We'd like to give you a lab demonstration of the plane wave, but there's a couple of reasons that's hard. First of all, plane waves are formally infinite in extent, and that makes it a little bit tricky. But perhaps less seriously, or more seriously, the interesting part about the plane wave is the relationship of the wavelength to the phase. And that's all small, on the order of a micron or smaller. And so it's kind of hard to get a camera down there and look at it. So instead, we've got here a little MATLAB simulation. And this is up on the website for the course. So you'll be able to play with this yourself. The reason we're doing this is we want you to have some good intuition for how the plane wave works, and particularly this concept of spatial frequency, which we're going to use in the next module of the course. So here I've shown a chunk of a plane wave. And I'm drawing the real part of the amplitude. And you can tell that it's the real part, because it's got variation. If I just drew the amplitude, Then it would be constant, right? The absolute value of the field is uniform. And I've drawn it versus two space dimensions, your x representing some boundary and z, some distance on propagating away from the boundary. So this right here represents a plane wave propagating directly away from the boundary. And so if I unfroze time, if I let time roll forward, these waves would be flying forward at the speed of light. I've normalized all coordinates by wavelength, because that's a convenient thing to do. So there's one bright fringe every wavelength, as you'd expect. Now what I want to do is I want to tilt this way a little bit. And I want to observe, as I do that, what happens to the field along x and along the z-axis, let's say? And I've already plotted those here. Along x, for example, well, the field is bright and constant, so it has a value of 1. And along z, it's a sine mu site. And the distance between the peaks is the wavelength and the material. Just what I'd expect. As I change that now by tilting it, I see now that my wave, if I just move along x, if I take that slice, is a sine wave. But it's a relatively large period sine wave. Before it was infinite period, now it's just large. And if you watched, let's do it again. I'll go back here, at the wave in the z direction, it's actually started out with the wavelength lambda, the wavelength in the material. But it's getting bigger, the period is increasing. And you can tell that by the waves are sliding to the right. So that's interesting. Somehow I'm decreasing my period, or increasing my spatial frequency, 1 over the spatial period, along x. And I started out with a small period, or high spatial frequency along z, and it's decreasing. Well, if we keep talking about frequency, that suggests we should think about Fourier transform. So over here I've plotted the Fourier transform of this function. Of course, it's a delta function, because this is a complex exponential. I've drawn just the real part so that we can see it. And really, I didn't tell you what this slider was over here I was moving, but it's the spatial frequency. So let's go over to, I don't know, one-half. And of course, there we go. So now we have a spatial frequency, units of 1 over distance, conveniently normalized by multiplying by the wavelength, equal to one-half. That says that we should have a spatial period that is, let's see, equal to 2. Look at that, right, spatial frequency one-half, spatial period 2, 0 to 2. That makes perfect sense. So what I'm doing here when I'm running my slider, is I'm changing the spatial frequency, which corresponds directly to an angle over here in real space. And we'll comment in just a minute on what that angle is. The smallest period I can have is lambda, the free space wavelength. Therefore, the largest spatial frequency I can have is 1 over lambda. Remember, this is normalized to lambda, so this point right here. So as I continue to tilt my wave, I can switch the x and z-axes in comparison to what I originally had, until eventually I have a wave that's traveling perpendicular to the boundary, instead of normal to it. And that's right there, look at that. So now I have a wave going this way. The period here is 0, yep. The period along x is one wavelength and the period along z is infinity. So you can see if these two trade. Well, how do they trade? Well, maybe the way to understand that, there's two ways. First, let's remind ourselves that spatial frequency here, 1 over lambda, 1 over wavelength. If I multiply that by 2 pi, that's k. That's the wave number which is radium spatial frequency. kx and kz are these, well, here's kx, and they're the 1 over these periods. And we derived, when we came up with the wave equation, that kx and kz, they're just components of the k vector. And the k vector itself has an overall magnitude, 2 pi over the material wavelength. In other words, the square root of kx squared plus kz squared is k. The kx and the kz are constrained. You can't pick one without the other, because you have to have, overall, just one wavelength, when you go along the beam. So last time I checked, an equation kx squared plus kz squared, the square root of it, is the equation for a circle. So that seems to me, and that's the last bit over here, that as, I'm going to go backwards here. As I change kx, I should find that kz, that's this direction, because fz times 2 pi is just kz, that it moves along a circular arc. This is just the magnitude of k, if I multiply it by this 2 pi, tracing out a circle. So what does that mean? If I tell you a spatial frequency on this boundary, then you can tell me the spatial frequency on this axis, because the k vector terms have to have, overall, a magnitude of k. A final way of thinking about this, and we're really now dipping our toes into Fourier optics, is what if I just took the two dimensional Fourier transform of this space? This plot here is the one dimensional Fourier transform of the field on the x boundary. But what if I took the two dimensional Fourier transform of the x/z space? Well, that's what this little dot is. And indeed, what we're finding is that it always moves along this arc. And that arc is the k vector having a magnitude of 2 pi over lambda. Now you might be thinking, what are these funny red lines out here? And what happens when I have a spatial frequency that's bigger than 1 over lambda? And we're going to leave that for you to explore on your own. So play with this tool. The key is we can understand a plane wave in a space, just by its spatial frequency on a boundary. Because once you tell me the spatial frequency on the boundary, I can paint the wave out into the space, because I know what the spatial frequency in z has to be. Because kx and kz are constrained, but the overall magnitude of k is 2 pi over lambda.