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.
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 803
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 poynting 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--
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 poynting 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 a 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 corrent-- 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,
There it is.
I don't even have to tell you
what the zero order is, do I?
The zero order is the same location
for all the colors.
So clearly, this is the zero order.
You may be surprised that the zero order
is so broad.
Why isn't it very narrow
because there are so many lines per inch.
The reason is of course that no matter how many 
lines per inch you have
your spectrum will never be narrower
then the actual size of the light source itself.
And so the reason why this is as big as it is,
is because the light source is big.
So all the colors, fall at the same location,
at the zero order that's why you see white light.
Here is the first order spectrum.
Notice that the red is farther away from the zero 
order then the blue.
Here is the first order,
on the other side.
Again the red is further away
from the zero order than the blue.
And you can even see here
the second order.
And again the red is further away
than the blue.
I now have a red laser which is pointing in exactly 
the same direction as the white light.
And so I make a prediction now that the zero order
of the red laser must fall in here
that the first order of the red laser must fall here
in the red.
That the first order of the laser, red laser,
must fall here in the red--
and it must fall here in the red
of the second order.
So I'm going to turn on the laser.
If you're a little patient.
There it is.
Exactly as predicted.
Notice, that the red laser always falls
at the location
where the white light
has the red in it's spectrum.
And that's what you expect of course.
And the zero order,
is right here.
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 is
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
which emit very discrete frequencies,
very discrete wavelengths.
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 803.
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 5400 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.
Aaah!
I'm beginning to see the dark lines.
Aaah!
The diffraction width is taking over.
Look at the center maximum.
Right there.
It's getting wider.
It's getting wider.
And I make the slit narrower.
It's getting wider.
It's getting wider.
And look how beautiful you see
these destructive interferences.
Where light plus light
gives darkness.
Wider.
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 is 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 closer 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 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 a 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.
In both cases and in what follows,
I have ignored the factor one point two.
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.
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.
