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Path of Light- Calculus of Variation

  1. Aug 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Let y(x) represent the path of light through a variable transparent medium. The speed of light at some point (x,y) in the medium is a function of x alone and is written c(x). Write down an expression for the time T taken for the light to travel along some arbitrary path y(x) from the point (a,yI) to the point (b,yF) in the form:
    T=[tex]\int L(y,y')dx[/tex]
    (between a and b, can't seem to add in limits on the integral sign)

    where L is a function you should determine.
    2. Relevant equations

    S = [tex]\int \sqrt{1+y'^{2}}dx[/tex]

    I derived this though, don't think its meant to be a known equation.

    3. The attempt at a solution

    Well using calculus of variation I know you can formulate a path length as:
    S = [tex]\int \sqrt{1+y'^{2}}dx[/tex]

    So the time for the light to travel along this path should just be:

    T= [tex]\int \frac{ \sqrt{1+y'^{2}}}{c(x)}dx [/tex]

    However, this has a dependence on x where the question states it must be only in terms of y and y'. This is because in the first part of the question I had to derive the Euler-Legrange equation and as light takes the path which minimises time, you must plug this function L into the equation.

    So, how do I get rid of the x dependence? If there's a trick or a technique involved could someone just inform me of the name so I can look it up for myself? I've read the classical mechanics textbook which is meant to accompany this course and can find nothing which helps with this.

  2. jcsd
  3. Aug 27, 2008 #2
    can x be written as a function of y?
  4. Aug 27, 2008 #3
    Well thats what has been running through my head the whole time, however everything that I have been given in the question I've posted. The only thing I can come up with is c(x) = c(y[tex]^{-1}[/tex](y)) as in the inverse function of y...of y. I don't think that helps...

    Sorry if I'm being dense and there's something obvious going on here.
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