# Relative Permittivity and Refractive Index

1. ### iLIKEstuff

25
in relating the index of refraction to the relative permittivity (dielectric constant/function). it is known that $$n = \sqrt{\epsilon_r}$$ for optical frequencies (i.e. $$\mu_r=1$$.

now this website
http://hyperphysics.phy-astr.gsu.edu/Hbase/tables/diel.html#c1

gives the relative permittivity of water as 80.4 i.e. $$\epsilon_r = 80.4$$
but we also know that the index of refraction of water is 1.33.

so it should be $$n = \sqrt{\epsilon_r} = \sqrt{80.4} = 8.9666$$ ??? am i missing something here?

i want to use the relative permittivity in an equation to calculate the electrostatic approximation of the scattering/absorption efficiencies of small spherical particles.

should i be solving for relative permittivity from the index of refraction? i.e. $${n}^{2} = {(\sqrt{\epsilon_r})}^{2} \Rightarrow \epsilon_r={1.33}^{2}=1.7689$$

what value should i use for the relative permittivity in this equation?

thanks guys.

2. ### mpolyanskiy

4
Permittivity is a function of wavelength (frequency). 80.4 value is valid for microwave diapason, not for optical one.

For optical frequencies you should calculate permittivity from refractive index, i.e. $$\epsilon_r = n^2$$

To find accurate value of n at a particular wavelength in optical diapason use, for example, RefractiveIndex.INFO

Last edited: Jul 25, 2008