Total internal reflection (TIR) occurs at the boundary from an optically denser into an optically less dense medium; the boundary condition requires and evanescent wave to enter into the second medium. This wave is propagating only along (x) the boundary, and exponentially decaying normal (z) to it. Also, there's no net energy transport in z since the average of the Poynting vector vanishes. All the energy is therefore in the reflected wave. So far, so good.

If I now bring a second interface, again with an optically dense medium, close to this first one, I can convert the evanescent wave into a propagating wave (frustrated TIR). The energy transported by this propagating wave depends critically on the distance between the two surfaces, i.e. on the amount of energy picked up from the exponentially decaying wave.

But: The evanescent wave did not transport any energy in z, only in x. Where is the energy now coming from? Even more: Energy conservation would require that, as I bring the second interface closer to the first, the energy in the reflected wave decreases at the same time as the energy in the transmitted wave increases. Yet, both processes happen at two different interfaces. How can the reflected wave possibly "know" that there will be a transmitted wave that it has to share its energy with? What am I missing?

I've read the answer to this question: Evanescent wave coupling and checked the referenced chapter in Born & Wolf, but I'm still not any smarter ...

  • $\begingroup$ Duplicate physics.stackexchange.com/questions/216892/… $\endgroup$
    – Farcher
    May 18, 2017 at 12:36
  • 1
    $\begingroup$ The answer to the question you linked states "Instead, the wave inside the barrier is a superposition of exponentially decaying and exponentially growing waves." Mathematically, I can accept that. But physically, the exponentially growing wave makes little sense. Also Born & Wolf discards the solution with positive exponent with the words "Naturally, only the negative sign in front of the square root of [... $exp(± omega_z/ny_2*sqrt(...))$]". I don't think the explanation given in that answer suffices ... $\endgroup$
    – Wiebke
    May 18, 2017 at 12:42

1 Answer 1


I found a nice list of other physics forums etc. in the meta thread here and finally found some pointers towards an answer on physicsforums.

Summary of the qualitative description found here:

Evanescent waves are non-propagating and decaying, as described by the $\mathrm{e}^{-kz}$ term (instead of $\mathrm{e}^{-ikz}$ for propagating waves). The decaying term is analogues to the decaying terms found in tunneling effects (hence frustrated TIR is also termed "optical tunneling" from time to time). There is an energy transfer, but it is non-radiative. The intensity of the reflected wave will be decreased as as the gap between the two interfaces is closed and more energy is contained in the propagating wave.

In a second thread, I found a link to this paper containing also a quantification of the transmission coefficients for the propagating wave. Assuming the gap is air ($n_2=1$) and the other two media are identical ($n_1=n_3=n$), they obtain:

$$\frac{1}{T} = 1 + \alpha \sinh ^2 \gamma$$ with $$\gamma = \frac{2\pi d}{\lambda}\sqrt{n^2\sin{\phi_1}-1}$$ (I'm omitting the values for $\alpha$ since they don't add to the understanding)

The authors note the similarity with quantum mechanical tunneling: $$\frac{1}{T} = 1 + \frac{V_0^2}{4E(V_0-E)}\sinh^2\gamma d$$ with $$\gamma = \sqrt{\frac{2m}{\hbar^2}(V_0-E)}$$

Obviously, these two equations can be converted into each other for appropriate values of $\gamma$ and $\alpha$ and frustrated TIR is just quantum tunneling observed for photons.

  • $\begingroup$ "...frustrated TIR is just quantum tunneling observed for photons." Could you elaborate on this? The derivation of evanescent waves can be done in purely classical electrodynamics, so I don't see how quantum effects are related apart from strictly formal analogies. $\endgroup$
    – Jono94
    Jan 23 at 8:58

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