Frustrated Total Internal Reflection

[mgkii]

I've been reading into - and watching videos on - FTIR as an explanation of Quantum Tunnelling. The articles and videos I've watched switch between classical and quantum systems so frequently its left me with a question I can't find an answer to - the texts that seem like they might answer it have been beyond me. I'm hoping someone can bring it down to my level!

So - parking all things Quantum for a moment and focussing just on the classical.

I set my experiment for TIR up so that my beam of light is reflected from the boundary with the maximum possible brightness - i.e whatever losses in transmission I get, whatever the losses due to the evanescent wave are, I adjust the angle so that the reflected beam is at it's maximum possible brightness.
Question: When I put a lens close to (behind) the TIR point and get a beam "out of the back" of the experiment, am I simply focussing the evanescent wave into something that I can now see, or does the TIR reflected beam dim? In other words am I actually increasing the amount of light that comes through the boundary, or just manipulating the evanescent wave?

Now switching to the Quantum Realm
When FTIR is being used to describe quantum tunnelling, is this just stating that the probability wave describing where my particle/photon will be found has non-zero values the other side of the boundary, so with enough particles/photons bouncing off a boundary, some number of them (presumably based on the size of the evanescent wave) will appear on the other side of the boundary?

Thanks
Matt
(Please remember this is a "B" question - even if the answer isn't, my brain is! Thanks in advance :-)

 

[hutchphd]

You don't need to use a lens for this. Any flat surface held sufficiently close will "bleed" intensity from the evanescent face. But the other face needs to be not coplaner somehow to actually let the light out. Wedge shapes are sometimes used. And the reflection will diminish. It is possible to use periodic gratings to make wavelength selections. This is a very robust field.
The formalism shares much with Quantum tunneling: much more than just "stuff bounces back". In fact the evanescent waves are very similar in form. A major practical difference is that light is a vector and so polarization sometimes matters.

 

[mgkii]

Thanks hutchphd; i have to say that's not what I had expected! How is the reduction in reflected beam intensity / increase in evanescent wave intensity explained in classical physics?

 

[hutchphd]

The TIR solution at the interface has a nonzero (decaying exponential ) tail into the lower index medium. When a higher index object is placed in that tail region (pretty close), the new solution is again wavelike in the new region and connects to the exponential tail in the appropriate manner. Then the maths give us the numbers.

 

[mgkii]

Thanks again Hutchphd. I've been doing some more reading based on what you've said and I have one final (I think it's final anyway) question on the tail that extends into the lower index medium.

From what I've read, this evanescent tail extends but does not propagate into the lower index medium - as you say above it decays exponentially. My question is - if I was to try and measure this tail, would I be able to detect an electrical / magnetic component of the tail in the lower index medium?

Even as I'm writing this I'm half expecting you to tell me that the act of measuring it will cause it to propagate into whatever device I'm using to measure it... If this is the case, should I be left with the understanding that left untouched, this tail carries no energy? (i.e no energy unless I do something like measure it, stick another medium close enough to let it propagate, etc)

Thanks again

 

[hutchphd]

Let me define the z axis to be normal (perpendicular) to the surface. The wave is still happily propagating in, say, the x direction an all the associated field are still oscillating. In the z direction, because of reflection, there will be an standing wave pattern in the high index medium and exponentially diminishing strength fields in the low index. All real and measureable and oscillating. There is no net energy flux in the z direction (anywhere) because of the total reflection but as you say any measurement device will choose the part it wants to see..

 

[mgkii]

Thank you - this has really helped.