0:04

So if we can Fourier transform a Gaussian beam at the front focal

point of the lens and find that we get the correct focus to Gaussian

beam at the back focal point of the lens, the next obvious function to

try would be a square wave at the front focal point of the lens.

So this is something you might make by illuminating an iris or

some sort of aperture.

Which is zero transmission and then unity, it's a whole, and then zero again.

And so that makes a function that's zero, 1, zero.

What does that look like when we focus it though a lens?

And this is what's called the Airy disk.

It is the classic focus that you get at the back of a lens, and

it's very representative of if you look at a star, for example.

The star is sending a spherical wave at us, but by the time it gets to us,

it's pretty much a plane because the stars are very far away.

The aperture there is the front of your telescope.

You let some light in, you focus the light through the object of the telescope,

this is what that star will look like.

So this is a pretty important equation and function to understand

because it really is the common focal shape that you get behind lenses.

And we now have the tools to direct that.

1:23

So at the front lens, I'm going to do this first for the one-dimensional case,

that is, I have no variation out of this plane.

And then we'll come back and do it for

the two-dimensional case where it's a circular hole instead of a slit.

But right now, we'll first do the slit.

And it's almost exactly the same solution, they're only slightly different.

So at the front focal point of the lens, I have a rectangle function.

If you're not familiar with this function, it simply means something that's zero, 1,

zero, with a total diameter or width, D.

2:09

And so the, when we do the Fourier transform of this rect,

the D here shows up there.

Notice again, we have units of distance, units of 1 over distance.

So this is the Fourier transform of the rect, but

it's in spatial frequency of this function over here.

We can't see spatial frequency in real space because it's got units of 1

over distance, and I don't understand what that means.

So the only way to understand how it's really painted in x prime and

real space here is we have to make the scaling substitution.

Fx equals the spacial coordinate behind the lens x prime,

over the focal length of the lens and the wavelength that we're in.

And that is units of 1 over distance.

Yay, that's cool, but now we have x prime here with units of distance.

So this is the Airy disk.

That's actually a slice, or very close to it, along this function right here.

Sine x over x has a peak at the origin, and then bumps off to infinity.

3:14

It's probably nice to have some measure of how big this is.

And the common measure is to go from the center to the first null.

That's arbitrary.

You could go center to half way down the peak, full width, half max,

or half width half max.

The Gaussian beam, we want the radius of the 1 over e point

of the field, because it was convenient.

The point is, there's a little bit of ambiguity here in

how big is this function, that's dependent on how you choose to measure it.

And you should hope be sure you know what that measure is.

But for a function that's got nulls,

as this one does, it's really common to measure the radius to the first null.

Well, that's really easy.

We set this function right here equal to pi.

And that tells us where that r naught is, that first radius, coordinate x prime.

And that's simply FD over lambda.

That bit right there.

We'll notice focal length over diameter is a quantity we've already run into.

That's called the F number of this lens.

Or we've also run into this and described it as numerical aperture.

That's just basically the diameter, sorry, radius of the lens, over the focal length.

So in the paraxial approximation, it's this angle right there.

Or, the sine of it in the non-paraxial approximation.

So, both of these expressions are really useful to remember.

The spot size measures radius to first null is F number times lambda.

This is why F number is an important and useful quantity.

It tells you the resolving power.

The size of your focal spot given in units of the wavelength.

So if you have an F1 lens, it's a powerful well corrective lens,

it's tiny foci that decides the wavelength.

If you have an F10 lens, your spots in radius to first null are 10 wavelengths.

5:18

Since F number is NA related directly to NA with a factor of 2 in it right here,

this expression is also handy to have.

So expressed as a numerical aperture,

the radius of the first null is 0.5 lambda over NA.

Notice as we've been talking about through this whole course now, light has a shape.

As you take the angle of this cone of light up, that's what the NA is,

then the spot size r naught goes down.

So that's this concept that Heisenberg's uncertainty principle.

The angle of the cone of light is the uncertainty and

momentum, r naught is the uncertainty in position, and they're inversely related.

If you want a tiny little spot, you need a large numerical aperture, and vice versa.

6:06

We'll come back to the two dimensional solution in just a minute, but

the point is, it's almost exactly the same.

This is formally an infinitely long slit, and this is a cylindrical lens.

That solution we did here.

If you solve this for a round lens with a round hole,

you get slightly different Fourier transform.

Has the same shape, and the answer is 0.6 lambda over NA instead of 0.5.

So pretty darn close.