In the last couple of slides, we have found the impulse response of an optical system illuminated by a plane wave. So we took a plane wave, we chopped it off with a rectangular or circular aperture, either 1d or 2d, and we Fourier transformed that to find the area disc. We did it first in 1d, now we're going to go back and do it in 2d, but we found sinex over x was the shape of the light that we got in the back focal point of the lens. We also then looked at the lens as something that limited the ray angle and thus the spatial frequency that got through the system. We defined the pupil function as looking into the system, what light gets through at low angles versus what light does not get through at high angles. And we realize that could be called the transfer function of the system. It could be used to describe the spacial frequencies that got off of the object, into that lens, and on through to our image plane. So we now have an impulse response and a transfer function. And of course, that's a happy pair to have, and linear systems theories tells us that those two should be related by a Fourier transform, so let's check that. There's that relationship. We should be able to take the spatial frequency transfer function of the system. Unlike temporal frequency transfer function, this is a function of two variables because we have two space coordinates x and y in one of our meters. And if we inverse transform that in x and y, we should find the impulse response, little h, in x and y. So, we did this before, where we took a Gaussian and we found that we had Gaussian point spread function. We did it with a one dimensional rect here as the transfer function system. And we found that the one dimensional sinex over x version of an airy disk. And we found that this radius to first null was f number times lambda. Now let's do it to the two dimensional version of a pupil, where this is rect r over d. Appropriately scaled to special frequency like we just did. It turns out the inverse Fourier transform of that function, instead of being sinex over x, is a special function x over x which looks a lot like sinex over x as a matter of fact. So this is actually the airy disc expression. This is that function right there as a function of radius, r. We had a lens of a particular diameter, d, that made it. Or particular distance, t, away from that lens and we're using light to that particular wave length lambda. Probably the first thing we'd like to know is, where is this null? Gives us a measure, a ruler for how big is our focus spot. So you look up in any text book, or online, what's the first null of the first order vessel function J, and it's a value of 3.83 and change. So we set this top thing equal to 3.83, right there. And we solve for r naught, and it turns out to be, instead of 1, lambda t over D, that's f number of times lambda that we had before, it's 1.22. So our expression before, for the one dimensional case was off by about 20%. And that's just the difference between one dimensional slit and a round hole in terms of how they Fourier transform. If you translate f number here to a numerical aperture, before we had 0.5 lambda over NA for the slit, now we have 0.6 lambda over NA for the round hole. So the point is we have beautiful intellectual framework now. This is probably the most important thing coming out of it, there's two. First is just knowing that there's a finite diffraction limited spot size. It's an airy disc, and the distance to first knoll is 0.6 lamda over NA. That's a phrase that should just roll of your tongue. The other concept, the equivalent one in spacial frequency is that we can look at the aperture of an optical system, think about it as spacial frequency low pass filter and the transfer function and the cut of frequency is NA over lambda. So we can think about these systems either through their impulse response, their diffraction limited spot size, or through their Fourier transform, and their transfer function, and that they are low pass filters with a cut off frequency, lambda over NA. Notice, and this is the same concept over, and over, and over here, that the spot size is 0.6 lambda over NA. And the cut off frequency is NA over lambda. So the other way I could write this is 0.6 over the cutoff frequency. In other words the cutoff frequency and the spot size are inverse, and that makes perfectly good sense. That's the same Heisenberg uncertainty relationship that we saw before.