I would have a quite stupid question: What is the connection between complex refractive index and the "normal refractive index" that I find in every high school books? Is that the real part of the complex refractive index (and thus corresponding to transmittance?) or is it something like "absolute value", that already contains both real and imaginary part? Thank you a lot for any answer, I´m really confused about the practical interpretation.


3 Answers 3


When using refractive indices we usually assume the materials are perfectly transparent and don't absorb any light. In this case we get a real refractive index that is just the ratio of the velocities.

However real materials always absorb some light, and when this happens happens we get a complex refractive index:

$$ \bar{n} = n + i\kappa $$

where $n$ is the phase velocity of the light and $\kappa$ is the extinction coefficient.

To get a feel for why this is and what it means we can do a simple calculation. The light wave is described by something like:

$$ E(x, t) = E_0 e^{i(kx - \omega t)} \tag{1} $$

This is just the usual equation for a plane wave with $k$ the wave vector and $\omega$ the angular frequency. To see how the refractive index is involved we note that the wave vector is given by:

$$ k = \frac{2\pi}{\lambda} $$

The wavelength, $\lambda$, is related to the wavelength in a vacuum, $\lambda_0$, by:

$$ \lambda = \frac{\lambda_0}{n} $$

and substituting for $\lambda$ gives:

$$ k = \frac{2\pi n}{\lambda_0} $$

Now we take our plane wave equation (1) and substitute for $k$. If we write $n$ as the complex refractive index, $n + i\kappa$, we get:

$$ E(x, t) = E_0 \exp\left(i \frac{2\pi (n+i\kappa)}{\lambda_0} x - i\omega t\right) $$

and rearranging this gives:

$$\begin{align} E(x, t) &= E_0 \exp\left(i \left(\frac{2\pi n}{\lambda_0} x - \omega t\right)\right) \exp(- 2\pi \kappa x) \\ &= E_0 \exp\left(i \left(kx - \omega t\right)\right) \exp\left(- \frac{2\pi \kappa x}{\lambda_0}\right) \end{align}$$

So what we've ended up with is a plane wave multiplied by the exponential damping factor:

$$ \exp\left(- \frac{2\pi \kappa x}{\lambda_0}\right) $$

And that's why $\kappa$ is the extinction coefficient.

  • $\begingroup$ Thank you for your detailed answer! But honestly, I still don´t understand. In the equation, you replaced "n" by "n + ik", but the "n" is not the same, is it? $\endgroup$
    – Anne
    Feb 23, 2018 at 0:00
  • $\begingroup$ Probably I did not formulate my question well, but my main problem is: I have some values of the refractive index as function of ω, that were determined experimentally. So, according to your answer, I can not suppose that these values are corresponding to the transmision? $\endgroup$
    – Anne
    Feb 23, 2018 at 0:09

Anne, It took me some time to understand this myself. But if you take a Newtonian system perspective, it makes this a little easier. User128785 as well as John Rennie have answer your question beautifully.

Let me add : Whenever we consider a moving wave in a medium, we (simplistically) assume forward motion. With random media of the type described above (for example glass slide which is partially transparent to a range of particular wavelengths), this simplistic assumption is challenged. Which means that the net motion may be forward , but the backward wave cannot be ignore (and it's intensity defines the net velocity of the system).These are the origins of the "effective medium theory". This means that for a particular set of wave (particle) the same media will behave one way and for another set of the things will be different. As a result we see dispersion ( of wave in a prism experiment). It takes time to understand that the prism experiment is just a model of what might be going inside a thin film (made up of small-small prisms...if you will).So a signal consisting of a range of wavelengths come out of a thin film differently, except we should be able to decipher the interaction with high precision to say if the 'meta-material" thin film is absorbing or desorbing type. You will have to think dynamically (a-la "riding on the boat on the wave analogy"...how so ever big or small the boat may be, effectively the system remains scalable). You realize that the Newtonian idea holds good here. With each particle(wave) of the media, facing forward (action) and backward (reaction) forces, as a result of the perturbation the problem reduces to the following statement : How does on represent this system Mathematically (this is what John Rennie has done)? At which point one must also consider the property of this interactive system a "net yielding" type or "net un-yielding" type, leading to the effective medium theory. To answer your two specific question :

  1. "n+ik" approach ,helps in the developing this understanding that each forward push also sees a lesser (or bigger) reactive pull. If we keep taking the observations of each wavelength point (rather difficult to do ,but nevertheless this is what they do, when they build spectrometers etc.,) our experimental reduces to a plot of "big-data-points" and Signal to noise plots over a range of wavelength. The noise data is also plotted and deciphered to add richness to our understanding.
  2. when you plot , you will see a exponentially decaying or rising curve depending upon the nature of the interaction (I have seen people going to the extent that they compare each interaction exothermically or endothermically, depending upon the nature of particles and media involved). \In the end you will see the data and realize if the media is "net" absorbing kind or "net " desorbing type for that range of wavelengths. The "n" represents the forward and the "k" represents the backward part of the interaction.

(HOPE this VERBOSE statement makes sense) Vinamra


The real part of the complex refractive index is the normal refractive index, and is related to the part of wave that transmits through the medium. Let's see what the imaginary part of the refractive index would do to a wave of the form $Ae^{\iota k x}.$ The refractive index is part of $k$ in the medium, so that being imaginary will make the wave decay exponentially. You have your answer there: the imaginary part of the refractive index is related to the part of the wave that gets absorbed in the medium.


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