Is the Fourier Transform of J0(x) a Rect Function or a Ring Disk?

[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?

 

[nicksauce]

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

 

[Count Iblis]

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?

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.