Just having some trouble with a question about Fabry-Perot intereferometers. Here's the first part of the question:

Consider a beam of light undergoing multiple reflections in a Fabry-Perot cavity between two surfaces, both with reflectance R, and with no absorption.

What fraction of the original beam power entering the cavity is left in the beam inside the cavity after a number N of reflections?

So my problem is incorporating N into my final answer. I've derived a general equation of the electric amplitude of the reflected waves inside the cavity (as shown in this diagram), which I found to be:

$E_N = A·τ·ρ'^N·e^{i(ωt-Nδ)}$,

where ρ is the Fresnel reflection coefficient and τ is the transmission coefficient [Not sure if the equation is entirely correct]

And by geometric series:

$Ec = \frac{A·τ·e^{i·ω·t}}{(1-R)^2·e^{-iδ}}$

Using a power density function for both inside the cavity Sc and the original beam S:

$Sc = \frac{1}{2}·n_f·c·ε(Ec^*·Ec)$,

where E* is a complex conjugate

$S = \frac{1}{2}·n·c·ε·A^2$

I get: $\frac{S_c}{S} =\frac{ τ^2 }{(1-R)^2 ·(1+C·sin^2 (\frac{δ}{2}))}$,

where C is $\frac{4R^2}{(1-R)^2}$

So, as you can see, this ratio doesn't have the number of reflections N, and I'm not sure where to go from here.

enter image description here

  • $\begingroup$ Maybe you can give some Refs on Fabry-Perot intereferometers that you were looking at in your context. It may help people to follow. $\endgroup$
    – wonderich
    Sep 30, 2016 at 1:07
  • $\begingroup$ This is very hard to read. You can do us (and yourself) a favor by using MathJax. Here’s a MathJax tutorial Without reading in detail (I can't!) I might suggest that perhaps you did not truncate the geometric series correctly. $\endgroup$
    – garyp
    Sep 30, 2016 at 1:38
  • 1
    $\begingroup$ You are on the right track. What changes with each pass? How do you sum the total loss out of the cavity for N passes? It is staring you in the face... $\endgroup$
    – Jon Custer
    Sep 30, 2016 at 1:47
  • $\begingroup$ Thanks for advice! I've tried to edit all formulas, should be easy to read now. $\endgroup$
    – Anthony
    Sep 30, 2016 at 4:30
  • $\begingroup$ This precise problem is addressed in Cavity-Enhanced Absorption Measurements Across Broad Absorbance and Reflectivity Ranges. Dasgupta, P. K., Bhawal, R.; Li, Y.-H. Anal. Chem. 2014, 86, 3727-3734. doi:10.1021/ac404251w $\endgroup$ Jan 6, 2017 at 19:48

1 Answer 1


The geometric series is described here as follows. If $r$ is a complex number such that $|r|<1$, then the sum of a finite number of powers of $n$ is: $$\sum_{k=1}^nr^k = \frac{r(1-r^n)}{1-r}.$$

In the limit as $n\to\infty$, we get $$\sum_{k=1}^\infty r^k = \frac{r}{1-r}.$$

The only I can see that you could lose the $n$ dependence is to take the infinite limit.


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