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?

1 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.

  • $\begingroup$ So the dielectric constant of water depends on the frequency of light? Is this a general thing or does this only apply to polarized molecules like water? Also, what does it mean for the frequency to be 0 Hz? How is this possible for light? And why is the dielectric constant taken at this value in literature? $\endgroup$
    – Stallmp
    Commented Dec 2, 2022 at 20:55
  • $\begingroup$ Yes, at $0\rm{Hz}, 10\rm{GHz}, 70\rm{GHz}$ is about $80, 60, 20$, resp. $0 \rm{Hz}$ means electro-static measurement. The higher the frequency the less the dipoles can follow it, so the effect of the biasing E field on them is smaller. $\endgroup$
    – hyportnex
    Commented Dec 2, 2022 at 21:00
  • 1
    $\begingroup$ Thank you for your response! I have some more specific questions about this: (i) Is the dependence of the dielectric constant on the frequency of light a general thing, or mainly for polarized molecules like water? I assume there isn't a strong dependence for diamond. (ii) What do you mean with electro-static measurement? From my understanding, it is the frequency of light, and how can that be 0? $\endgroup$
    – Stallmp
    Commented Dec 2, 2022 at 21:05
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    $\begingroup$ What causes dispersion is quite complicated and you should ask a separate question on that, there are many and more knowledgeable people than I on this site who will be able to respond on that subject. Measuring at $0\rm{Hz}$ just means "dc", get an aquarium filled with water between two metal plates and connect them to a battery and measure capacitance. $\endgroup$
    – hyportnex
    Commented Dec 2, 2022 at 21:10
  • 4
    $\begingroup$ @Stallmp, Yes dependency of refractive index on wavelength is a general thing, so it applies not only to water: $$n(\lambda )=A+{\frac {B}{\lambda ^{2}}}+{\frac {C}{\lambda ^{4}}}+\cdots ,$$ where $A,B,C$ is material-dependent coefficients. $\endgroup$ Commented Dec 2, 2022 at 22:01

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