 where I'm gonna integrate over all the transverse components. And of course I need to give them some different amplitudes to each one. So I'm thinking like Fourier again. You can decompose anything into things that oscillate. I just need to find what they are. So I'm gonna put something here. Depends on these transverse things here. And I'm gonna integrate only over d new x and d new y. Why not the c component? Why only x and y? Because the c one is determined by the x and y. But if I write this again, substituting this form, then this is the same as the integral. And it's from minus infinity to infinity, a of new e to the minus i 2 pi new dot. And this part that I can write as e to the i. Let me write this part as h of new d, d new y, where h of new is that exponential. e to the i 2 pi. And then I can bring that lambda inside of the square root. So this gives me one over lambda squared minus new x squared minus new y squared. Is that okay? And it is assumed that when this is bigger than one, this is positive. When this is smaller than one, the square root is positive imaginary. So that this whole thing, the case, times c. Times c. Now, how do I find this a? Four minutes. How do I find this a? So this h depends on new and depends on z. What happens if I let z equals zero? What is the exponential equal to? One. So notice that h of new comma zero is equal to e to the i zero equal to one. So if I evaluate the field at the initial plane, u at x comma y comma, let me say z equals zero, then what I have is the integral from minus infinity to infinity of a of new e to the minus i to pi new dot x. d new x, d new y, dh goes away. And what is this formula? I was seeing this before. It's the inverse Fourier transform. The inverse Fourier transform of a is equal to the field at the first plane. So I can call this, let me call this the initial plane of x comma y. That means that a is the Fourier transform of the initial plane. So therefore, this a of new, which is called the angular spectrum, is the Fourier transform as x goes to new x, let's say y goes to new y of the initial field. So if I have a field and I know what it is at the initial plane, I Fourier transform that and that tells me what the weight of the plane waves is. From there to propagate, I just let these plane waves travel. In other words then, the field at x comma y comma z is now, this is the inverse Fourier transform of what? Of this, no, this whole expression looks now like an inverse Fourier transform of the product of this times h. So that is, and this is the Fourier transform of the initial field. So this is the inverse Fourier transform as new x goes to x, new y goes to y of the transfer function times the Fourier transform as x goes to new x and y goes to new y of the initial field. So to propagate the field from a known initial plane, the plane waves of the property tells me I just have to do the Fourier transform, multiply it by this simple transfer function and the inverse Fourier transform. So it's again our signal and response. The signal is now what? Is the initial field and the response is the final field after propagation and transfer function is this phase that accumulates in the propagation or the exponential decay. So the plane waves give us this connection between solving the Hamel's equation and Fourier theory. So I'm gonna start from this tomorrow and try to interpret again the uncertainty relation and all these properties in terms of this and show you on the computer how this can be used to propagate and all the properties and then we'll go into measuring the resolution of an imaging system, et cetera. I just wanted to get to the punch line before we leave. All right, so coffee break and at four, please be here for professor, sorry. Okay, so about that, green fluorescent proteins. Very good.