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Light beam path

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    In plane parallel plate the refractive index is a function of coordinate ##z##, so that ##n=n_0 -{n}'z^2## for ##{n}'>0##. The origin of the coordinate system is in the middle of the layer, and ##z## is parallel to the normal of the layer. In paraxial approximation calculate the path of a light beam.

    2. Relevant equations
    If we use parametrization ##s##:

    ##\frac{d}{ds}(n(z)\frac{d\vec r}{ds})=\nabla n##

    3. The attempt at a solution
    For ##z## coordinate: $$\frac{d}{ds}(n(z)\frac{d z}{ds})=\frac{dn}{dz}=-2{n}'z$$ $$\frac{d}{ds}([n_0-{n}'z^2]\frac{d z}{ds})=-2{n}'z$$ Now I hope I can use approximation that ##dz\sim dx## this would bring me to $$[n_0-{n}'z^2]\frac{d^2z}{dx^2}=-2{n}'z$$ and finally since ##z## is very small, than I can forget about ##z^2## term. $${z}''+\frac{2{n}'}{n_0}z=0$$ which brings me to my final solution $$z(x)=Asin(\sqrt{\frac{2{n}'}{n_0}}x)+Bcos(\sqrt{\frac{2{n}'}{n_0}}x)$$
    Or is this completely wrong?
     
  2. jcsd
  3. Apr 20, 2015 #2

    TSny

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    Reconsider this. Paraxial means that the ray always makes a small angle to the z-axis.

    Note that this is a vector equation. So, you will need to think about its components.
     
  4. Apr 20, 2015 #3
    Hmmm, ok firstly, let me apologize for a mistake in my first post. There should be ##ds\approx dx## and not ##dz\approx ds##.
    But you got me a bit confused now. I think you are trying to say that I should use approximation where ##ds\approx dz## but... this is in contrary with the problem. The problem says that the ray is only travelling close to the centre of the layer (where ##z## is very small). Or...?

    The other two components only give me $$n(z)\frac{dx}{ds}=n(z)\sin\vartheta _x=C_x$$ and $$n(z)=\sin\vartheta _y=C_y.$$ Do they have any physical interpretation?
     
  5. Apr 20, 2015 #4

    TSny

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    OK, I guess I misinterpreted the question. Since they only mention a z-axis, I assumed that paraxial meant that the ray is traveling approximately in the z-direction as it passes through the plate. So, yes, I was thinking ds ≅ dz.

    [EDIT: I agree with your solution.]
     
    Last edited: Apr 20, 2015
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