Let $n_1$ be the index of refraction of the first medium and $n_2$ be the index of the second medium.

When $n_2>n_1$, then for an incident angle of $90^{\circ}$ we get a refracted light at a maximum angle (since the incident can't be more than $90^{\circ}$, the refracted can't be more than that maximum angle). But, at a $90^{\circ}$ incident wave, Frensel's equations give us that the transmition coefficients are equal to zero, so there is no EM field there (please correct me if I am wrong).

But, my problem with this is with the reversibility principle. When we have $n_1>n_2$, then when the incident angle is equal to the critical angle we have total reflection and we have a refracted evanescent wave. Now, if we put a second boundary, after which there is a third medium that is the same as the first as this picture shows:

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then when the evanescent wave arrives at the third medium, it creates another refracted traveling wave (and remember that $n_3>n_2$ so $n_1>n_2$ at the second boundary). This is the same case with that of my first paragraph, in which a wave arrives at an angle of $90^{\circ}$ but here it gets refracted while (following from the first paragraph), Frensel's equations predict that when the angle of incidence is $90^{\circ}$, then the refracted wave has zero intensity.

So, clearly there is something wrong with my reasoning.

Does it have to do with the fact that I am considering an evanescent wave? So, for an evanescent wave there can indeed be a refracted wave of non-zero intensity while for a normal wave the refracted wave has zero intensity?

  • $\begingroup$ Because I know that I might have not explained the situation very well, if something needs clarification, please point it out to me. $\endgroup$ – TheQuantumMan Nov 6 '15 at 1:24
  • $\begingroup$ Yes, what is a refraction coefficient? Do you mean the transmission coefficient at the interface? $\endgroup$ – Rob Jeffries Nov 6 '15 at 7:23
  • $\begingroup$ I edited the question. $\endgroup$ – TheQuantumMan Nov 7 '15 at 10:31

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