1
$\begingroup$

How does the intensity of light scattered by a medium depend on the refractive index and the size of the particles dispersed in the medium?

I found plenty of literature about the dependence of intensity on particle size. Scattering of light goes from Rayleigh scattering to Mie scattering (Tyndall effect) for larger particles in colloidal solutions.

But why do solutions with much larger particle sizes (suspensions) not show Mie scattering? Why is the intensity of scattered light less in such cases?

Also, does refractive index also play a role in this? Specifically, if we were to increase the difference between the refractive index of the dispersed phase and the dispersion medium, would it have any effect on scattering? (not just on the intensity of scattered light, but any effect whatsoever)

$\endgroup$
  • $\begingroup$ Do you mean by "But why do solutions with much larger particle sizes (suspensions) not show Mie scattering?" that the cross section approaches the "physical" cross section of the particle, because Mie theory shows this. Or do you mean that the scattering is not very strong for some colloids. The scattering amplitude goes like $\Delta (n^2)$ - the difference between the particle and steeping fluid indices, so it is naturally much weaker in liquids. $\endgroup$ – WetSavannaAnimal Oct 28 '14 at 4:28
  • $\begingroup$ @WetSavannaAnimalakaRodVance: I mean that the scattering is not very strong for some colloids. $\endgroup$ – Gerard Oct 28 '14 at 4:48
2
$\begingroup$

Refractive index manifestly plays a role in Mie scattering: if the suspended colloids have the same refractive indexand characteristic impedance as the surrounding fluid, the whole system is electromagnetically homogeneous, and there is no scattering. For nonmagnetic materials, this statement is the same as that of a homogeneous refractive index.

So I believe your observations will mostly be explained by the dependence of Mie scattering on refractive index. The following is a summary of Chapter 13 of Born and Wolf, "Principles of Optics". You calculate the power scattered from a spherical, dielectric particle of refractive index $n_s$ steeped in a medium of index $n_0$ through its effective scattering cross section:

$$\bar{Q}(\frac{n_s}{n_0},\,q) = \frac{2}{q^2}\operatorname{Re}\left(\sum\limits_{\ell=1}^\infty\,(-i)^{\ell+1}\,(\ell+1)\left(\mathscr{E}_\ell\left(\frac{n_s}{n_0},\,q\right)+\mathscr{M}_\ell\left(\frac{n_s}{n_0},\,q\right)\right)\right)$$

and you use this quantity by multiplying the actualy cross sectional area of the sphere presented to an incoming plane wave by $\bar{Q}(\frac{n_s}{n_0},\,q)$ and the reflexion intensity calculated from the normal incidence Fresnel equations. The size parameter for the sphere is:

$$q=\frac{2\,\pi\,n_0\,r_s}{\lambda}$$

i.e. the sphere's radius expressed in the corresponding radian delay in the suspending liquid. $\mathscr{E}_\ell$ and $\mathscr{M}_\ell$ are the complicated expressions:

$$\mathscr{E}_\ell(\rho,\,q) = i^{\ell+1}\,\frac{2\,\ell+1}{\ell\,(\ell+1)}\,\frac{\rho\,{\rm Re}(\zeta_\ell^\prime(q))\,{\rm Re}(\zeta_\ell(\rho\,q))-{\rm Re}(\zeta_\ell(q))\,{\rm Re}(\zeta_\ell^\prime(\rho\,q))}{\rho\,\zeta_\ell^\prime(q)\,{\rm Re}(\zeta_\ell(\rho\,q))-\zeta_\ell(q)\,{\rm Re}(\zeta_\ell^\prime(\rho\,q))}$$

$$\mathscr{M}_\ell(\rho,\,q) = i^{\ell+1}\,\frac{2\,\ell+1}{\ell\,(\ell+1)}\,\frac{\rho\,{\rm Re}(\zeta_\ell(q))\,{\rm Re}(\zeta_\ell^\prime(\rho\,q))-{\rm Re}(\zeta_\ell^\prime(q))\,{\rm Re}(\zeta_\ell^\prime(\rho\,q))}{\rho\,\zeta_\ell(q)\,{\rm Re}(\zeta_\ell(\rho\,q))-\zeta_\ell^\prime(q)\,{\rm Re}(\zeta_\ell(\rho\,q))}$$

where $\zeta(z) = \sqrt{\frac{\pi\,z}{2}}\,H_\ell^{(1)}(z)$ and $H_\ell^{(1)}(z) = J_\ell(z)+i\,Y_\ell(z)$ is the Hankel function of the first kind. I was interested in Mie scattering by conductive spheres which means $\rho = \frac{n_s}{n_0}$ is complex, and I used the Lentz-Thompson recurrence method to calculate the complex argument Bessel functions stably. Here are my results:

Mie Scattering

The asymptotic normlised cross section as $q\to\infty$ in all cases is $2$. This is multiplied by the simple minded Fresnel co-efficient:

$$|\Gamma|=\left(\frac{n_s-n_0}{n_s+n_0}\right)^2$$

So not only does the asymptotic strength of the reflexion $|\Gamma|$ gets very small for $n_s\approx n_0$, the colloids need to be very big relative to the light wavelength for the reflexion to approach its asymptotic value: for very small index difference, the Rayleigh behaviour (i.e. scattering is small and varies like $1/\lambda^4$) prevails even for very large colloids.

$\endgroup$

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.