# Fourier transformation of J0(x)

1. Feb 15, 2010

### KFC

In wiki (http://en.wikipedia.org/wiki/Fourier_transform), there said the fourier transform of the Bessel function (zeroth order J0) is a rect function (window). But I also saw a text (about optics) that the Fourier transform on a ring slit (or ring disk) is zeroth-order Bessel function, so which one is correct? If wiki is correct, what is the fourier transform on a ring disk?

2. Feb 15, 2010

### nicksauce

Well presumably it is the same thing, but the first one is in x,y space, and the second in is radial space.

3. Feb 16, 2010

### Count Iblis

It is a rectangular window divided by sqrt(1-omega^2). In case of the 2 dimensional Fourier transform, you are considering the function J0(sqrt(x^2 + y^2)) as nicksauce said.