So let's work out what fuzziness, what blurriness diffraction applies for a telescope.
So let's say we have the mirror or lens or aperture of a telescope here, of diameter D.
And assuming the telescope is correctly built, if you get light coming in here, it will all be brought
to a focus at some particular point. Now if it was an ideal telescope
the light from any other position would not contribute to what you're picking up here on that particular pixel.
So if the light was coming from any other angle at all, it would land somewhere
else on the detector. So now let's consider light coming from a different angle, the light
you would like to go somewhere else. So we'll say this angle here is theta.
And you want none of that light to get to that point. Now, if the light can be straight, all the waves
add up in phase, for example if it's a peak here, it'll be a peak all the way here
if it's a trough, then a trough all the way to here, and your optics is designed so all those
peaks will add up, and if you in-phase it on here. So that's coming from an angle, though.
The waves are going to be coming in something like this. And you see that... let's say these
are the peaks. When there's a peak here, it's not a peak over there. It's a little bit different
from a peak. So let's say for example that, that length there that we're going to call L.
Now if L was a wavelength, then when it's a peak here, it'll also be a peak there
but a different peak. So it might be like the peak of this one and the peak of that wave.
So in the middle it'll be a trough. So, peak trough, middle peak. If it was lambda over
two, then when it's a peak here, it'll be a trough there and sort of halfway between
a peak and a trough in the middle, and so on. What that means is, if you had all peaks
along here, it all adds up in phase over there. If it's got a peak and a trough and a peak
then it absolutely won't add up in phase over there. It'll cancel out. So generally speaking
if L, this distance here, if L is about a wavelength, then the things are not going
to add up in phase there. If L is much smaller than a wavelength, like a tenth of a wavelength
you're still going to get maybe not quite all peaks, but fairly close. So you'll see
you'll still get some light over there. So that's our requirement.
If the path difference from one end to the other is about a wavelength, then you're not going to get light from this
angle contributing over here. Which means for example if this is light from the star
and that was the planet, you could separate them nice and clearly. If L is much less than
a wavelength of light, then there's going to be some contamination. Some of the light
from both will contribute, and so it will be very hard to tell the planet apart from the star.
Ok. So what we need to do is work out what L is. So let's zoom in now.
So we've got the parallel waves. We've got the waves coming in at an angle, theta, and we'll do
a right angle wavefront there, and what we're going to work out is what L is. We know the
diameter D of the telescope mirror. Now if this is theta, that's a right angle, so this must
be 90 minus theta. So that must be theta. So from trigonometry, this is a right-angle
triangle, so the opposite is L and the hypotenuse is D. So sine theta equals opposite over hypotenuse
over D. But now we can play a trick. If you remember the small angle approximation for
theta less than about 10(or tan?) degrees. Sine theta is very close to tan theta, and
it's very close to theta. They're all about the same thing. Only works for very small
angles, but in this case the different between a planet and a star is going to be a tiny
fraction of an arcsecond or maybe an arcsecond or two, so it's a very small angle. So it's
a good approximation here. So instead of writing sine theta we can just write theta.
As long as we just measure it in radians. So what we actually have is theta equals L over D.
But if you remember, the requirement to get the light not contributing was, L is about
the wavelength, so that's going to be the wavelength over D. That in fact is the equation
for diffraction limit. Roughly speaking if the angle is in radians equal to the wavelength
divided by the diameter, you're ok. If the angle is much smaller than that, you're in
trouble, you're not going to see things very clearly. Technically speaking for a perfectly
circular (???) you could add up your peak, slightly off a peak, slightly more of a peak
all the way down to trough all the way over there, and it turns out it comes out as one
point two lambda over D. But just straight lambda over D will work fine for the purposes of this course.
So what is that for the Hubble Space Telescope? For the Hubble Space Telescope it works at
optical wavelengths, so lambda is about say, naught point five micrometres.
Diameter of the Hubble Space Telescope is two point four metres, so the diffraction
limit, the angular limit is going to be naught point five times 10 to the minus six to convert
from micrometres into metres, over two point four, which is two by 10 to the minus seven radians
which is naught point oh four arcseconds. Pretty good, considering from the ground you
might get ten times worse than that even in the best sites. But still not perfectly sharp.