Physics Asked on October 12, 2020
Wikipedia’s Mie scattering Mathematics discusses the scattering amplitudes $S_1(theta), S_2(theta)$ for each outgoing polarization of an incident EM plane wave on a uniform sphere.
It defines the scattered intensities as
$$i_1(theta) = |S_1(theta)|^2$$
$$i_2(theta) = |S_2(theta)|^2$$
and then defines the scattering intensity
$$I(theta) = I_0 frac{lambda^2}{8 pi^2 r^2} left(i_1(theta) + i_2(theta)right)$$
where r is the radius of the sphere.
The incoming intensity $I_0$ would likely have units of power per unit area, whereas I’d expect the scattered intensity or radiance $I(theta)$ to have units of power per unit solid angle, but if that were true the units don’t work.
Could $I_0$ possibly be the power incident on the geometrical cross-section $pi r^2$?
I’m doing an approximate calculation of something similar to what’s described in this question. I’ve found a python script that generates the scattering amplitudes $S_1(theta), S_2(theta)$, I’ll have a laser pointer with $I$ watts/m^2 and a detector at some angle $theta$ with solid angle $Delta Omega$ and I’d like to calculate the scattered intensity reaching that detector for a given single particle, but I’m stuck on how to correctly convert from $S$ to scattered power into $Delta Omega$.
The answer there says:
Wikipedia just pointed me to an English translation of the original paper by Mie, which I didn’t know existed. I haven’t yet had a chance to read it, so I don’t know how useful it is.
but I’m not able to access that link.
I think the intensity here in the radiometry context is called irradiance and it is power per unit area. When your detector is far enough from the scatter, power per unit area is also power per unit area per solid angle (radiance). To calculate ΔΩ, we can integrate intensity over a solid angle. There might be a cos factor, depending on whether your detector is perpendicular to the scattered light.
Answered by xmq on October 12, 2020
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