle <phase difference> gives the <destructive interference intensity> minimum. <Wave amplitudes> add before their squares determine <wave intensity>. https://openstax.org/books/university-physics-volume-3/pages/3-1-youngs-double-slit-interference[Ling, Sanny and Moebs, University Physics Volume 3, §3.1: coherent waves and interference]{ref}.
Consider two overlapping beams in a homogeneous medium, with the same <polarization of an electromagnetic wave> and individual local time-averaged <light intensities> $I_1,I_2$. Their <phase difference> $\delta$ is measured in <radians>. In the scalar detection convention, the combined local time-averaged <wave intensity> is
$$
I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta.
$$
The last term records <wave interference>; adding $I_1+I_2$ alone misses it. The formula describes <irradiance> when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared <electric field> response, proportional to <scalar wave intensity>; a specified component of the <Poynting vector> also requires the propagation geometry. The ideal crossing <plane waves> below make that distinction explicit.
For equal individual <light intensities> $I_1=I_2=I_0>0$, the <cosine> half-angle identity gives
$$
I=2I_0(1+\cos\delta)=4I_0\cos^2(\delta/2).
$$
The <constructive interference intensity> is $4I_0$ at $\delta=0$ modulo $2\pi$, and the <destructive interference intensity> is zero at $\delta=\pi$ modulo $2\pi$. At <phase quadrature>, $\delta=\pi/2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary <wave intensity> unit, the maximum is $2.25$, the minimum is $0.25$, and the <phase quadrature> value is $1.25$: unequal <wave amplitudes> cannot cancel completely.
\Image[/interference-intensity-versus-phase.png]
{title=Two-beam <wave intensity> versus <phase difference>: equal beams reach 4 and 0; unequal beams reach 2.25 and 0.25}
Both curves use the same <wave intensity> unit and a horizontal <angle> in <radians>. The dashed levels are the averages over a full <phase difference> cycle. Equal beams have <interferometric visibility> one; the unequal example has <interferometric visibility> $0.8$.
A spatially varying <phase difference> turns the same law into <interference fringes>. Take two ideal <plane electromagnetic waves> with common <wavelength> $\lambda$, equal individual <light intensities> $I_0$, and propagation directions at <angles> $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta<\pi/2$. Both have <linear polarization> along $y$, perpendicular to both propagation directions. Choose equal initial <wave phases>; their <wavevectors> and <phase difference> are
$$
\mathbf k_\pm=k(\pm\sin\theta,0,\cos\theta),\qquad k=\frac{2\pi}{\lambda},\qquad
\phi_\pm=\pm kx\sin\theta+kz\cos\theta-\omega t,\qquad
\delta=\phi_+-\phi_-=2kx\sin\theta.
$$
Here $k$ is the <wavenumber> and $\omega$ the <angular frequency>. The common $z$ and time terms cancel, so the <interference fringes> stay at fixed $x$ for every $z$. Their <interference fringe spacing> is
$$
\Lambda=\frac{2\pi}{2k\sin\theta}=\frac{\lambda}{2\sin\theta},\qquad
\frac{I(x,z)}{I_0}=4\cos^2(kx\sin\theta).
$$
Set the length unit to $\lambda$, so that $\lambda=1$, and take $\theta=30^\circ=\pi/6$: then $\Lambda=1$. The <constructive interference intensity> bands lie at integer $x/\lambda$; the <destructive interference intensity> bands lie at half-integer $x/\lambda$.
\Image[/two-beam-interference-fringes.png]
{title=Ideal crossing <plane waves> at $\pm30^\circ$: stationary <interference fringes> with spacing one <wavelength>}
The grid covers $-3\leq x/\lambda\leq3$ and $0\leq z/\lambda\leq4$; both axes are dimensionless lengths in <wavelength> units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal <plane wave> superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized <plane electromagnetic waves> in <vacuum>, the longitudinal time-averaged <Poynting vector> obeys $\langle S_z\rangle=I\cos\theta$; the plotted scalar $I$ therefore differs from the <irradiance> through a plane normal to $z$ by that common factor. This relation follows by adding the <magnetic fields> as well as the <electric fields>.
The bright peaks do not create <energy>. Averaging over one <interference fringe spacing>, or equivalently over one full <phase difference> cycle, gives
$$
\frac1{2\pi}\int_0^{2\pi}I(\delta)\,d\delta=I_1+I_2,
$$
because the <integral> of the <cosine> is zero, with <sine> as its <antiderivative>. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal <energy flux> is correspondingly $2I_0\cos\theta$. Bright and dark regions redistribute the incoming <energy>; <conservation of energy> concerns the full <energy flux>, rather than the height of one peak.
<Wave interference> is useful because <phase differences> convert small changes in <optical path length> into measurable changes of <wave intensity>, as in a <Mach-Zehnder interferometer>. Its beauty is that the same <superposition principle> produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a <diffraction grating>. Feynman's treatment connects this many-source pattern directly to addition of <complex amplitudes>. https://www.feynmanlectures.caltech.edu/I_30.html#Ch30-S1[Feynman, Leighton and Sands, The Feynman Lectures on Physics, I §30–1: the resultant amplitude of identical oscillators]{ref}.
A short proof outline uses local real fields $E_j(t)=A_j\cos(\omega t+\phi_j)$ with parallel <linear polarizations>. Write $I=C\langle(E_1+E_2)^2\rangle_t$ and $I_j=CA_j^2/2$ for the same positive conversion factor $C$. Averaging over many <wave periods> while the <phase difference> stays fixed uses $\langle\cos^2(\omega t+\phi_j)\rangle_t=1/2$ and $\langle\cos(\omega t+\phi_1)\cos(\omega t+\phi_2)\rangle_t=\cos\delta/2$, giving the stated law. Equivalently, add the <complex amplitudes> and take the squared <complex modulus>. https://www.feynmanlectures.caltech.edu/I_29.html#Ch29-S5[Feynman, Leighton and Sands, The Feynman Lectures on Physics, I §29–5: the mathematics of interference]{ref}.
For different fixed <linear polarizations> in a common transverse plane, the cross term acquires the <dot product> of their real <unit vectors>, equal to $\cos\alpha$ for relative <angle> $\alpha$. Orthogonal <linear polarizations> remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common <polarizer> can select overlapping components and recover <wave interference>. Likewise, rapid random variation of the <phase difference> removes the observed cross term only when the detector's time average has $\langle e^{i\delta(t)}\rangle_t=0$. Slowly varying <wave phases> can instead move resolved <interference fringes>, and biased or incompletely sampled <wave phases> can leave residual contrast. Finite <optical coherence> therefore depends on the observation time as well as the fields.
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