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Destructive interference occurs when coherent contributions partially or completely cancel their complex amplitudes. For two equal contributions, a phase difference of $\pi$ modulo $2\pi$ gives zero resultant wave amplitude. The missing local light intensity is consistent with conservation of energy because the full spatial interference pattern redistributes power.

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.