Converting indices of refraction If I know that the index of refraction of a given substance is $1.4$ for the average wavelength (say ${550\rm\ nm}$), and I would like to know what is the index of refraction with a wavelength of ${832\rm\ nm}$, how would I go about computing this?  I'm not looking for an exact solution, just a rough estimate that is better than $1.4$.
 A: You can calculate the refractive index using an empirical formula, such as the Sellmeier equation. The Sellmeier coefficients are different for each material, you can find those for common materials on sites like https://refractiveindex.info/
Once you know the Sellmeier coefficients of the material, you can use this online calculator.

Update
I understand from the comments that you're specifically looking for the refractive index of the crystalline lens of the eye. I found this paper: Development of a human eye model incorporated with intraocular scattering for visual performance assessment, where the authors state in Equation (4a) that
$$
\begin{multline}
n(\lambda) = 1.389248 + 6.521218 \times 10^3∕\lambda^2 \\− 6.110661
\times 10^8∕\lambda^4 + 5.908191 \times 10^{13}∕\lambda^6,
\end{multline}
$$
with $\lambda$ in nm.
A: You're really not going to like this. You have no choice but to find out exactly what the "substance" is and then look up dispersion curves for it. Either that or you're going to have to measure it yourself at 832nm.
In short - dispersion is not a fundamental property of matter, there is no simple law that works for all substances.
If it's a very "simple" substance (simple make up, simple crystal structure), there may be a quantum optic model it will tell you the dispersion behaviour. Otherwise it's most likely that you will have to go to an experimental curve. Optical glasses manufacturers publish detailed description of their wares' dispersion: they will usually do so as coefficients in either the Sellmeier or Schott models (these are models that have the form that arises from a series of resonances, but the coefficients are fitted experimentally).
Try also https://refractiveindex.info to see whether your "substance" is there.
If you have to resort to data from optical material manufacturers and if very high accuracy is important to you then beware of the following: optical designers define the refractive index of AIR AT STANDARD TEMPERATURE AND PRESSURE to be 1.0. In this definition, therefore, the vacuum has a refractive index slightly less than one. This is altogether crazy, I know, but optical designers really do mean something different from the rest of us when they say "refractive index"!
If you really stuck and you can get hold of the Abbe number for the substance, then this will give you enough data to find the coefficients in the Cauchy model:
$$n(\lambda) = A + \frac{B}{\lambda^2}$$
The Abbe number is defined by either :
$$V_D = \frac{ n_D - 1 }{ n_F - n_C }$$
where $n_D$, $n_F$ and $n_C$ are the refractive indices of the material at the wavelengths of the Fraunhofer D-, F- and C-spectral lines (589.3nm, 486.1nm and 656.3nm respectively) (more common in Japan and US) or
$$V_e = \frac{ n_e - 1 }{ n_{F^\prime} - n_{C^\prime} }$$
where $n_e$, $n_{F^\prime}$ and $n_{C^\prime}$ are the refractive indices at the green mercury e-line (546.073 nm), and blue and red cadmium lines at 480.0 nm and 643.8 nm, respectively (more common in Europe). Again, these are optical designers' refractive indices as warned above.
