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Color vision distinguishes distributions of visible light across wavelengths through the different spectral sensitivities of visual receptors. A perceived colour need not correspond to one wavelength: different optical spectra can produce the same receptor responses. Consequently the broad spectral filtering of thin-film interference need not produce pure monochromatic colours.

At the same output in air, use incident intensity $I_{\mathrm{inc}}\propto |A{|}^2$. The two specified output amplitudes, including their common reflection sign, give the thin-film interference resultThe output is a cosine with mean $0.0234I_{\mathrm{inc}}$, period $2\pi$, and extremaThe unequal amplitudes prevent a dark minimum; their interference visibility is $0.009/0.0234=5/13\simeq 0.385$. This is the stipulated two-beam model: the supplied amplitudes already include propagation and transmission factors, and internal re-reflections are excluded by the stated assumption. They need not be recomputed from the later index value.

In division of amplitude, one incident beam is split into two or more parts at a partially reflecting interface; they follow distinct paths and are recombined. The Michelson interferometer and thin-film interference use this principle. In division of wavefront, spatially distinct parts of the same wavefront are selected and allowed to overlap, as in double-slit interference. Both require adequate optical coherence; wavefront division particularly exposes the need for spatial optical coherence between the selected regions.

In thin-film interference, coherent beams reflected at the two faces of a transparent film overlap. Let the film have refractive index $n_f$, thickness $t$, and internal ray angle ${\theta }_f$, determined by Snell's law. For vacuum wavelength ${\lambda }_0$, the propagation part of the relative phase isAlthough the internal round-trip ray length is $2t/\cos {\theta }_f$, the two emergent rays must be compared on the same outgoing wavefront. Subtracting their external lateral path contribution gives the effective optical path difference $2n_{f}t\cos {\theta }_f$, explaining the cosine rather than its reciprocal.

Reflection also changes phase. At normal incidence in lossless positive-index media, the electric-field Fresnel equations give $r_{ij}=(n_i-n_j)/(n_i+n_j)$: reflection towards higher index reverses sign, whereas reflection towards lower index does not. If the two interfering reflections differ by one such phase reversal, their total relative phase is ${\delta }_{\mathrm{prop}}+\pi$ modulo $2\pi$. Then reflected maxima and minima obeyIf there are zero or two reversals, the conditions interchange. These bright/dark conditions assume overlapping beams with adequate temporal optical coherence; complete cancellation further requires equal amplitudes. In transmitted light, a lossless complete film has complementary intensity variations because reflected plus transmitted power is conserved.

For multiple internal reflections at normal incidence, summing the geometric series gives the exact film reflection amplitudeThis includes both the amplitude mismatch and repeated-beam interference. A normal-incidence antireflection coating on substrate index $n_s>n_0$ can cancel reflection at one wavelength when $n_f=\sqrt{n_0{n}_s}$ and $t={\lambda }_0/(4n_f)$: both interface reflections reverse sign and the film's round-trip propagation adds $\pi$, with equal interface amplitudes. A varying thickness or angle selects different maxima for different wavelengths, producing the colours of transparent films in white light. Broad bandwidth and large incidence angles generally prevent perfect cancellation across all wavelengths.

A bubble is a thin water layer surrounded by air. At a locally flat patch, let its thickness be $d$, its refractive index be $n$, and the internal angle of refraction be ${\theta }_t$. The two principal reflected light contributions are the reflection at the first surface and the contribution that enters the layer, reflects at its second surface and leaves again. The first reflection, from lower to higher refractive index, introduces a $\pi$ phase shift; the second does not. Their optical path difference is $2nd\cos {\theta }_t$, so their phase difference is, modulo $2\pi$,where ${\lambda }_0$ is the vacuum wavelength. Constructive interference in reflected light therefore occurs approximately whenDestructive interference occurs at $2nd\cos {\theta }_t=m{\lambda }_0$. In the weak-reflection approximation, the reflected light intensity at each wavelength is proportional to ${\sin }^2(2\pi nd\cos {\theta }_t/{\lambda }_0)$. This is a smooth spectral filter, with broad maxima and minima: it does not select three pure wavelengths corresponding to display primaries. Ordinary illumination contains many visible wavelengths, and the eye combines their remaining contributions through color vision. The result is a mixture of colours, often pastel or desaturated.

The layer thickness varies over the bubble because of drainage and flow; the bubble's curvature also changes the local angle of refraction and the viewed optical path difference. The spectral filter therefore changes continuously across the surface and in time. Very thin regions become dark because the two reflected contributions nearly cancel across the visible optical spectrum. Thin-film interference explains both the changing colours and this dark limit.

The ray sketch shows a local patch and is not to scale. The plotted curves use a chosen constant $n=1.33$ at normal incidence and illustrative thicknesses; they show the spectral filter rather than calibrated perceived colours.

Reflection and central spot. At a boundary from lower to higher refractive index, the near-normal electric-field reflection coefficient is negative, $r=(n_1-n_2)/(n_1+n_2)$. This gives a $\pi$ wave phase reversal on reflection from the glass plate through the air film. Reflection at the lens's glass-to-air underside has no such reversal. These are consequences of the Fresnel equations for the reflected electric field.

At ideal contact the film thickness is zero, so the two reflected contributions have zero geometrical optical path difference but a relative $\pi$ phase reversal. They produce destructive interference, and the central spot is dark. This conclusion concerns the film reflection; other lens surfaces can add a weak background.

Rings and their radii. At radial distance $r$, let the air thickness be $t$. The two relevant rays reflect at the upper and lower film interfaces. The lower reflection accumulates an extra round-trip optical path length $2t$, plus the relative $\pi$ reflection shift. As $t$ increases away from contact, alternating constructive and destructive thin-film interference forms Newton's rings. At near-normal incidence, a bright ring satisfiesThe extended source can illuminate these localized film fringes: each incident contribution interferes with its own two reflected paths; unrelated source points need not be mutually coherent.

For the spherical lens surface, geometry gives $r^2+(R-t{)}^2=R^2$, so $r^2=2Rt-t^2$. Since $t\ll R$, $t\simeq r^2/(2R)$. Combining the equations yieldsThe $m=0$ bright ring is the first bright ring around the dark contact point.

Curvature from ring diameters. Writing $D_m=2r_m$, subtract the squared diameters to eliminate the unknown initial order:Using SI units,

Liquid-filled film. With film refractive index $n=1.4$, still smaller than both glass indices, there remains exactly one relative reflection phase reversal. The round-trip optical path length is now $2nt$, soHere $\lambda =500\mathrm{nm}$ is the illuminating vacuum wavelength, approximately its air value, rather than its shorter value inside the liquid. The ring labelled $m=2$ therefore has diameterFilling the gap contracts each ring radius by $1/\sqrt{n}$ at fixed order and curvature.

Thin-film interference combines coherent light contributions reflected or transmitted at the two surfaces of a layer. For a film of refractive index $n$, thickness $d$ and internal angle ${\theta }_t$, the round-trip optical path difference is $2nd\cos {\theta }_t$. The relative reflection phase shift must also be included. For a nonabsorbing higher-index film surrounded by air, one of the two leading reflected contributions has a $\pi$ phase shift and the other does not. Reflection maxima therefore satisfy $2nd\cos {\theta }_t=(m+1/2){\lambda }_0$. A smooth reflection spectrum filters many visible wavelengths at once; varying thickness and angle produce spatially varying mixed colours.