Evanescent light waves, energy flow and tunneling

[Joker93]

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).

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 traveling wave?
3) When the evanescent wave reaches the second interface and produces another traveling 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?

 

[mfb]

1) So, how can the evanescent wave which does not transfer any net energy produce another wave at the second interface?

It does not transfer net energy if it is fully evanescent and time-independent. In your setup it is not.

2) Is evanescent light wave a standing or traveling wave?

I don't think those two categories are useful here.

3) When the evanescent wave reaches the second interface and produces another traveling 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)?

Sure, reflection will go down a bit. Why: because the structure of the wave-function changes. That change spreads at most with the speed of light, of course, so short light pulses can lead to a different response than longer ones.

 

[Joker93]

It does not transfer net energy if it is fully evanescent and time-independent. In your setup it is not.
I don't think those two categories are useful here.
Sure, reflection will go down a bit. Why: because the structure of the wave-function changes. That change spreads at most with the speed of light, of course, so short light pulses can lead to a different response than longer ones.

Thanks for the reply! But, how would you categorize evanescent waves and why are you saying that in my setup it is not an evanescent wave?

 

[mfb]

But, how would you categorize evanescent waves

I would call them evanescent waves.

and why are you saying that in my setup it is not an evanescent wave?

They transfer energy.

 

[Andy Resnick]

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).

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 traveling wave?
3) When the evanescent wave reaches the second interface and produces another traveling 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?

My responses:

2) Evanescent waves are non-propagating waves, the surfaces of constant phase are not normal to the propagation direction, but parallel to it. They are not standing waves, either. Mathematically, evanescent waves do not have the form exp(-ikz) but exp(-kz).

1) Energy does not propagate in an evanescent wave; the analogy for FTIR is quantum tunneling through a barrier. The wavefunction is analogous (exp(-ikz) vs. exp(-kz)) as well.

3) FTIR does involve energy transfer from object 1 to object 2, but it's non-radiative energy transfer. The intensity of light in object 1 is decreased in proportion to the amount of energy transferred to object 2.

4) Are you referring to FTIR fluorescent microscopy? The excitation field is extended into the sample via evanescent waves, acceptor atoms are excited by the excitation field and fluoresce; this fluorescence is a propagating wave that is collected by the microscope objective.

Does this help?

 

[Joker93]

My responses:

2) Evanescent waves are non-propagating waves, the surfaces of constant phase are not normal to the propagation direction, but parallel to it. They are not standing waves, either. Mathematically, evanescent waves do not have the form exp(-ikz) but exp(-kz).

1) Energy does not propagate in an evanescent wave; the analogy for FTIR is quantum tunneling through a barrier. The wavefunction is analogous (exp(-ikz) vs. exp(-kz)) as well.

3) FTIR does involve energy transfer from object 1 to object 2, but it's non-radiative energy transfer. The intensity of light in object 1 is decreased in proportion to the amount of energy transferred to object 2.

4) Are you referring to FTIR fluorescent microscopy? The excitation field is extended into the sample via evanescent waves, acceptor atoms are excited by the excitation field and fluoresce; this fluorescence is a propagating wave that is collected by the microscope objective.

Does this help?

Yes, thank you for your answer!