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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: May 3, 2017
  2. jcsd
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