= Solution
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 $2\pi$,
$$
\Delta\phi=\frac{4\pi nd\cos\theta_t}{\lambda_0}+\pi,
$$
where $\lambda_0$ is the <vacuum> <wavelength>. <Constructive interference> in reflected <light> therefore occurs approximately when
$$
\boxed{2nd\cos\theta_t=(m+\tfrac12)\lambda_0,\qquad m=0,1,2,\ldots.}
$$
<Destructive 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.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2021/paper-2-a2-film-interference.png]
{title=Two reflected paths through a thin film and the broad spectral filtering that produces mixed colours}
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.
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