Two-Wave Interference and the Interference Term

The interference term is the sensing element of precision optics. Amplitude-splitting interferometers — Michelson, Mach–Zehnder, Sagnac — turn a path difference into a fringe count, measuring displacement, refractive index, rotation, and, in Fourier-transform spectrometers, the entire spectrum of a source from the way its visibility decays with path difference. Gravitational-wave detectors are kilometre-scale Michelson interferometers reading a phase shift $\delta$ corresponding to a strain of $\sim10^{-21}$. Thin-film interference, the same cross term mapped onto film thickness, is engineered into the anti-reflection coatings on every camera and eyeglass lens and the high-reflectors of every laser mirror. In metrology the term underlies laser interferometers that define the metre, calibrate machine tools, and test optical surfaces to nanometre flatness; in astronomy, stellar interferometers recover the angular sizes of stars from the visibility of fringes formed by widely separated apertures — the direct experimental route from this two-line formula to the diameters of distant suns.

Represent the two fields as phasors of length $\sqrt{I_1}$ and $\sqrt{I_2}$ separated by an angle $\delta$. The resultant intensity is the squared length of their vector sum, and the law of cosines hands you $I_1+I_2+2\sqrt{I_1I_2}\cos\delta$ immediately. When the phasors point the same way they add head-to-tail into a long resultant (bright); when opposed they nearly cancel (dark). Because intensity is the square of a sum, it is the amplitudes that combine, not the intensities — which is why two equal beams give $4I_0$ and $0$, not a flat $2I_0$.

The whole pattern lives or dies by whether $\cos\delta$ survives the time average. If $\delta$ drifts randomly faster than the detector integrates (incoherent light), or if the two fields are orthogonally polarized, the average of the cross term vanishes and the fringes flatten into a featureless $I_1+I_2$. When the beams are unequal, the contrast survives but weakens — and that weakening has a clean formula, the punchline.

Equal beams give the sharpest fringes, $V=1$; a four-to-one intensity ratio already drops the contrast to $0.8$; a very weak second beam ($r\gg1$) gives $V\approx2/\sqrt r\to0$ and the fringes fade. The visibility is the bridge from this elementary law to coherence theory, where a partial-coherence factor multiplies the cross term.

We start from the two recalled electromagnetic facts and the superposition principle. First, what the detector measures — recalled from electromagnetism — is the time-averaged flux, so a single wave of amplitude $E_0$ has intensity:

where: $\mathbf S=\tfrac1{\mu_0}\mathbf E\times\mathbf B$ — the Poynting vector; $\langle\cdot\rangle$ — cycle average; the $\tfrac12$ from $\langle\cos^2\rangle=\tfrac12$. Both the Poynting vector and this cycle-averaged intensity are owned by electromagnetism (aq-em-energy-density-poynting-vector, aq-em-intensity-time-averaging); optics recalls them and superposes. Two scalar monochromatic waves of the same frequency, sharing a polarization, are written:

where: $E_{0i}$ — real amplitude; $\mathbf k_i$ — wavevector; $\omega$ — common angular frequency; $\varphi_i$ — intrinsic phase. By the superposition principle (recalled from the previous article) the total field is $E=E_1+E_2$. Squaring the summed field and time-averaging (Derivation 1) yields the two-wave interference law, the central result:

where: $I$ — total time-averaged intensity; $I_i\propto E_{0i}^2$ — the intensity of each wave alone; $\delta$ — phase difference at the observation point. The third term is the interference term; the first two are the naïve sum that would hold if the waves were incoherent. The phase difference collects the geometry and any intrinsic offset:

where: $\Delta=n\,\Delta\ell$ — optical path difference; $\lambda$ — vacuum wavelength; $\Delta\varphi$ — intrinsic offset (for example a $\pi$ shift on reflection). Bright fringes occur at $\delta=2m\pi$, dark at $\delta=(2m{+}1)\pi$, with $m$ an integer. The extremes of $(3)$ are read off at $\cos\delta=\pm1$:

where: $I_{\max}$ — bright-fringe intensity; $I_{\min}$ — dark-fringe intensity. For equal beams $I_1=I_2=I_0$ these become $4I_0$ and $0$, and the whole pattern collapses to the tidy form:

where: $I_0$ — intensity of each equal beam; the half-angle form makes the $4I_0$ peak and the zero minimum explicit. The contrast of the fringes is quantified by the visibility (Michelson contrast), derived in $(5)$:

where: $V$ — fringe visibility, $0\le V\le1$; it equals $1$ when $I_1=I_2$ and falls as the beams become unequal, incoherent, or cross-polarized. Finally, energy is conserved because the cross term averages away over a fringe:

where: $\langle I\rangle_\mathrm{fringe}$ — intensity averaged over one fringe period; the interference term integrates to zero, so no energy is created or lost.

1 · Equal beams: the 4I₀ fringe and where the energy goes (confidence builder)

Two equal coherent, co-polarized beams of intensity $I_0$ each overlap. Find the bright- and dark-fringe intensities, and show the pattern conserves energy.

Step 1. With $I_1=I_2=I_0$ the law $(3)$ becomes $I=2I_0+2I_0\cos\delta=2I_0(1+\cos\delta)=4I_0\cos^2(\delta/2)$. The amplitudes, not the intensities, are what add.

Step 2. At a bright fringe $\cos\delta=+1$: $I_{\max}=(\sqrt{I_0}+\sqrt{I_0})^2=\boxed{4I_0}$ — four times a single beam. At a dark fringe $\cos\delta=-1$: $I_{\min}=(\sqrt{I_0}-\sqrt{I_0})^2=\boxed{0}$ — total darkness.

Step 3. Average over one fringe: $\langle I\rangle=\tfrac1{2\pi}\int_0^{2\pi}4I_0\cos^2(\delta/2)\,\mathrm d\delta=\boxed{2I_0}$, exactly the sum of the two inputs. Energy is conserved: the $2I_0$ removed at the dark fringes is exactly the $2I_0$ of excess brightness gathered at the maxima.

Step 4. The visibility is $V=2\sqrt{I_0I_0}/(2I_0)=\boxed{1}$ — maximum contrast, the signature of equal, coherent, co-polarized beams.

2 · Unequal beams and the Young fringe spacing (exam level)

Two coherent beams have an intensity ratio $I_1/I_2=4$. Find the visibility. Then, for a Young double slit of separation $d=0.20\ \mathrm{mm}$ illuminated at $\lambda=550\ \mathrm{nm}$ with a screen at $L=1.0\ \mathrm{m}$, find the fringe spacing.

Step 1. Visibility from $(7)$: with $I_1=4I_2$, $V=\dfrac{2\sqrt{I_1I_2}}{I_1+I_2}=\dfrac{2\sqrt{4I_2^2}}{5I_2}=\dfrac{2\cdot2I_2}{5I_2}=\boxed{0.8}$. Even perfectly coherent light gives less than unit contrast once the amplitudes are unequal.

Step 2. For the double slit the two wavefront-split beams acquire a path difference $\Delta=d\sin\theta$, so $\delta=2\pi d\sin\theta/\lambda$ and the pattern is $I=4I_0\cos^2(\pi d\sin\theta/\lambda)$ (equal beams). Bright fringes fall at $d\sin\theta=m\lambda$.

Step 3. For small angles $\sin\theta\approx y/L$, so adjacent bright fringes are separated on the screen by $\Delta y=\dfrac{\lambda L}{d}=\dfrac{(550\times10^{-9}\ \mathrm{m})(1.0\ \mathrm{m})}{0.20\times10^{-3}\ \mathrm{m}}=\boxed{2.75\ \mathrm{mm}}$.

Step 4. So the same cross term draws millimetre-scale fringes from a sub-micron wavelength — the geometry $L/d$ magnifies $\lambda$ into a visible spacing — while the amplitude ratio sets how sharply those fringes stand out. This is the elementary two-wave law feeding directly into Young's experiment and every amplitude-splitting interferometer.

The protagonist here is a single algebraic object — the cross term — that recurs across all of two-beam optics, and it repays being looked at directly.

Texts. Hecht, Optics, 5th ed., ch. 9 (Interference), Eqs. 9.11, 9.14 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 2 (Interference of Light) · Loudon, The Quantum Theory of Light, 3rd ed., ch. 3 (coherence and the degree of coherence) · Born & Wolf, Principles of Optics, §7.2 (two-beam interference) · Foot, Atomic Physics (interference background).

Provenance. That intensity is the time-averaged Poynting flux $(1)$ is recalled by name from aq-em-intensity-time-averaging and aq-em-energy-density-poynting-vector (grep-verified: the Poynting vector, energy density, and the cycle-averaged intensity with its $\tfrac12$ factor are derived and owned there); superposition of the fields is recalled from aq-optics-superposition-principle-em-waves. Optics owns the two-wave interference law, its visibility, and the coherence/polarization conditions. verify_two-wave-interference-term.py reproduces: $\langle\cos^2\rangle=\tfrac12$ and $\langle\cos A\cos B\rangle=\tfrac12\cos(A-B)$; the law $(3)$; $I_{\max}=(\sqrt{I_1}+\sqrt{I_2})^2$, $I_{\min}=(\sqrt{I_1}-\sqrt{I_2})^2$; $I=4I_0\cos^2(\delta/2)$; $V=2\sqrt{I_1I_2}/(I_1+I_2)$ with $V=0.8$ at $I_1/I_2=4$ and $V=1$ at $I_1=I_2$; the energy integral $(8)=I_1+I_2$; the uniform-random-phase averages $=0$; and the Young spacing $\Delta y=\lambda L/d=2.75\ \mathrm{mm}$ for $\lambda=550\ \mathrm{nm}$, $d=0.20\ \mathrm{mm}$, $L=1.0\ \mathrm{m}$. Follows Hecht §9 Eqs. 9.11, 9.14; the visibility matches Michelson's contrast; the coherence/polarization averaging follows Loudon ch. 3 and the Fresnel–Arago laws.

Builds on: Superposition Principle of Electromagnetic Waves · Optical Path Difference and Phase Retardation · Conditions for Observable Interference. Leads to: Young's Double-Slit Experiment · Fringe Visibility and Contrast · Thin-Film Interference · Mutual Coherence Function and Degree of Coherence.