Last time we discussed the interference pattern of two
openings in the screen, which established the wave
character of light. Today, I am going to expand
that to capital N. And we will make N thousands of
openings in a screen. And, when it comes to optical
light, we call those gratings. Suppose here each one of those
dots is a small opening in a screen.
It could be a little hole or it could be a slit perpendicular to
the blackboard. And imagine that plane waves
are coming in. And so each one of those
openings are going to be Huygens sources and they are going to
radiate waves. And the question now that we
want to answer is if we look in a direction theta away from the
normal to that screen, what is then the light
intensity that you will see as a result of the interference of
all these Huygens sources? Suppose that the separation
between two adjacent Huygens sources is d.
In other words, here is one and here is the
other. This is a blow-up of what you
see here. Here you see hundred,
maybe thousands. Here I have only two.
And, if that separation is d, then I can calculate the phase
difference between those two. If this is theta,
this is no different from what we did last time for the
double-slit, then the past difference from this spherical
wave to a point far away at an angle theta and this spherical
wave is this. And this is d sine theta.
And so, we will introduce, just like we did before,
a phase angle delta, which is the phase between the
spherical wave from this point and that point.
Two neighboring points. And here are a thousand.
So this is only between the two neighboring sources.
And so, first we want to know how many times can we fit a
wavelength on this path? And, for each time that we can
fit a wavelength on there, we have a phase difference of
2pi. So this delta is the phase
difference. And we had the same equation
last time for double-slit interference.
No difference. And, if this delta is a
multiple times 2pi, then you will have constructive
interference. My goal today is way more
ambitious. I want to know what the light
intensity is for any angle theta.
It is going to be an extremely complicated function.
I first want to revisit some of your high school geometry.
If you have a triangle, and this side is A and this
side is also A, the length of the triangle,
the length of the sides, and if you want to know what
this one is, and if this angle is delta then this side here is
2A times the cosine of one-half delta.
You can revisit that in your high school geometry,
but I need that today. And now I am going to do the
vectorial superposition of capital N E vectors that come
from these various sources. And the neighboring ones are
off by delta. That is the phase angle between
them. I will raise this later,
but I first want to work above my head so that you can see what
I do. I start with a circle.
You will see shortly why I do that.
The radius of the circle is unimportant.
You will see that it will cancel.
This radius arbitrarily chosen is R.
This length here is also R. I will call this point C,
this point O and this point P. And I am going to add three
vectors, which are all offset relative to each other over an
angle delta. But I will make the calculation
as if there were capital N. Capital N could be a million.
There is no limit. But I am going to make the
drawing only for three. This is one vector.
And let's say it has a length A, but it is really the electric
vector that is going to be added vectorially to the other
electric vectors. And so here is the second one.
And here is the third one. And so, the angle between the
second and the third one, this angle here is delta.
And this angle here is also delta.
That is the delta between the neighboring sources,
which we just derived. It follows from the geometry of
a circle that this angle here is delta, this angle here is delta
and this angle here is also delta.
So that means if I have capital N of these vectors that this
angle here is then N times delta.
And that means that the angle here is then pi minus N times
delta. I will draw one more line right
through the middle of this vector.
I think of this as length A but it is the electric vector.
You can also think of it as E zero.
And I draw a line straight through there.
This angle is 90 degrees. And so, that means I know this
angle now is one-half delta. And this is all the geometry
that I need to calculate the incredibly complicated light
intensity as a function of theta.
My goal is to find the magnitude of the vector OP
because that is the result, in this case,
of these three vectors. But I will make my calculation
as if they were N. First, look at this triangle
here, which has a one-half delta angle here.
This is 90 degrees. And so, this part here is
one-half A. I just give it the length A.
That means that one-half A divided by R is exactly the sine
of delta over two. That is exact.
It is not a small angle approximation.
This is exactly the sine of that angle.
In other words, my radius R is one-half A
divided by the sine of delta over two.
I am really interested in OP. And now I am going to use this
knowledge. I have a triangle.
I know two sides and I know this angle then I can calculate
this one. I know two sides,
this is R and this is R. I want to know the third side.
And I know this angle which is pi minus N delta.
And, therefore, this length here is 2R times
the cosine of half that angle. Pi minus N delta divided by
two. Pi over two is 90 degrees,
and 90 degrees minus the angle is the sine of the angle.
So I can also write for this 2R times the sine of N delta over
two. Now, I take this R here and I
pop that in here to eliminate my R.
And so, now I get that the lengths of that vector OP,
which is really my goal, that is the vectorially sum of
N E vectors. The adjacent ones off by phase
angle delta. That is going to be two times
R. And so, that gives me an A
upstairs. And then I get the sine of N
delta divided by two divided by the sine of delta divided by
two. And this is really the key of
what is following. This is now the length,
the magnitude of the E vector. If you think of this A as being
E zero, it is the magnitude of the electric field.
So the light intensity, the pointing vector obviously
goes by the square. And so, now comes the very
famous equation that I, as a function of delta,
is I0 times the sine of N delta divided by two divided by the
sine of delta divided by two and this whole thing squared.
And you could call this the grading equation.
You will see that this is a very complicated function.
We will beat it to death together.
This is an exact derivation. This is no approximation.
Delta can be anything from zero to 10 million pi.
There is no approximation made here.
10pi, 20pi, 30pi, anything for delta is allowed.
It is not an approximation. Now, the intensity,
we think in terms of watts per square meter.
And the meaning of I0 is that if there were only one opening
in the screen instead of N then this is the intensity that you
would see I0. Now, if the upstairs here is
zero, you would think that the intensity is zero.
That is not always the case. Because, if the upstairs is
zero and the downstairs is also zero, you get zero divided by
zero. And now you need l'Hopital to
calculate what that ratio is. And that ratio then becomes the
maximum value possible. I will write that here.
The maximum value possible, if you use l'Hopital,
you will find that that ratio is N squared.
And so, this becomes N squared times I0.
That is the case when you get zero divided by zero in that
equation. Before I will show you how
dramatic this function is, I want to remind you that for N
equals two, which is what we covered last time,
you can use this equation. This holds for any capital N.
It holds for any value for delta.
If you substitute in this equation that you have there,
N equals two, and you do a little bit of
massaging of the algebra, I will let you do that,
and this is for the double-slit interference,
you will find that I is 4I0 times the cosine squared of
delta over two. And, for those of you who have
a good memory, remember that I derived this in
class. When you only have two vectors
that you add that light intensity goes with the cosine
square delta over two. Now, you also have this here.
You also know that at the maxima with two slits you see
four times more light than if there were only one opening.
And you can do this, of course, on your own by
substituting in this equation N equals two.
The first thing that I want to do now is to make a drawing,
a plot of that function. And I will do that for N equals
four. And then we will discuss all
the consequences also for cases that N is much larger than four.
I am going to plot this for N equals four.
We only have four openings now. And I always plot only sine
theta. The reason why I like to plot
always things in terms of sine theta, theta is a real
geometrical angle. Here is this screen with the
openings. And theta is an actual angle in
the lecture hall. Theta is something that I can
immediately relate to. This is 10 degrees.
This is 20 degrees. This is 30 degrees.
Delta is a phase angle. That is not a real angle in
phase. And so, I always like to plot
the intensity in terms of sine theta, but you can also do it,
if you want, in terms of delta.
If sine theta is zero then you get zero divided by zero.
And you are going to get a maximum.
If sine theta is lambda divided by D, you are going to get a
maximum. If sine theta is lambda divided
by D, you see that delta is 2pi. That means you get a maximum.
You get a maximum here. If sine theta is two lambda
divided by D, you get a maximum.
And, of course, on the other side minus lambda
over d, you also get a maximum. And, according to the equation,
if you really believe that equation verbatim,
then all these peaks would have the same maximum.
It would be 16I0 because this is the N squared.
And I will show you, but first I will plot it,
that if there are four slits that in between the prime maxima
there are N minus one locations whereby you have completely
destructive interference. There is zero light.
N minus one, in this case,
is three. You will shortly why it is N
minus one. There are three locations here
whereby there is zero light. I put them in here and then I
will draw the curve. That is the zero.
And so, I am going to make an attempt now to draw the light
intensity. It is prime maxima when zero
divided by zero is N squared. Here is another one where zero
divided by zero becomes N squared.
And here is another one where by zero divided by zero becomes
N squared. If you wanted to know what the
delta is here, well, the delta here is zero,
of course. And the delta here is 2pi.
And the delta here is 4pi. I first want to show you,
or at least draw your attention to the fact that there is a
wavelength dependent lambda. And what that means is that if
you take 650 nanometers, which is red,
the red would have a maximum here, the red would have a
maximum here, the red would have a maximum
here and the red would have a maximum there.
But if now you have 400 nanometers, which is violet
light, it would have a maximum at different locations.
Here at zero it would always have the same location as
maximum, but the wavelength is shorter for blue,
for violet. Here would be the maximum for
violet, here would be the maximum for violet and roughly
here would be the maximum for violet and roughly here.
The reason why the red and the blue there almost coincide --
I mentioned that also last time for the double-slit
interference. Two times 650 is roughly three
times 400. So they live a life of their
own. Let's now address the issue of
the N minus one zeros. I first want to calculate what
the location is here where you have your first zero,
completely zero. Well, you would have your first
zero when the upstairs is zero but the downstairs is not zero.
Because, if they are both zero, you are at what we call a prime
maximum. What is the first time that
this one becomes zero and the downstairs not zero?
It is when this is pi. When it is zero they are both
zero, but when that is pi then, of course, I would have my
first zero. Let's do that here.
I call that my first zero. That is the case when N delta
divided by two is pi. And so, that means when delta
is 2pi divided by N. Now I go to this equation,
and I put in here for delta 2pi divided by N.
And what you see then, that sine theta becomes lambda
divided by sine theta, is now lambda divided by Nd.
In other words, this point here,
in terms of sine theta which, of course, is the same as angle
theta in radians because these angles are small.
This is an angular dimensional plot.
This here is lambda divided by Nd.
And then, of course, the second one will be twice
lambda divided by Nd and the third one will be three times
lambda divided by Nd. You will again have completely
destructive interference. You may be interested in what
the magnitude is, what the light intensity,
I should say, is of this little mini-maxima.
Well, that is very easy to calculate.
If we know that the sine of theta here is lambda divided by
Nd and we know that here it is twice that much then all I have
to ask that equation is what is your light intensity when the
sine of theta -- Now we go to that mini-max,
the first mini-max right here. I ask the equation,
what is your intensity when the sine theta is now 1.5 times
lambda divided by Nd? Then I am right in between
these two zeros. Now, whether I am exactly on
that maximum, I don't know.
I am really not interested, but I surely am close.
When sine theta is 1.5 lambda divided by Nd then delta is
going to be 3pi divided by N. That is easy because you take
this equation, 1.5 lambda over Nd,
you put in sine theta, 1.5 lambda divided by Nd.
The 1.5 times 2.0 becomes 3pi. You get N downstairs,
so you get the delta of 3pi divided by N.
And now you revisit this equation and you just put in
there N equals four, you know what capital N is,
you know now what delta is, so you calculate your upstairs,
you calculate the downstairs. And you will find now that I is
approximately 1.17I0. That is low compared to 16.
This is only some 7%, 6.8%.
This height here is only some 7% of that height,
so it is very low. You have C now,
the bizarre consequences that if you get zero divided by zero,
you get these maxima at 16I0. Then you get N minus one point
whereby you get complete zeros here.
But even these mini-maxima don't mean very much.
They are very low in light intensity.
And, even if you go to an N of 1,000.
And today we will even go beyond that with our
experiments. We will have capital N.
We go up to 2,000. If you make capital N 1,000,
you can redo all this. And you will see that this
mini-maximum is roughly 4.5% of this maximum.
That maximum now becomes a million times I0.
If you have a thousand of these openings, they will add up at
the maxima. You get a million times the
intensity that one alone will do.
The reason is obviously you get a thousand times the E vector,
and they are all in phase with each other.
And the pointing vector is the square of the amplitude of the E
vector. You get the million.
All these maxima have the same width.
And if this is lambda divided by Nd then this here on this
side in terms of angular distance is, of course,
the same. There is complete symmetry.
And so the width here, if I take the width roughly
without being very precise, the width of each one of those
peaks must be roughly lambda divided by Nd.
So I take half this distance, angular distance.
All these things are angles in radians because sine theta is
very much smaller than theta. Now, you can see that the
larger N is, the more of these openings you have,
the narrower these lines are going to be, if I think of these
as being a line. And that means your ability to
distinguish two neighboring frequencies from each other,
two different lambdas increases.
And that is what we call spectral resolution.
The larger N is, the better spectral resolution
you have, your ability to separate two neighboring
frequencies then increases. There is a very easy way that I
can convince you, without any math,
well, little math, why the width of these peaks,
the width of these maxima must be proportional to one over N.
And that is purely an energy conservation argument.
And follow me closely. If I have N of these openings
in the screen, they will let N times more
light through than one opening. That is straightforward.
You can tell that to your kid brother.
If you have N openings in the screen, you get N times more
light through than if you had one opening.
That is nonnegotiable. But if each maximum is N
squared times higher, the only way that you can
conserve energy is if you make the maximum N times smaller.
Then you know that N times more light went through.
The argument once more, you have N openings that gives
you N times more energy than one opening.
But, if each maximum gives you a light intensity that goes with
N squared, the only way that you can conserve energy is if you
make the maxima N times narrower.
And so that is a very easy way to see that the width here must
go down with increasing N. A very powerful argument.
Now I will make you see in another way.
If we have four of these openings in the screen,
why there are only three minima.
And all these methods that I am using, in a way,
are complimentary. It is all the same thing,
but I just want you to see it in different ways.
It helps me enormously to look at it in different ways.
We have N equals four. And I start with delta equals
zero. There is no phase difference
between adjacent sources. This is the case that this is
the E vector of source number one, this is the E vector of
source number two, number three and number four.
All four E vectors line up in the same direction.
Otherwise, delta could not be zero.
Therefore, I get 16 times the light because I square 4E and I
get 16. This is your factor of 16.
And you get a maximum. Now we go to delta equals pi
over two, 90 degrees. I don't have to look at that
equation. I don't need that one.
I know what 90 degrees is. I went to high school.
I am educated. This is one vector.
This is 90 degrees. That is the second vector.
This is 90 degrees. That's the third vector.
This is the fourth vector. 90 degrees.
What do I end up with? Zero.
If all four relative two neighbors 90 degree phase angle
then clearly you have zero here. Now I go delta equals pi.
I know what pi is. One, two, three,
four. What do I end up with?
I add four vector, 180 degrees,
flip, flip, flip, flip.
What is the net result? Zero.
So it is dark. This one is this one.
And the second one is this one. Now I am going to do this one
for you. Delta equals three-half pi.
Three-half is 270 degrees. Well, this is one.
This is 270 degrees, this is 270 degrees,
and that is 270 degrees. What is the net result?
Zero. That's this one.
Now I am going to do 2pi. When I do 2pi I am back here.
I have another maximum, and that is this one.
And so, you can see, purely by playing vectors,
very simple, high school level you can see
that there will be N minus one minima, exact minima between the
prime maxima. The first thing that I want you
to see, when I take the grating that you have and I use my laser
pointer, is incredible impact of using many, many lines.
My laser beam has about a diameter of about three
millimeters. Here is this laser beam.
And this I estimated to be roughly three millimeters.
Your grating, believe it or not,
is a super grating. That grating has 13,400 lines
per inch. Imagine how anyone can put
grooves in your plastic. 13,400 per inch.
That means that the separation d between two groups is about
1.9 times 10 to the minus 6 meters.
That is only two microns. How anyone can do that beats
me, but it can be done. I can calculate now how many of
those lines I have here in the three millimeters,
and I end up with N about 1,600.
When I shine my laser beam through my grating I use
effectively 1,600 of those lines.
And I can calculate now what the angles are,
where on the screen there on the wall, the maxima will fall,
those maxima. And, by the way,
there will be 1,599 of these zeros in between.
And these minima are so small that you won't even see them.
You will only see the maxima. Let's calculate at what angles
we would then see the first. By the way, these things have
names. We call this zero order.
And we call this first order. This is first order.
And we call this second order. And this is also called first
order, but it is on the other side, of course.
We call these orders of the spectra.
You can now calculate that the sine of theta of N is N times
lambda divided by D. And so, when N is zero,
that is your zero order, you get, of course,
a maximum. So the zero's order is at theta
equals zero. Now, your first order is when
the sine of theta one is lambda divided by D.
And I can take my lambda which, in my case, of my laser I will
have to tell you, my lambda is 532 nanometers.
It is green. And so, I can calculate what
theta one is. And I find 16.3 degrees.
And then I can go to the second order, theta two,
for that color. It is different for different
colors. And I find 34 degrees.
And I can go to the third order, and I find theta three is
then 57 degrees. And there is no fourth order
that would make the sine of theta larger than one.
There are only zero order, and then there are first,
second and third order. The grating that you have,
I always carry that with me no matter where I go.
It is easy to put in your calendar.
And so, I will show you now, by simply shining through this
grating, I will show you there on the wall the zero order would
fall right smack in the middle so-to-speak.
First order 16 degrees away. If you know my distance to the
wall, you can calculate how far that is.
It is probably 1.5 meter. And so, you will see these
maxima, you will see them extremely narrow,
high value for N. And, well, it speaks for
itself. I have to rotate my gratings to
make sure that the grooves are in this direction.
If the grooves are in this direction, the spreading is out
in this direction. The one that you see right now
on the right side of the screen there is my zero order.
Now, you cannot tell that it is zero order because they have
only one color. And then you see,
I will move it a little, the first order now on the
right side of the screen. And that angle should be very
accurately about what I calculated, the 16 degrees.
And then you see on the blackboard here,
we saw it right here, you see the second order.
And then you see here the third order, if you have good eyes.
And the fourth order does not exist.
And imagine that between those maxima, if I really use 1,600
lines, there would be 1,599 points with exact zeros,
and then all these silly mini-maxima that you don't even
see. That is the power of grating
when you use many lines. If I use 1,600 lines,
I use 800 times more lines than when I have double-slit
interference. The double-slit interference
pattern will be extremely different from this.
That is why those locations are so narrow because,
if it were double-slit interference,
you would get the cosine square function.
And so you would see the maxima would be 800 times broader than
this. That is the power of using
1,600 lines. Now I want you to get your
gratings out. And I want you to look simply
at a desktop plan. And the reason why I want you
to do that is that I want you to be disappointed because you may
not see what you erroneously expected, that the zero order is
incredibly narrow. It is not.
It is the lamp itself. Of course, if you light source
in angular size is way larger than this angular size you
cannot expect to see the light source get smaller.
In other words, clearly the limiting factor of
seeing zero order very narrow is only if the light source itself
is small enough in size. And so, when you look at this
light now, you will see all colors at zero order.
That is one thing that is important.
You just see the lamp. That is your zero order
maximum. That is the red,
the blue, the green, the yellow, the violet.
That is all of it. And then you will see,
on either side, blue appear.
First the blue. That has the lowest shortest
wavelength. I do not think there is much
violet in this lamp. And then you will see the blue,
the red. Make sure that you line it up
so that the grooves are vertical.
That is so that you get the spread in the horizontal plane.
And then you see the green, and you see the red.
And it really looks like you have a continuous spectrum.
Even at second order, it still looks like a
reasonable spectrum. It starts in blue and it goes
to red. But, when you go to the third
one, you already see that the blue and the red are going to
interfere with each other there. Then they really do not look
any more clearly like separate spectra.
I want you to appreciate the fact that if your light source
is huge that each little location of the light source
gives you a line which is this narrow, but if you have many,
many of those locations it smears it out,
of course, and you see a very broad line, if I call this a
line, this zero order maximum.
We have here a grating that works not in transmission.
Your grating works in transmission.
The light goes through it, like what we discussed here.
But we have here one that is metal, whereby grooves I put on
the metal. And this is called a reflection
gradient. And I can show you,
with this reflection grading that has a large light source
just like this, it is large,
the spectrum when we project it here on the screen.
If Markos can give me a hand? Thank you, Markos.
We have white light that goes onto this reflection grating.
It is a big spot of white light.
And you will see them the first order will not be narrow because
the light source itself is so big.
And then you will see, on either side,
something similar to what you saw here.
You will see the white. The zero order is always the
light of the source itself which is, in this case,
white. And then on both sides you will
see first order, second order.
And I think you see up to four or five orders.
I do not remember how many lines there are per inch.
It is not important because it is really the effect that I want
to show you. I was quantitative on my own
demonstration with my green laser.
Let us turn this on. I think that is it.
And we turn all the lights off so that you can enjoy this
fully. Here you see the size of that
zero order maximum. It is not narrow at all because
the light source is not narrow. And then you see here nicely
the blue, the green. And you see here the red.
Blue, green, red.
You begin to see spectra, and then the whole thing sort
of fettles out. And you see the same on this
side. So, this is a reflection
grating. If we turn on a red laser that
is much narrow then you get the advantage of your many,
many lines. And then you get,
of course, an extremely narrow zero order maximum.
And that is this laser, I believe.
We will know very shortly. This is the 633 nanometer
laser. And so, now I take advantage of
the many lines that I cover. Now my light source is narrow
enough to let me benefit from the grating, the many slits that
I use. And you see,
as you expect, that the red zero order always
falls at the same location. And then the red here,
the 633 coincides, of course, with the red from
the white light and so on. And you even see here 633.
And this is first order, second, third order.
There is no fourth order. The separation in terms of
angle is only a function of d, the spacing between the grooves
and, of course, lambda.
And so, if the spacing of a double-slit interference pattern
is the same as the spacing of a multiple one,
the separation and angle is the same.
But, of course, the gain is that you make them
narrow and they go up with N squared.
That is what you gain. Now I would like you to take
full advantage of your grating. And we have prepared light
sources that are narrow enough so that you come very close,
maybe not exactly, to seeing the width of the
light source about lambda over Nd.
If you put the grating in front of your eye, your pupil itself
has a diameter of about three millimeters.
You will only use, effectively with your eye,
about 1,600 lines, just like my laser beam did.
Because it is the same three millimeters.
You look through about 1,600 of these lines.
And you know what the d is, the separation of the grooves.
And so, you can now look at a line source, which we have
prepared. We have prepared line sources
which are very narrow, not like this one,
but very narrow, which are in here.
Now, if you look at that light source, which is helium,
and now you use your grating, then you begin to understand
the idea of spectral resolution. You will now begin to see the
individual atomic lines nicely separated through your grating.
And so we will turn this on. And we will make it completely
dark. And then I want you to
appreciate this and spend some time looking at these lines.
It is really quite remarkable. And this, of course,
you could never do with the two slit interference.
You need these many, many lines.
You look through about 1,600. This very strong yellow in the
helium is a well-known helium line, and it has a wavelength of
587 nanometers. It is the brightest,
the strongest one in helium. And you can see them in first
order, you can see them in second order.
And then, gradually, when you go to higher orders,
the various colors begin to overlap with each other because
they each live their own lives. The angles are only dependant
on lambda over d. And now I can show you neon,
which has even more lines.
You can look at your grating. Amazing.
I mean, the grating does not cost more than maybe a dollar.
It is absolutely stunning. And it has an incredible
spectral resolution already because you need a prepared line
source to take advantage of that spectral resolution.
And, as I said, you will probably approach,
certainly the audience all the way in the back of 6120,
you will approach this angular resolution.
The ones that are closer may not approach it because they see
the line source, of course, wider.
The angle at which they see the line source may well be larger
than this width, but the ones in the back of the
audience are there for a little [better off?].
The angle that you see to the line source, the widths of the
line source become smaller the farther you are away.
This is remarkable. Absolutely incredible.
I think this is a great moment to rest and to digest this
wonderful experience and to have a four-minute break.
Thank you very much. What I want to discuss now is
the logical consequence of this whole concept of Huygens sources
where spherical waves come from each point in the aperture.
And we are now going to extend it to a single opening.
Not multiple but one single opening.
And the opening is now d. This separation is now this
opening. It is d.
Think of them as being a slit which has a width d.
It is open, single-slit. And we have plane waves coming
in like this. And now the question is if I
look in various directions, and that is my famous angle
theta, what will I see now on a screen that I place very far
away? Well, each point in this
aperture can now be considered, according the Huygens-Fresnel
Principle as a source of spherical waves.
And they are going to interfere with each other.
Strangely enough, for reasons beyond me,
we call this diffraction, but it is exactly the same
phenomenon as interference. We draw a strange distinction
in physics between interference which was the grating and the
double-slit, we say double-slit interference.
No one would ever say double-slit diffraction,
but it is the same thing. Somehow, when we deal with
individual openings we call that diffraction.
It is the same. You can use it any way you want
to. You can call it interference.
But the individual Huygens sources, there they are,
they are all going to do their own thing.
And I pick this one at the top, number one, and I pick this one
number two right in the middle. Why don't I do that?
You will see why I do that. Now I can calculate what the
past difference is between the Huygens source right in the
middle and the Huygens source right at the top.
Well, this past difference here is clearly one-half d times the
sine of theta. We did that before.
We had a little d here. If I make that one-half lambda,
I claim that in that direction there will be darkness.
Why will there be darkness? Because, if source number one
can kill number two because they are 180 degrees out of phase,
then the source just below one can kill this one below two and
the one below there can kill this one.
So I can always identify two pairs which kill each other.
That means darkness. This must be a criterion for
destructive interference. Now, it may not be the only
angle for which there is destructive interference.
But I want to convince you that there is at least one for which
you see no light. I will now introduce a phase
angle beta, which is the phase difference between source one
and source two. If this is the slit,
one side of the slit and the middle of the slit,
it is not the phase angle between two neighboring sources.
Because the neighboring sources touch each other.
There is an infinite number of Huygens sources.
It is a continuous Huygens source.
It is the phase angle between the edge of the slit and the
center of the slit. That is the way I define beta.
And so my beta now becomes 2pi divided by lambda times one-half
d times the sine of theta. And so, you see the one-half
eats up the two so you get pi d divided by lambda times the sine
of theta. And now I will not derive for
you as I did. Precisely for the gratings,
I will not derive for you what the light intensity is at
function of angle. Again, it is a matter of adding
vectors. But you can look that up in
Becefi and Barrett. It is done in Section 8.7.
And so, I will only give you the answer.
I did it for the grating. You do this one.
And you can show now that the intensity of the light,
as a function of that phase angle beta, is I0 times the sine
of beta divided by beta. And, of course,
no surprise that you get a square there because that has to
do with the pointing vector. And this function is very
different from a grating function.
And, before I plot it, let me make a few calculations.
I will do it here on the center board so that you can still
compare with this equation. I am going to write down here
what the sine of theta is. And then here my column comes
beta. And then here comes the sine of
beta. And here comes the intensity I.
I first take the sine of theta as zero.
Well, if the sine of theta is zero, it is immediately obvious
that beta is zero. And it is also obvious that the
sine of beta is zero. So you get zero divided by
zero, you use l'Hopital, and this ration becomes one.
And now the light intensity is I0.
Well, that is not so surprising that you have there a lot of
light. Because, of course,
if you have an opening here and you shine light through it,
then you expect that you see right on the wall there,
in the middle of the slit you expect to see a lot of light.
So that is not so surprising. This, by the way,
is the way that we define I0. I0 is defined as that maximum
that you will see when you so-to-speak look straightforward
at angle theta zero. That is the way we define I0.
Now let's take sine theta as lambda divided by D.
Now beta is pi. We put in here sine theta is
lambda divided by D. Here it is.
So you see that beta is pi. The sine of beta is now zero.
And, therefore, the intensity is zero because
the upstairs is zero but the downstairs is not zero.
That is this. Because, look,
if you take this half away and you take this half away,
you get that the sine of theta is lambda divided by D.
I already predicted that you would have destructive
interference. That is exactly this case,
of course. And now I have two lambda
divided by D. That gives me a beta now of
2pi, that gives me again a zero here, and there is another zero.
There are more locations in space where there is complete
darkness. Not just one.
In fact, there is an infinite number of them.
You can go on like this. I will first plot the curve,
and then we will discuss it in a little bit more detail.
And I will only plot curves in terms of sine theta because that
is an angle that I can relate to.
I can tell my mother about theta.
I cannot tell my mother about phase angles,
but I can tell her about theta. Mom, this is theta.
10 degrees. 20 degrees.
30 degrees. That is a real angle in my
laboratory. That is that theta.
That I can relate to. I always plot things in terms
of sine theta. And so, here is my zero.
And let this be lambda divided by that capital D.
That is the slit. This is two lambda divided by D
and here is three lambda divided by D, and the other side minus
lambda divided by D and so on. I will put one more in.
And so, now the curve that you see, not so obvious but when you
plot it you will see that, you get here this maximum,
which we define to be I0. It is the central maximum.
And then you will get here your first zero.
You get here some kind of a mini-maximum.
And then you get another zero here.
And you get an infinite number of zeroes every time.
You get three lambda over D, four lambda over D,
five lambda over D, on this side here and so on.
The central maximum, the width in terms of angles,
this of this as angles. Right sine theta is close to
theta in terms of radians, so this is an angular size.
The linear size depends on how far you are away from the screen
on which you show it. Then you have to multiply this
by L. L is the distance to the
screen. This is the angular side.
This width here, very crudely,
is about half this. And so that width is about
lambda divided by D. Let's take an example which is,
I think, the demonstration that I have lined up for you anyhow.
We will take a laser light, which is about 600 nanometers.
The fact that it is 633, of course, is not so important.
I give you easy numbers. And suppose we have d which is
about 0.1 millimeters. We have a slit which has a
width opening of only 0.1 millimeters.
And we will put a screen there, L at a distance of about three
meters. We can calculate now what
lambda divided by D is. That would give you the angle
in radians. That is six times 10 to the
minus 3 radians.
Sine theta is very close to theta.
And so, now you can calculate the linear size of this central
maximum, as I am going to show you there on the screen.
And the linear size is now L. The linear dimension is L times
lambda over D. That is how wide that central
maximum will be. And that, in this case,
will then be about two centimeters.
Now, think about it. Think about the absurdity.
We have a slit which has an opening of one-tenth of a
millimeter. And because of Mr.
Huygens, it will show up there with a width of two centimeters.
Two hundred times broader than the actual opening.
Whereas, if you would think high school, then you would say
if you have light going through a tenth of a millimeter,
you look, what you see on the wall would be a tenth of a
millimeter. No, you see two centimeters.
And that is the result of diffraction.
That is the result of the fact that each one of those sources,
in this aperture, are going to radiate spherical
waves. They are going to interfere
with each other. And they then cause this huge
broad center. And the smaller you make D,
the more you tighten the nuts on the slit, the wider it is
going to be. Because, look,
if you make this smaller d then this angle will become larger.
Very known intuitive. Before I show you this,
I want to know roughly what that maximum is here.
That mini-maximum. Well, that is easy.
You could do that now on your own, because all you have to do
is substitute in this equation sine theta which is sort of
half-way in between. If I do that,
halfway in between, that is 1.5 times lambda
divided by D. That is right in between these
two. That gives me then a beta of
1.5 pi. I can go to this equation and
put in for sine theta 1.5 lambda over d.
The lambda and the d cancel, you get 1.5 pi.
Straightforward. Just turning the crank.
Now you have beta. You can calculate what the sine
of beta is, which is minus one. And so, now you know what the
sine of beta divided by beta is. And you will find 0.045 times
I0. This mini-maximum is 4.5% of
the central maximum. And when you go further out,
these maxima are even smaller. But, when I show you this
phenomenon, which I will, you will see distinctly these
zeros. You will see actually very nice
dark locations. And you will see the central
maximum and then a little bit of light, but not very much light
on this side. And, of course,
again, this is wavelength dependant.
So whether you do this in red light or in blue light,
you will see something very different.
If you do it in red light, you will see this white.
If you do it in blue light, you will see it narrower.
And also those locations, of course, are then further in.
And I have a slide which shows you the idea in different
colors. Maybe we can make it a little
darker. Here you see it in three
colors, red, green and blue. Notice how very different this
is from a grating. You see some broad central
maximum. That is that central maximum
that you see here on the blackboard.
And then you see the sharp black locations where there is
almost no light here. And you see infrared farther
apart than for green. And for green farther apart
than for blue. And then, when you do it with
white light, of course, then it becomes always more
difficult. You see the sharp dark areas
because the colors overlap. All right.
Now I want to demonstrate this to you.
And the way that we are going to do this is with a slit that
we can vary in size. Where was my calculation?
I made a calculation here. I am going to do it with a 633
nanometer laser light. That the beam of that laser is
about three millimeters, and then we have a slit here.
This is the opening. This is the d.
And we can make the d smaller. And so, we already made a
prediction that the linear dimension then,
if the opening is only a tenth of a millimeter,
you would expect that the central maximum on that screen,
which is about three meters away, that is why I chose the
three meters, will be two centimeters wide.
But I can make it way wider because I can make d way smaller
than a tenth of a millimeter. What you see now is that the
slit is very large, very open.
I don't know, maybe a millimeter or so.
And I am going to tighten it. And I will stop at one point
here where it is a nice point to stop.
Here you see that central maximum.
See how powerful and overwhelming that is in terms of
its brightness? We understand now why,
because of this crazy function sine beta divided by beta
squared. Already now it is here,
oh, I would say five centimeters.
Already now this slit width must be less than a tenth of a
millimeter, because it would be two centimeters if it were a
tenth of a millimeter. And I hope you can see
distinctly those dark locations. And you see there are a lot of
them. But, keep in mind,
that this mini-maxima next to the broad maximum is only 4.5%.
And it gets smaller and smaller, lower and lower as you
go further away. Now I go way beyond one-tenth
of a millimeter, way smaller.
Now, keep in mind that when I make the slit width smaller less
light will go through. I cannot help that.
The whole image will become fainter.
That is the price I pay for letting less light through.
But what I gain is to show you the absurdity that the center
maximum gets wider and wider and wider as I make the slit smaller
and smaller and smaller. The slit is now narrower and
narrower and narrow and narrows, and the central maximum that
you see there is almost a foot. The center, the opening of my
slit must now be something like maybe only ten microns or so.
This is a very highly accurate device, whereby,
we have the option of making the slit with,
indeed, as small as ten microns.
And you see this is the result that you get.
I will now make the slit open and open and open and open.
And here we have the point. If I make the center point
about two centimeters then the slit width, which is about now,
is about one-tenth of a millimeter.
I now have to make an important confession about the grating
equation. Some of you who were very
observant may have noticed that the maxima of the grating that I
showed, also when I did the experiment with my own laser,
were not all exactly the same brightness.
There was a difference. And no one asked me about it,
and I was hoping that no one would ask.
And the reason for that is that each one of those grooves has a
finite size opening. And each one of those openings
acts this way. They cause diffraction.
We call that diffraction. It is just semantics.
And so, superimposed on the grating equation this causes
diffraction. And the net result then is that
you get the product of the two. If I amend here now,
and I am removing this beta now because this is for a grating,
but the beta is defined this way, this d is the opening of
each groove in your grating. And little d is the separation
between the grating. Then I can write down now here
times the sine of N delta over two divided by the sine of delta
over two. And now I put here the square.
And now I have the real grating equation.
And so, what you see now is that since little d is always
larger than capital d, you are going to see that the
maxima, which comes from this equation, are being modulated by
this one. And so, if this were a grating
whereby capital d was the opening of each individual
groove -- And if the grating wanted,
of course, the zero or the maxima is always here,
and if the first order maximum grating would fall here and the
second order would fall here and the third here then this is the
price you pay for the fact that these grooves have an opening.
And so, you see a modulation in the strength of your first,
second, third, fourth order and so on.
And so, when you look carefully at the grating lights --
And I will demonstrate that to you.
They are not 16 times I0 if you have N equals four.
And if you have N equals 1,000, they are not all a million
times pi zero, but they have this overall
envelope which modulates it. And that is the result of the
finite opening of the grooves. To make sure that you
understand the difference between the two D's --
If this is my grating and this is the open area,
this is, say, where the light cannot go
through, then the definition of D is this and the definition of
this is d. And that d shows up in here and
this capital D shows up in there.
So this is the single-slit diffraction and this is the
multiple-slit interference. There you see again we make the
distinction in wording, but that has no meaning because
it is all diffraction, of course.
If somehow D were approximately d divided by five,
and it just so happens we have a grating here for which that is
the case, then the fifth order maximum of the grating is going
to be killed. Because that is when this
function becomes zero. So you would see then zero
order one, two, three, four,
five would be killed here. And then they would build up a
little again. And then, ultimately,
of course, they would all peter out.
And so, if I show you a spectrum of a grating,
you can actually roughly estimate what the ratio capital
D over little d is by seeing sort of that modulation pattern.
And so that is what I want you to see now.
It is not so exciting. Many of you who were observant
may have seen it anyhow, because it was every time
there, even when I showed my own grating.
We are going to make it quite dark for this.
This is the wrong switch. I have so many switches here.
There we go. This is a grating that I
purposely offset. I purposely offset it so that
here is the zero order. This is the zero order.
I think it is that one, actually.
It is easy to test where my zero order is.
This is the zero order, so we aim exactly here.
And so, this is the first order, second order,
third order, fourth order.
Look at this sucker. It is almost gone.
That is the result of the fact of the single-slit interference.
And then it comes up again here.
And the reason why it comes up again is because now you enter
this little mini-maximum in the single-slit interference.
You see it here quite well. Sometimes with gratings you can
see it remarkably well. Other times it is harder to
see. It depends, of course,
on how many maxima you have. But here you see quite well the
modulation. You see it comes up here again.
And this little point here would then be somewhere here in
this maximum. This is then the complete
equation that combines single-slit diffraction with
multiple-slit interference. If we change a single opening
from a slit to a circle, your eye is a circle,
your pupil is a circle, then very little change,
except of course if you have a circular opening everything is
now axial symmetric. And so, you will get circles.
These things become circles, which is not so obvious.
And then, which is not so obvious, this minimum does not
fall at lambda over two in terms of angular dimension,
but 1.22 times lambda over two. And, if you want to use [1.2?]
for an approximation that is fine enough.
It is a little larger for a circular opening than it is for
a slit. I have here one of those
pinholes. So this is now a circular
opening for which this relation has to be used now.
This central is then a little wider.
And we are about four meters away from the screen.
It is this one, four meters away from the
screen. And I am going to do this with
a wavelength lambda of 594 nanometers.
It is a circular opening. We will call it a pinhole.
And lambda is about 594 nanometers.
It is also a laser. And the distance through the
screen L is about four meters. And what you are going to see
is a ring, which is this ring, and then you see this light
inside, which is very difficult for me, this is the very high
maximum. And then you will see more
rings outside. This ring is quite
well-defined. You are going to see that very
sharply defined. And if this ring,
which I measured, is about five centimeters in
diameter, you should be able to tell me what the diameter of the
opening is. Of course, because you know now
what the angle is. And so, you can calculate what
D is. And, when I did that,
I came up with something like, I think, an eighth of a
millimeter or so, but you can confirm that.
A very small opening of an eighth of a millimeter,
if I did that correctly, would then give you a central
maximum, which is from zero to zero, five centimeters wide.
And so let's take a look at that.
These single pinhole diffractions are always very
difficult because the pinholes have to be so small to see it,
and that means very little light will go through.
And so, here you see it. For those of you who are
sitting close, you will clearly see that
central circular maximum. And then you clearly see the
first ring, the dark ring. My pinky is right on it.
This is about five centimeters across.
I see a second dark ring, but if you are far away in the
audience you may not see that so well.
This is a nice example of circular single-slit
diffraction. To make you see it even better,
we have handed out cards. And those cards have a small
pinhole in one location and they have double-slits in the other
location. And so, I am going to aim at
you, very slowly I am going to scan this over the audience a
light-emitting diode, bright light.
And, as it passes you, you only get one shot at it,
and you look through it, maybe we can make it darker.
If you look through the pinhole, you will really see
beautifully this ring structure with the dark lines and the
center maximum. But, if you look through the
other opening, you will see the beautiful
double-slit interference. Notice that the widths of the
dark lines and the widths of the bright lines is about equal
because you only have two slits. Remember, you get this cosine
square function. It is not as dramatic as a
grating. Now I am going to rotate this
through the class so that each one of you get a chance to look
through both openings one at a time.
If I go too slowly let me know. And you can keep these cards.
You need a very bright light. You need a very small like,
too, because if the light is too large in size then,
of course, your dark and your bright areas are going to merge
with each other. You wash it out.
Your light source always has to be carefully thought through in
terms of its dimension, in terms of angular dimension.
If the angular dimension of your light source is too large
you kill all the phenomenon. I am going to rotate it back.
Who has not seen it? You have not seen it.
How could I do that to you? But now you can see it even
longer than others.
This single-slit diffraction or single-opening diffraction has
major consequences, even for our daily experiences,
because it ultimately determines our ability to
separate two light sources in the sky.
If you have a telescope and the telescope has a lens or it has a
mirror which has a diameter D then there is a limitation to
which it can separate two stars in the sky.
Let's assume there are two stars at roughly equal strength.
Here is star number one and here is another star which is
start number two. And the angle between them is
delta theta. Then somewhere here,
on the photographic plate or, in your case,
on your retina, there will be an image.
And that image will be like this for one star.
And there will be another image a little bit displaced from the
other star. And if those two blurs are too
close together, you don't see two stars anymore
but you see only one star. And so, now comes the question,
how small can this angle be so that you still say there are two
light sources? A car comes toward you with two
headlights, how close does the car have to be that you still
say there are two and not one? Well, there is a criterion,
which is a little bit arbitrary, called the [rally?]
criterion for angular resolution.
And that is we want the angle between the two lights larger or
equal to this angle so that the maximum of the second light
would fall here. And so, you would clearly see
then that this thing is broadened.
And you may even see a little dip in that [curl?].
On your photographic plate you would really be able to say,
yes, there are two sources and not just one source.
And so, the angular resolution would then be,
in terms of angle, 1.2 times lambda divided by D.
Delta theta would have to be larger than 1.2 times lambda
over D for you to be able to say, yes, there are two stars.
Which is the ultimate limit of angular resolution for you,
for me, but also for optical telescopes.
Suppose we take the Hubble Space Telescope,
HST, that has a mirror D, which has a diameter of 2.4
meters. And it is prepared so carefully
that the claim is made that it is really diffraction limited.
And so, that means if I take an average wavelength in the
optical spectrum of about 500 nanometers, I realize that it is
all the way goes from 400 to 650, but if I take this as a
representative wavelength then 1.2 times lambda divided by D
translates into about one-twentieth of an arc second.
That is an incredible resolution.
One-twentieth of an arc second. If two stars of equal
brightness are one-twentieth of an arc second apart,
the Hubble Space Telescope can see there are two stars.
The same telescope on earth would do no better than half an
arc second, this is an arc second, to maybe even two arc
seconds. Why is it so much worse for a
telescope on the ground than it is for Hubble Space Telescope?
Any one of you know that? Atmosphere.
The earth's atmosphere is in turbulence, is always in thermal
motion. And it is that thermal motion
that is the problem. That makes your imagine on your
photographic plate or on your CCD move around.
They work like little lenses. And so it broadens it,
but it broadens it in an incredible way.
We call this the seeing. If the seeing were one arc
second then Hubble's Telescope, which is above the earth's
atmosphere, has an angular resolution that is 20 times
better linearly. That means over a surface.
It has 400 times more resolution elements because it
is two-dimensional, so it is highly superior in
terms of angular resolution than any ground based observatory.
And now comes your human eye. Your human eye,
the opening of your human eye depends on the time of the day.
At night, when it is dark, your pupil opens.
During the day it goes down a little.
If we take about four millimeters, a reasonable
number, and we take again 500 nanometers as our representative
wavelength then we can calculate what 1.2 times lambda over D is.
And that translates to half an arc minute.
That is 30 arc seconds. That is 600 times worse than
the Hubble Space Telescope. You cannot do better than this.
This is Mother Nature. You cannot beat diffraction.
On your retina, when you look at a light
source, the image on your retina will look like this,
will look exactly what you saw there.
That is what your retina will see.
And, if those two lights are too close together,
your brains will say, sorry, I do not see two lights.
Now, in practice, Mother Nature did not desire
our eyes, at least most of us not really down to diffraction
limitation. In practice,
I think it is more like one to two arc minutes.
This is my symbol for arc minutes.
The angular resolution of your eyes is not quite as good as it
could be, but it is close to that.
And I am going to test that with you.
I have here a screen, and this screen has holes in
it. And I will give you the code of
the holes. Here is that screen.
Two holes, two holes, two holes, and we repeat them
three times. This is one millimeter apart,
two millimeters apart, three millimeters apart,
four millimeters apart. Students who are two meters
away, very few are, but if you were --
We take here a student who is at a distance of two meters.
If the student looks at the one millimeter separation of these
slides then the angular separation is 1.7 arc minutes.
You should be able to see them as two lights.
If you look at the two millimeter separation then
obviously it is about 3.4 arc minutes.
You should have no problems. In other words,
the students who are close should be able to see this as
two lights, this as two lights and this as two light.
But let's now go to the students who are five meters
away. These are the students who are
five meters away in the audience.
If we go to the one millimeter separation, there is no hope on
earth that you will see that when you are sitting there.
You will not be able to see the upper two as two light sources
because the separation is only 0.7 arc minutes.
And I do not think that any one of you can see lights that are
0.7 arc minutes apart. The two millimeter slots would
be 1.4 arc minutes and the three millimeter slots would be about
two arc minutes. And so, I am going to make it
dark now in the room. And I want each of you to just
look at these pinholes. And I am going to rotate this
so that all of you get a chance. And then I want you to raise
your hand if you can see the top two as separate ones.
And only those who will raise their hands will be very close
to me. And then we will slowly go
farther into the audience. Look closely at the upper one.
And then also try to see the one below there,
the two millimeters which are here, the three millimeters and
the four millimeters. Can you see the upper one?
Even the upper one you cannot. You're only three meters away.
Now I will just rotate this. The upper one,
again, is the one millimeter separation.
You see it repeat three times. And then two millimeters.
And then below that is three millimeters.
And then it is four millimeters.
Raise your hand if you can see the upper one at two light
sources. No one.
Oh, boy. Well, if you really can,
then your resolution is very close to 0.6 arc minutes because
the two meters was 0.7 arc minutes.
That is really remarkable but possible.
Who can see the second row clearly as two distinct sources?
Now people are coming in. Even the ones that are close
can only do it. You see that?
No one in the audience there in the back is raising his hand.
Now we are talking about a resolution of the two
millimeters, that 3.4 arc minutes for the ones that are
two meters. You are talking about 2.5 arc
minutes. Who can see the third row
separate? Now the hands go up.
And who can see the fourth row as separate?
I think the whole class. The ones in the back there,
can you not see? Really?
Boy, you have to go to an eye doctor.
You cannot see the bottom? I was there this morning where
you were and could see the bottom one distinctly as two
light sources. And I could kid myself that I
even saw this one as two. But I was really kidding myself
because I knew it, I think.
But this one I could clearly see.
Really, none of you can see the bottom one as two differences?
Well, that shows that your angular resolution is no better,
and you shouldn't be ashamed of that, it is not your fault,
than about two arc minutes. With that idea in mind, 1254 1:23:51,320 have a good weekend.