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Paraxial wave equation

  1. Dec 1, 2008 #1
    paraxial wave equation - Solved

    1. The problem statement, all variables and given/known data

    When a laser beam travelling is traveling in one direction, we can make the paraxial approximation.

    Question: Find an expression for the surfaces with constant phase in the beam.

    2. Relevant equations

    From a previous part of the question, I had to work out the paraxial wave equation
    which
    [tex]\nabla^2_{x,y}G - 2ik\frac{\delta G}{\delta z}[/tex]
    and confirm that the below was a solution
    [tex]G(r, \omega) =\frac{1}{s^2(z)}exp[-\frac{x^2+y^2}{s^2(z)}][/tex]
    where [tex]s^2(z) = w_0^2 - \frac{2iz}{k}[/tex]

    3. The attempt at a solution

    I think I have to find an equation of the form.

    [tex]G = Re^{i\phi}[/tex]
    [tex]phase = \phi[/tex]
    where R = radius of curvature

    The complex parts are confusing, the solution/equation isn't spherical so I'm a bit stuck on where to start.
     
    Last edited: Dec 1, 2008
  2. jcsd
  3. Dec 1, 2008 #2

    turin

    User Avatar
    Homework Helper

    I considered the more general algebraic problem of finding the expression for the constant phase of

    [tex]
    G=\frac{1}{a-ib}\exp\left(-\frac{c}{a-ib}\right)
    [/tex]

    I first do the conjugate multiplication thing, so that I'm dealing with sums of real and imaginary parts.

    [tex]
    G=\left(\frac{a}{a^2+b^2}+i\frac{b}{a^2+b^2}\right)\exp\left(-\frac{ca}{a^2+b^2}-i\frac{cb}{a^2+b^2}\right)
    [/tex]

    Then I factorize, with my eye on the prize.

    [tex]
    G=\frac{1}{\sqrt{a^2+b^2}}\exp\left(i\arctan\frac{b}{a}\right)\exp\left(-\frac{ca}{a^2+b^2}\right)\exp\left(-i\frac{cb}{a^2+b^2}\right)
    [/tex]

    Finally, I collect factors so that I have the standard polar form.

    [tex]
    G=\frac{1}{\sqrt{a^2+b^2}}\exp\left(-\frac{ca}{a^2+b^2}\right)\exp\left(i\left(\arctan\frac{b}{a}-\frac{cb}{a^2+b^2}\right)\right)
    [/tex]

    So, a constant phase would require the transcendental equation to be satisfied.

    [tex]
    \arctan\frac{b}{a}-\frac{cb}{a^2+b^2}=\textrm{constant}
    [/tex]

    Perhaps the paraxial approximation allows the [itex]\arctan[/itex] to be approximated so that the equation becomes solvable ...
     
    Last edited: Dec 1, 2008
  4. Dec 1, 2008 #3
    Thanks. Very clever piece of factorising there :). The lecturer said not to be concerned with a long equation.
     
  5. Dec 1, 2008 #4

    turin

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    Homework Helper

    I wouldn't call that result "a long equation" (even though it is unsolvable). Are you looking for something simpler?
     
  6. Dec 1, 2008 #5
    Shouldn't of said long without a reference if I'm studying physics really :). The paraxial approximation allows the small angle approximation to be used.
     
  7. Dec 1, 2008 #6

    turin

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    Homework Helper

    I believe you mean
     
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