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Far Field Approximation After a Prism

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    A wave with an initial profile ##U(z)=Ae^{ik_0z}e^{-\frac{z^2}{2\sigma^2}}## is traveling in the z direction (yes, the Gaussian profile and the optical axis are not perpendicular). It then passes through a prism with apex angle ##\alpha## and refractive index ##n=n_0+\gamma(k-k_0)## as in the drawing.
    Assume the prism is infinite in the axes perpendicular to z.
    a) What is the refraction angle (from the z axis); assume small angles.
    b) Define the above angle as the new optical axis. With respect to that axis, find the field on a screen of distance d, using the small angle and far field approximations.

    2. Relevant equations
    Fraunhofer approx., Snell's law.

    3. The attempt at a solution
    The first part is a breeze, using Snell's law to get $$\phi=\theta_2-\alpha=\sin^{-1}(n\sin(\alpha))-\alpha$$ and then Taylor etc.
    The second part is more of a snow storm with Gaussian blur. It might not be that bad, it's just that I can't figure out how to handle the "infinite" prism.
    We actually solved a similar setup in one of the problem sets, where the prism was finite, with width D and height L (corresponding to the apparent sides of the triangle in the picture), and the wave had the form ##Ae^{-\frac{\rho^2}{2\sigma^2}}##. Writing the amplitude right after the wave exits the prism, we have $$U\left(x,y,0\right)=Ae^{-\frac{x^{2}+y^{2}}{2\sigma^{2}}}e^{ikD}e^{ik\left(1-n\right)\left(L-y\right)\alpha}$$
    after approximating the distance traveled through the prism as ##d(y)=(L-y)\tan(\alpha)\approx(L-y)\alpha##.
    Then, using Fraunhofer approx. $$-\frac{ik}{2\pi z}e^{ik\left(z+\frac{x^{2}+y^{2}}{2z}\right)}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dx'dy'U\left(x',y',0\right)e^{-i\frac{k}{z}\left(xx'+yy'\right)}$$
    it's just algebra to get to something like $$U\left(x,y,z\right)=-\frac{ikA}{2\pi z}\left[\frac{1}{\sqrt{2\pi\sigma^{2}}}\right]e^{ikD}e^{ik\left(n-1\right)\alpha L}e^{-\frac{\sigma^{2}}{2}\left(\frac{2\pi}{\lambda z}\right)^{2}\left[x^{2}+\left(y+\left(n-1\right)\alpha z\right)^{2}\right]}$$ (I might have erred along the way... nvm...)
    We can also deduce the angle of the new optical axis and compare it to the geometric analysis, by noticing we have the center of our new Gaussian, so it's just quick trig to get the angle.
    But now for this setup, I can't make much progress. First, all my answers were up to a constant phase, because I don't have D, so I can't really tell how much phase the wave in different places along the "y" axis had accumulated. So I can only make it relative. Second, Using Fraunhofer method doesn't make sense, because since it doesn't take z elements into account, it predicts that for all "z-only" profiles we would get the same diffraction pattern. So my next two guesses are:
    1) simply claim ##\Phi=Ae^{ik_0d}e^{-\frac{d^2}{2\sigma^2}}e^{-i\omega t}## because it's the same beast but in a different direction. But that's kinda lame, I don't buy it myself....
    2) assume the profile remains in the z direction, and just calculate using trig for a given d and refraction angle ##\phi## from the z axis, what's the corresponding z value for all x,y. But that also seems incomplete, as it makes no use of some of the data, specifically the refractive index and proposition to assume far field approx.

    Sorry for the length.
    Any thoughts?
     

    Attached Files:

  2. jcsd
  3. Feb 1, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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