a) Convince yourself that the brightness pattern of light on the screen is a cosine function. (HINT: Think about the conditions for constructive and destructive interference of the light waves emerging from each slit).
b) Now imagine a second set of slits placed just inward of the rst set. How does the second set of slits modify the brightness pattern on the screen?
c) Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?
d) Notice that this continuous set of slits forms a "top hat" transmission function. What is the Fourier transform of a top hat, and how does this compare to your sum from the previous step?
e) For the top hat functions FT, what is the relationship between the distance between the first nulls and the width of the transmission function (HINT: it involves the wavelength of light and the width of the aperture)?
f) Take a step back and think about what I'm trying to teach you with this activity, and how it relates to a telescope primary mirror.
The two-slit experience is a deceptively simple one. We've got a single-wavelength beam of light shining on a screen with two minuscule slits in it. For now, we're assuming that these waves are coming in at exactly 90 degrees to the screen, and are perfectly straight and parallel (a star, for instance, is emitting light in spherical shells in every direction, but after traveling light-years to get to us, the shells are essentially flat planes--or lines if simplified into two dimensions). When it hits them, waves of light spread out of each one, then interfere constructively and destructively with each other, leaving an interference pattern on another screen in front of it. I've drawn this setup about a thousand times over the past week and a half, so rather than subjecting everyone to my somewhat questionable artistic skills yet again, here's a much better illustration.
As you can see, the light is emitted isotropically from each slit, and each set of waves interferes with the other. Wherever crests line up (along the blue lines), we end up with a bright spot on the right-hand screen (constructive interference). Directly in between is where the two lines are exactly half a wavelength out of phase with each other, giving us dark spots (destructive interference), with the intensity of the light waxing and waning in between to make a sinusoidal pattern--a cosine rather than a sine, since we have a maximum directly in the center at the "origin," with brightness on the y axis and distance from the central maximum on the x-axis.
So what happens when we add a second slit in between? To figure that out, I'm afraid I'm going to have to subject you to my drawing skills, so we can work out a mathematical relationship between D, the distance between the slits, and the spacing between the bright spots on the detector screen.
This is a pretty confusing diagram, so let's work out what's going on here. Light is coming in from the left, just like in the last image, and passing through the slits, which are $D$ apart. $x_1$ is the distance from each slit to the central maximum. The next maximum is $d$ away. It is also $x_2$ away from the top slit and $x_2 + y$ away from the bottom slit.
However, since maxima will only occur an when crests line up, light has to be traveling $x_2+\lambda$ from the bottom slit to the first maximum (if the next maximum is $x_3$ away from the top slit, it will be $x_2+2\lambda$ from the bottom slit, and so on), so we can tell that $y = \lambda$. We can also work out some angle equalities as well--we know that the two angles labeled $\phi$ are equal by the vertical angle theorem, which gives us $\phi + \theta = 90$. Bearing in mind that $L$, the distance between the two screens, is much larger than $D$ (yeah, my drawing isn't exactly to scale), we can also approximate that the two lines $x_2$ and $x_2 + y$ are parallel, and that the bottom line $x_1$ and $L$ are parallel, giving us the two right angles shown (and the "rectangle" redrawn below, which tells us that $x_1 \approx L$). This means that the angle between the line $x_2 + y$ and the left-hand screen is also $\theta$. The topmost angle of that right triangle (also redrawn below), then, is $\phi$.
Phew. That was a lot of geometry just to get everything labeled. But if we apply some trig to it, it can tell us what we need to know (and give us the answer to part (e) for good measure). That small triangle on the left tells us that $cos(\theta) = \lambda/D$. Meanwhile, we also know from the triangle made by $d$, $x_2 + y$, and $x_1$ that $cos(\theta) = d/x_1 = d/L$. This allows us to construct a key equality.\[cos(\theta) = \lambda/D = d/L\] We can rearrange this to give us a function for $d$ in terms of $D$: $d(D) = \frac{\lambda L}{D}$. Thus, the closer together our slits get (the smaller $D$ gets, the bigger $d$ will be--in other words, closer slits will give us fatter cosine functions. So if $D=1$ gives us cosines that look like this:
Then $D=2$ will give us cosines that look like this:
And $D=3$ will give us cosines that look like this:
We could keep doing this with steadily decreasing values for $D$, until the limit where $D=0$, we have only one slit, and there is no interference pattern, just an even distribution of light. However, more interesting than looking at all of these individual cases is looking what happens when we combine all of them, superimposing every possible cosine and giving us an entire aperture rather than a pair of slits. That gives us what's called a sinc function:
As you can see, superimposing all the slits gives us a huge, bright central maximum, with much smaller maxima at regular intervals off to the sides. Since the y-axis is brightness and the x-axis is distance from a central point, that translates to an image with a bright center surrounded by evenly spaced dimmer rings:
Anyone who has looked through a telescope will recognize this as the Airy disk one sees when looking at a bright point source. This is really cool--we've taken the results of the two-slit experiment and generalized them to apply not to two slits, but to an entire aperture.
And if we go deeper, there's even another layer to all of this. Here's the transmission function for our first set of slits:
Where the x-axis is position and the y-axis is the amount of light passing through (zero meaning none, one meaning all). The two lines are the locations of our two slits, $D$ apart. Thus, the transmission function for a set of two closer-in slits just has the two lines moved slightly closer:
And if we move the two slits even closer together, the lines follow suit:
Thus, if we take every possible spacing of slits from our first set all the way down to zero, we end up adding every possible pair of lines until they all run together into a rectangle:
That's called a top hat function, and it models the transmission of an aperture. It also just so happens to have a Fourier transform that looks a little familiar:
What this means is that whenever you point your telescope at an object in the sky, it's actually taking the Fourier transform of the light it's receiving across its aperture, going from planar waves of light spread across a mirror or lens to a single, defined image. So when you look through a telescope, you're not just seeing photons that have traveled through space for years to reach you, you're seeing some pretty complicated (and elegant) math, too.
Awesome job Tom! You've really worked on conveying the key concepts clearly, well done. Maybe try making your drawing larger so it's easier to see. Also the Fourier transform of a top hat is called a 'sinc' function, not a 'sink'.
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