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Law of Refraction with changing index of refraction

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A light ray enters the atmosphere of the Earth and descends vertically to the surface a distance h = 101.2-km below. The index of refraction where the light enters the atmosphere is n = 1.00 and it increases linearly with distance to a value of n= 1.000293 at the Earth's surface.


    Over what time interval does the light traverse this path?

    2. Relevant equations
    v1/v2=n2/n1

    v1=3e8
    n1=1
    v2=3e8/n2
    D=distance from atmosphere barrier

    3. The attempt at a solution
    So the increase in n2 per m = 1/345392491

    Therefore n2=(D/345392491) +1

    v2=3e8/(1+(D/345392491))

    So v2 changes with distance. I'm not sure where to go from here to get time.
     
  2. jcsd
  3. Sep 24, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    See if you can write an expression for v as a function of altitude. That is, if y is the vertical height then v(y) = ?. It will rely on also having an expression for n(y). Then knowing that v is dy/dt you should be able to write a differential equation.
     
  4. Sep 26, 2015 #3
    Thanks I got the answer!
     
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