Fabry-Perot Cavity: Energy and Round Trip Phase Dynamics

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SUMMARY

The discussion centers on the dynamics of energy and phase in Fabry-Perot cavities, specifically addressing the conditions for constructive and destructive interference. It establishes that even when destructive interference occurs, energy is not lost; instead, in a lossless cavity, all incident light is reflected, maintaining the intensity. The maximum reflected intensity formula is provided as (I_r)_{max} = I_i (4r^2)/(1+r^2), where r is the amplitude reflection coefficient. For lossy cavities, some energy is transmitted and some remains within the cavity, leading to broader resonance peaks.

PREREQUISITES
  • Understanding of Fabry-Perot cavity principles
  • Knowledge of multiple-beam interference
  • Familiarity with optical path difference concepts
  • Basic grasp of amplitude reflection coefficients
NEXT STEPS
  • Study the mathematical derivation of the Fabry-Perot cavity resonance conditions
  • Explore the implications of damping in optical resonators
  • Investigate the role of dielectric films in interference patterns
  • Learn about the applications of Fabry-Perot cavities in modern optics
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Optical physicists, engineers working with optical systems, and students studying advanced optics concepts will benefit from this discussion.

cepheid
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Hi,

I know this is dumb question, and I should be able to figure it out and move past it, but I can't. The optics book I have begins its treatment of Fabry-Perot cavities by considering the case of an undamped optical resonater (perfectly reflecting mirrors). Here's a quote from it:

In order for a plane wave to exist in the cavity, it must return to exactly the same phase after one round trip through the cavity. In this case, it will constructively interfere with itself. Otherwise, the phase will precess on each successive round trip, and eventually lead to destructive interterference.

If that occurs, what happened to the energy in the wave?
 
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I found my own answer to this question. Looking at the *results* of the treatment of multiple-beam interference within a thin dielectric film in Optics by Eugene Hecht, (which is not the same book as the one I was referring to in my first post), it turns out that the reflected intensity (the intensity of the light that ends up coming out of the same side of of the dielectric film as the incident light, after the multiple reflections occur within the film), is a maximum when the optical path difference in the cavity is a half-integer multiple of the wavelength (the condition for destructive interference and absolutely NO cavity resonance). This maximum reflected is given by

(I_r)_{max} = I_i \frac{4r^2}{(1+r^2)}​

where I_i is the incident intensity, and r is the amplitude reflection coefficient. If it's a lossless cavity, r = 1, and the reflected intensity is equal to the incident intensity. So just because destructive interference occurs in the cavity doesn't mean that energy disappeared. In the steady state, all of the incident radiation is reflected, and none is in the cavity or is transmitted out the other side.

Edit: For a lossy cavity, some is reflected, some is transmitted, and some is inside the cavity. This is consistent with damping of the optical resonator causing the resonance peaks to broaden, so that there is some non-zero intensity in the cavity even at wavelengths other than those of the Fabry-Perot modes. I hope you guys agree with my resolution of my problem.
 
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