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Homework Help: Refraction through a nonuniform medium

  1. May 13, 2010 #1
    1. The problem statement, all variables and given/known data

    A plane electromagnetic wave refracts in a nonuniform medium. The ray trajectory is known,
    [tex] y = a \left[ 1 - \left( \frac{x}{2a} \right)^2 \right] [/tex]
    where a=const.

    a) Plot y(x)

    b) From the slope of the curve find the angle [tex] \theta \left( y \right) [/tex] which the k-vector makes with respect to the surface normal (parallel to the y-axis)

    c) Using Snell's law, find the refractive index as a function of height, [tex] n\left(y\right) [/tex]

    2. Relevant equations

    [tex] n_i sin \left( \theta_i \right) = n_t sin \left( \theta_t \right) [/tex]

    3. The attempt at a solution

    a) simple downward facing parabola with y-intercept at a, x-intercepts at -2a and 2a

    b) It hinted at what to do by stating the slope of the curve which is the derivative of the trajectory.

    [tex] \frac{dy}{dx} = \frac{d}{dx} a \left[ 1 - \left( \frac{x}{2a} \right)^2 \right] [/tex]
    [tex] = a \left[ -2 \left( \frac{x}{2a} \right) \left( \frac{1}{2a} \right) \right] = \frac{-x}{2a}[/tex]

    The slope is thought of as "rise over run" so if we're looking for the angle with respect to the y-axis the triangle formed from the tangential to the trajectory and the y-axis the sides of the triangle given from the slope is adjacent and opposite.

    [tex] cot \theta = \frac{-x}{2a} [/tex]
    [tex] \theta \left( y \right) = tan \frac{-x}{2a} = -tan \frac{x}{2a} [/tex]

    c) This is where I'm stuck, Snell's law is given in the earlier section. I can foresee the need to go to infinitesimal levels as the index of refraction changes over the minute values of y through the trajectory but I cannot see how to begin this task.
     
  2. jcsd
  3. May 13, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Good...

     
  4. May 13, 2010 #3
    Ha thank you for the catch, I haven't been completely here lately.

    [tex] \cot \theta = \frac{-x}{2a} [/tex]

    [tex] \theta \left( \textbf{x} \right) = \cot^{-1} \left( \frac{-x}{2a} \right) [/tex]

    [tex] = \cot^{-1} \left( - \frac{2a \sqrt{1-\frac{y}{a} } }{2a} \right) = - \cot^{-1} \left( \sqrt{1-y/a} \right) = \theta \left( y \right)[/tex]
     
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