# Relative Permittivity and Refractive Index

by iLIKEstuff
Tags: index, permittivity, refractive, relative
 P: 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...s/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.
 P: 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

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