Understanding the transmission of an etalon

[HermanTheGerman]

Hi there,
I have a question concerning the etalon (One would think that I find an answer for this in every standard physics textbook or this forum, but I actually did not):

An etalon can be built from a solid block of glas (for example) with dielectric high reflectivity coatings (let's say R=99 %) on both sides. For the right wavelength their transmission can get close to unity. I understand that constructive interference of the beams reflected inside the cavity is the reason for this. Here is the part that I don't understand: If the outer surfaces reflect 99 % of the light, I would assume that only 1 % of the light can enter the etalon and undergo constructive/destructive interference. The other 99 % are reflected back to the laser at the very first surface and thus cannot interfere with something. So how can the etalon transmit overall almost 100 % of light (if it has the right wavelength) if 99% are reflected away from the etalon?

Or to cut a long story short: I don't understand why most light is transmitted at the surface when coming from the outside, but mostly reflected when coming from the inside. In my understanding dielectric reflectivity surfaces should work the same in both directions.

I would be glad if someone can help me understanding this effect.
Greetings, HermanTheGerman

 

[sophiecentaur]

Hi and welcome.
Both faces of the etalon are involved and not just the input face. In the same way that Blooming of a lens works, the phases of the waves inside the etalon result in most of the light being transmitted and not much being reflected.
I think its true to say that the etalon in which a laser operates is different because the 'source' is inside the etalon. Your comment seems just about right to me:

thus cannot interfere with something.

exactly.
This constitutes a high Q resonator with a very high energy density inside and a relatively small proportion of that energy emerging from the output. It's the high Q factor of the resonator that controls the frequency of the spontaneously emitted photons and produces such a high coherence. You can pull the laser frequency by small changes in the etalon spacing.

 

[HermanTheGerman]

I am sorry but I thought about it again and still don't get it completely. You are talking about the waves inside the resonator. But what I meant are the waves that don't enter the resonator because they are reflected at the outer surface of the etalon. In textbooks no one talks about how much of the incoming light enters the resonator, they only consider the light that already entered the resonator. In papers about experimentally realized etalons transmission values of well above 50 % are claimed. But how is this possible if 99 % of the light does not even enter the resonator and gets reflected back to the light source? Even if it destructively interferes with the 1 % that enters the resonator, the reflected part is still verly large (since its amplitude is much higher than the amplitude of the 1 % of light inside the resonator).
I hope someone can help me with understanding this issue.
Thanks in advance,
Greetings, Herman

 

[sophiecentaur]

the waves that don't enter the resonator because they are reflected at the outer surface of the etalon.

OK. When a wave reaches a boundary, there has to be continuity across the it (all the energy has to go somewhere and the phases have to be righted.). The incident, transmitted and reflected wave vectors have to add to zero (the reflected wave 'makes up the difference', which is quite significant for just a single 'reflecting' boundary. In an etalon, there is another wave involved which has been reflected from the far wall. So you have four waves to consider. If the phase of this extra wave is just right (i.e. the distance between plates is right), then the difference that has to be made up is tiny (notionally zero) and that means the transmitted wave is more or less the same amplitude as the incident wave. In an arm waving sort of way, you could say that this extra wave from the other side presents the right conditions for full transmission; it sort of pulls the energy in. However, the incident wave is is the wrong phase for it to encourage that extra wave to pass through the surface so the energy stays inside.
A problem I had with this was that there are many internal reflections and the above explanation only deals with one - hence the question "how does it know?". But it's a general principle in such situations (Resistor networks, mechanical structures, water in pipes etc.) where we just consider the state of things once everything has settled down and the "how does it know?' answer is that the element is just behaving according to the surrounding conditions. In an etalon, there will be a time during which the energy within the etalon builds up until equilibrium (power in = power out + power dissipated) is reached - just like an LCR resonant circuit.

 

[tech99]

I am sorry but I thought about it again and still don't get it completely. You are talking about the waves inside the resonator. But what I meant are the waves that don't enter the resonator because they are reflected at the outer surface of the etalon. In textbooks no one talks about how much of the incoming light enters the resonator, they only consider the light that already entered the resonator. In papers about experimentally realized etalons transmission values of well above 50 % are claimed. But how is this possible if 99 % of the light does not even enter the resonator and gets reflected back to the light source? Even if it destructively interferes with the 1 % that enters the resonator, the reflected part is still verly large (since its amplitude is much higher than the amplitude of the 1 % of light inside the resonator).
I hope someone can help me with understanding this issue.
Thanks in advance,
Greetings, Herman

I am answering you from an electrical engineering perspective.
I don't know if you are familiar with transmission lines and resonances but it seems to be the same thing. First imagine a matched transmission line carrying energy from a generator to a load. Now cut the line and insert a piece of line having greatly different characteristic impedance. There will be strong reflections at the junctions with the new piece, but if it is half a wavelength long, all the reflections cancel out and it looks transparent. There are standing waves on the new piece and it is resonant. The sharpness of its resonance curve defines its Q or finesse.
If the half wave line is many half waves long, the frequency response will have many resonances, like the glass block.
If we couple the half wave line more lightly to the matched line, the Q increases but theoretically the transmission loss at resonance is still zero.This effect is achieved my raising the characteristic impedance to a very high or low value or, as in the case of the glass block, by inserting further intentional mismatches, equivalent to capacitors, at the junctions.
The figure you mention for 99% reflection seems to be under a test condition using just one interface and where there is no resonance taking place, but when used with the resonant block, the mismatches at the two interfaces are canceled out and transmission loss falls to a low value. The fields within the block then rise to very high values at resonance.

 

[tech99]

Sophie Centaur: Sorry I just missed your response before replying myself.

 

[sophiecentaur]

@tech99
I always prefer the EE explanation, where possible. You can stick a meter on and measure everything - inside and outside. That's something that the optics guys find a lot more difficult.

 

[HermanTheGerman]

OK. When a wave reaches a boundary, there has to be continuity across the it (all the energy has to go somewhere and the phases have to be righted.). The incident, transmitted and reflected wave vectors have to add to zero (the reflected wave 'makes up the difference', which is quite significant for just a single 'reflecting' boundary. In an etalon, there is another wave involved which has been reflected from the far wall. So you have four waves to consider. If the phase of this extra wave is just right (i.e. the distance between plates is right), then the difference that has to be made up is tiny (notionally zero) and that means the transmitted wave is more or less the same amplitude as the incident wave. In an arm waving sort of way, you could say that this extra wave from the other side presents the right conditions for full transmission; it sort of pulls the energy in. However, the incident wave is is the wrong phase for it to encourage that extra wave to pass through the surface so the energy stays inside.
A problem I had with this was that there are many internal reflections and the above explanation only deals with one - hence the question "how does it know?". But it's a general principle in such situations (Resistor networks, mechanical structures, water in pipes etc.) where we just consider the state of things once everything has settled down and the "how does it know?' answer is that the element is just behaving according to the surrounding conditions. In an etalon, there will be a time during which the energy within the etalon builds up until equilibrium (power in = power out + power dissipated) is reached - just like an LCR resonant circuit.

Thanks, that's a good explanation. I also had problems with understanding the "how does it know?", but your answer for this seems very intuitive.