So last time we discussed the interference patterns due to two
coherent light sources. Today I will expand on this by
exploring many many light sources.
Suppose instead of having two slits through which I allow the
light to go I have many. I have N, capital N.
And let the separation between two adjacent ones be D,
and so plain parallel waves come in and each one of these
light sources is going to be a
Huygens source, is going to produce spherical
waves. And so now we can ask ourselves
the same question that we did before, and that is look at a
long distance far away at certain angle theta.
Where will we see maxima and where will we see minima?
And then we can put up here a screen at a distance L and we
will call this X equals zero, and then we can even ask
the question where exactly on that screen will we see these
maxima? You will have constructive
interference exactly the same situation that we had with the
double-slit interference pattern, when the sine of theta
of N equals N lambda divided by D.
And if you're dealing with very small angle theta,
you should all remember that the sine of an angle is the same
as the angle itself, provided that you work in
radians. So for small angles,
you can always use this approximation,
if you remember that it is in radians.
And that's only in the small angle approximation.
And so the conclusion then is if we work in radians for now
that theta of N for the maxima is then at N lambda divided by
D, N being zero right here, N being one right here,
N being two right there. And if you want to express that
in terms of a linear displacement from zero then X of
N again for small angles is L times that number.
And so now you get displacement here in terms of centimeters or
in terms of millimeters. So you will say well big deal,
it's the same result that we had for uh the double-slit
interferometer. We had exactly the same
equation. There was no difference.
And D now is the separation between two sources here.
It is obvious that it is the same because if these two are
constructively interfering then these two will too and these two
will too and these two will too so all of them will,
so it's not too surprising that you get exactly the same result.
But now comes the big surprise. We haven't discussed yet the
issue where the locations are where light plus light gives
darkness. We haven't discussed the
destructive interference. And to derive that properly is
very tricky. In fact if you take eight oh
three you will see a perfect derivation.
But I will give you the results.
What is not so intuitive, that if you have N sources,
that between two major maxima, that means between this maximum
at N equals zero and a maximum at N equals one,
there are now N, capital N, minus one minima.
And minima means complete destructive interference.
So if capital N is two, which we did last time,
two minus one is one, exactly, that was correct.
We had only one zero in between the two maxima.
But that's not the case anymore when capital N is much larger
than two. And so let me now make you a --
a sketch whereby I plot the intensity of the light as a
function of angle theta and this is the intensity,
so that's in watts per square meter, remember that's the
pointing vector, and let this be zero,
and let the angle theta one be here, and for small angles then
that's lambda divided by D, and here you have theta two,
which is two lambda divided by D, and so on.
I take the small angle approximation.
So this angle is now in radians.
What you're going to see now is the following intensity,
as a function of theta. You see here a peak,
and you're going to see here a peak, and you're going to see
here one, and so on, and the same of course is true
on the other side. And here in between you're
going to see now N minus one locations whereby you have total
destructive interference. And the same is the case here.
And this can be huge. N can be a few hundred.
So we have many many locations where you have a hundred percent
destructive interference. Now this point,
this first location, where we hit the zero,
that now is at the position lambda divided by D divided by
capital N. And I will call that angle from
the maximum to that zero from this maximum to this zero,
I will call that angle for now delta
theta. Because that delta theta is a
measure for the width of the line, here is at maximum,
here it is zero, and so that angle delta theta
in terms of radians is lambda divided by D times N,
which then is approximately theta one divided by N,
because theta one itself is lambda divided by D.
And so you see that it is N times
smaller than this distance. And so if N is large,
these lines become extremely narrow, and that's the big
difference between two-slit interference and multiple-slit
interference. And the larger N is,
the higher these peaks will be. The height of these peaks,
the intensity here, is proportional to N squared.
And you may see gee why -- why not -- is why is it not linearly
proportional to N? Well that's easy to see.
Suppose I increase capital N, the number of sources,
by a factor of three. Then the electric field vector
where there are maxima is three times larger.
But if the electric field vector is three times larger the
pointing vector is nine times larger.
So you get nine times more light.
Now you may say, gee that's a violation of the
conservation of energy. Three times more sources,
nine times more light, how can that be,
well you overlook then that if you make N go up by a factor of
three that the lines get narrower by
a factor of three, because of this N here,
and so they get higher by a factor of nine,
and they get narrower by a factor of three,
and so you gain a factor of three in light.
Of course you gain a factor of three.
You have three times more sources.
You get three times more light. So you see there's no violation
of the conservation of energy here.
And I want to demonstrate this to you using a -- a red laser
which we have used before. And I will use what we call a
grading, a grading is a plate which is
specially prepared, a transparent plate,
which has grooves in it, and the one that I will use has
tw- twenty-five hundred grooves, we call them lines,
per inch. That means the separation D
between two adjacent grooves in my case is about two point one
six microns. A micron is ten to the minus
six meters. And the wavelength that I'm
going to use is our red laser, which is about six point three
times ten to the minus seven meters.
And I'm going to put the whole thing there.
I'm going to make you see it there at a distance L.
Which is about ten meters. And so this allows me now to
calculate where the zero order will fall, where the first order
and where the second order will fall.
We call when N is zero, we call that zero order,
so this is zero order, when N is one we call that
first order, and when N is two we call that
second order. And you have of course the
first order also on this side and the second order also on
this side. Everything that you have here
you have to also think of it as being on the other side.
So I can predict now where the zero order will be when N is
zero. That is zero degrees.
That's immediately obvious. I use that equation.
If N is zero the zero order is always right at the center,
provided that all these sources are in phase.
And they will be in phase because I use plain waves.
So Huygens will tell you that they're going to oscillate
exactly at the same time they produce the same frequency,
they produce the same wavelength, and they're all in
phase with each other. So there will be a maximum at
theta one equals zero. And then there will be a
maximum which I calculated to be at three point five five
degrees. I calculated that from this
equation and then theta two will be at roughly seven point one
degrees. If you want to know how wide
the width of this peak is going to be,
then you have to know how many lines of my grading I will be
using. Well, my grading is like so.
Here I have these lines not unlike the grading that you have
in your optics kit. There are twenty-five hundred
of those lines per inch. And my laser beam is roughly
two millimeters in size. So this is about two
millimeters. And that tells me then that I
cover about two hundred lines. And if I have two hundred lines
I can now calculate how wide that line is going to be.
Because this factor of N enters into it here.
And if I express that in terms of that angle delta theta,
then the angle delta theta going back here so delta theta
is then the three point five five degrees divided by two
hundred, and that's an extremely small
angle, that angle is approximately one arc minute,
which is sixty times smaller than one degree.
And if you want to translate that in terms of how wide that
spot will be, if I see it on the screen ten
meters away from me, and if you want to call that
delta X, then you would naively predict that delta X is
something like three millimeters, and the reason why
I say naively because you will not see that it is
three millimeters, it will be extremely narrow,
but it will be more than three millimeters, because the
limiting factor is always the divergence of my laser beam.
And so the divergence of my laser beam is more than one arc
minute, and so I don't get down to the one arc minute narrow
beam. I'm not too far away from it
though. So this is what I want to show
you first. I will turn on the laser first
and then make it very dark because we do need darkness or
this has to come off because that
would obviously -- oh, I turned off the wrong laser,
but that -- I turned on the wrong laser, but that's OK.
That's a second demonstration which I do with uh with the
green laser, this is the one that I need, this is the red
laser, will come on very quickly.
There it is. Tom, if you can turn that off,
maybe that will help, although everyone can see it
but it would help. So you see here very clearly
the um the zero order is at the m- at
the -- right in the middle. This one, so theta is zero,
and this theta one is my three-and-a-half degrees,
this is also three-and-a-half degrees.
This is the seven point one degrees and so on.
And so you see this whole pattern of -- uh of uh
interference as a result of multiple slits.
And so this is a -- these are grooves in a piece of plastics.
And notice how small they are, how narrow they are compared
with the double-slit interference.
So they don't have that theoretical minimum with this
one over N, but they approach that, and the reason why they
are not that narrow is because the divergence of the laser beam
itself is larger than that one arc minute that I calculated.
And so you can never beat that of course.
We did this experiment with white,
with uh red light, but keep in mind that if I take
red light, here is a maximum, at zero order,
here is a maximum, provided that this lambda is
lambda for red, here is a maximum,
provided that this is the lambda for red.
But if I have white light, then of course I deal with
other colors. And if I have blue light in my
white light, it will also have a maximum here,
that's nonnegotiable, but it has its first order
maximum here, because the wavelength is
shorter. And it will have its second
order maximum here.
This will be the same distance. And so when you do this with
white light you're going to see always at zero order white
light, because all the colors have their maximum at zero
order, but at first and second and higher orders,
the cor- the -- the -- the colors uh walk at their own pace
so to speak. And then the smaller the
wavelength is the closer it will be to the zero order and then
the spacings between first and second will also be smaller than
in the case of the long wavelength,
in this case red. And this is something that I
also want to demonstrate to you. It is not so easy to get a very
strong powerful source of white light.
I'm using for this a -- a reflection grading.
You can also use gradings in reflection.
You take metal and then the -- the grooves are made on the
metal. And you get a reflection,
which we will have there on the wall.
This reflection grading has a spacing which is four times
smaller than the one we have here.
It's only two-and-a-half microns and so the angle of
theta one will not be three-and-a-half degrees but it
will be four times larger. The main purpose wh- why I want
to show you this is I use white light that the zero order is
white. And then we will see also of
course the first and the second order if we have good eyes
because the f- the -- the whole thing is not so -- so very
bright. Make sure that I have the -- my
light, flashlight, so your eyes
may have to adjust a little bit to the darkness.
The effect is not overpowering because our light source is not
very bright. But you see the where is my
laser pointer, this is the zero order maximum,
and the zero order maximum is very wide, the reason is the
divergence of the white light beam, it's not this factor of N
that I gain, I gain much less, uh I can turn on
also a laser which I use at the same time, and you will see that
the red laser will give its own th- there they come,
so the zero order of the red laser is of course also here,
all colors are here, and then you see here the first
order of the red, that's a large angle,
but the D is very small, you see here the first order of
the red, second order of the red, you see here the first
order of the blue, first order of the blue,
you see a big difference, between the separation of the
red and the separation from the zero order and the blue,
it's a big difference. There is actually a much better
way that I can make you see all this and that is if I ask you
which I think I'm going to do now to use your grading,
but hold -- hold it for a second.
Before you get your -- your own gradings out.
Our equations that we have derived so far
only hold if we look very far away.
These angles of theta are only true if you go very far away.
For reasons that we discussed last time we can however do
something very clever. We can use a lens and if we
have a lens we can bring the image very close without
disturbing the angles. If this is a grading,
this is the number of sources that I have,
and so the light comes in in this direction and if I put here
a lens and this is the screen, the focal point of the lens,
then if the angle theta -- if the angle theta for which I
would have expected in this direction my first order
maximum, and here of course my zero order maximum,
then the lens will not change that angle.
We never discussed lenses so it may not be so obvious to
you. But the lens will always
maintain the integrity of angles.
So the angle theta that we derived here is the correct
angle, but of course in terms of X that is enormously reduced if
this distance is very small. So then we have the option that
we don't have to allow for very large distances like now ten
meters. And so your eye is a perfect
tool to use for that because you have a lens in your -- in your
eye, that's the whole idea. And so now I want you to get
your gradings out.
And I want you to hold the gradings in front of your eyes
and manipulate the gradings a little bit so that you get the
lines vertical. And you will easily be able to
do that, this is your light source.
Your lines, your gradings, have a thousand lines per
millimeter. That means the spacing of your
grading is one micron. One micron is ten times smaller
than this number. So the angles are huge in your
case. They're way larger than what we
have there. I will make it completely dark
and then I want you to rotate your gradings such that you get
the spectra on either side, on the left and on the right.
That means your grooves are then in vertical direction.
And what you see now way better than what I could show you in my
previous demonstration, you see the zero order is the
lamp itself. All the colors are right at the
center. That's the lamp.
That's your zero order maximum. And then you see if you go to
the right you see the blue coming in beautifully first
because that's the smallest wavelength, you're going further
to the right, you see the first order red,
you're going further to the right, you see the second order
blue, you go further to the right, and you see second order
red, but since D is so amazingly small in your case,
there may not be too many maxima in the red.
In fact I will ask you on one of the homework assignments
which is the optional one how many maxima in the red you will
see. So essential here is that you
see that the f- zero order maximum at the center is white
light. And so it's not until you get
to first and second order that you begin to see the colors
separate. Now these gradings can be used
to do atomic physics. There are atoms and molecules
uh which emit very discrete frequencies, very discrete
wavelength. And when you look at them with
a grading you can see very distinctly where these lines
fall. Where these wavelengths fall.
And that's the next thing that I want to do.
I will show you now, I will turn off the white light
source and I will turn on for you neon.
And so I want you to look now at the neon and if you give
yourself some time you will see that the neon is not a
continuous spectrum like you saw with the white light bulb
but you see very distinct locations where you see maxima.
You see many in the red, and I think you see several in
the orange, and I noticed this morning that I see two lines in
the green. You have to look very carefully
because these lines in the green are very faint.
And so the whole purpose then is that with these gradings you
can not only find out which wavelengths are emitted by these
atoms and these molecules but you can also find the relative
strength, and that is of course entering the
domain of atomic physics. And these gradings are
extremely powerful to do that. And I would advise you to carry
these gradings with you at least for the next few weeks and when
you're outside and you get a chance to see some bright lights
out on highways or on the street, to look through the
grading and see whether you can see these emission lines,
uh mercury if you get mercury lamps, they are very beautiful.
You see many many different colors very discrete
frequencies, very discrete wavelengths,
are emitted by mercury in the same way, the same kind of
physics that you see that here with um with neon.
So now comes something that may come as a surprise to you.
Because now I would like to discuss with you um the
interference pattern if we have only one slit.
We discussed two, we discussed capital N is a
large number, but what now if we have only
one slit? Even if you have only one slit
there will be directions in space whereby light plus light
gives darkness, and there will be directions
whereby the light constructively interferes with each other.
And strangely enough this is given a different name.
We call this diffraction. It's exactly the same physics.
There is no difference. It should have been called
interference, but it's in the literature you
will see it under the name diffraction.
It all comes down again to Huygens'
principle. So let me here now have a
single opening and this opening is a slit perpendicular to the
blackboard and the opening is A in size.
A single one. And the plain waves come in and
Mr. Huygens says that all these
sources here are all going to be emitting the light at the same
frequency, the same wavelength, and they will all be in phase
with each other. Because these plain waves
arrive here all at the same time.
And so I could now ask myself the question at what angle of
theta will I see maximum, what angle of theta will I see
minimum, and you can also put a screen then at large distance L,
you can call X equals zero here and you can ask yourself where
will I see these maxima and where will I see these minima?
To derive this in this case is not easy.
Again I refer to eight oh three.
It's as difficult as deriving in this case this
whole structure in between the maxima.
One thing is obvious and that is that you got to get a maximum
that is nonnegotiable when theta equals zero.
That's simply a matter of symmetry of the problem.
If all these sources are in phase, clearly you're going to
get a maximum here. No one will question that.
The minima is very tricky. And the minima will fall at the
following locations. The sine of theta of N equals N
times lambda divided by A. And for small angle
approximation, this is the same as theta N in
radians. And when you see that equation
your first reaction should be that maybe I goofed by a factor
of two. Because your first reaction
will be that's the same equation that we have there and then we
have D there, and so how can we have minimum
here where we have maxima there? Well, the situation is
different. The best way that you can see
that that equation is not wrong is
perhaps the following. Suppose you take the angle
theta one. So that's for N equals one.
Then the relation that you see there will tell you that this
source here, this Huygens source, and this Huygens source
have a difference in path length of lambda.
And so you will then say aha that's constructive
interference, that's true,
but that means this Huygens source and this Huygens
source will then have a difference in path of half
lambda. So they will kill each other.
And that means this Huygens source and this Huygens source
will have a difference in path length of one-half lambda.
So they will also kill each other.
And so in this upper half there is always one Huygens source
which will kill the one at the bottom.
And you can do a similar reasoning for the angle theta
two. And so that is indeed the
correct equation. You will find complete
minima when theta one in terms of radians is lambda divided by
A and theta two is two lambda divided by A and so on.
That's where you find your minima.
And if you convert that into X where they actually fall on a
screen, well then X one will be for small angle approximations L
times lambda divided by A. That's no different.
And so now what I owe you is a pattern.
What will the pattern look like?
When I look on the screen there,
what will I really see? Well, it's looking,
it's going to look very different from what you may
think. I will plot it now in terms of
X. I could have plotted it in
terms of theta. But I decided to plot it in
terms of X. So here at X equals zero there
is unmistakable, unnegotiable,
there is a maximum, that is -- that coincides with
theta equals zero. And then here at lambda divided
by A, completely destructive interference,
that's that angle theta one uh times L.
And then here L two lambda divided by A complete
destructive interference and the same is true of course on the
other side. And what you're going to see
now in terms of the intensity that I showed you there,
watts per square meters, is a curve that looks like
this. You get an enormously broad
maximum, absolutely zero here, very small maximum,
absolutely zero there, very small maximum and zero
there, and this continues for a long time.
And this is very different from anything we've seen with
multiple sources. If the intensity here is I
zero, then the maximum here which I didn't even calculate
where it is, it is somewhere in between these two minima,
but I didn't calculate precisely where it is,
that maximum is very low, it's only four point five
percent in strength of I zero. And this is even lower.
This is only one point six percent of I zero.
So when you look at a diffraction pattern,
we call this a diffraction pattern like this,
you will see a very broad center maximum and then you see
these dark spots on either side, and you see light coming up
again in between them, sort of submaxima.
And so the width of this center maximum, this width,
which is really X one, that width is then L lambda
divided by A and if you want to be picky and you say well the
center maximum is really twice that much, fine,
be my guest, but this is clearly a measure
for how wide that center spot will be when you see it on a
screen. And now there comes something
that is completely not intuitive for you as well as for me.
And that is if you make A very small that means you let the
light go through an extremely narrow slit,
then this what you will see on the wall is extremely wide.
The smaller A is the wider it will be.
It's exactly opposed to what you would predict.
You would think if you make the opening through which you put
light very small, you would think that what you
see on the screen is also very small.
It's exactly the opposite. And that's is what I want to
demonstrate to you. I have here a demonstration
with a variable slit. I can vary A.
And we will use the brightest laser that we
have, a beam of green laser light, about five thousand four
hundred Angstroms. And what I will do is I will
make this opening narrower and narrower and narrower.
And as I make the opening I start very large.
I start with a large opening of maybe five millimeters.
At a large opening A is so large that this is negligibly
small because A is very large, then this is not very large,
this is very small. But as I make A smaller and
smaller and smaller and smaller there comes a time that
the diffraction width is going to be dominating the whole scene
and what you will see then on the screen there that the bright
spot will get wider, wider, wider,
wider. And that for me is always so
enormously fascinating because it goes so strongly against our
intuition. And so this is what I have next
on the menu. I have here this variable slit.
I must make sure that I have my flashlight and I'll turn on this
laser that I earlier accidentally turned on.
I hope it's coming on. Yes it is.
And so you see there the slit is now very wide.
And so the size of that slit that you see there -- the size
of the -- of the bright spot is now entirely dictated by the
divergence of my laser beam. The diffraction width is
negligibly small because A is very large.
But now watch. Now I'm going to
tighten, make A smaller. And I'm doing that now.
I'm making it smaller and smaller.
The diffraction width is still negligible.
Making it smaller. Ah.
I'm beginning to see the dark lines.
Ah. The diffraction width is taking
over. Look at the center maximum.
Right there. It's getting wider.
And I make the slit narrower.
It's getting wider. It's getting wider.
Look how beautiful you see these destructive
interferences. Where light plus light gives
darkness. Wider.
Right now it is at least ten
times wider than it was before. And now it is twenty times
wider. And it gets very faint.
Of course it gets very faint because if I make the slit very
narrow not much light goes through, there's nothing I can
do about it. But notice how incredibly
impressive this is. How wide that center maximum
becomes. And that is very characteristic
for diffraction. The narrower the slit the wider
the diffraction pattern at the center maximum.
Now I did this in monochromatic light, monochromatic light means
that you have practically only one wavelength.
And there is a way that I can make you see this from your own
seats. And that is what we're going to
do with the um with the cards that you have.
So if you can get the cards out now, we did this in one color,
right, in green light, almost one color,
almost monochromatic, so you see a beautiful pattern
here, very well-defined dark lines.
You now can use this little slit that you have,
put it in front of your eye, and I'm going to make you see
this white light and you would see all the features that you're
supposed to see but it's even more interesting in white light
because with white light you have a little red,
you have a little blue, you have a little yellow,
and so these minima will fall at
different locations of course. And so you don't see it as
beautiful as I showed you, as distinct as I showed you in
green, but you see very distinctly the center maximum
and you see the dark lines on either side.
But the main reason why I want you to see this is that if you
manage to manipulate the size of the slit, the size of A,
if you manage to manipulate that, when you make it narrower,
notice that the diffraction pattern gets broader and not the
other way around. So first make sure that you get
it, that you begin to see the dark areas, and then try to make
it a little narrower and then you see that it opens up.
This is precisely what I did with the variable slit here.
So give it a d- b- bit of time. What helps me that actually you
don't have to pull it open but yet you can move one piece of
the card behind the other card. So your one thumb goes to the
back and the other your thumb comes forward.
That works very well for me. Who can see clearly the
diffraction pattern? Unmistakably.
Very good. Take the card with you,
impress your parents. And look at home at very bright
street lights and you will still see the same diffraction
pattern. Although not as ideal as you
see it here because our source is a line source and that helps
of course if your slit if vertical.
OK. If I don't have a slit as an
opening but if I have a circular opening then the pattern that
you would expect is the same that you see here but you have
to rotate it about this line because you now have axial
symmetry. So you don't have a -- a long
slit, but you have a circular opening.
And indeed that is approximately correct.
If you had a circular opening you would see a center maximum
which would be very bright and then you
would see rings around it of zeroes.
And so if I try to make you see it this would be the center
maximum. It would be a ring around it
here. Complete darkness and then
again a little bit of light, not very much,
because remember that this maximum is only four-and-a-half
percent of that one. And then you would see again a
ring with complete darkness and so on.
So you have a little pinhole and this is the image
that you would get on a screen. From that pinhole.
And you would think now that this angle from here to here is
this theta equals zero and we call that theta one.
This is the theta one where you see your first zero,
you would think that that is lambda divided by A if A is the
diameter of your pinhole. Well, it's almost that.
It is a little larger. Because a
pinhole, a circular geometry is different from a line.
And so take my word for it that it comes not at lambda divided
by A but it comes at roughly one point two lambda divided by A.
If you want to be picky it's really one point two two times
lambda divided by A and this of course is in radians again.
I work now exclusively in terms of radians, small angle
approximation. And so this raises the issue of
what we call in physics angular resolution.
Suppose I have a pinhole and I look
at the images of two light sources.
One light comes in from there and the other comes in from
there, could be the headlights of a car, could be two stars in
the sky, well, each one of them will give a
diffraction pattern, that's nonnegotiable,
you can never bypass diffraction.
So one star will give a diffraction pattern here,
or one headlight, and the other will give a
diffraction pattern here. You would have no problem to
say oh, yeah, they're all,
there are two light sources, there's one star and there's
another star.
OK. Now make the angle between the
two sources smaller and smaller and smaller and smaller so that
these two diffraction patterns come close and closer and closer
and closer. How close can you now bring
them so that we, you and I, will still say,
yeah, there are still two sources?
We call that angular resolution.
And so how we define angular resolution is that both light
sources have exactly the same strength and let's assume for
now for simplicity that they're monochromatic,
so that there's only one wavelength that they emit.
Then the criterion that is generally accepted so that we
can still decide that there are two light sources,
that this one, the center of this one,
is no closer than the location where this one has complete
darkness. In other words,
the spot of the second star should fall right where the
other one has darkness. If you bring them closer your
brains will say no. No, no, that's not two sources.
That's really only one source. And we call this the Rayleigh
criterion of resolution. And that Rayleigh criterion of
resolution therefore is that the separation between the two light
beams, stars or the headlights from a car, the separation has
to be larger than this angle. And it is a function of A.
If A is larger, then that angle can
be substantially smaller. And this is what we call the
diffraction limitation on angular resolution.
It doesn't matter whether you have a pinhole or whether you
have a lens a c- circular lens that we use with telescopes.
Or whether you have concave mirrors, which we use with
telescopes. In all cases are you always
stuck with the idea at best, that's the best you can ever
do, that is the angular separation that theta
I call it here the theta one, is that one point two lambda
divided by A, and that is then in radians.
If you take a lens which has the diameter A of about twenty
centimeters then that translates into a theta one of about half
an arc second, for five thousand Angstroms.
So I take lambda five thousand Angstroms, remember an Angstrom
is ten to the minus ten meters.
So theta one then becomes half an arc second.
Oh point five arc seconds. An arc minute is one-sixtieth
of a d- arc degree, and an arc second is sixty
times lower than that. And so if A were two point four
meters, telescope, two point four meter telescope,
that's a real biggie, then theta one would be
approximately one-twenty-fifth of an arc
second. So the larger you make your
telescope the better angular resolution you would have.
This angular resolution is twelve times better than this
one. Because A is twelve times
larger. So now you may think that if
you take a two point four meter telescope on the ground and you
look at stars that two stars equally bright you would be able
to tell them apart if they are farther away from each other
than one-twenty-fifth of an arc second.
That is not true. The contrary.
It is very bad. You can't even tell them apart
often when they are half a second apart.
And the reason for that is not that the diffraction limitation
is going to kill you, but the reason is that the air
is always turbulent. And it is the turbulence on the
air that makes the image, the diffraction-limited image,
which itself is very small, move around on your
photographic plates, it moves it around in a matter
of ten seconds over an area which
itself could be as large as one second.
Astronomers call that seeing. And so when you look at your
picture, the whole star is smeared out over an area which
is in angular size one arc second or maybe half an arc
second at best. So you can never achieve this
from the ground. So all telescopes from the
ground without exception can do at best half an arc second.
They cannot do much better because of the air turbulence.
And this is now the great thing about the Hubble space
telescope. Hubble is up there,
or maybe down there, whichever it is,
I don't know where it is, maybe Jeffrey knows where it
is, but it is somewhere. And Hubble has no air.
And so Hubble doesn't suffer of the air turbulence.
And so Hubble's mirrors are indeed diffraction-limited.
And Hubble has a mirror which is two point four meters
diameter. And indeed at five thousand
Angstrom, I checked that yesterday with people at Hubble
space telescope, indeed Hubble is
diffraction-limited, and Hubble has an angular
resolution at five thousand Angstroms, which is about
one-twenty-fifth of an arc second.
And at shorter wavelengths, it's even better,
and at longer wavelengths, it is a little worse.
And so I would like to show you at least one picture of Hubble
without going into the details of what you are seeing of the
astro- of the astronomy. And that's the one that is
coming up. It's a very famous picture that
Hubble made several years ago. It is a picture of a supernova
explosion. John if we can have the slide,
there it is. You're looking here at an
explosion, it's called Supernova nineteen eighty-seven A,
which occurred in February of nineteen eighty-seven.
This object is a hundred fifty thousand light-years away from
us. That means the explosion really
took place -- took place a hundred fifty thousand years
ago. But we saw it for the first
time in February eighty-seven. And without going into the
details of what you're looking at, which is of course very
fascinating, but that's not the objective today,
I want you to appreciate that this
distance here is one arc second.
And look at the incredible detail that Hubble can show you.
If you took a picture like this with a ground-based observatory,
this whole part would just be one smudge.
You would not be able to resolve that.
And that is the power that you see in front of you now of a
diffraction-limited telescope which -- which has a diameter of
two point four meters. You get an angular resolution
which is very close to four-hundredths of a an arc
second. The amount of detail that you
see is incredible. That's the big power,
the big reason why this telescope was put in orbit,
to do away with the um air turbulence, what the astronomers
call as seeing. Which is always a limitation of
your angular resolution. So in the remaining five
minutes I want to address the issue of the angular resolution
of your own eye. You can now calculate what the
ultimate angular resolution is of your own eye.
Because you can estimate what the diameter is of the pupil.
The opening of your eye. Three millimeters,
maybe five millimeters, a little bit larger at night
when it is dark. Pupil opens up.
But we can calculate what this is.
Uh if I take four millimeters, so I put in for A four
millimeters, and if I take lambda five thousand Angstroms,
it's not an unreasonable value, then I find that the best
angular resolution of a human eye is half an arc minute.
Cannot be any better. There's just no way around it.
You're always stuck with the diffraction limitation.
I think though that most of you will not be able to see with an
angular resolution of one-half arc minute.
Most of you are probably in the domain of one arc minute.
It's a little larger than diffraction-limited.
But it's very close to that. And that is something that I
would like to test. Not to see how good your eyes
are, but for yourself to get a feeling for angular resolution.
And the way I'm going to do that is as follows.
We have prepared a box which Marcos is going to wheel in very
shortly which has two pinholes at the top.
And these two pinholes are two-and-a-half millimeters
apart. And then there are two pinholes
which are five millimeters apart.
And then there are two pinholes which are ten millimeters apart.
And then there are two pinholes which are fifteen millimeters
apart. So maybe we can take a look at
that. There it comes.
Thank you Marcos. Here are two pinholes which are
two-and-a-half millimeters apart.
We repeat the whole thing three times.
And the reason why we do that is so that the different angles
in the audience you can probably always see two pinholes well.
What am I going to do now to test the angular resolution of
your eyes? If I make the assumption that
your angular resolution is one arc minute, no worse,
no better, remember it can never be better than half an arc
minute, that's nonnegotiable.
But it could be worse than one arc minute.
That would mean that all students who are closer than
nine meters from me should be able to see this as two indep-
independent light sources. Those who are farther than nine
meters away from me will not see these as two sources.
If they did their angular resolution would be better than
one arc minute. And all students which are
closer than seventeen meters will be able to say yeah,
I see these as two sources, and all students which are
closer than thirty-four meters should be able to say yeah,
I see these as two sources. And so we're now going to make
it a little darker. And you don't need your
gradings, you don't need these cards, I just want you to look
at the lights, these light sources,
and then tell me which you're going to see as two separate
light sources. And that then allows me to tell
you very roughly what your angular resolution is.
So try to look at them. And so now my first question is
who can see the upper, either here or there or there,
they're the same, who can clearly see those as
two light sources? Come on.
You don't see them as two?
Something wrong, is the lights not on?
You must be kidding. Are all of you sick or
something? Oh man I have no difficulties
at all. The upper one is two light
sources. You've all got to see an eye
doctor. OK, who sees the second line as
two? Who sees the third line as two
but not the second? You see how interesting,
just look around you, you see we're moving back in
the audience. Who sees the -- who can see
none of them at two -- two light sources?
OK, so maybe not -- no eye doctors are needed then.
Who can only see the third line as two sources,
only the third line? OK, well, I expect that,
you see, no -- yeah, maybe your angular resolution
is not very -- do you wear glasses?
So you -- yeah, I'm asking you,
so you see the third line as double?
And not the second line? Yeah, there's nothing to worry
about, maybe two arc minute resolution.
So now you have tested your angular resolution.
When you think of diffraction it's really an incredibly
fascinating thing because what this diffraction actually means
that it is a limitation that is put upon us,
on everyone, also God, no one can bypass
diffraction. No matter how hard we try we
can never undo our chains and handcuffs that are imposed upon
us by diffraction. And remember it's all Huygens'
fault. But let's forgive Huygens
because after all he was Dutch.