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The special frequency tool in MATLAB,

that we just explored,

hopefully has given you some intuition that we can think

about plane waves as Fourier transforms or through Fourier transforms.

And that if we know the spatial frequency on a boundary,

let's say the X boundary we were looking at,

that corresponds to a term of the wave vector, the KX term,

and since K is constrained to be

equal overall in magnitude to two pi over the wavelength,

if we know the KX term,

we then know the KZ term and we know the angle of propagation.

Those concepts are fundamental to a classical Fourier optics,

not surprisingly, but we need a little bit of that here,

and that's why we're walking through this particular module.

The reason we need it is we need to be able

to predict the light at the focus of a system,

and systems that are different than just Gaussian beams.

We need arbitrary distributions of light.

And it turns out the way to do that is this Fourier transforms.

That's what optics studio does.

And next, we'll go look at how these two

terms and how these techniques look in optics studio.

But we need to understand where the plots are coming from.

So that's what this part of the class is about,

is to bring Fourier concepts into our geometrical ray tracing class.

And the relationship is,

that spatial frequency explore,

that the K vector can be thought of as our ray vector and therefore on a boundary,

the K vector corresponds to a spatial frequency and now we have the Fourier concepts.

So we're going to walk through the key concepts of the Fourier optics class that we need,

it's very powerful stuff.

And to really understand diffraction in greater detail,

I encourage you to go look at that class.

So, let's think about an imaging system,

actually not an imaging system,

a one F system where we have a single lens.

And on the front boundary,

which is going to be the front focal plane,

we're going to paint that same complex exponential,

drawn here through its real part,

that we have in the spatial frequency Explorer.

And remember that if it's got a particular period,

we'll call that lambda X, because it's in the X direction,

we could relate that to a wave going at a particular direction theta

with a local wavelength

between the waves along the direction of the propagation length naught.

One over wavelength is frequency.

And so, the spatial frequency is sin theta over lambda.

Now, let's walk through what we expect

a special frequency on this front boundary to do when we get to the back boundary.

And we don't really have the ray tracing tools for that.

Well, it turns out we do and we're going to go see those.

So, here is that plane wave that corresponds to that period on the boundary.

It's just the same as my spatial frequency Explorer that you played with last.

It's got a vacuum wavelength,

or a vacuum here of lambda naught.

It's going in a direction theta.

And so, notice my little dotted lines here that I have

peaks of this wave along X that are separated by lambda X,

and peaks of this wave along the direction of propagation separated by lambda.

And the two of those through a little bit of

trigonometry tells you that the direction of wave is theta,

and the relationship is line theta.

So, great.

Now we can take any arbitrary complex exponential on the boundary,

that it's a periodic function here and we can relate

that to the plane wave that must be propagating off towards the lens.

And we know that then can be described in

geometrical optics by a ray or a bunch of parallel rays if you like.

Awesome. That means, those parallel rays hit the lens and of

course they converge to a single point in the focal point here.

That's simple graphical ray tracing.

And a converging cone of rays corresponds to a converging spherical wave.

Awesome roll over of that.

And of course the single ray I traced here is our world's simplest,

which is the one that goes through the center of the lens,

so that the angle theta here is the same as the angle theta here.

Good so for far, doing great. So, even the

geometrical approximation where this comes to an infinitely sharp focus,

I have a delta function right here.

In the position, in the electric field,

we've got nothing and all right here to focus on.

The position of that delta function would be given by the focal length times sin theta.

You might be thinking it should be 10 theta and it actually is sin theta.

That's a subtlety will get to in the third class.

Turns out the lens is focused on the spheres, not on the planes.

So, but in the practical approximation,

it doesn't matter because everything would be just theta whether it's in sin or 10.

So, I know the position of my focus focal length times sin theta. Oh, hey?

I know sin theta over here,

so I could substitute.

And the key result is right here,

is that the position of this delta function is the focal length,

the wavelength in this material,

times this spatial frequency,

the one over here.

That is important because now I can draw all this picture.

I have a sinusoid or complex exponential

on this side with a particular spatial frequency ethics.

On this side of the lens,

I want to program it like behind it,

I have a delta function,

complex exponential on this side,

delta function on this side.

It's a Fourier transform relationship.

And conveniently, we find that the position of this delta function is linearly

related to the spatial frequency of

the sin wave and the scale factor because there has to be a scale factor,

Fx is in one over meters.

And I don't know how to measure distance in one over meters.

So you better multiply by something with quantities unit.

I better multiply this by something that has

the units distance squared because I'm measuring the distance here in real space.

There's one distance, second distance on axis.

So there's scale factor, focal length, and free space wavelength.

But after I remove that scale factor,

what I have on the right- hand side is that

Fourier transform of what I've on the left-side.

So there's nothing but a proof.

You can prove this with final equations derived from the wave equation.

But hopefully, I have motivated the fact that if I

have a lens and I paint an electric field on the front side,

I find the Fourier transform of the electric field on the back side where

the coordinate system here is scaled by F times the wavelength.

That is called the Fourier transform geometry.

There are others, but this is the kind of classic geometry.

And we'll now use that to understand if we

have an arbitrary field distribution on this side,

what's the actual size and shape of the focus spot we find over here?

So, in summary, the optical system consisting of one F propagation distance,

a lens, and another F for propagation distance is a Fourier transform.

We paint an electric field on the front plain here X.

It's got a spatial frequency FX.

We find on the back focal plane X prime a delta function,

if we have complex exponential over here.

But in general,

the Fourier transform variable is one over a distance of spatial frequency.

And since this is in real space,

there must be some sort of scale factor. There's the scale factor.

So, the black box system of this lens

performs a Fourier transform of the electric field with this particular scaling.

That's how we're going to take arbitrary fields into the front of the lens and

understand what the actual shape and size of the focus is on the back of the lens.