If the spherical approximation is good enough, you should be able to convert the surface areas into radii. I mention that because in my work with light scattering, all the equations are usually written in terms of the radius of the scatterers. It also looks like all the Mie scattering tables are written in terms of the radius of the scatterers as well.
The best I can do in terms of actual formulas is that the scattering is proportional to $1/a^2$, times an intensity factor that you have to look up in a table. $a$ is the radius of the scatterer. (If it seems weird that scattering could go down as radius increases, see below for an explanation.) Knowing that you're only looking at one angle eliminates the angular portion of the intensity factor, but it's a complicated function of the size of the scatterer due to resonance effects when $a/\lambda \approx 1$. I found two tables of Mie coefficients. The first only has values for $n=1.40$. The second has more indexes of refraction, but might have access restrictions (it initially told me I had access thanks to my university library).
Section 10.4 of Jackson derives some equations for the scattering of electromagnetic radiation from spherical particles, beginning from Maxwell's equations. He eventually leaves off without discussing the full problem. That might be a useful starting point for the theory.
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.
Everyone seems to refer back to Kerker as the first textbook that contains a full treatment of the Mie problem, but it's difficult to find, and very expensive. I would consider it only if you can get it through your university's library. I have a copy of the Dover edition of van de Hulst, which focuses specifically on the light scattering problem. It appears that there is a Dover edition of Stratton, now, as well.
Why does the scattering intensity go down as the particle's radius increases? The first caveat is that the scattering isn't just proportional to $1/a^2$; the intensity factor is also a function of particle radius. I'm only marginally familiar with the general Mie problem, so I don't fully know all the complications that introduces.
The other issue, and one that I am comfortable with, can be explained with reference to the Rayleigh scattering problem. In the experimetns I'm used to, we plot $1/I$ on the y-axis, and $\sin^2(\theta/2)$ on the x-axis ($I$ is the scattering intensity, and $\theta$ is the scattering angle). Under a set of approximations, that gives a straight line. The y-intercept is proportional to 1 over the molecular weight of the particle. The slope is proportional to the square of the radius of the particle. So for a given molecular weight, increasing the particle's radius will increase the slope of that line. So for a given angle, you have to increase $1/I$, which means that $I$ must go down. I think the underlying explanation for that that the density of the particle decreases, because you have increased the volume while keeping the same mass.
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:
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.
Best Answer
Creative thinking with this question.
Rayleigh scattering is the scattering of a plane wave from a single small particle well-separated from other particles. If a second small particle is somewhat nearby, it will scatter coherently with the first and create interference. The far-field will not be Rayleigh anymore; you cannot just add the intensity, the phases also matter.
Say enough small particles aggregate very close together to form a large sphere, with the spacing between particles much less than the wavelength of the light. Then the combined interference effects from each scattering point produce the Mie solution. The idea is nicely illustrated in this method, for example, http://en.wikipedia.org/wiki/Discrete_dipole_approximation. The interference from each discrete dipole slowly builds up to give the scattering from arbitrary scattering-particle shapes. (Mie is for spheres.)
The molecules in a glass of water are separated by a distance much less than the wavelength of the light, so Rayleigh is not applicable. The water is effectively a homogeneous medium with a uniform index of refraction.