Relative Permittivity and Refractive Indexby iLIKEstuff Tags: index, permittivity, refractive, relative 

#1
Jul2408, 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.phyastr.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. 



#2
Jul2508, 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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