Optical Fourier Transform for Propagation

[bryverine]

Homework Statement

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) $f(x,y)=cos^2(\pi y / 2 \lambda)$

Homework 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):
$g(x,y)=h_0exp(\frac{i\pi(x^2+y^2)}{\lambda d})F(\frac{x}{\lambda d},\frac{y}{\lambda d})$

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 propagation 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!

 

[blue_leaf77]

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.