EM Theory: Refractive index of water

[samreen]

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?

 

[lepton5]

Note: this relation n \simeq \sqrt{K_{\epsilon}} (Maxwell Relation) only holds for simple gases (air, Helium, Hydrogen).

For water, this relation doesn't work well because K_{\epsilon} 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:

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


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