1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook