I'm reading through the book Introduction To Laser Technology and just got to the section on Fabry-Perot interferometers. The book talks about the seemingly paradoxical behaviour of the optical power passing through a pair of 95% mirrors, separated by an integer multiple of the half-wavelength, not being 90.25% but instead ~100%.

I understand that this is due to wave interference, and that a resonant Fabry-Perot transmits almost all of the light, whereas a non-resonant Fabry-Perot reflects almost all of the light.

What has thrown me a little is what the book says about the optical energy in a resonant Fabry-Perot:

For a Fabry-Perot to be resonant, the separation between its mirrors must be equal to an integral number of half-wavelengths of the incident light. Such an interferometer is shown in Figure 4.12. Note here that because the mirrors are separated by an integral number of half-wavelengths, the light is exactly in phase with itself after one round trip between the mirrors. Thus, all the waves traveling in one direction (say, left to right) are in phase with each other. And the waves moving right to left will likewise all be in phase.

In this case, all the individual waves between the mirrors add together and result in a substantial amount of power bouncing back and forth between the mirrors. For the interferometer shown in Figure 4.10, about 20W will circulate between the mirrors, even though only 1W is incident on the interferometer.

Those 20 watts are constantly reflecting off the second mirror, which transmits 5%. That is where the 1W of transmitted light, shown in Figure 4.10, comes from.

A recreation of Figure 4.10 is as follows:

Fabry-Perot interferometer constructed from two 95% mirrors, with 1W of optical energy flowing through it.

A recreation of Figure 4.12 is as follows:

A diagram showing a wave travelling forward, back, and forward again in a resonant Fabry-Perot

The book explains that Figure 4.12 shows the same wave bouncing left-to-right, right-to-left, and back again.

What I'm not confused about is the transmission of 1W to the right, and I also (mostly) understand why no optical power is reflected back to the source. What I am confused about is how the 20W power figure arises; the book doesn't give much of an explanation for this. I inferred that 20W comes from the 95% reflectivity and 1W input, but the depiction and explanation don't explain how 20W of optical power appear from 1W of input.

I can almost get my head around it by thinking more along the lines of power density than just power, with a kind of energy storage in the space (cavity?) between the two mirrors, but I can't quite fully grasp it. Can someone explain this?


1 Answer 1

  1. You know that the reflectivity of the right hand mirror is 95%, therefore to achieve 1 W travelling away to the right there must be 20 W incident on that mirror from the left.

  2. You have assumed the system is lossless and at steady state (the energy contained within the interferometer isn't increasing or decreasing with time), and therefore the power travelling away to the right must be equal to the power incident from the left.

So if there is 1 W incoming from the left, there must be 1 W exiting to the right. And for that to be true there must be 20 W incident on the left side of the right mirror.

  • $\begingroup$ I got that far, in terms of "there must be this much power for the numbers to make sense", but the part I'm not quite following is the conservation of energy. When the system turns on it cannot possibly have 20W of optical power flowing between the mirrors, otherwise we'd be creating power (and thus energy) from nowhere. $\endgroup$
    – Polynomial
    Sep 5, 2022 at 4:58
  • 1
    $\begingroup$ @Polynomial, yes there will be some transient behavior after you turn on the power source, and it will take some time (maybe a few nanoseconds or 10's of nanoseconds, assuming the interferometer is less than 0.5 m long) for the power to build up in the cavity. It would probably be easiest to explore that with a simple numerical simulation. $\endgroup$
    – The Photon
    Sep 5, 2022 at 5:01
  • $\begingroup$ Got it. So the power in the cavity increases with every reflection up to the point of 20W of optical power being present, at which point it reaches a form of equilibrium where 1W is flowing into the cavity and 1W is flowing out? $\endgroup$
    – Polynomial
    Sep 5, 2022 at 5:04
  • 1
    $\begingroup$ @polynomial Yes, You are correct the factor of 20 comes from the mirror being 95% in steady state. Yes the reflections are adding up if the round trip ends up being in phase. With the math you write it as a summation. Sometimes it is hard to judge how much math, or back ground, or what book someone is using may be different. $\endgroup$
    – UVphoton
    Sep 5, 2022 at 17:49
  • 1
    $\begingroup$ @Polynomial problem, A really good book on lasers (I think you can find on the web) is Joseph Verdeyen, 'Laser Electronics' It written for upper level undergrads and first year grad students, and goes into more detail and math than the book you are using. It has a whole chapter on optical cavities and explains them well if you want to get into the gory detail. The book you are using covers more types of lasers and optics, and is good but when I looked at it kind of hand-waves stuff sometimes. $\endgroup$
    – UVphoton
    Sep 5, 2022 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.