EM Theory: Refractive index of water

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samreen
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Homework Statement




Problem: Sea water has k = 80 in the low frequency limit. Yet its refractive index is around 1.34. Explain the discrepancy



Homework Equations



For a non magnetic dielectric medium, the absolute refractive index in the low frequency range, is given by : n = √k where k = Є/ Єo is the dielectric constant of the medium. Є and Єo are the permittivities of the medium and free space, respectively.




The Attempt at a Solution



No idea. Does it have anything to do with the fact that water shows a wide variety of behaviour in various frequency ranges?
 
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Note: this relation [tex]n \simeq \sqrt{K_{\epsilon}}[/tex] (Maxwell Relation) only holds for simple gases (air, Helium, Hydrogen).

For water, this relation doesn't work well because [tex]K_{\epsilon}[/tex] and then n are actually frequency-dependent, known as 'dispersion'.
You can consult to your Optics book for the dispersion eqn. I only summarize dispersion eqn, as:

[tex]n^2(\omega) = 1 + A (\frac{1}{\omega^2_0 - \omega^2})[/tex], A is constant value.


you see, if the frequency ([tex]\omega[/tex])is low (as your question) than resonance [tex]\omega_0[/tex], the refractive index will be greater than 1. so in case for sea water, n is about 1.34