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Fabry-Perot Cavity

  1. Apr 6, 2008 #1

    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:

    If that occurs, what happened to the energy in the wave?
     
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  3. Apr 7, 2008 #2

    cepheid

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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

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

    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.
     
    Last edited: Apr 7, 2008
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