The index of a mixture
There is another prediction of our theory of the index of refraction that we can check against experiment. Suppose we consider a mixture of two materials. The index of the mixture is not the average of the two indexes, but should be given in terms of the sum of the two polarizabilities, as in Eq. (32.34). If we ask about the index of, say, a sugar solution, the total polarizability is the sum of the polarizability of the water and that of the sugar. Each must, of course, be calculated using for $N$ the number per unit volume of the molecules of the particular kind. In other words, if a given solution has $N_1$ molecules of water, whose polarizability is $\alpha_1$, and $N_2$ molecules of sucrose (C$_{12}$H$_{22}$O$_{11}$), whose polarizability is $\alpha_2$, we should have that \begin{equation} \label{Eq:II:32:37} 3\biggl(\frac{n^2-1}{n^2+2}\biggr)= N_1\alpha_1+N_2\alpha_2. \end{equation}  We can use this formula to test our theory against experiment by measuring the index for various concentrations of sucrose in water. We are making several assumptions here, however. Our formula assumes that there is no chemical action when the sucrose is dissolved and that the disturbances to the individual atomic oscillators are not too different for various concentrations. So our result is certainly only approximate. Anyway, let’s see how good it is. We have picked the example of a sugar solution because there is a good table of measurements of the index of refraction in the Handbook of Chemistry and Physics and also because sugar is a molecular crystal that goes into solution without ionizing or otherwise changing its chemical state. We give in the first three columns of Table 32–2 the data from the handbook. Column A is the percent of sucrose by weight, column B is the measured density (g/cm$^3$), and column C is the measured index of refraction for light whose wavelength is $589.3$ millimicrons. For pure sugar we have taken the measured index of sugar crystals. The crystals are not isotropic, so the measured index is different along different directions. The handbook gives three values: \begin{equation*} n_1=1.5376,\quad n_2=1.5651,\quad n_3=1.5705. \end{equation*} We have taken the average. Now we could try to compute $n$ for each concentration, but we don’t know what value to take for $\alpha_1$ or $\alpha_2$. Let’s test the theory this way: We will assume that the polarizability of water ($\alpha_1$) is the same at all concentrations and compute the polarizability of sucrose by using the experiment of values for $n$ and solving Eq. (32.37) for $\alpha_2$. If the theory is correct, we should get the same $\alpha_2$ for all concentrations. First, we need to know $N_1$ and $N_2$: let’s express them in terms of Avogadro’s number, $N_0$. Let’s take one liter ($1000$ cm$^3$) for our unit of volume. Then $N_i/N_0$ is the weight per liter divided by the gram-molecular weight. And the weight per liter is the density (multiplied by $1000$ to get grams per liter) times the fractional weight of either the sucrose or the water. In this way, we get $N_2/N_0$ and $N_1/N_0$ as in columns D and E of the table. In column F we have computed $3(n^2-1)/(n^2+2)$ from the experimental values of $n$ in column C. For pure water, $3(n^2-1)/(n^2+2)$ is $0.617$, which is equal to just $N_1\alpha_1$. We can then fill in the rest of Column G, since for each row G/E may be in the same ratio—namely, $0.617:55.5$. Subtracting column G from column F, we get the contribution $N_2\alpha_2$ of the sucrose, shown in column H. Dividing these entries by the values of $N_2/N_0$ in column D, we get the value of $N_0\alpha_2$ shown in column J. From our theory we would expect all the values of $N_0\alpha_2$ to be the same. They are not exactly equal, but pretty close. We can conclude that our ideas are fairly correct. Even more, we find that the polarizability of the sugar molecule doesn’t seem to depend much on its surroundings—its polarizability is nearly the same in a dilute solution as it is in the crystal.