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When considering an evanescent wave travelling in a region between two regions where the solution takes the form of a travelling wave from the maths we have a forwards travelling wave that decreases in amplitude and a second solution that increases in amplitude. Does this second solution give us a reflected wave or would the reflected wave be something more along the lines of $e^{-\alpha(b-x)}$ where alpha is some constant related to the decay and b the far side of the evanescent region.

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  • $\begingroup$ Could you give a clearer description of the configuration you are considering. For instance, what is the direction in which the wave guide is oriented with respect to the wave you are interested in and what do you mean by a reflected wave? $\endgroup$
    – Crimson
    May 24, 2016 at 9:29
  • $\begingroup$ @Crimson Well the situation much more general than just the waveguide problem I was thinking about at the time. I have edited the question to reflect this removing the reference to a waveguide. In any situation where we have evanescent waves we have two solutions for k normally we can discount the second positive solution but with a finite thin barrier reflections could have significant amplitude from the secondary interface. $\endgroup$
    – Goods
    May 24, 2016 at 9:39

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from the maths we have a forwards travelling wave that decreases in amplitude and a second solution that increases in amplitude

An evanescent wave doesn't travel forward or backward: it's a non-propagating solution. In particular, its wavenumber is purely imaginary.

Does this second solution give us a reflected wave

Yes, the reflected wave is indeed directly related to the growing solution.

Suppose we have a Dirichlet boundary at $x=0$. This means that our stationary wavefunction $\psi(x)$ (regardless of the exact form of our equation — be it Schrödinger's, Maxwell's, vibration of membrane or whatever else) must vanish at this point. Then, for a uniform medium, it must have the following form in some neighborhood of $x=0$:

$$\psi(x)=A\sin(kx).\tag1$$

If the conditions of the actual eigenproblem (like e.g. potential in the Schrödinger's equation, or refractive index in Maxwell's equation) make $k$ purely imaginary, i.e. $k=i\kappa,$ $\kappa\in\mathbb R,$ then we can rewrite $(1)$ as

$$\psi(x)=A\sin(i\kappa x)=Ai\sinh(\kappa x).\tag2$$

Notice that we've obtained a hyperbolic sine function, which is a linear combination of exponentials:

$$\sinh(x)\equiv\frac{e^x-e^{-x}}2.\tag3$$

This means that, if our wave, in the conditions where it appears to be an evanescent wave, comes to a reflecting boundary, both growing and decaying exponentials will appear in the solution of the wave equation, while without this reflector we'd have only the decaying solution.

Moreover, if the wave were actually a propagating wave, i.e. it had $k\in\mathbb R,$ then we'd indeed could view $(1)$ as a combination of the incident wave $\propto\exp(-ikx)$ and the reflected wave $\propto\exp(+ikx)$.

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I presume that when you talk about travelling evanescent wave, you are referring to surface plasmons waves. There was an interesting paper(I can provide you the paper if you cannot find it) on whether growing waves could have a physical significance :

"Surface polaritons like waves guides by thin lossy metal films "

In a word, it is mentioned that a growing wave could be reflected away from a waveguide if the wave also had a limited extent along the waveguide axis. In this case the energy is finite and the far field of this wave can be obtained with near to far field transformation. However, even though this situation was theoretically plausible, the way to excite experimentally such a wave was still questioned in the paper.

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