In general, scattering intensity is written as $I = I_{0} \frac{\pi a^2 Q_{sca} P(\theta)}{r^2 4 \pi}$

It is also written as $I = I_{0} \frac{i_1 + i_2}{2 k^2 r^2}$

I am sort of confused, given the expressions of $i_1$ and $i_2$, is the phase function $P(\theta)$ ONLY dependant on the scattering angle $\theta$?

It is my understanding that the scattering efficiency $Q_{sca}$ incorporates depends on index of refraction, sphere size and wavelength, but the phase function only depends on the scattering angle. Is my understanding correct?


1 Answer 1


Be careful what you mean by phase here. There is an angle at which intensity is maximum

$$I = \underbrace{|{\boldsymbol E}|^2}_{\mathrm{total\,\,\, measured \,\,\,field}} = I_0 \frac{\pi a^2 Q_{\mathrm{sca}} P(\theta)}{4 \pi r^2} $$

which is the mod squared of the electric field for one particle with boundary conditions within the Mie regime of the quasi-static scattering theory.

However, when you have multiple particles, as in a disperse medium, you will have near-field scattering effects and have to incorporate the interference from scattered electric fields at different phases relative to one another. This phase information is stored in the Bessel functions of the analytical solution.

$$ I_{\mathrm{measured}} = \left| \sum\limits_k {\boldsymbol E}_k + {\boldsymbol E}_{\mathrm{res}}\right|^2 $$

for $k$ scatter centers, which makes the problem more computationally taxing. I don't remember if the incoming beam also interferes, but there is some residual effects $({\boldsymbol E}_{\mathrm{res}})$ in the environment. Perhaps this answer will help.


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