The index of refraction which represents how much light gets refracted when entering a medium is defined as

$$n = \frac{c}{v}$$

I have seen it stated in several places, such as here, that we can replace the index of refraction with a complex index of refraction

$$\tilde{n} = n + ik$$

I don't understand how we can do this and what it actually represents. Either the index of refraction is complex valued or not, I don't see how you can just arbitrarily replace a real valued variable with a complex valued variable. So how is this valid?

Also, in the link above it states that $k$ represents the absorption loss. Why is a complex number needed to represent this?

  • 4
    $\begingroup$ How much do you know about complex numbers? e.g. That $\exp [i(n + ik)x]$ would be the product of an oscillatory function and an exponential decay. Closely related physics.stackexchange.com/q/108580 $\endgroup$ – Rob Jeffries Oct 18 '15 at 16:26
  • $\begingroup$ If the general solution to a differential equation is $y = \exp (\alpha x)$, there is nothing to stop you making $\alpha$ a complex number. You may have met an example in damped harmonic motion? $\endgroup$ – Rob Jeffries Oct 18 '15 at 16:29

Electromagnetic theory combined with optics gives an electromagnetic interpretation for the refractive index: $$ v = \frac{1}{\sqrt{\epsilon_0\mu_0}\sqrt{\epsilon_r\mu_r}} = \frac{c}{\sqrt{\epsilon_r\mu_r}} = \frac{c}{n} \quad\Longrightarrow\quad n^2 = \epsilon_r\mu_r $$

If we ignore magnetic properties, ie $\mu_r\approx 1$, then we have: $n^2 = \epsilon_r$. If you solve maxwell equations for an waving electromagnetic field in vacuum, you will get this result: $\mathbf k\times\mathbf B = -\omega/c^2\epsilon_r\mathbf E$. Taking those values and placing into the electromagnetic waves, it forms, well, a wave: $$\mathbf E = \mathbf E_0 \exp(-i(\omega t - \mathbf k\cdot\mathbf r)$$.

But, then, when you solve for an electromagnetic wave inside a conducting material, you get: $$ \mathbf k\times\mathbf B = -\frac{\omega}{c^2}\left(\epsilon_r + \frac{ig}{\epsilon_0\omega}\right)\mathbf E $$

Then, in comparison with previous result, we defined a complex dielectric constant: $$ \hat\epsilon_r = \epsilon_r + \frac{ig}{\epsilon_0\omega} $$

And then, we define a complex refraction index: $\hat n^2 = \hat\epsilon_r$. And again, placing into the wave, you now get a damped wave, where the complex part of the refraction index $k$ makes the wave decrease its electric field exponentially: $$ \mathbf E(\mathbf r, t) = \mathbf E_0\exp\left(\frac{-k\omega r}{c}\right) \exp\left(-i\omega\left(t - \frac{nr}{c}\right)\right) $$

  • $\begingroup$ @sammygerbil The first link you've sent, the answers has hardly any equations. The second link is this question itself. Well.. Out of curiosity.. what is the procedure when an answer is a duplicate? =D. $\endgroup$ – Physicist137 Mar 15 '17 at 16:15
  • $\begingroup$ Sorry, I posted to the wrong question. Reviewers (rep > 3k) vote on whether one question is a duplicate of another. Usually the newer question is closed, unless it has more and better answers. I did not intend this question to be closed. I made a mistake switching between windows. $\endgroup$ – sammy gerbil Mar 15 '17 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.