This question already has an answer here:

An ideal lens in geometrical optics will have a refractive index $n$. However, other materials can have complex refractive indices $n^{\prime}=n+i k$. What are the physical effects of the complex part of the refractive index? Would it result in a shifting of the focal length? Would it attenuate the light passing through the lens?


marked as duplicate by sammy gerbil, David Hammen, ZeroTheHero, Yashas, John Rennie Mar 16 '17 at 7:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Absorption, so yes, it attenuates the light. $\endgroup$ – Jon Custer Mar 14 '17 at 23:05

You can see the math and the physics pretty easily. Since $n = c/v$, with $v$ the velocity in the medium, and the frequency does not change with refraction. Physically it's because the microscopic process has the EM wave causing movement of the electrons in the medium, and they vibrate and radiate at the same freq at which altered; it is the interference of those with the incident wave which causes the resulting phase velocity to be smaller. Since v = f x wavelength in media, as v is smaller than $c$, the wavelength in the media goes down by the same factor n, from that in vacuum.

ie, $\lambda \rightarrow \lambda/n$

To not be confusing with k usually used as the wavenumber, I'll use $b$ for your (the OPS) k, and $a$ for the real part of n. So $n = a+ib$ for a complex index of refraction. [but please note that in the literature n$^\prime$ = n + ik is normally used, as you did in your question. I'm just trying not to confuse the wavenumber symbol too much, and simplify the use of too many primes with just an easier set of symbols]

Now the wave's amplitude goes as $e^{i(kz-wt)}$. $k$ is the wavenumber, z the distance into the medium, wt remains unchanged as we go from vacuum to the medium, but kz changes to be

$$kz = \frac{2\pi}{\lambda}z \rightarrow \frac{2 \pi n}{\lambda} z = \frac{2 \pi (a+i b)}{\lambda} z = (k_r + k_i)z$$

And so $$e^{ i \left(kz-\omega t \right)} \rightarrow e^{i(k_r z-\omega t)} e^{i^2 k_i z} = e^{i(k_r z-\omega t)} e^{- k_i z} $$

From that last expression you can see that the imaginary part of the index of refraction causes the amplitude to decrease, i.e., attenuation, with a skin depth of $\delta=2\pi/k_i$. The imaginary part of the index of refraction is used to describe the attenuation, ie, absorption.

The Wiki article on refraction has a similar explanation, and more. See https://en.wikipedia.org/wiki/Refractive_index


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