Lec 21: Diffraction, Angular Resolution | 8.03 Vibrations and Waves (Walter Lewin)


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, have a good weekend.

10 Comments

Add a Comment

Your email address will not be published. Required fields are marked *