3
$\begingroup$

I'm trying to understand what Mie scattering means. In spherical particles that are larger than the wavelength of the light, the direction of scattering depends on the size of the particle. To me, what happens now seems incredibly counterintuitive. Larger particles scatter more light forward, while in smaller particles, the light is scattered into many directions? If anything, I would have assumed that a small particle affects the beam path less than a bigger one.

In some diagrams that I have seen to visualize this, it looks like the light is transmitted through the particle, then scattered when exiting it. Is that really what happens, or does it rather glance off the side of the particle?

If it is transmitted through the particle, does that mean that the refractive index of the material of the particle also has a large effect on how the light is scattered, and not just the size of the particle?

$\endgroup$

3 Answers 3

1
$\begingroup$

You should provide the source of your information. According to my source (M. Born, E. Wolf, Principles of Optics, fourth ed., Pergamon, 1968), "Larger particles scatter more light forward" when they are still smaller than the wavelength. When a particle is much larger than the wavelength, it mostly reflects, in accordance with geometric optics.

As for intuition, one can compare optical paths for rays diffracted from opposite edges of the sphere: their difference is greater for larger spheres, so the interference condition is satisfied for smaller forward angles.

And yes, the refractive index has a significant effect for example: if the refractive index equals 1 and there is no conductivity, there is no scattering.

$\endgroup$
4
  • $\begingroup$ Thanks! That helps. So with a refractive index of 1, there is no scattering at all, because the light can't pass through the particle - but what about backscatter, can't that still be happening? $\endgroup$ Nov 29, 2021 at 11:18
  • $\begingroup$ @protoperidinium : if the refractive index is 1, the light passes through the particle alright, but what one sees in the forward direction is unscattered light, and there is no backscattering. $\endgroup$
    – akhmeteli
    Nov 29, 2021 at 14:21
  • $\begingroup$ I see! Sorry, I got a bit muddled there. So by conductivity you mean conductivity of the material, and if the material is conductive, that means it interacts with the light to some degree = refracts and scatters it? $\endgroup$ Nov 29, 2021 at 15:14
  • $\begingroup$ @protoperidinium : That is right, with one caveat: what matters is the conductivity (or absorptivity) at the frequency of the light (see the chapter on metallooptics in Born,Wolf). $\endgroup$
    – akhmeteli
    Nov 29, 2021 at 15:40
1
$\begingroup$

Generally, forward scattering by spherical particles is dominated by two effects:

  1. Diffraction of the incident wave that propagated around the particle (or, for a complementary view, diffraction of the particle shadow),
  2. Transmission through the particle.

This is easy to see if we examine the examine the Debye series, which is a decomposition of the Mie solution into the terms labeled by variable $p\ge0$. Each term of this series corresponds to $p-1$ internal reflections of the wave occurring before the wave gets out of the sphere ($p=0$ being external reflection+shadow).

Here's a plot of the two Debye terms (corresponding to the two above mentioned effects) for scattering of a wave with wavelength $650\,\text{nm}$ by a sphere with size $10\,\text{μm}$ and refractive index $1.33$:

The black line shows the full Mie solution for the scattered far field, scattering angles changing from $0°$ to $90°$. The green line is the reflection+shadow term of the Debye series, whose main contribution is in the small angles $\theta\lesssim5°$. The red line is the transmitted wave through the sphere. As you can see, the transmitted wave scatters very broadly, but mainly in the forward direction.

Other terms of the Debye series add to the oscillatory behavior of the phase function, but don't change the large-scale behavior for $\theta<90°$.

$\endgroup$
0
$\begingroup$

Sorry but I think you are a bit confused. A transparent medium different from vacuum space, can simultaneously refract and scatter visible light but this does not mean that these two phenomena are related. For visible light spectrum you can have a transparent medium which refracts light but practically does not scatter light like clear glass but also a medium like air which practically does not refract visible light (or tiny little) and only scatters light.

A homogeneous medium either refracts monochromatic light or in your case mentioned scatters light or both. If the material is refractive to a specific monochromatic light it will homogeneously twist the path of a non vertical incident beam of monochromatic light when passing through it. In case of a spectrum of light like for example white light it will cause due to different refractive indices to different wavelengths, frequency dispersion of the incident white light at an angle and split it to its color components and a colored spread out spectrum of light will exit the medium (like a rainbow) but this has nothing to do with the scattering phenomenon but the light gets spread out to its spectrum components due the different refraction indices.

Scattering is a total different phenomenon where incident monochromatic light gets deflected (bounced of) at multiple random angles to the forward direction and all different form the incident angle, from particles of a medium. In this case a monochromatic beam will be much spread out when it exits such a medium. In Rayleigh scattering the size (diameter) of the particles must be less than about ~$1/10$ David J. Lockwood the incident monochromatic wavelength of light for this to occur. When a light spectrum is incident to such a medium only a specific wavelengths components matching the above condition are scattered by the medium and the rest of the spectrum components get either refracted or Mie scattered or reflected or absorbed according the the medium particle's size and incoming wavelengths of light. In case of air particles and white mid-day sunlight, blue light component is mostly forward scattered and intensified and directed towards the Earth surface therefore the blue of the sky. During dawn or sunset light travels a much more thicker layer of atmosphere to reach to us therefore there is larger probability to blue light to be scattered to directions different from your straight line of sight and thus mainly the red (largest visible wavelength) which is not scattered by air reaches your eyes. Similar the mid-day sun disk on a clear sky appears yellowish because the dominant percentage of blue scattered light and red unscattered light mix, reaching your straight line of sight.

So Rayleigh scattering acts like a dynamic color filter.

In Mie scattering particle size must be larger than the incident wavelength. Deflection angles opposite to Rayleigh scattering are in average more shallow creating a color halo effect around the particle when irradiated with monochromatic light. In the case of the white light spectrum, a Mie scattering medium does not act selectively to specific color components but all wavelengths get more or less the same scattering amount effect (i.e. beam of white light gets spread out inside the medium without any coloring from the initial white). Examples of Mie scattering is milk, clouds and inhaled cigarette smoke. Mie scattering in the visible spectrum is usually related with white.

Actually there is a very nice demonstration of Rayleigh and Mie scattering by Prof. Walter Lewin of MIT:

https://www.youtube.com/watch?v=gitR_ikNOfM&t=5031s

Again, light refraction and light scattering are different phenomena and not related.

refraction vs scattering

image source: link

In the above figure the laser monochromatic light is refracted by an angle $Θ$ and simultaneously spread out by light scattering effect of the medium inside the vial. Refracted photons pass through the particles and scattered photons bounce of the particles.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.