# Mie theory: Interpretation in terms of intensity

I'm trying to understand Mie theory. For this I'm reading the book "Absorption and Scattering of Light by Small Particles" by Bohren and Huffman.

The derivation of the formulas is fine, but I'm stuck at interpretation. I'm expecting the following:

Take a ray hitting a sphere. Then three things are possible:

a) The ray is absorbed

b) The ray is scattered

c) No interaction between ray and sphere

Let's assume the ray has unity intensity. Then some fraction of it should be absorbed, some fraction of it should be scattered, and the remaining part has no interaction with the sphere. Note that the scattered part should be given by an angle-dependent function.

I'm unable to compare this expectation with the formalism presented in the book - they talk mostly about cross-sections

So here is my question:

Hopefully, this question makes sense.

Any finite object illuminated by an infinite planar wave scatters a part of its energy. The ratio of the overall scattered energy to the power density of the incoming wave has the dimensions of meter squared, and thus is described as scattering cross-section.

A large mirror has scattering cross-section A similar to its geometrical dimensions. However, in mostly transparent objects A would be less, and interestingly, in resonant objects it can be more than would be deduced from the geometry.

Mie theory uses the extinction cross section, $$\sigma_e$$, which is equal to the sum of the scattering cross section and the absorption cross section:

$$\sigma_e = \sigma_s + \sigma_a$$

For radiation (e.g., light, radar) propagating in the $$z$$ direction, the flux will decreases as:

$$\Phi(z) = \Phi e^{-n\sigma_ez}$$

where $$n$$ is the number density $$(L^{-3})$$ of scattering centers, $$z$$ is length $$(L)$$, and the cross section has dimensions $$L^2$$, making the exponent dimensionless.

Note that in the Rayleigh region, the total scattering cross section is proportional to $$d^6$$ ($$d$$ is the diameter of the scattering sphere), which is why rain radar reports results in $$dB_Z$$, where $$Z=1$$ has units mm$$^6$$ per cubic meter, corresponding to one single 1 mm raindrop per cubic meter in the radar's resolution cell.

• So the number of scattering areas in the volume is 'd to the 6th'? Sep 7, 2020 at 17:48
• no. $d^6$ is the diameter of the raindrop to the sixth-power, so in real rain radar the reflectivity goes as the 6th moment of the drop-size distribution, while the total volume of water goes as the 3rd moment, and then the rain-rate (which is what end-users want to know) also depends on how fast those drops fall, which a function of $d$.
– JEB
Sep 7, 2020 at 18:34