The refractive index is given by:

$$ n = \sqrt{\mu_r \epsilon_r} $$

This equation is symmetric about wavelength and is same for all wavelength of light i.e. since $\mu_r$ and $\epsilon_r$ are dependent only on material $n$ is independent of wavelength of light chosen. But this contradicts the Cauchy formula. Why is this?

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    $\begingroup$ $\mu_r$ and $\epsilon_r$ are functions of wavelength not constants. $\endgroup$ Aug 7 '18 at 16:20

Permittivity and permeability are not just constants, but instead are complex functions that depend on a number of other quantities, including the wavelength of the light. In fact, the refractive index $n$ is only half the story - there is also a related quantity, the extinction coefficient $k$, that describes the absorption in a medium. To put it more accurately, the refractive index $\tilde n$ is a complex function that depends among other things on the wavelength.

$$\tilde n (\omega) = n(\omega) + i k(\omega)$$

As stated in the question, the refractive index is related to the permittivity and permeability. In fact, they contain the same information about the material, and which one is chosen largely depends on convention and convenience. Also the permittivity ("dielectric function") and permeability have real and imaginary parts. For a non-magnetic material ($\mu_r = 1$, valid for most common materials), dielectric function and refractive index are related as follows: $$\varepsilon' + i \varepsilon'' = (n + ik)^2$$

With the individual components being: $$\varepsilon' = n^2 - k^2$$ $$\varepsilon'' = 2nk$$

And in reverse: $$n = \sqrt{\frac{\sqrt{\varepsilon'^2 + \varepsilon''^2} + \varepsilon'}{2}}$$ $$k = \sqrt{\frac{\sqrt{\varepsilon'^2 + \varepsilon''^2} - \varepsilon'}{2}}$$

As can be seen from the formula for $n$, the refractive index becomes imaginary if the real part of $\varepsilon(\omega)$ becomes negative and the extinction coefficient becomes large. This corresponds to a large absorption in the medium. The absorption coefficient $\alpha(\omega)$ is also again dependent on the frequency:

$$\alpha(\omega) = \frac{2 \omega}{c} k(\omega)$$

There are a number of different models to describe the dielectric function and permeability of materials. A good example is a Drude metal, in which the real part of the permittivity is given by

$$\varepsilon'(\omega) = 1 - \frac{\omega_p^2}{\omega^2 + \gamma^2}$$

where $\omega_p$ is the plasma frequency of the metal (the characteristic oscillation frequency of the bulk of the electrons), $\omega$ is the frequency of the EM wave, and $\gamma$ is a damping constant.

Here for example is the real and imaginary part of the dielectric function (permittivity) of gold at different photon energies (= wavelengths):

enter image description here

This means permittivity and permeability, and consequently also the refractive index, depend both on the materials and the wavelength. Same also goes by the way for the absorption spectrum, which mainly depends on the imaginary part of those functions and also depends on the material and wavelength.

If the real parts of both permittivity and permeability are negative, then the refractive index of the material is also negative - it is then a so-called negative-index (left-handed) metamaterial. It has further interesting properties, for example that light is refracted in the other direction at an interface. This can be used for planar lenses etc.

enter image description here

  • $\begingroup$ This would be a great answer if you could expand a bit on the imaginary part and how it enters in the index $n$, and how one reconciles negative values of the real part (as per your graph) with a real index of refraction as per the OP question. $\endgroup$ Aug 8 '18 at 0:57
  • $\begingroup$ does this means that er and ur not only depend on the the material but also the wavelenght of light present there ? does this also mean that blue light bend more because ur and er has changed there and ur and er changed there because blue light is present there ?? means the wave slow down and reason the of that is itself that the wave is present there ?? how will you explain this ? $\endgroup$ Aug 15 '18 at 16:30
  • $\begingroup$ @ParthGoyal Exactly: $\varepsilon_r$ and $\mu_r$ are different for different wavelengths. Imagine it like this: you have a thick layer of snow. A light animal like a bug will sink only very little into the snow. But a human for example very much. The property "how much do you sink into the snow" depends not only on the snow itself, but also on the one who is stepping on it! In a similar way, $\varepsilon_r$ and $\mu_r$ depend not only on the material, but also the wavelength. So, blue light is refracted differently than red light, even though the material is the same. $\endgroup$
    – ahemmetter
    Aug 15 '18 at 18:39

The relative permittivity and permeability depend on wavelength.


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