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Relative Permittivity and Refractive Index

by iLIKEstuff
Tags: index, permittivity, refractive, relative
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iLIKEstuff
#1
Jul24-08, 05:18 PM
P: 25
in relating the index of refraction to the relative permittivity (dielectric constant/function). it is known that [tex]n = \sqrt{\epsilon_r}[/tex] for optical frequencies (i.e. [tex]\mu_r=1[/tex].

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

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

so it should be [tex]n = \sqrt{\epsilon_r} = \sqrt{80.4} = 8.9666[/tex] ??? 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. [tex]{n}^{2} = {(\sqrt{\epsilon_r})}^{2} \Rightarrow \epsilon_r={1.33}^{2}=1.7689[/tex]

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

thanks guys.
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mpolyanskiy
#2
Jul25-08, 11:58 AM
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. [tex]\epsilon_r = n^2[/tex]

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


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