Refractive index of water

Question

The index of refraction is given by the following formula: $$ n = \sqrt{\frac{\epsilon \mu}{\epsilon_0 \mu_0}} = \sqrt{\epsilon_r \mu_r},$$ where $\epsilon_r$ is the relative permittivity/dielectric constant and $\mu_r$ is the relative permeability of the material (source: Introduction to Electrodynamics by Griffiths).

Let's look at the case of diamond. The dielectric constant of diamond is approximately 5.8 (source: Griffiths).

For most materials, $\mu_r \approx 1$, so $n \approx \sqrt\epsilon_r$ (source: Griffiths). Using the dielectric constant of diamond, we get $n_{diamond} = \sqrt{\epsilon_r} = \sqrt{5.8} \approx 2.4$, which agrees with the literature value for diamond.

However, when using this formula for water, the result does not agree with the literature value. The dielectric constant of water is 80.1 (source: Griffiths) and the relative permeability of water is also approximately 1 (source: Wikipedia), so we can use $n \approx \sqrt{\epsilon_r}$. But then, this gives $n = \sqrt{80.1}\approx 8.9$, which definitely does not agree with the literature value ($n_{water}\approx 1.33$).

1. Could someone explain why this formula does not work for water? In Griffiths, they mention the first equation at the top is for linear homogeneous mediums. Does that mean that diamond is a linear homogeneous medium, but water isn't? Is that the reason why the formula does not work for diamond?
2. If so, then why is water not a linear homogeneous medium? And how does the dielectric constant exactly depend on the properties of linear homogeneous media in that case?

Answer 1

Water is a strongly polarized molecule and indeed at $f=0 \rm{Hz}$ it has $\epsilon_r=80$ but it is also very dispersive and $\epsilon_r$ drops precipitously beyond $f > 10 \rm{GHz}$ all the way through the optical region as you can see it in any rainbow.