0
$\begingroup$

Rayleigh scattering occurs when the dimensions of the scatter is much smaller than the wavelength of the incident electromagnetic radiation.

Mie scattering occurs when the dimensions of the scattered is much larger than the wavelength of the incident electromagnetic radiation. An example is when light is scattered by small water droplets in clouds.

So does Rayleigh or Mie scattering occur in liquid or bulk water?

It seems the answer is Mie scattering. But in Raman spectrometry. The incident light say 532nm is rayleigh backscattered in liquid. This is why the equipment needs to have notch filter, etc. So why does Rayleigh scattering (?) occur in liquid/bulk water when the particles are much larger than the wavelength of light? Or is it Mie Scattering that the raman device is filtering in the incident light backscattered?

$\endgroup$
1
  • 2
    $\begingroup$ I am not sure that Mie is the right model in a continuous homogeneous medium. Raman works at molecular basis or so. Mie needs interfaces. For theoretical treatment you can ask on Physics as well. A solution or a pure liquid in different from a suspension of solid or liquid in a gas is what I wanted to tell. You are looking for abrupt change from a phenomenon to another where things can go smoother or coexist. See my other comment. $\endgroup$
    – Alchimista
    Nov 1, 2019 at 8:32

1 Answer 1

2
$\begingroup$

So does Rayleigh or Mie scattering occur in liquid or bulk water? So why does Rayleigh scattering (?) occur in liquid/bulk water when the particles are much larger than the wavelength of light?

What is the approximate size of water molecule? It is on the order of 2.7 Angstroms or 0.27 nanometers. When we are talking about visible light, the wavelength is 500 nm/0.27 times larger than the scatterer! In case of pure water and with visible light, Rayleigh scattering is observed along with Raman effect. If you disperse larger "particles" in water such as milk, one sees Mie scattering.

$\endgroup$
17
  • 1
    $\begingroup$ Those are either ice particles or large (compared to wavelength) droplets suspended in air. $\endgroup$
    – AChem
    Oct 29, 2019 at 4:19
  • 1
    $\begingroup$ "However, the water vapor will not readily condense without help from other particles. The air making up our atmosphere is full of microscopic floating particles of dust, soil, smoke, sea salts, and other matter. These particles are called condensation nuclei when they assist in cloud formation. Just as water particles condense on grass to form dew, the tiny airborne particles of water vapor condense into liquid or ice on the surfaces of dust particles in the air. As more water vapor condenses into water droplets, a visible cloud forms." ssec.si.edu/stemvisions-blog/what-are-clouds $\endgroup$
    – AChem
    Oct 29, 2019 at 4:20
  • 1
    $\begingroup$ I don't remember the specific details but keep in mind those are all classical theories. At that time quantum mechanics was developing. The classical version of scattering is that the incoming electric field oscillates the charge in the molecule at the same frequency. An oscillating charge then emits light of the same wavelength. $\endgroup$
    – AChem
    Oct 29, 2019 at 4:57
  • 1
    $\begingroup$ @Jtl also the phenomena can be concomitant. Certainly the Raman spectrum of the water droplet is the same and involves Rayleigh sc. What gives the Mie sc. would be the droplet of water at its interface rather than water molecules. $\endgroup$
    – Alchimista
    Oct 29, 2019 at 9:00
  • 1
    $\begingroup$ @Jtl, No there is nothing like that. If you hold a glass of water in sunlight, it will also do Rayleigh scattering. Read Wikipedia on Rayleigh scattering. $\endgroup$
    – AChem
    Oct 30, 2019 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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