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YEN-JIE LEE: OK.
So welcome back,
everybody, to 8.03.
So before we start
the lecture today,
we will give you, as usual,
a short review on what we
have learned, and also, an
introduction about what we
are going to learn today.
So last lecture,
we were discussing
an interesting phenomenon, which
is seeing film interference
pattern.
As you can see
from this slide, we
were wondering why the
soap bubbles are colorful.
And in the end of the
class, we actually
recognized that the reason why
the soap bubbles are colorful
is because of the
interference phenomenon
between the refracted
light on the bubble.
One puzzle path is that
the light goes into the--
goes refracted directly from
the surface of the soap film.
The other possible
optical path is
to get refracted by the
inner surface of the film.
Therefore, the interference
between these two paths
actually created a colorful
pattern on the bubble.
So we also learned about
how thick is the soap film.
And I think just a quick
reminder, actually,
we concluded that in order
to see a colorful pattern,
the thickness of the
wall or, say, the film,
should be something like in
the order of 100 nanometer.
So that's actually
pretty remarkable,
because that's already in the
order of the size of a virus.
OK.
OK that's actually pretty cool.
So what are we
going to do today?
What we are going to do today
is to continue the discussion
that all kinds of
different phenomenons,
which can be explained
by interference.
We will learn interference
phenomenon with a double slit
experiment and using, for
example, laser or water,
and which I have
a water tank here,
which I will show you
the interference pattern.
And also, the second thing
we are going to learn today
is how does a phased
radar actually works.
OK?
So by the end of the
lecture today, you
should be able to learn
why we should construct
the radar in the way
and how to actually
focus on the
electromagnetic wave
to work one specific direction.
So that's essentially what
we are going to learn today.
The third goal is
that we are going
to make a connection to quantum
mechanics from the lecture
today.
All right.
So let's immediately
get started.
So before we start the
discussion of a double slit
experiment, I would
like to remind everybody
about Huygens' Principle, which
you may already learned it
from 8.02 or in the
high school days.
So what essentially
is this principle?
So this principle
is saying that if I
take a look at all the points
in the wavefront, basically,
you can treat all those points
on the wavefront a point
source.
And this point source,
essentially, a point source
of a spherical wave.
And it's immediate from all
the points on the wavefront.
So you can see from
this slide, basically,
if we choose to focus on the
yellow point on the wavefront,
you can see that from
each yellow point,
you can actually treat that as
a spherical wave point source.
And then what you actually
need to do in order
to calculate what would be
the total electric field,
for example, is to add
up all those contribution
from each point.
And then you will be
able to actually explain
the interference pattern,
which we see in the experiment.
You may wonder where is this
Huygens' Principle coming from?
And although we are not
going to derive that directly
in the lecture today,
but I can actually safely
tell you that essentially,
it can be derived
from Maxwell's equation.
OK?
I will link some
document, which actually
shows the proof of the
principle on the website
and for your reference.
The other thing which you
may or you may not know
is that we are really lucky so
that we can use this Huygens'
Principle in our universe.
Why is that?
Because if you look at
the mathematical proof
of this principle, it is because
the number of dimension, number
of spatial dimension
is odd, which
is three in our universe--
Or in my universe
also is yours, OK,
[LAUGHS] happened to
be yours, as well--
such that the Huygens'
principle actually works.
On the other hand, if the
number of dimension is even,
there's no Huygens'
principle, actually.
So that's pretty
interesting in that we
are really lucky that it
actually works in our universe.
But I will not go
into detail in 8.03.
So let's get started with
a concrete example, which
we would like to further
investigate to understand
the interference phenomena.
And those will prepare
ourselves to the understanding
of the design of the
radar, for example.
All right.
So suppose I have
experimental set
up here which contain a
wall where on the wall,
there are two slit, A and a B.
The upper one is A. The lower
one is B, as designed here.
And from the left inside
there's an insert plane wave
with a wavelength lambda,
which is showing here.
And this plane wave, plane
electromagnetic wave,
or can be water wave,
et cetera, essentially
approaching the wall with
these two slits there.
And we were wondering what
would be the resulting
pattern on the screen.
This screen is actually
pretty far away
from the experimental setup,
the wall on the left-hand side.
How far is that?
The distance between the screen,
which shows the resulting
interference pattern, and that
the wall is actually defined.
It's actually given here.
It's actually
called L, capital L.
And in this experimental setup
L essentially pretty, pretty
large and is much, much larger
than the d, where d, small d,
is the distance
between the two slits.
OK?
So our job now is to
understand what will be--
and to predict what is
going to be the interference
pattern coming from the
electromagnetic wave which
pass through point
A and point B,
and what is going
to happen over, say,
what would be the result which
we will observe on the screen.
OK?
So the first thing
which we can do
is that we can now
assign observer,
which is called P, one
of the point of interest
on the screen, which
is located here.
And then we can
link or, say, the
connect the point A, which is
the location of the first slit
and the location of
the second slit, which
is called B. We can link those
points together by a line.
And that is actually denoted by
AP and the BP, these two lines.
Since we are talking about L,
which is essentially very, very
large, assuming that
the distance, the length
scale of the distance between
the wall and the screen
is much, much larger
than the length
scale of the distance between
the two slit, which is d.
Therefore, I can safely
assume that AP and the BP
are almost parallel
to each other.
Right?
And I can also try to express
the location of the P point
by using the angle between BP
and the horizontal direction.
OK?
And the horizontal
direction is actually
showing there's
a dash line here.
And the angle between BP
and the horizontal direction
it's called theta here.
OK.
So since AP and the BP are
almost parallel to each other,
I can now calculate what would
be the optical path length
difference between
AP and the BP.
Right?
So in order to
actually calculate
the phase difference between
the electromagnetic wave coming
from slit A compared to slit
B, I need to calculate--
again, like what we did last
time-- optical path length
difference.
OK?
In this case, I can call the
distance between A and P, rA.
And then I can also call
the distance between B
and the P, rB.
Then the optical path
length difference
is called rB minus rA.
And then we can
actually calculate
that because we have already
given you the angle between BP
and the horizontal direction.
And basically, we
can safely conclude
that the path length difference
is actually this line here.
Therefore, I can actually
calculate and get
the optical path
length difference,
the difference between rB and
the rA to be d sine theta.
OK?
Once we have that, it's
actually pretty straightforward
to calculate what would
be the phase difference.
The phase difference between the
field coming from slit A, which
I will call it EA
here, and the field
coming from the slit B,
which I will call it EB here.
The phase difference, as
you define a lot of time
to be delta, delta can be
calculated by the optical path
length difference,
d sin theta, divided
by lambda, which essentially
telling you how many period
have passed when the light have
to actually overcome this--
or say have to pass through this
optical path length difference.
And, of course,
these things need
to be modified by 2 pi
in order to translate
from a number of period
to a phase difference.
Therefore, you get the phase
difference between AP and BP
to be delta equal
to d sine theta
divided by lambda times 2 pi.
OK.
So you can see that
all those calculations
are pretty straightforward.
Maybe you have already seen
that before in an earlier class.
But what I want to say
is that it is actually
because of Huygens'
Principle, such
that you can't expect something
which will show up at point P,
right?
If you don't have
Huygens' Principle
what is going to happen?
What is going to happen
is that the light
passing through this slit
will just go straight.
And they will never
overlap each other.
OK?
So that's actually why, because
of the Huygens' principle,
all the points on the wavefront
are treated as a point
source of a spherical wave.
OK?
So that is essentially why you
can expect that something will
hit the P point, which
is because, in this case,
we have two points,
two point source.
And they are emitting
spherical waves
coming from these two points.
OK?
So it is really because of
Huygens' Principle, which
applies here, such that we can
actually observe the phenomenon
at the P. And now,
we have managed
to calculate the
phase difference,
which is delta, presented here.
So what the next question is,
what would be the intensity?
Since we have already calculated
the phase difference delta,
what would be the
intensity observed at P?
So for that, we have
already prepared ourselves
from the last few lectures.
So now, we can
actually calculate
what would be the total
E. The total E will
be equal to EA plus EB.
And here, I'm going to
use complex notation just
for simplicity.
And basically,
you can rewrite EA
and the EB as E0 exponential
i omega t minus k times rA
plus E0 exponential
i omega t minus k rB.
The first term is actually
telling you the contribution
from the first slit, slit A.
And the second term is actually
telling you the contribution
coming from slit B.
In this set up, I'm telling
you that I have the plane
wave coming from the left
hand side of the experiment
and actually hitting the wall.
And you can see that
from the drawing.
Actually, the
wavefront, essentially,
actually telling you that the
direction of the electric field
is actually in the Z direction
in my coordinate system shown
on the board.
So basically, the Z direction is
actually pointing to you guys.
And that means the
electric field is actually
oscillating in this direction.
OK?
So therefore, I have to be
careful of those vectors.
So therefore, I need to
give it other direction.
And in this case, it's
actually the Z direction.
And also, you can see that
the amplitude is actually
denoted by E0 because I always
assuming that both slit have
the same finite width.
For the moment, ignore
the width of the slit.
And also, they are coming
from the same plane wave.
Therefore, the amplitude
is all denoted by E0.
OK?
So now, I have the
expression here.
And I can now go ahead and
simplify this expression
and rewrite that in this form.
So I can now extract the E0.
And also, I extract the
common factors here,
which essentially the
exponential i omega t
and also, minus k rA.
I can actually
factorize some part
of the exponential function out.
So the choice I made
is that I actually
could factorize out exponential
i omega t minus k times r.
Basically, I take these out.
And I get this term showing
here, omega t minus k rA.
I take this out.
Then basically, what you
are doing to get inside
will be 1 plus exponential
minus i delta, actually.
times z.
OK?
Why is that delta?
Because once you factorize out
or take out exponential i omega
t minus k rA,
basically, you are left
with something proportional
to exponential i minus k
rB minus rA, right?
And that is actually the optical
path length difference here.
And also, of course, you
can always rewrite lambda
over 2 pi, right?
Basically, you write this
to be k times d times theta.
Right?
So therefore, you can
actually immediately identify
the second term is to
essentially exponential
minus i delta.
OK?
Any questions here?
OK.
Because d sine theta
essentially is just rB minus rA,
therefore, I safely
replace that by delta here.
OK?
All right.
So since everybody's
on the same page,
I can now, again, factorize
out not only the omega t
minus kA term,
but I can actually
do a trick to factorize
out, also, exponential
minus i delta divided by 2 out.
And basically, what
I'm going to get
is exponential i delta over 2
plus exponential minus i delta
over 2.
This reason why I'm doing
this is because, huh, now,
I have this term identified.
And this is actually
just 2 times cosine delta
divided by 2.
All right?
OK?
So now, I'm really pretty
close to the intensity.
So what would be the
intensity coming out
of this electric field?
That is actually going
to be average intensity,
as we discussed last
time in the lecture.
The average intensity is
proportional to square
of E vector.
Right?
In the complex
notation, how do we
evaluate the absolute
value of E vector square?
In the complex
notation, basically, you
get basically, E times E
star, where E is actually
the amplitude, which is the size
of the E vector, the magnitude
of the E vector.
Then, basically, you
will see that this
will be proportional to cosine
square delta divided by 2.
Right?
Because you can see that
if I calculate EE star,
then all the terms with related
to exponential i something
actually got cancelled.
Right?
So therefore, you can see
the "aha" very, very quickly.
We can show that
the intensity will
be proportional to cosine
square delta divided by 2,
where delta is the
phase difference
between the first path
and the second path.
OK?
Any questions so far?
OK.
So we can see that the intensity
essentially changing really
rapidly as a function of delta.
Right?
So when I have a situation
where delta is equal to 0--
let's actually stop here a
bit and enjoy what we have
as you learn from here.
All right?
So if you have delta equal
to 0, what does that mean?
That means there's
no phase difference
between the first and
second electric field.
Therefore, when you
add them together--
just a reminder about the
notation we were using before.
So if you draw the vector
in a complex frame, what
you are doing is that you
are actually adding EA
and the EB together in the
most efficient way, right?
Because the delta is equal
to 0, the phase differences
is equal to 0.
Therefore, you are actually
adding them in a straight line.
OK?
So that actually will give
you the maxima intensity.
Because when delta is
equal to 0, cosine 0 is 1.
Right?
Therefore, you are reaching
the maxima in the intensity.
So now, I can always
increase my delta
until a number which
is actually pi.
What is going to happen
is that if I still
use the notation which I was
using for the complex frame,
what it does this is that, huh.
Now, I am actually completely
cancel the electric field,
because the phase
difference now is pi, right?
So therefore, in
the complex frame,
you are adding the two
vectors in way such
that they completely
cancel each other.
The magnitude of
the two vectors are
the same, as shown here,
which is actually E0, right?
Therefore, what you are
going to get, as you expect,
is going to be 0, because
they completely cancel.
OK?
You can also see that from
this formula we did right here.
When delta is equal to pi, then
essentially, cosine pi over 2.
Then you get
intensity equal to 0.
OK?
Everybody accept this?
All right.
Now, I can still continue
and increase the delta,
for example, until
delta is equal to 2 pi.
Then you are getting this again.
Basically, you have EA and the
EB, again, line up each other.
And the difference is
that this EB actually
rotated maybe 360 degree.
And basically, you will see
that, again, the intensity
become the maxima again.
OK?
So that is actually
how we can actually
understand this result.
And, of course, you
can also go ahead
and plot or simulate
this result in the computer
and really draw the amplitude,
really draw the intensity
as a function of angle here,
or, say, the delta here.
As you can see from
here, that the intensity
is actually reaching the
maximum in the center.
Why is that?
In the center, if I
have observer here
in the center, what
is going to happen
is that the path
length, optical path
length between AP
prong and the BP prong
is going to be the
same by symmetry,
because it's actually
in the optical center.
Therefore, you will expect that
delta is actually equal to 0.
OK?
So that's essentially why
you see the maxima there.
And if you start to
move away from there,
you will see that the
delta start to increase.
And at some point, you'll
reach a minima, which
you can see that on the plot.
And that is actually
because now,
due to the increasing
optical path length
difference and the
phase difference,
the two electric
field is starting
to cancel each other,
which actually produce
the black pattern there.
And finally, after it
pass delta equal to pi,
then these two electric fields
start to work together again.
All right?
They're collaborating again.
And you can see that again.
You would get another
maxima afterward.
OK?
And here, you can see that
is actually my calculation.
And, of course, I can do
a demonstration to you
to really show that
this is actually
what we are going to see
based on the demonstration we
are going to show here.
So now, I am going to
turn the light off.
And here, I have a device which
actually contain a water tank.
And I need to actually
turn this thing up.
On the water tank I have two
vibrator, which is actually
acting as a point source.
So basically, those vibrator
vibrating up and down
to create waves in this tank.
OK?
So basically, you can
see that, huh, really,
you have two point-like source.
And you can see spherical waves
is actually really generated
and is really propagating
away from the point source.
OK?
And what I can do
now, you can see
that this picture is
really dynamic, because we
can see that wavefront
essentially moving
as a function of time.
So what I'm going to
do is to really change
the frequency of the
light, which is actually
shining on this water,
so that you can actually
see the fixed pattern here.
And now, I am going to
change the light frequency.
You can see now I only
shine the water tank
at the specific time which match
the speed of the propagation
of the water wave.
And you can see, aha,
I've actually managed
to freeze the wavefront.
We see?
OK.
So you can see, now, really, you
can see coming from the source,
they are circular
wavefront, which
actually mimicking the result
from Huygens' Principle.
And you can see that
they are complicated
interference pattern forming.
You can see that
at some point they
have constructive interference.
If you focus on
the central part,
you can see that the maxima
is actually reach there.
On the other hand, if you
move away, a little bit
away from the center, you can
see that really, the intensity
drop.
And at some point, you will
also see that, OK, again, I
am changing the
procedure in such
that the phase difference
between the contribution
of our source A and the B
essentially equal to 2 pi.
In that case, you will be able
to see that another maxima is
actually created again.
So now, we can
actually also show you
that a lot, in fact, based
on this glorious pattern,
let's actually take a look
at the projector here.
So if I look at on the
individual slide, which
I have here, you can see
that those are actually
a point-light source and is
creating a circular pattern.
And now, I can actually overlap
with two patterns together.
And you can see that when I have
the center of the two circles
pretty close to each other,
you can see that really, you
have very small d.
In this case, you have
very small distance
between source number
one and number two.
Then basically, based
on our expression,
so you can see that delta is
equal to d sine theta divided
by lambda times two pi, right?
And you can actually
calculate sine theta
will be equal to delta
divided by k times t.
OK?
When delta is equal to pi,
that is going to give you
a minima where, essentially,
also showing here,
the minima is shown as
the black pattern here.
OK?
You can see from on here.
So what this says, your
formula is showing you
that when I have d,
which is very small,
what is going to
happen is that I'm
going to get sine theta to be
very large when d is actually
very small.
And that can be shown here.
When I have d, which is the
distance between the center
of these two point
source, very small,
you can see that the
place you get the minima
is really far away from the
center, which is actually here.
OK?
Now, what I'm going to do
is to increase the distance
between these two source.
According to our position,
what is going to happen
is that the central
maxima will decrease.
The position where you get a
minima will be moving closer
to the center, according
to that formula,
because it's
proportional to 1 over d.
And we can do this
really carefully
to see if I can succeed.
And you can see that
really, when I am moving
these two slides
away from each other,
you can see that the
pattern is changing, right?
And the center maxima, or, say,
this Gaussian-like curve there
becoming narrower and narrower.
OK?
So that essentially what we
can actually observe form here.
And our calculation really
works very well here.
Very good.
So do we have any
questions regarding
the demonstration we have here?
OK.
So all those things seems to be
pretty straightforward to you.
And what we are actually
now is seeing a position
where we can actually
discuss how we actually
can understand the radar, which
is how actually radar works.
So here is actually
how radar works.
Suppose you have
some unknown object,
which is like an airplane, OK?
And you would like to
know where is this object.
What you do, actually, is to
shoot whatever radio waves
toward some direction
and see if there
are something coming back.
Right?
Then you know there's
something on the sky
because you can detect
the refracted wave.
Right?
So we shoot this airplane.
And then something is
going to come back.
And now, we can say OK.
In that direction I have
something coming back.
That means there's
something there.
And I can also measure
the time it takes
for the wave to come back.
Then I know where it's
actually that object.
Right?
So that's actually a pretty
straightforward thing to do.
However, there's one difficulty.
So this is actually
the radiation pattern
of oscillating dipole which
we actually learned before.
So the problem is that,
OK, what we really
need is electromagnetic wave,
which is actually very, very
narrow in angle and pointing
to some specific direction.
And then I would
like to see if I
can get some refractive wave
coming from that direction.
OK?
The problem is that, look!
if I oscillate some
charge up and down,
the radiation I'm getting
is really, really broad.
Right?
So it's going toward all
kinds of different direction.
So if you use this
to detect things,
you are always going to
get something coming back,
because it's actually shooting
the electromagnetic wave
into random direction.
And you are not
sure any more where
is actually this object
you are trying to detect.
OK?
So that's actually
apparently a problem.
And what we can actually do is
to make use of the interference
phenomenon, which
we can actually
learn from here to
actually try to make sure
that the electromagnetic
wave is actually
pointing to some specific
direction we want.
So let's actually go ahead
consider a three slit
experiment.
I have this setup changed.
Originally, I have two slits.
And now, I drew it in
three holes on the wall.
And, again, I have the distance
between the slits to be d.
And I call this slit
number 1, 2, and 3.
And we were wondering what
would be the interference
pattern on the screen,
which is actually
far away from the wall,
as a distance of L.
And I'm interested
in their intensity
at the point P on this screen.
OK?
So what I am going to do
is to basically repeat
what we have done in
the previous example.
I'm trying to connect
1 to the P, 2 P,
and the 3 P, basically,
connect the slit
to the point of
interest on the screen.
And I can actually also--
you know this angle, this 1 P
to the horizontal direction,
this angle is called
theta in my notation.
Then clearly, I can
go ahead and calculate
what will be the
optical path length
difference between
of the light coming
from slit number 1, slit
number 2 and the slit number 3.
OK?
And in this case,
what I'm interested
is delta 1, 2 and delta 1, 3.
Right?
Since the screen is really
far away from the wall,
therefore, I can actually
savor the assurance
that these two angle is actually
theta because the three lines,
due to the large
distance, this L
is actually really,
really large.
Therefore, they are actually
almost parallel to each other.
OK?
So what is going to happen
is that delta 1, 2, which
is the phase difference
between light
from the first slit
and second slit,
is actually going to
be equal to delta 2, 3.
It's going to be
equal to the phase
difference between the
second slit, the light
from second slit and third slit.
And what is actually
that number?
This number is going to
be equal to d sine theta
divided by lambda times 2 pi.
It's exactly the same
as what we actually
get from the first example.
OK?
Therefore, what
is going to happen
is that no matter
what theta I choose,
the phase difference
between nearby slit
is actually a constant,
which is actually this one.
And I will call this phase
difference to be delta.
I would like to ask
you a question now.
The question is, how
do we choose the delta
here such that I have completely
destructive interference?
Now, I have three vectors,
vector E1, vector E2,
and the vector E3.
The phase difference
between E1, E2, and E3,
the nearby phase difference
is actually delta.
So the question is, how do
I actually completely cancel
the electric field so
that I have completely
destructive interference?
Can somebody help me here?
The hint is that you can
actually use this vector sum
idea in the complex frame.
STUDENT: [INAUDIBLE]
PROFESSOR: Yes, very good.
To form a triangle in
the complex frame, right?
So what we can do is now
choose the phase difference
delta to be such that
E1, E2, and E3 actually
form a triangle.
You see what I mean?
Therefore, you can
actually already get
what would be the
required delta value.
The required delta value is
going to be 2 pi divided by 3.
Right?
OK?
So very good.
So now, we are not
afraid anymore.
So how about four
slit experiment?
I just add another slit,
d essentially the distance
between the fourth slit
and the third slit.
What will be the
delta required to have
destructive interference?
Anybody can help me?
STUDENT: [INAUDIBLE]
YEN-JIE LEE: Very good.
So if you have four slit,
based on this intuition,
which we developed from
the complex notation vector
sum, what is going to happen
is that if you have four slit,
the delta will be equal
to 2 pi divided by 4.
OK?
So what does this tell us?
So remember, the
sine theta, sine
theta is telling you
the location where
you get the minima.
OK?
So this is actually the
power profile, or, say,
the intensity profile.
OK?
And this is actually equal to 0.
And this is actually delta.
OK?
The place which you
get zero intensity
is actually becoming
closer and closer to zero.
Right?
Because sine theta,
which is the angle
between horizontal direction
and this observer P,
is proportional to delta.
When you have destructive
interference at angle
which is smaller, smaller,
and smaller, that means what?
That means the central
Gaussian-like structure
is going to be becoming
narrower and narrower.
Does that make sense?
Very good.
So at least we found
something interesting now.
That means, ha, one
idea to get very narrow
electromagnetic wave
pointing to some direction
is to have a huge number
of point light source
and slit experiment
such that I can actually
construct something which is
actually very narrow in angle.
And I can use that to
shoot the object which
I would like to detect.
You see what I mean?
Does that make sense?
OK?
All right.
So that's very good.
So now, let's actually consider
an N slit interference pattern
OK?
So suppose, now, I have not
only 1, 2, 3, and then many
more until N slit.
All right?
I can now go ahead and
calculate the E total, which
is the total electric field
coming from all of the slit
we have.
Basically, this will be equal
to E0 exponential i omega t
minus kR where I define r1
is roughly capital R. OK?
That's essentially
the contribution
from slit number 1 OK?
And this contribution
from slit number 1
is going to be looking like
exponential i omega t minus kR
minus delta, right,
because there is a phase
difference between
the light coming
from first slit and the second
slit, which is actually delta.
All right?
So what would be the third term?
So these actually coming
from slit number 2.
What would be the third term?
Exponential i omega t
minus kR minus what?
STUDENT: 2 delta.
YEN-JIE LEE: 2 delta,
yeah, because you
can see that coming
from here, seems
the distance between theta
as constant, which is d.
Therefore, the phase
difference between nearby slits
is actually a constant.
Therefore, I accumulating
the phase difference now.
I get 2 delta here.
And this is actually
a contribution
from the first slit.
And the et cetera, et
cetera, until the Nth slit,
which is actually
going to be exponential
i omega t minus kR
minus N minus 1 delta.
And summing all those things
together, and all of them
are in the Z direction.
OK?
So I'm now going to
calculate this dimension.
So basically, you are getting E0
exponential i omega t minus kR.
I can actually factorize
these factor out.
And what am I going to get is 1
plus exponential minus i delta
plus exponential minus
exponential minus i
2 delta plus blah, blah, blah.
And basically, you will
get exponential minus i
minus 1 delta in the first term.
And all those things are
pointing to the Z direction.
And this, I know how
to actually calculate.
Right?
Just a reminder, basically,
if you calculate summation
N equal to 0 to N
minus 1 r to the Nth.
And these will
give you 1 minus r
to the N divided by phi minus r.
OK?
So basically, I can now go
ahead and calculate this.
And this will basically
give you 1 minus--
OK, so the small r here has
been replaced by exponential
minus i delta, right?
So therefore, what
I'm going to get
is 1 minus exponential minus
i delta N for the upper part.
And then I have 1
minus exponential
minus i delta in the lower part.
OK?
So that actually make use of
this formula, which are here.
And, again, it should be
simplify these series.
All right?
As usual, what
I'm going to do is
to use the trick similar
to what I have done there
to actually get
cosine function out
of the exponential functions.
All right?
So what I'm going to
do is to factorize out
exponential minus i delta N
over 2 for the upper part.
So basically, I get
exponential minus i delta
N divided by 2, exponential
i delta N divided by 2
minus exponential minus
i delta N divided by 2.
OK?
This is actually
divided by exponential
minus i delta over 2
exponential i delta
over 2 minus exponential
minus i delta over 2.
All right?
The reason I'm doing
this is because I
would like to actually make
this a cosine function.
OK?
Any questions so far?
OK.
So if no question,
then basically,
this expression can
be, again, rewritten
as exponential minus i
delta N minus 1 divided
by 2, because I have this
denominator nominator
exponential i delta N over 2 and
that exponential minus i delta
divided by 2.
OK?
Therefore, I can combine
them all together
and then get this
expression here.
And this is actually
exponential minus exponential.
Therefore, I am going
to get sine out of it.
And basically, I
get sine and delta
divided by 2 divided
by sine delta over 2.
OK.
So now, I can actually
go ahead and calculate
what will be the
resulting intensity.
Right?
The resulting intensity
is going to be
proportional to the square
of the electric field.
Right?
So basically, the intensity will
be proportional to E square.
And that is actually
equal to E times E star.
E and the E is a
complex conjugate.
And basically, you
will see that this
will be proportional to sine
and delta divided by 2 divided
by sine delta over 2 square.
Therefore, the intensity
will be equal to i
0 times sine and
delta divided by 2
divided by sine delta over 2.
And then square that.
Any questions?
So after all this work, we have
arrived at expression which
is very hard to understand.
Right?
[LAUGHS] So what I'm
going to do to help you
is to really plot the result
as a function of delta
on the screen.
You can see there
are four plots here.
The first one is N equal to 3.
The upper left one
is N equal to 3.
So you can see that the
pattern looks like this.
So at delta equal to
0, surprise nobody,
you are going to get maxima.
Right?
Because delta is equal to
0, you are adding N vectors
the most efficient way.
Therefore, you are going
to get the maxima, which
is i equal to i 0.
OK?
And if you move away from
the center, delta equal to 0,
and you see that is a
small bump in between.
Then you can continue
and continue.
And you see that there's
another big peak again.
You see?
So that's essentially
the structure
if you plot this result, i
equal to something proportional
to sine square this
expression there.
And that's essentially what you
will get when N is equal to 3.
OK?
And this is essentially how
I remember this pattern.
OK?
So when N is equal to
3, you have a family
of two adult and one child.
[LAUGHTER]
Right?
So basically, you
have two big peak.
And between them,
there's a small peak.
OK?
That's actually how I
remember this pattern.
And I think it's
pretty nice, right?
So you can have N equal to 4.
It's a bigger family.
You have two adults.
The adults are slimmer, OK?
All right?
[LAUGHTER]
Because they have a
lot of work to do.
Then they have two child.
All right?
N equal to 5, how many
children do we have?
STUDENT: We have three.
YEN-JIE LEE: Three.
Therefore, the adults
are really frustrated.
So they are even slimmer in a
happy way, making it positive.
And N equal to 6, woo.
Oh my god, I have four
children in the family.
All right?
So there are two things
which we learned from here.
The first one is that the
number of big peak, which
I would call it principal
maxima, the number
of principal maxima
is actually pretty
similar as a function of delta.
But the number of
secondary maxima
increase as a
function of N value.
N value is actually
telling you how many slits
you have in the experiment.
And also, you can see that the
delta is actually becoming--
the first minima,
the delta value
is actually decreasing
as a function of N value.
Right?
So the parents are
getting slimmer.
All right?
So therefore, you
can see that if I
would like to have a radar
which is actually pointing
to a very specific
direction, what
essentially the choice of
N value which we will need?
Infinity or a very large number.
OK?
For sure in your life,
we cannot do infinity.
But now, we have found
a way to actually design
our radar since sine theta is
actually proportional to delta.
Therefore, what we
actually really need
to do is to really maximize
the number of slits
we have so that actually
we can create a radar which
would really point toward the
direction of the enemy, which
is shown there,
invading the earth.
OK.
[LAUGHTER]
And we can actually detect it.
OK.
So we will take a five minute
break before we actually
go to the last part of the
course, which is the connection
to quantum mechanics.
So we come back at 35.
[SIDE CONVERSATIONS]
[SIDE CONVERSATIONS]
YEN-JIE LEE: OK so welcome
come back from the break.
So before we move
to the connection
to quantum mechanics,
I would like
to talk some more about
what we have learned
from the design of the radar.
OK?
So this essentially
what we actually get.
The position of the minima that
required the phase difference
delta is actually equal to
2 pi divided by N value,
because it was this delta value.
The N vectors is going
to cancel each other.
And you are going to form
something like a circle
if you choose delta equal to
2 pi divided by capital N. OK?
And don't forget why
this is actually delta.
The delta is actually d sine
theta divided by lambda.
Right?
OK?
And times 2 pi.
OK?
Right?
So therefore, you can see that
the sine theta is actually
proportional to lambda
divided by N times d.
OK?
And in this case, you can
see that if you increase
N value, the
resolution or the width
of the central
principal maxima is
going to be decreasing
as a function, though,
N value you're putting.
So in short, how do I actually
design a high-resolution radar?
What I really need is to
have lambda to be small.
OK?
So that means I need to use
high-frequency electromagnetic
wave.
I can maximize the N value.
I can actually
make d very large.
That means I'm going to have
a very large radar design.
Right?
Then I can have a
very good resolution.
OK.
So we are almost
done with radar.
But there's a problem.
The problem is that if you look
at this, if this is actually
the position of the
principal minima,
you can see that
is always pointing
to the center of the radar
where the delta is equal to 0.
OK?
And then that means I can
only scan in one direction.
There is a reason why those
radar are called phased radar.
That is because
now I can actually
change the relative phase of
all those point source emitted
from the radar so
that I can shift
the direction of the
central principal maxima.
OK?
So what is actually
done here is like this.
So basically, I
can have introduced
before emitting the
electromagnetic wave,
I can introduce a zero
additional phase difference.
And for the second one, I
introduce additional phase
difference of phi.
OK?
And for the third
one, I introduce
additional phase difference
between the third slit--
or say the third emitter and
the first emitter by 2 delta.
And for N's emitter, I
introduce a phase difference
of N minus 1 phi.
OK?
If I add this phase difference
into the setup, what
I'm going to get is like this.
So basically, delta will become
2 pi divided by lambda d sine
theta minus phi angle.
All right?
And this phi is actually
the artificial eddy phase
difference between those source.
OK?
And that means I will require--
and this will be equal to 2
pi divided by N value, such
as you have completely
destructive interference.
OK?
I can now make this phi
to be time-dependent.
For example, it's
increasing as a function
of time, phi times t, right?
Then what is going to happen
is that as a function of time,
I'm going to change
the sine theta value
so that I can get a complete
cancellation, 2 pi over N.
Right?
So effectively, I'm
changing the angle
of the central principal
maxima by introducing
additional artificial phase
difference between all
those point source.
OK?
And this is actually
the way we can actually
rotate the place we are
scanning up and down
and get a very nice result
to detect the enemy.
OK?
Any questions?
No?
OK.
So now, I'm going to
move on and discuss
a very interesting experiment.
So this is very exciting
experiment content,
billiard balls
and the two slits.
OK?
And we will wonder,
then, what is
going to happen when
those balls especially
pass through the slit.
Can anybody actually tell me
what she is going to happen?
And what will be the
statistics, or say,
the count, which I am going to
go on to get in the receiver
later?
Anybody can actually tell me?
If I actually shoot a lot
of balls through this slit--
don't be shy, right?
It's easy.
No?
Nobody wants--
STUDENT: They make [INAUDIBLE]
YEN-JIE LEE: Yeah, that's right.
Right?
Doesn't surprise nobody, right?
[LAUGHS] Yeah, too afraid
of answering questions.
OK, you can see that they
make two path, right?
No?
Right?
OK.
Very good.
So now, this is
the exciting part.
Now, instead of shooting
billiard balls, what I'm going
to do is to shoot electrons.
So I can actually prepare
an electron source
and heat it up, such that
it start to emit electrons.
And I have two slits and have
them pass through this slits.
And I have a screen,
which actually
have an electron
detector to count
the number of electron which I
am going to get on the screen.
The reason why I call it
single electron source
is because each time I control
my experiment such that it only
emit one electron every time.
OK?
The question I'm trying
to ask is, will I
see some pattern,
which is actually
light like billiard balls, and
they form two piles in a pack?
That's actually
option number one.
Or I'm going to see
really something crazy?
It's the electron is
going to be interfere--
it's going through the
interference with itself.
And that essentially
option number two.
OK?
The lure of 8.03 is that
everybody had to choose one.
OK?
So how many of you think what is
going to happen is number one?
Come on.
I have only one
electron each time.
Nobody think so?
Wow.
Maybe all of you are wrong.
[LAUGHS] How about
the second option?
STUDENT: [LAUGHS]
YEN-JIE LEE: Hey,
some of you actually
didn't raise your hand.
Come on.
Come on.
[LAUGHTER]
OK, everybody.
Wow.
What is actually
happening to you brain?
[LAUGHTER]
My brain is not
functional like this.
OK.
So I really hope that I can
bring the experiment to here.
But unfortunately,
that's actually
going to be difficult. OK?
So what I'm going
to do is that I'm
going to show you the
experimental result,
this video.
And we are going to see
what is going to happen.
You see that there
are dots popping out.
What are those?
Those are the detected
electron one-by-one on screen.
OK?
So basically, you can see
that the number of dots
are increasing as
a function of time.
And I actually-- I mean,
speeding up things a bit
so that actually you can
see the pattern quicker.
OK.
So you can see that there
are more and more dots.
And each time, you can
see that I only get
one electron per image here.
Right?
So you can see now
there are more and more
and more and more, and
accumulating more data,
like what we actually done
in The Large Hadron Collider.
We wait there,
collect more data.
And we are speeding things up.
And you can see that,
wow, something's
actually developing.
What is that?
Can you see it?
Now, you are speeding up
like 1,000 times faster.
You can see what pattern?
STUDENT: Interference pattern.
YEN-JIE LEE:
Interference pattern.
What is going on?
You are not surprised?
STUDENT: No.
YEN-JIE LEE: Oh my god.
What is going on?
[LAUGHTER]
I'm so surprised.
Look at this.
So I have emission of
one electron each time.
And that is actually the
four snapshot which I took--
which actually this
experiment, Hitachi Group
actually did this experiment.
You can actually click on
this link to the more detail.
And they took four
snapshots of the experiment.
And you can see that
in the beginning,
you can see clearly
each time you only get
one electron out of the source.
OK?
But as a function
of time, you're
accumulating more and more.
And you see that
clearly, there's
a pattern forming, which, is
actually consistent with what
we see in this calculation.
OK?
So I think that's
actually truly amazing.
And what does that mean?
That means the electron
is playing with itself.
It's interfering with itself.
Right?
That's really strange.
What is going to happen?
What is going on?
So one single electron pass
through both slit, which is
actually the option you choose.
Surprise me.
And then they
interfere like waves.
And they produce the pattern
which we see on the screen.
That is actually
really crazy to me.
What is actually even more
crazy is this situation.
So now, if I make measurement in
front of the slit, OK, so now,
I puts on a little device.
When the electron pass
through one of the slit,
I say, send me a signal.
OK?
So now, I can clearly
know that which
slit the electron is
actually going through
in the experiment.
OK?
And the crazy thing is
that if I do that, then it
becomes two piles.
OK?
Of course, maybe there are
some diffraction pattern.
But it really
changes the pattern
of the experimental result.
And that is actually
really very strange.
And we are going to
talk about that briefly
in the next lecture.
So before the end,
I'm going to show you
an additional
demonstration which
motivate the discussion
what we are going
to have in the next lecture.
So now, I can actually turn off
the light again and also hide
the image.
OK.
I hope I can find the pattern.
[LAUGHS] All right.
So here, I have two laser.
So I'm going to turn
up the first laser.
And this laser is going to
pass through a two slit--
a two really nearby slit and
form an interference pattern.
As you can see on the wall--
I hope you can see, I don't
know if you can see clearly--
that you can see there are
many, many dots, nearby dots,
which actually shows
you the position
of the principal maximas,
right, because are actually
two slit experiment.
Therefore, how many children
do we have in the family?
Zero, right?
Because they are--
OK, they just got
married, maybe.
[LAUGHS] All right.
So therefore, you
will see only adults.
And that is actually
the principal maximas.
You can see many,
many nearby dots.
They are almost equally bright.
OK?
But there's something happening
to this pattern as well.
And you can see that--
wait, wait, wait a second.
In the calculation we
get the principle maxima
to have the same height, right?
That means you are going to get
exactly those same intensity
for all the maximas.
But you don't see that here.
You can see that if you move
away from the center too much,
the intensity is decreasing.
You see at the edge?
It actually even goes to zero.
Right?
What is actually happening?
Something clearly is actually
missing in our calculation.
And that missing
part is actually
diffraction, which we will talk
about that in the next lecture.
So if you compare this
pattern to the second demo,
you can see in the
right hand side
setup, which I have here,
which I should give you
a projection on the
wall, which is actually
lower part of the demo, you can
see that this laser actually
pass through a single slit.
But this slit is
actually pretty wide.
OK?
And you can see that indeed,
you see the laser coming out,
but essentially,
not a single spot.
And it has some kind
of pattern, which
is actually popping out there.
And this is also
related to interference
between infinite
number of source.
OK?
And you can see that the
pattern seems to really pretty
similar to the pattern
we see in the upper demo,
except that upper demo have
individual similar structure,
which is the principal maxima
from the two slit interference.
And we are going to solve
the mystery in the lecture
next time.
OK.
So thank you very much.
And if you have any questions
related to the lecture today,
I will be here to
answer your questions.
So this is a demo
which we would like
to show you, Single Slit and
the Double Slit Interference
Pattern.
OK?
So the first scene is the setup.
So we have a laser
beam, which is actually
passing through this either
single slit or double slit
experiment.
And then the laser beam
will be going through this
and interfere and show
interesting pattern
on the screen.
And there are two setup.
The left-hand side one is two
slit interference experiment.
And right-hand side is a single
slit diffraction experiment.
So you can see left-hand side
one, I already turned it on.
Laser beam passed
through two slits.
And they form complicated
pattern on the screen.
And you can see there are
two kinds of structure here.
The first one is the
very fine structure,
which you can see that
it's like some row of dots
in the center of the pattern.
And there are larger
scale pattern as well,
which you can see that the
overall intensity of all
those little dots
are also variating
as a function of distance
with respect to the center.
So during the lecture, we
were wondering what actually
cause this kind of pattern.
And the answer is
that this is actually
coming from the effect of
single slit interference.
The reason why we
have this pattern
is because the two
slit is actually not
infinitely narrow in my setup.
Therefore, within a
single slit, there
is already a interference
pattern coming out of it.
Therefore, the compound effect,
results in a very complicated
structure we see on the screen.
So to demonstrate
this effect, now, I'm
going to turn on the
right-hand side setup.
In the right-hand
side setup, I am
going to have the
laser beam, which
you see emitting from here,
pass through a single slit.
I actually set it
up so that they
have the same width between
the single slit experiment
and double slit experiment.
And then you can see
after I turn it on,
you can see that now, we
have two sets of pattern.
The lower set is actually coming
from a single slit interference
experiment.
And you can see very nicely
that first of all, it
has a similar pattern, like
what we see in the double slit
experiment.
Secondly, you can see that
basically, we carefully tune
these two experiments
so that the distance
between the slit and the
screen is roughly the same.
Finally, we also set it up,
as I've mentioned before,
such that the width
of the individual slit
in the double and the single
slit experiment are the same.
And you can see that with
single slit experiment,
we also see a very
similar pattern
that you have a central maxima.
You have a high-intensity
light going toward the center
of the pattern.
And the intensity actually
decrease dramatically really
quickly as a
function of distance.
And also, you can see
that the pattern actually
matches with what you see in
the double slit experiment
very well.
And that is actually
pretty remarkable.
And from these
two experiment, we
understand why we have also
a complicated structure
in the double slit
experiment, not
just like many,
many little maximas,
many, many little dots.
But also, you have
this overall modulation
in the light intensity.
And that is actually mainly
coming from the single slit
diffraction pattern.
