How does Snell's law work with a complex refractive index?

Question

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?

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.

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$\mathbf{IN\ DEPTH\ EXPLANATION}$

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).