So now we've seen for a Gaussian beam and for a rectangular aperture that when you go through a lens that takes a Fourier transform, the shape of the focused spot, defraction limited focused spot, is given by the Fourier transform of the light at the front focal plane. We're now going to take that concept a little further, or perhaps expressed a little different or might be a better way to say it. And anyone that's very familiar with Fourier transforms, as soon as we found the impulse response of the system, because that's what those focus spots are, you might expect to find a transfer function soon after. And that's what we're going to do. This is, again, the concept stolen from the course Fourier optics, just pulling the bits that we need. And encourage you to go and think about that course, ff you want to know more about this. We're going to get just enough that we can understand how optic studio will calculate properties for us, and particularly, how we understand the limits of the geometrical optics approach. Not going up the universe thing. So let's imagine we have an object here at a front focal plane. And that object is going to be relatively small in comparison to the diameter of the lens. Relatively small is defined by the Fresnel integral which is taught in Fourier optics. We don't really need to to worry exactly what [INAUDIBLE] fit means here. I'm going to imagine that I want to think about a particular spatial frequency being launched off of this object. And remember, we can take any spatial frequency with a, well there's a spatial frequency, so its period is lamda x. The spatial frequency is 1 over lamda x, and would be given by the angle of the plane wave being launched off of the axis over the pre-space wavelength. And that makes sense, right? Because let's see, if sine of this angle were zero, that would be a wave going right down the axis, that would give me a spatial frequency of zero or a period of infinity. Okay, check, that's what we saw in that spatial frequency explorer. Conversely, if the angle here was 90 degrees, that would correspond to a wave going right up the boundary. Sine theta would equal 1, and the spatial frequency would equal 1 over lambda. And the spatial period would equal lambda. That's just what we'd think, this period here would get tilted over and be going straight up the boundary. So the range of spatial frequencies that we can launch is between 0 and 1 over lambda. The question we'd like to ask is, but which of those actually get through the system? Notice that there's some metal mounting my lens here. And way up here, for high angles, I don't get through. Way down here, for low angles, I do get through. What's the angle or spatial frequency which we cut off, where we just hit this edge and we'd expect to suddenly see the transmission of the system drop as we tilt our array up here? Well, that's relatively easy. We have our expression for the spatial frequency here, and it's got sine alpha in it, which is my angel right here. Well, in the paraxial limit, that's easy to calculate. That's this over that, so it's D over 2 over the distance between object lens t. Notice the sign doesn't really matter here. All the systems are symmetric, so positive spatial frequency and negative spatial frequency are the same thing. So I'm not worrying about the sign. But I've done it and written it down in our sign convention. So here's my approximation for sine alpha D over 2 over t. Which, notice that the quantity D over 2t is the numerical aperture, or in other words, sine theta is the numerical aperture. That's not very surprising. And that we're going to call the cutoff spacial frequency. We're now thinking about this object as a sum of spatial frequencies. That is its Fourier transform. And we've just found that for low spatial frequency, you get through the system. And other things may get screwed up along the way. But at least if we're worried about what can possibly get through, we find any frequency below NA over lambda gets through. And any frequency above does not. This is another way of thinking about what numerical aperture means. In units of 1 over lambda, it's the cutoff frequency of the system. That's a pretty important concept. And what you can see here is let's imagine we do have an imaging condition between x and x prime for example. So the object has been painted over here along x prime, perhaps with inversion, perhaps with some scaling. But it's not a perfect copy, because it's undergone a low pass spatial frequency with a cutoff frequency of f0. That's why it's blurry. It's got some finite resolution, because it's gone through a low pass spatial filter.