In order to calculate Fresnel coefficients for layered media, we often need to calculate the angle that light travels inside a material with complex refractive index. Naturally, this is related to the original angle of incidence in air, but how does Snells law work with a complex refractive index? If I come in at $\theta_1$, what is $\theta_2$? Is it still calculated by $n_1 \sin\theta_1 = n_2 \sin\theta_2$ where $n_1$ and $n_2$ are the real parts of the refractive indices? If I treat $n_2$ as complex, I can get a complex angle $\theta_2$. Does this imply there is an evanescent wave alongside a propagating wave?


1 Answer 1


Your last statement is correct. At least in the case where only one refractive index is complex, namely, when an incident wave coming from some medium enters a conductor, then, the refracted wave is evanescent. This is because Snell's law remains unchanged; just that the transmitted refractive index is complex and hence so is the transmitted angle.


One of Maxwell's equations is $$\nabla\times \mathbf H=\frac{4\pi}{c}\mathbf j+\frac{1}{c}\dfrac{\partial\mathbf D}{\partial t}.$$ For linear, isotropic and homogeneous dielectrics, $\mathbf j=0$ and $\mathbf D=\varepsilon \mathbf E$, and therefore, the equation becomes $$\nabla\times \mathbf H=\frac{\varepsilon}{c}\dfrac{\partial\mathbf E}{\partial t}.$$ Moreover, if we take the curl on both sides and interchange the temporal and spatial differential operators, we have $$\nabla\times(\nabla\times\mathbf H)=\nabla(\nabla\cdot\mathbf H)-\nabla^2 \mathbf H=\frac{\varepsilon}{c}\frac{\partial}{\partial t}(\nabla\times\mathbf E).$$ Using the other Maxwell equations, $$\nabla\times\mathbf E=-\frac{\mu}{c}\frac{\partial\mathbf H}{\partial t}\ \ \ \text{ and }\ \ \ \nabla\cdot\mathbf B=0,$$ and knowing linear media also satisfy $\mathbf B=\mu\mathbf H$ we arrive at the following wave-like PDE in $\mathbf E$ and $\mathbf H$, $$\nabla^2\mathbf E=\frac{\varepsilon\mu}{c^2}\frac{\partial^2\mathbf E}{\partial t^2}.$$ Upon comparison with the wave equation, we find $$v=\frac{c}{\sqrt{\varepsilon\mu}}\implies n=\frac{c}{v}=\sqrt{\varepsilon\mu}.$$ If the material isn't ferromagnetic, $\mu=1$ and thus $n=\sqrt\varepsilon$.

Returning to the PDE, one of its solutions is the typical plane wave solution $$\mathbf E=\mathbf E_0\exp\left[{i(\omega t-k\,\mathbf r\cdot \mathbf s)}\right]=\mathbf E_0\exp\left[{i\frac{\omega}{c}(c t-n\,\mathbf r\cdot \mathbf s)}\right].$$ Similarly, in the simple case of non-ferromagnetic ohmic conductors, knowing that the current density is proportional to the electric field, $\mathbf j=\sigma\mathbf E$, the original equation becomes $$\nabla\times \mathbf H=\frac{4\pi\sigma}{c}\mathbf E+\frac{\varepsilon}{c}\dfrac{\partial\mathbf E}{\partial t}.$$ And if we assume a plane wave solution like in the dielectric case, then $$\frac{\partial\mathbf E}{\partial t}=i\omega\mathbf E\iff \mathbf E=-\frac{i}{\omega}\frac{\partial\mathbf E}{\partial t}$$ and so, $$\nabla\times\mathbf H=\frac{1}{c}\left(\varepsilon-i\frac{4\pi\sigma}{\omega}\right)\frac{\partial\mathbf E}{\partial t}\sim \frac{\varepsilon}{c}\frac{\partial\mathbf E}{\partial t}.$$ In view of the resemblance with the dielectric case, we can recycle the same solution but with different refraction index or dielectric constant if you will, but this time they're complex: $$\tilde\varepsilon\equiv \varepsilon-i\frac{4\pi\sigma}{\omega}\implies\sqrt{\tilde\varepsilon}=\tilde n\equiv n-i\kappa\implies \mathbf E=\mathbf E_0\exp\left[{i\frac{\omega}{c}(c t-\tilde n\,\mathbf r\cdot \mathbf s)}\right].$$ This way Snell's law will still hold but we'll have to add the new refractive index: $$n_i\sin\theta_i=\tilde n\sin\theta_t\ .$$ Since $n_i\sin\theta_i\in\mathbf R$ and $\tilde n\in\mathbf C$, then it must mean $\theta_t\in\mathbf C$, like you said. The fact that it is complex will mean a change in amplitude rather than only phase so the transmitted wave will be evanescent.

To demonstrate this we'll pose a specific problem:

Let's suppose the incident wave is found on some medium with refractive index $n_i$ and WLOG the incident plane is $y=0$. The ray enters the conductor with angle $\theta_i$ wrt to the normal of the conductor surface ($z=0$). The positive $y$ axis is pointing towards us, the positive $x$ one rightwards, and the positive $z$ one downwards. With this convention, the incident propagation vector will then be $\mathbf s_i=(\sin\theta_i,0,\cos\theta_i)$ and the incident electric field $$\mathbf E_i=\mathbf E_{0,\,i}\exp\left[{i\frac{\omega}{c}(c t-n_i(x\sin\theta_i+z\cos\theta_i))}\right].$$ As for the transmitted wave, it enters the conductor with an angle $\theta_t$ wrt to the same normal and its transmitted propagation vector is $\mathbf s_t=(\sin\theta_t,0,\cos\theta_t)$. The transmitted electric field will be $$\mathbf E_t=\mathbf E_{0,\,t}\exp\left[{i\frac{\omega}{c}(c t-\tilde n(x\sin\theta_t+z\cos\theta_t))}\right]=\mathbf E_{0,\,t}\exp\left[{i\frac{\omega}{c}(c t-xn_i\sin\theta_i-z\tilde n\cos\theta_t)}\right].$$ What's left now is finding $\tilde n\cos\theta_t$: $$\begin{aligned}\tilde n\cos\theta_t&=\pm\sqrt{\tilde n^2-\tilde n^2\sin^2\theta_t}\\ &\hspace{-0.5cm}\overset{\text{Snell's law}}{=}\pm\sqrt{\tilde n^2- n_i^2\sin^2\theta_i}\\ &=\pm\sqrt{n^2- n_i^2\sin^2\theta_i-\kappa^2-i2n\kappa}\\ &=\pm(\rho\cos\alpha+i\rho\sin\alpha)\equiv a-ib, \end{aligned}$$ where $$\rho=\sqrt[4]{(n^2- n_i^2\sin^2\theta_i-\kappa^2)^2+4n^2\kappa^2}\ \ \ \text{ and }\ \ \ \alpha=-\frac{1}{2}\arctan\left(\frac{2n\kappa}{n^2- n_i^2\sin^2\theta_i-\kappa^2}\right)$$ Finally, the definitive transmitted electric field will be $$\mathbf E_t=\mathbf E_{0,\,t}\exp\left[-\frac{\omega b}{c}z\right]\exp\left[{i\frac{\omega}{c}(c t-xn_i\sin\theta_i-az)}\right],$$ whose amplitude decays really rapidly with $z$ so it is an evanescent wave. Notice we'll always pick the sign in $\pm$ so that $b>0$ and the amplitude doesn't increase to $\infty$ (which doesn't make sense physically).

  • 1
    $\begingroup$ This is an excellent answer! I always struggle with the math when curl is used, and I can’t seem to implement this for the case that you didn’t mention: complex refractive index in dielectric (applicable in e.g. food dyes, or even just water in infrared). I will try again later, but perhaps your answer can be improved by adding that case. $\endgroup$
    – W_vH
    Commented May 2 at 6:58
  • $\begingroup$ I see, I had only seen complex refractive indexes in conductors. I'd really like and it'd be really interesting if you added an answer about these other cases as to complement my answer. $\endgroup$
    – Conreu
    Commented May 2 at 7:30
  • $\begingroup$ Thanks for making this so clear! For the following question, I believe no matter what the origin of the complex part, all of this should still work, i.e. for conductors its just ohmic loss, but if you are absorbing light due to electronic transitions in a semiconductor, it is still all n + iK. $\endgroup$ Commented May 3 at 2:06

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