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Fiber optics: using continous variation refractive indexes hold light?

  1. Mar 26, 2013 #1
    My teacher said that instead of using a fiber optic made of just one material we could use a set of materials with progressively low refractive index to turn the light back in, like so:

    2013-03-27 01.11.36.jpg

    In the case of a continous variation of refraction index, how can you do the math. Can you please show me, the right way for it?
     
  2. jcsd
  3. Mar 26, 2013 #2

    Drakkith

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    Staff: Mentor

  4. Mar 27, 2013 #3
    It was a good help, thanks.
    The math at the bottom is the fermat principle

    [itex] \int_{0}^{end} n ds=minimum[/itex]

    this is because n=c/velocity, if you take the c out (factorize) you are summing the infinitesimal times of the path the light travels.
    However, I can't get anywhere from here, I tried doing the following:

    [itex]\int_{0}^{end} n ds=minimum \Leftrightarrow \int_{0}^{maximumheight} n(y)\sqrt{\frac{dx}{dy}^{2}+1} ds=minimum \Rightarrow \frac{d}{dt}\int_{0}^{maximumheight} n(y)\sqrt{\frac{dx}{dy}^{2}+1} ds=0[/itex]

    we can pass the derivative to the inside of the integral and assume that n(y) varies linearly with y to simplify calculations but anyway, it doesn't like very nice.. Any suggestions?
     
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