- #1
bananabandana
- 113
- 5
Is there some way to - from an intuition standpoint - justify the fact that there should be a factor of ##a^{6}##, (where ##a ## is the particle diameter) in the Rayleigh Scattering formula? I've seen a few sources hint that there should be. I can follow the derivation from e.g a Lorentz atom, but I don't see why I should immediately be thinking of the factor of ##a^{6}##? [Is it somehow related to a dipole moment?]
Rayleigh Scattering Formula:
$$ I \propto I_{0} \lambda^{-4} a^{6} $$
Rayleigh Scattering Formula:
$$ I \propto I_{0} \lambda^{-4} a^{6} $$