1. Not finding help here? Sign up for a free 30min 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!

Parabolic light-path's function for index of refraction

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data

    The setup is a planar cross-section of a medium. The horizontal direction is 'x' and the vertical direction is 'y'. I fire a beam of light into this plane, starting at some positive y, at x = 0, and traveling horizontally in the x+ direction. I need to find a function n(y) describing the index of refraction that would make the beam of light follow the path s(t): x(t) = t, y(t) = y_0 - t^2 , where y_0 is the initial y position of the beam.

    2. Relevant equations

    L = integral{A to B} of [ n ds ] = integral{A to B} of [ n sqrt( (dx/dt)^2 + (dy/dt)^2 ) dt ] . Here L is the optical path length along the path the beam follows. It is a path integral of the index of refraction along the path of the beam.

    f(x,y,x',y') = n(x,y) * sqrt(x'^2 + y'^2) is an extremum function that can be constructed from the above integral. x' is dx/dt and y' is dy/dt .

    dn/dx - (d/dt) * (df/dx') = 0 and dn/dy - (d/dt) * (df/dy') are the Euler-Lagrange equations.

    Please note that all of these equations have been taken from the website "http://www.mathpages.com/rr/s8-04/8-04.htm" [Broken]. The equations there are for a general n which is dependent on both x and y. However, in my problem, n is a function of y only, so there is probably a simpler version of the above.

    3. The attempt at a solution

    I have tried many times to solve the above equations for n. My latest attempt involves first of all defining y = -x^2, and then defining a function F = n(y) * sqrt(1 + (dy/dx)^2), and then using F in the equation dn/dy - (d/dx)*(dF/dy'), where y' = dy/dx . I think a formulation of y(x) instead of x(t),y(t) is also valid since x(t) = t and y(t) = y_0 - t^2. After several pages of algebra, I end up with the equation: ((8y + 6) / (16y^2 - 8y + 1)) dy = dn / n . Using the Wolfram Integrator for the LHS, I end up with: n(y) = e^(0.5 * ln (1-4y) + 2/(1-4y) + C) = sqrt(1-4y) * e^(2/(1-4y) + C).

    I am certain that I have made an error somewhere, as each time I solve the equations I get a different answer. I am also pretty sure I have made a more fundamental error in setting up the equations somewhere, as the exponential answer I get does not seem to make much sense - I do not think an exponentially increasing index of refraction would be necessary to cause a light beam to bend in a parabolic trajectory, plus also the answer becomes imaginary as soon as y > 0.25
    Last edited by a moderator: Apr 23, 2017 at 8:48 AM
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?

Similar Discussions: Parabolic light-path's function for index of refraction
  1. Hyperbolic functions (Replies: 0)