= Rayleigh scattering wavelength dependence
{c}
{title2=$\sigma\propto\lambda^{-4}$}
For a fixed <sphere> and fixed <refractive index>, both the <Rayleigh scattering cross-section> and <scattering efficiency> are proportional to $\lambda^{-4}$. Halving <wavelength> multiplies either by $16$, as long as the shorter <wavelength> still satisfies $ka\ll1$. If <optical dispersion> changes $m$, the material factor changes too; the pure fourth-power comparison then needs correction.
For equal incident <light intensities> at blue <wavelength> $450\,\mathrm{nm}$ and red <wavelength> $650\,\mathrm{nm}$, the same toy <sphere> gives
$$
\frac{\sigma(450\,\mathrm{nm})}{\sigma(650\,\mathrm{nm})}=\left(\frac{650}{450}\right)^4=\frac{28561}{6561}\simeq4.35315.
$$
Blue <light> therefore produces about $4.35$ times the scattered <power>. This compares the particle's response at two <wavelengths>; it does not assert that an actual sky has this ratio of observed blue and red <light intensities>.
\Image[/rayleigh-wavelength-scaling.png]
{title=<Rayleigh scattering wavelength dependence> for a fixed small <sphere>, normalized at $650\,\mathrm{nm}$}
{description=The curve is $(650\,\mathrm{nm}/\lambda)^4$ over $350$ to $750\,\mathrm{nm}$. Blue and red comparison points are marked at $450$ and $650\,\mathrm{nm}$. Fixed <radius> and nondispersive <refractive index> are assumed, with $ka\ll1$ throughout. This is a particle response curve, not a simulated sky <optical spectrum>.}
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