# Optical Fourier Transform for Propagation

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1. Oct 9, 2015

### bryverine

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).