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We know that the reflective coefficient in total internal reflection is 1, which means that all energy is reflected. But we also know that for evanescent wave, Poynting vector in the direction parallel to the reflecting surface is not zero. Isn’t this contradictory to the statement that all energy is reflected?

Also, energy in the direction perpendicular to the surface exponentially decays, it seems that there is still energy here just that it’s not propagating?

Edit: the major confusion I have is that why doesn’t the evanescent wave’s propagating parallel component and non-propagating perpendicular component break energy conservation?

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  • $\begingroup$ The Poynting vector in an evanescent wave is zero. Or more precisely it time averages to zero. $\endgroup$ Commented Feb 15 at 9:59
  • $\begingroup$ Only the component perpendicular to the surface is 0, the component parallel to the surface is not $\endgroup$
    – km12180202
    Commented Feb 15 at 22:30
  • $\begingroup$ OK, yes I see your point. I have voted to reopen. $\endgroup$ Commented Feb 16 at 5:21

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Those are two different approaches to look into the problem.

We know that the reflective coefficient in total internal reflection is 1, which means that all energy is reflected.

This is true in the ray optics regime:

From Wikipedia

This is true for the model in which electromagnetic radiation is a ray, and therefore, the evanescence is technically "not important here". This is in quotation marks because essentially since you assume that the analysis is in dimensions much larger than the wavelength $l >> \lambda$ then it holds true that, at TIR condition, all light will be reflected. This analysis is useful, for example, when analyzing optical systems such as tabletop optical tables, large telescopes, etc.

But we also know that for evanescent wave, Poynting vector in the direction parallel to the reflecting surface is not zero. Isn’t this contradictory to the statement that all energy is reflected?

In the ray optics regime, it is not contradictory, since the evanescent portion of the radiation is much smaller than the reflected ray, therefore, if you do a rigorous wave-regime analysis, you will find that there is actually some evanescent field through the reflecting interface, but it will be much smaller than the reflected light.

However, note that this also arises from the energy conservation laws and the boundary conditions of the problem.

Also, energy in the direction perpendicular to the surface exponentially decays, it seems that there is still energy here just that it’s not propagating?

The wave may still have a non-zero instantaneous component in the propagation direction of the wave, even if the average is zero (meaning that energy does not necessarily cross the interface or travel along with the propagating wave, but if you measure at a specific point you may measure the evanescent wave, or for example, use the evanescent wave as it is done for evanescent excitation for plasmons or prism coupling).

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  • $\begingroup$ It is not just true in ray optics. It is true in the Maxwell equations, that is, in the Fresnel equations. $\endgroup$
    – my2cts
    Commented Feb 15 at 10:41
  • $\begingroup$ The rest of the explanation discusses the caveats of the comment. In Fresnel equations which deal with the wave behavior it is possible to calculate the evanescent wave: brown.edu/research/labs/mittleman/sites/… As noted in the presentation, the evanescent penetration length is much smaller than the wavelength and the time-averaged power is 0. Nonetheless it is possible to transfer energy even in the TIR case (e.g. coupling theory for resonant rings exploits this concept). $\endgroup$
    – ondas
    Commented Feb 16 at 14:07
  • $\begingroup$ Thank you so much for your explanation, but I am still confused about why energy exists in transmitted wave and R=1, and it does not break energy conservation. Even if the energy in evanescent wave is small, it’s still extra energy? $\endgroup$
    – km12180202
    Commented Feb 16 at 22:17
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    $\begingroup$ There is a whole paper that deals with exactly your question: mmrc.caltech.edu/FTIR/Literature/ATR/… a very, very narrow summary is that even if in general TIR transfers a beam that reflects off an interface, all reflections have a phase and therefore at instantaneous moments, there may be a slight energy imbalance, which is dealt with through the instantaneous energy of the evanescent wave. I highly recommend to read the paper I suggested as it walks you through a rigorous analysis that is quite comforting. $\endgroup$
    – ondas
    Commented Feb 20 at 9:58

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