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Relative Permittivity and Refractive Index

  1. Jul 24, 2008 #1
    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/Hbase/tables/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. jcsd
  3. Jul 25, 2008 #2
    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
     
    Last edited: Jul 25, 2008
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