# Real part of refractive index

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.

## 2 Answers

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.

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.

• 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? – Anne Feb 23 '18 at 0:00
• 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? – Anne Feb 23 '18 at 0:09