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Path difference in Fabry-Perot etalon

  1. Jan 13, 2009 #1
    I'm trying to understand how the Fabry-Perot etalon works for an experiment regarding the Zeeman effect.
    Looking at the image could anyone tell me why the path difference in the etalon is [tex]\Delta l = 2t \cos{\theta}[/tex]? In my head, the path difference is [tex]\Delta l = 2t/\cos{\theta}[/tex], in other words, just the length of two more reflections.

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  2. jcsd
  3. Feb 19, 2011 #2
  4. Feb 20, 2011 #3
    Wow, I posted that question two years ago. Thanks for taking the time to answer it though.. :)
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