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I know that in a vacuum,

$$c=\frac{1}{\sqrt{\varepsilon_0\mu_0}}.$$

Does this apply to any medium, so that

$$c=\frac{1}{\sqrt{\varepsilon\mu}},$$

and its index of refraction is related to its dielectric constant and relative permeability by

$$n=\sqrt{\kappa\mu_r}.$$

I tried this for water, which has a dielectric constant of about $80$ and a relative permittivity of about $1$, but then the index of refraction is about $8.9$ when its actually $1.33$. Where did I go wrong in my assumptions/calculations?

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The dielectric constant is a function of frequency. The value of 80 is for DC regime (zero frequency). You need the dielectric constant at the frequency of visible light if you want to find the index of refraction for visible light. You can see how the dielectric constant drops with increasing frequency in this figure. The frequency of visible light is much higher, not contained in the figure. (https://images.app.goo.gl/QvodnBTuZ4JFVRMh9)

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  • $\begingroup$ So the dielectric constant at the frequency of visible light is much lower than the DC dielectric constant? Is this generally because the molecules have less time to polarize in an alternating electric field? $\endgroup$ – BumbleBlast Apr 28 at 18:23
  • $\begingroup$ @BumbleBlast Actually the big difference for water is not usual (usually the difference is smaller). This is, because water is a polar liquid, so the polarisation is there already but unordered. With an electrical field the water molecules rotate to form a polarisation. For light there is too much intertia (and more traditional polarisation forms) $\endgroup$ – lalala Apr 28 at 18:28
  • $\begingroup$ @lalala Yes I was referring to water specifically. Thanks! $\endgroup$ – BumbleBlast Apr 28 at 18:30

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