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Optical Fourier Transform for Propagation

  1. Oct 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The complex amplitudes of a monochromatic wave of wavelength ##\lambda## in the z=0 and z=d planes are f(x,y) and g(x,y), redprctively. Assume ##d=10^4 \lambda##, use harmonic analysis to determine g(x,y) in the following cases:
    (a) f(x,y)=1
    ...
    (d) ##f(x,y)=cos^2(\pi y / 2 \lambda)##

    2. Relevant equations
    Part (d):
    ##f(x,y)=cos^2(\pi y / 2 \lambda)=.5(1+cos(\pi y / \lambda)=.5(1+.5(exp(+i\pi y / \lambda)+exp(-i\pi y / \lambda)) ##
    Fourier Transform Equations:
    ##F(\nu_x,\nu_y)= \int_{-\infty}^\infty f(x,y)exp(-i2\pi (\nu_xx+\nu_yy))dxdy##
    ##f(x,y)= \int_{-\infty}^\infty F(\nu_x,\nu_y)=exp(+i2\pi (\nu_xx+\nu_yy))d\nu_xd\nu_y##
    Transfer Function of Free Space (Fraunhofer Approximation):
    ##g(x,y)=h_0exp(\frac{i\pi(x^2+y^2)}{\lambda d})F(\frac{x}{\lambda d},\frac{y}{\lambda d})##
    3. The attempt at a solution
    ##f(x,y)=1\\
    F(\nu_x,\nu_y)= \int_{-\infty}^\infty exp(-i2\pi (\nu_xx+\nu_yy)dxdy=\delta(\nu_x-0)\delta(\nu_y-0)
    \\
    g(x,y)=\int_{-\infty}^\infty F(\nu_x,\nu_y)H (\nu_x,\nu_y)exp(+i2\pi (\nu_xx+\nu_yy))d\nu_xd\nu_y\\
    Assume\;Fraunhofer\;Approx\; \lambda <<d\\
    g(x,y)=h_0exp(\frac{i\pi(x^2+y^2)}{\lambda d})F(\frac{x}{\lambda d},\frac{y}{\lambda d})\\
    g(x,y)=h_0exp(\frac{i\pi(x^2+y^2)}{\lambda d})\delta(\frac{x}{\lambda d})\delta(\frac{y}{\lambda d})\\
    g(x,y)=(i/\lambda d)exp(-ikd)exp(\frac{i\pi(x^2+y^2)}{\lambda d})\delta(\frac{x}{\lambda d})\delta(\frac{y}{\lambda d})\\
    g(x,y)=(i/\lambda d)exp(-ikd)[exp(\frac{i\pi x^2}{\lambda d})\delta(\frac{x}{\lambda d})][exp(\frac{i\pi y^2}{\lambda d})\delta(\frac{y}{\lambda d})]\\##

    I think because the function is f(x,y)=1, there should just be propogation through free space. This should go to g(x,y)= some phase shift. My guess is that having a delta function at the end is wrong and it should "disappear" somehow.

    I'm using the Fundamentals of Photonics, Bahaa E.A. Saleh (either edition).

    Thank you for your help!
     
  2. jcsd
  3. Oct 9, 2015 #2

    blue_leaf77

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    Science Advisor
    Homework Helper

    Mathematically, your calculation does not show any fault. The FT of a constant is a delta function, however the physics involved should not allow your calculation to be true because Fraunhofer approximation is actually a further approximation within Fresnel approximation, in other words, by assuming Fraunhofer approximation, Fresnel approximation must be implied. And this Fresnel approximation does not allow the input field to be unbounded like it is in this problem.
     
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