In the full internal reflection case where we have a refracted evanescent wave, If another object is nearby, then we could have wave tunneling phenomenon(frustrated total internal reflection). I am looking for an answer that addresses all my following concerns:

1) So, how can the evanescent wave which does not transfer any net energy produce another wave at the second interface? What happens with the transfer of energy before the evanescent wave reaches the second interface and what happens to it at the exact time that it reaches it? Does energy spontaneously flow from one interface to another to produce the wave at the second interface?
2) Is evanescent light wave a standing or travelling wave?
3) When the evanescent wave reaches the second interface and produces another travelling wave, then does anything happen to the reflected wave of the first interface in order to not have problem with conservation of energy or there is no such problem(and why)? And it something does happen to the reflected wave, does it happen spontaneously(when the evanescent reaches the second interface)?
4) If energy is not transferred by the evanescent waves, then how does its EM wave excite the atoms(or molecules or whatever) at the second interface?

NOTE: I think the evanescent wave is a travelling wave that travels parallel to the boundary between the two media but its wavefronts are exponentially decaying. Please enlighten me.

  • 1
    $\begingroup$ Evanescent waves are generalizations of reactive energy in a lumped element circuit, and if you think of it in these terms it is not surprising that they can be coupled "to" or "out". If you place an obstacle in a waveguide that can propagate in its fundamental mode only then the obstacle will scatter the incident energy into three parts, a reflective term propagating backwards, and transmitted term propagating forward (both these are in the fundamental mode) and a reactive stored energy that stays around the obstacle. The latter is the "evanescent" nonpropagating piece. $\endgroup$ – hyportnex Nov 5 '15 at 23:23

So yes, if you compute the Poynting vector (energy flux density), $\vec E \times \vec H$, for an exponentially decaying evanescent wave, you indeed find zero time-averaged energy flux perpendicular to the reflecting plane. Ask you say, this leads to a conundrum --- how do evanescent waves transfer energy across barriers? For sure, we know they can transfer energy.

The simple answer is that the frustrated total reflection case is not described by a single evanescent wave, and so what I just said above does not apply. Instead, the wave inside the barrier is a superposition of exponentially decaying and exponentially growing waves.

If you compute the energy flux density for that total wave, you find a nonzero value! This is due to the "interactions" of the two waves, each of which on its own would not have any energy flux. Moreover since one wave is exponentially decaying and one is exponentially growing at the same rate, you find that the energy flux density is a constant value throughout the barrier. Energy conservation is satisfied and the world goes on to live another day.

(Math left as exercise to reader because I am lazy.)

  • 1
    $\begingroup$ From where did the exponentially growing wave come? Maybe I did not understand something well, but if there where an exponentially growing wave, then we would have an electric field that has a huge intensity at a long distance from the interface, while I have read that the evanescent wave that reaches the second interface is very weak. So, an exponentially growing wave does not mean that the evanescent wave reaches the second interface very weakly. $\endgroup$ – TheQuantumMan Nov 5 '15 at 21:35
  • 1
    $\begingroup$ I think I understand now. Is it because the (only)exponentially decaying wave is due to the fact the we had a boundary condition of E=0 at x=inf. and now that there is a second interface the boundary conditions change and so we can not rule out the exponentially growing factor in the general solution? $\endgroup$ – TheQuantumMan Nov 5 '15 at 21:40
  • $\begingroup$ Yes exactly, the exponentially growing wave doesn't grow forever because there is the other interface, in the case of frustrated total internal reflection. It really is all about the boundary conditions as you say. $\endgroup$ – Nanite Nov 5 '15 at 21:42
  • $\begingroup$ (If you solve the Maxwell equations then the amplitudes of exponentially growing/decaying waves fall out naturally, once you assert the boundary conditions.) $\endgroup$ – Nanite Nov 5 '15 at 21:43
  • $\begingroup$ ok, i understand. But the exponentially growing factor(as well as the decaying) are standing waves in the direction perpendicular to the two interfaces(I think, correct me if I am wrong), so how can any of these waves transfer(net) power? $\endgroup$ – TheQuantumMan Nov 5 '15 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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