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Take $u>0$ to mean that the mirror recedes along the incident plane electromagnetic wave. Put $\beta =u/c$, with $|\beta |<1$. The moving-mirror Doppler shift can be obtained directly by matching wave phase on the moving surface. Write the incident wave phase as $k_{i}z-{\omega }_{i}t$ and the reflected wave phase as $-k_{r}z-{\omega }_{r}t$, where $k_i={\omega }_i/c$ and $k_r={\omega }_r/c$ are positive wavenumbers. At the mirror, $z=z_0+ut$, the two phases must have the same time dependence; their constant offset incorporates the reflection phase shift. ThusConsequently the observed reflected angular frequency and ordinary frequency areThis is a redshift for a receding mirror and a blueshift for an approaching mirror. At small velocity, ${\omega }_r\simeq {\omega }_i(1-2u/c)$.

Equivalently, the relativistic Doppler effect gives ${\omega }_i^{\prime }=\gamma (1-\beta ){\omega }_i$ in the mirror rest inertial frame. Reflection there reverses propagation and preserves frequency. Transforming the backward-going reflected wave to the laboratory gives another factor $\gamma (1-\beta )$, so ${\omega }_r={\gamma }^2(1-\beta {)}^2{\omega }_i={\omega }_i(1-\beta )/(1+\beta )$. A single Doppler factor would miss the effect of the return path.

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.

Away from the shell, the reduced radial equation is free. Regularity at the origin and the definition of the S-wave phase shift giveThe reduced radial wave function is continuous at $R$; otherwise its second derivative would contain a delta derivative not present in the equation. Integrating the equation across the shell gives the jumpDivide by the common shell value and substitute the two free solutions. This yieldsIf the shell value vanishes, use the undivided matching equations or a continuous limit rather than this logarithmic-derivative division. For an explicit phase shift of a delta-shell potential, write $x=kR$ and $g=2m{V}_{0}R/{\hslash }^2$. The cotangent addition formula givesThe phase is defined modulo $\pi$; choose its branch continuously as energy or coupling varies. At zero coupling the physical phase shift is zero.

The scattering length is $a_0=-{\lim }_{k\to 0}\tan {\delta }_0/k$. Since $\sin (kR)\sim kR$ and $\cos (kR)\sim 1$, the exact phase shift gives, away from the pole,Thus plot $a_0/R$ against $g$, so that both axes are dimensionless.

For repulsion, $g>0$, the curve rises from zero to the horizontal asymptote $a_0/R=1$, staying between zero and one. For weak attraction, $-1<g<0$, the scattering length is negative and tends to $-\infty$ as $g\to -1^{+}$. For stronger attraction, $g<-1$, it is positive and greater than $R$, tending to $+\infty$ as $g\to -1^{-}$ and to $R$ as $g\to -\infty$. Its only zero is $g=0$.

The pole at $g=-1$ is the zero-energy threshold of an S-wave bound state. To see this directly, use $u\propto \sinh (\kappa r)$ inside and $u\propto e^{-\kappa r}$ outside for energy $-{\hslash }^2{\kappa }^2/(2m)$. The same derivative jump givesThis approaches $-1$ as $\kappa \downarrow 0$ and admits a positive $\kappa$ for $g<-1$. At the threshold the scattering length diverges; this is a zero-energy scattering resonance, not a divergence of the finite-energy cross-section beyond its partial-wave unitarity bound. These are the main features of the scattering length of a delta-shell potential.

A phase shift is a change in an oscillation's wave phase, specified modulo $2\pi$. It can arise from propagation, reflection or an imposed response. With the time convention $e^{-i\omega t}$, a delay $\tau$ multiplies the complex amplitude by $e^{i\omega \tau }$. A sign reversal is a $\pi$ phase shift.

For $V(r)=V_0\delta (r-R)$, put $g=2m{V}_{0}R/{\hslash }^2$. The regular reduced radial wave function is proportional to $\sin kr$ inside the shell and to $\sin (kr+{\delta }_0)$ outside. It is continuous at $R$, while its radial derivative jumps by $(g/R)u(R)$. Dividing the jump condition by the common value gives $k\cot (kR+{\delta }_0)-k\cot (kR)=g/R$, which implies the displayed phase shift modulo $\pi$. At nodes, the undivided matching equations give the continuous limiting interpretation. For $g\ne 0$, an equivalent formula is $k\cot {\delta }_0=-k\cot (kR)-k(kR)/[g{\sin }^2(kR)]$. The exact S-wave scattering amplitude is $[k\cot {\delta }_0-ik{]}^{-1}$.

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.