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YEN-JIE LEE: So I
hope you can hear me.
Welcome back, everybody.
I hope you have the
full energy back
from Thanksgiving vacation.
Everybody, welcome back.
So before we start, we'll
talk about where we are now.
So this is actually the goal,
which we said, for 8.03,
at the beginning.
So what we have
been discussing is
to learn about how to
translate physical situations
into mathematics, simple
harmonic oscillator, coupled
oscillators, et cetera.
And we tried to put together
infinite number of oscillators.
And we found waves from
this interesting exercise.
And of course, we learned about
Fourier decompositions of waves
and also learned about how to
put together physical systems.
In order to do that, you need
to define boundary conditions.
And those conditions
need to be satisfied,
so that you can describe
multiple physical systems
all together.
And the third part
of the course, we
have been focusing on
many, many applications,
for instance, for the
phenomenon related
to electromagnetic
waves and also
many practical
applications in optics.
And we are pretty close to
the discussion between wave
and vibrations and the
future course, which
is 8.04, the connections
to quantum mechanics.
And if we have time, we will
talk about gravitational waves
if we manage to do that.
It depends on how
fast we progress.
So let's start
the lecture today.
So first, we will give
you a short review.
Before Thanksgiving,
we were talking
about polarizer filters.
And we have been researching
how to make a very good photo so
that you can post it
on Facebook, right?
So that's essentially
what we learned.
So if you want to take a picture
of the sky, which is deep blue,
then you need to
use a polarizer.
And the reason?
We also understand that
it's because, if you
look at the sky, which is
actually, roughly a 45 to 90
degree wave from the
direction of the sun.
Basically, what you get is
that all those light fronts
scattering, between the
sunlight and the molecules
or little dust in the
sky, polarize the light.
Therefore, you can actual
filter them using the polarizer.
Of course, you have to tune
your polarizer carefully,
so that you can
actually minimize
the light from the
sky, so that you get
a sharper image in your photo.
And also, we discussed about,
with a polarizer, polarization
filter, we can actually filter
out also the reflected light,
for example, from the water
or from the window of a car.
And that is because something
which is closely related
to the boundary condition, which
we learned about from Maxwell's
equation in matter.
And this is actually
the four equations,
which we discussed last time.
And in that issue, we were using
that to explain the incident
light, from air to
something which is denser,
for example lighter gas.
And then we found that, if we
start with unpolarized light--
this incident wave
unpolarized light--
what we found is that the
transmitted wave, which
is actually in the
bottom of this diagram,
is actually still pretty
close to unpolarized
light but slightly polarized
because of the transmission
and the boundary condition.
And the reflected light,
something very interesting
happens.
Only the component,
which is actually
polarizing the direction, such
that the electric field is
oscillating in a
direction perpendicular
to the surface of this
light, will survive.
And that actually gives
you polarized light,
which is actually
reflected from the surface.
And this interesting
phenomenon reaches the maxima,
where you get the fully
polarized light, when
you actually set the incident
angle of the unpolarized light
at so-called Brewster's angle.
And this Brewster's
angle is happening
when the reflected light
and this transmitted light
direction actually are
orthogonal to each other.
And that actually gives
you the maxima effect
we are looking for.
So that's actually
what we have learned
from electromagnetic
wave in matter
and, also, matching
the boundary conditions
between the electromagnetic
waves inside the material
and in the air, such that
we actually learn about all
those interesting phenomena.
And basically, we
have learned how
to describe
electromagnetic waves, how
to add electromagnetic waves
together, how they propagate
from one position to
the other position,
and how the boundary
condition works
and your equation of motion,
et cetera, and something
related to dielectric material.
And today what
we're going to do is
to put all the things
we have learned together
and see if we can
actually explain
a very interesting phenomenon.
So before I start,
I will show you
a demonstration, which I'm not
sure if I will be successful.
It's very difficult, actually.
And of course, during
the break, you're
welcome to come over and
play with all those demos.
And here, I have two sticks.
And I am going to create a
soap bubble from this soap
and water.
And let's see if I
will make it or not.
So basically, I put this
into the soap water.
And I will try to open it to
see if I can create a bubble.
Yeah.
You can see.
You can see that there's a
colorful soap bubble created.
You can see that it
is not always easy.
Oh, it's getting very messy now.
I'm trying to destroy
the classroom.
But it's OK, because we are MIT.
You can that it's
really beautiful.
It's colorful.
It live for a while,
then it breaks.
And of course, during the break,
you are welcome to do this.
And it's actually non-trivial
to create this size of bubble.
So the success rate is like 50%.
So as you can see from
this demonstration,
we see something
really beautiful.
This bubble is colorful.
And I didn't actually
shine this soap bubble
by all kinds of
different preset colors.
So it appears automatically
and just shiny
with all kinds of
different-- whatever,
wavelengths I get from the
lights that are in here.
Pretty bright light,
there, on my face.
And you can see that
it becomes colorful.
And we are going to
understand what is going on
and where this color
is coming from.
And the good news is that,
based on the knowledge we
have learned, we are
in a very good position
to understand this phenomenon.
So before we start to
explain this phenomena,
I would like to talk
about a phenomenon,
so-called interference.
So suppose we have two
electromagnetic waves.
We can actually
add them together
because of
superposition principle.
So what we can do
is that, suppose
I have two electric fields.
E1 is actually defined as
A1 cosine omega t minus kz
plus phi 1 in the x direction.
So by now, you should
know that this is actually
the electric field propagating
at angular frequency omega,
with a wave number k, going
toward positive z direction.
And the electric field is
perpendicular to the direction
of propagation in
the x direction.
And also, this electric
field have a phase of phi 1.
So that's actually what
we know already by now.
What does this mean,
this expression mean?
And it's actually the harmonic
oscillating electric field.
And of course, since I am
talking about interference,
basically, I can add this,
the first electric field
and the second
electric field together
and see what will happen.
So now, if you define
the second electric field
to be A2, which is the
amplitude, cosine omega
t minus kz, basically, they have
the same wavelengths and also
the angular frequency--
plus phi 2.
But they have different phase.
And of course, in
this setup, I asked
them to be pointing to
the same direction, which
is the x direction.
So we were wondering,
what is going
to happen if I consider the
superposition of these two
electric fields.
And the total electric field,
which is called the E vector,
is actually E1 plus E2.
So before that, I would
like to remind you
about pointing vector and
also the so-called intensity.
So pointing factor
is actually defined
as this S vector,
pointing vector,
is equal to 1 over mu 0 E
cross B. So this is actually
the directional flux of
energy per unit area.
So that should be the
pointing vector which
we have been using for a while.
And also, another reminder is
that, given the electric field,
which is actually a
harmonic progressing wave,
the corresponding
B field would be
equal to 1/v. v is actually
the speed of the light
in some specific material.
And k hat is actually
the direction
of propagation across
E. This will give you
the magnitude and
also the direction
of the corresponding
magnetic field
for the electromagnetic waves.
So now I'm interested
in what would
be the resulting intensity, I,
if I try to superimpose this,
to try to put together
these two electric fields.
As you can see, these
two electric field
have different phase.
The first one has phase phi 1.
The second one, has phase phi 2.
So this means that they
may reach maxima or minima
at different position in space.
In this case, it's
in the z direction.
And what I'm actually defining
here is two plane waves.
And so it really
depends on what would
be the relative phase
for the first and second
electromagnetic wave.
In order to quantify how
much they cancel each other
or how much they enhance each
other, what I'm going to do
is to evaluate the
intensity of the resulting
electromagnetic waves.
And what is actually intensity?
Intensity is actually
the amplitude
of the pointing vector.
So I write this as the length of
the S vector, pointing vector.
And I can, of course, calculate
what will be the value of this
or, say, the length of
the pointing vector.
This will be equal
to 1 over mu 0,
based on these equations here--
just a reminder.
And I would like to know what
would be the length of the E
cross B field.
That will give you the length
of the pointing vector.
And basically, what you're going
to get is 1 over mu 0 times--
since the B field is
actually highly related
to the electric field, and
it takes a hit of 1/v, right,
in terms of the size
of the amplitude.
So basically, you're going
to get 1/v, E squared.
And this cross product
actually is OK,
because it become E squared,
because B field and the E field
are always orthogonal
to each other.
And I can now rewrite
this v. v is actually
the velocity of the
speed of light in matter.
So basically, what
I can rewrite is
that will become
1/v will become c/n,
which is actually the refractive
index of a specific material.
And still I have E squared here.
Finally, I can
rewrite this formula,
since c is equal to 1 over
square root mu 0 epsilon 0.
Therefore, I can
rewrite this expression
in terms of epsilon 0.
And what I'm going to get
is c times n time epsilon 0,
There E squared.
So basically, what I do is
I multiply both numerator
and the denominator
by c, and also
I actually use this expression.
Then I can actually
cancel the mu 0
and then write everything in
terms of c, n, and epsilon 0.
So until here, there
was basically no magic.
Basically, it's just
rewriting the length
of the pointing vector
in terms of n and also
the electric field.
So now, what I am going to do
is to plug in this expression
into that formula and see
what we are going to get.
So let me evaluate what
would be the E squared,
the length of the E squared.
So basically, the definition
of the E is shown here.
It's the superposition
of E1 and E2.
Therefore, I can now
quickly write down
what would be E squared here.
Basically, you are going to
get A1 squared, cosine squared,
omega t minus kz plus phi 1.
Basically, that is coming
from the E1 times E1.
The second term, which
I'm going to get,
is actually E2 times E2.
E2 times E3, you are
going to get A2 squared,
cosine squared, omega
t minus kz plus phi 2.
Basically based on this
equation, and I square it,
and then I get the
second part here.
And finally, what
I am going to get
is the third term, which
is actually E1 times E2.
Basically, you are going to
get 2 times A1 A2 cosine omega
t minus kz plus phi 1
cosine omega t minus kz
plus phi 2, which is actually
the cross term of this E
vector squared.
Any questions so far?
I hope this is pretty
straightforward to you.
And of course, I can
now rewrite this product
with 2 cosine, cosine
times cosine, right?
Basically, I can
rewrite this using
the formula, which
we have related
to a cosine times a cosine.
Basically, I can rewrite this as
1/2 cosine 2 omega t minus 2kz
plus phi 1 plus phi 2 plus
1/2 cosine phi 1 minus phi 2.
So basically, the
first term is actually
collecting the content
of the two cosine
and their length together.
The second term is
actually calculating
the difference between the
content of the cosine function.
And what I'm going to get is
actually phi 1 minus phi 2.
Now, based on this
definition, intensity,
I is actually equal to
the magnitude of that
pointing vector.
So remember, our
goal is to evaluate
what will be the resulting
average intensity.
So now I can calculate what
would be the average intensity
over one period.
So this will be equal to
1/T, integration over 0
to T, one period, and the
instantaneous intensity, I, dt.
And what I am going
to get is that--
so we have three terms.
The first term is here,
which is actually A1 squared,
cosine squared.
It's actually
related to omega t.
And the second term is here.
It's also proportional to
cosine squared omega t.
And finally, we
have two terms here,
which is actually proportional
to cosine 2 omega t.
And finally, the last term is
actually independent of time.
So what I'm going to do is to
evaluate the individual terms.
For the first
term, basically, A1
squared, cosine
squared, omega t.
So by now, it should be
pretty straightforward for you
if I integrate cosine squared
over one period of time,
basically, what you are
going to get is 1/2.
So is actually done
several times in the p set.
So basically, what
you're going to get is--
I am going to collect all
of those constants from here
and copy here.
So basically, you have c
times n times epsilon 0,
which is actually coming from
the definition of intensity.
And then for the first
term, what I am going to get
is A1 squared divided by 2.
This 1/2 is actually
just an integral
related to cosine squared.
Similarly, you are going to get
the same result, a very similar
result, for the second term.
The second term is going to give
you A2 squared divided by 2.
Finally, you can
have the third term.
The third turn is going
to give you what value?
Can somebody help me?
AUDIENCE: 0.
YEN-JIE LEE: Yes, 0, right?
Because this is actually
cosine 2 omega t, right?
So if you integrate
over one period,
you are going to get 0 plus 0.
Each period will
give you 0, right?
So 0 plus 0 is 0, so
therefore you get 0.
Very good.
How about the last term,
anybody can help me?
AUDIENCE: It should
remain as it is.
YEN-JIE LEE: That's right.
Because it's a constant.
So the average of a constant is
a constant, which is actually
giving you 1/2 times 2
times A1 times A2 cosine
phi 1 minus phi 2.
Of course, I can
cancel this 1/2, which
is actually coming
from here, and the 2,
which is coming from here.
And basically, you are getting
A1 A2 cosine phi 1 minus phi 2.
And I need to
close this bracket.
Any questions so far?
So what we have been
doing is that I evaluated
the total electric field.
I basically calculated
the superposition
of the two fields.
And then I am interested in what
would be the average intensity
coming from this field.
And I write down E
squared explicitly.
There are four terms and only
three of them actually survive.
And basically, the expression
I'm getting is like this.
Basically, you have
some constant multiplied
by A1 squared over 2
plus A2 squared over 2
plus A1 A2 cosine
phi 1 minus phi 2.
You can see that the intensity
depends on phi 1 and phi 2,
right?
So this actually would change
the resulting intensity.
So in order to get some idea
about what does that mean
and also how does the
average intensity change
as a function of phi 1 minus
phi2, what I'm going to do
is to define phi 1 minus
phi 2 to be delta, which
I will call phase difference.
Then I would like to plot
the averaging intensity, I,
as a function of delta and
see what's going to happen.
So this is actually
the result. So if I
have the x-axis
to be delta, which
is actually phi 1 minus phi 2,
and the y-axis is intensity.
Of course, I would like to
take out the constant, which
is c times n times epsilon 0.
So I am plotting the y-axis'
average intensity divided
by c times n times epsilon 0.
What I'm going to
get is something
which is actually oscillating
up and down, like this.
The maxima value happens
when delta is equal to 0.
When delta is equal to 0,
what is going to happen?
This means that cosine
delta is equal to what?
, Therefore what you are going
to get is A1 squared over 2
plus A2 squared
over 2 plus A1 A2.
This is actually when
there delta is equal to 0.
And the intensity,
as you reach maxima,
and the maxima values
is actually 1/2
A1 plus A2 squared, based
on these calculations.
On the other hand,
you can expect
that that intensity
will reach a minima when
delta is equal to which value?
Anybody can help me.
AUDIENCE: Pi.
YEN-JIE LEE: Pi, yes.
When delta is pi, what
is going to happen?
Cosine pi is minus 1.
So therefore, what you
are getting is 1/2 A1
squared plus A2
squared minus 2 A1 A2.
And that will give you
1/2 A1 minus A2 squared.
You can see that, when
the filter is equal to 0
or when the filter is
equal to 2 pi, for example,
if you increase the phase
difference large enough,
or the filter is actually
4 pi, all of those number
will keep you maxima
constructive interference.
So what does that mean?
That means you are adding
these two electric fields
in the most efficient way.
On the other hand, when
the value of the filter
is equal to pi or equal to 3
pi or equal to 5 pi, et cetera,
the intensity, the average
intensity reaches a minima.
That means, instead
of adding them,
you are actually canceling them.
You are canceling
the electric field
of the first and the second
electromagnetic wave.
And now I give you a maxima
intensity, which is 1/2 A1
minus A2 squared.
Just a reminder, this
A1 and A2 is actually
the amplitude of the first
and second electric field.
So what will happen if
I set A1 equal to A2?
If I set A1 equal to A2,
that means the minima
would be equal to what?
AUDIENCE: 0.
YEN-JIE LEE: 0, yeah, very good.
How about the maxima?
AUDIENCE: [INAUDIBLE]
YEN-JIE LEE: It will
be A1 plus A2, right?
So basically, you are going
to get four times larger value
compared to the intensity
before you add them together.
So individual intensity is I.
And after adding them together,
with delta equal
to 0, you are going
to get four times
larger intensity
if the amplitude of the first
and second electric field
is the same.
So very good.
So that's actually the
result of the calculation.
And you can see that the amount
of intensity we can get out
of this highly depends on
the filter, which is actually
the phase difference
between the first
and the second electric field.
Can we actually get some more
feeling about this addition?
So what I am going
to do is to, again,
write everything down in
terms of imaginary number
or, say, a complex number.
So if I rewrite the
electric field, E1,
as a real part of A1
exponential i phi 1,
exponential i omega t minus kz.
In the x direction.
And I can also
rewrite the expression
for the second
electric field to be
the real part of A2 exponential
i phi 2, exponential omega
t minus kz, again,
in the x direction
So if I add these two fields
together, what I am doing
is like in the complex plane
I have an imaginary number
contribution in the y direction.
And the real part is
actually in the x direction.
Suppose omega t minus kz is 0.
At some instant of time omega
t minus kz is equal to 0.
So what I'm doing is I have the
first vector, which is actually
presenting the contribution
of the first electric field.
And this electric field is going
to be pointing to a direction
phi 1 away from the real axis
with amplitude equal to A1.
This is actually what we
learned from the first lecture.
And if I add the second electric
field, what I'm going to get
is another vector, which
is actually A2 in length.
And the angle is phi 2, here.
So the resulting
amplitude is actually
when I take the real part of
the first and second expression,
adding them together.
Basically, I am taking a
projection to the real axis.
And that is actually
the resulting amplitude
of the electric field, which
is actually the superposition
of the first and second field.
So you can see
that, when phi 1 is
equal to phi 2, what is going
to happen is the following.
So basically, what you're
actually going to get
is that you are
increasing the length
of the routing vector, which
is the addition of the two
vectors.
You are actually getting a
maxima out of this addition.
Because phi 1 is equal to phi
2, therefore, these two vectors
form a straight
line, therefore, you
can actually add and
get maximum amount
of the amplitude out of this.
On the other hand,
when phi 1 minus phi 2
is pi, which I define as
delta, when this happens,
what we are doing is like
addition of two vectors,
in a complex plane,
but they are pointing
to the opposite direction.
So the first one
will be like this.
And the second one will
be looking like that.
And they are actually
trying to cancel each other.
So that's actually how you can
understand what is happening
with different delta values.
In this case, delta
is equal to 0.
The phase difference
is equal to 0.
And in the second case, phase
difference is equal to pi.
And then what happened
in between is like this.
You are adding them
sort of together but not
in the most efficient way
or the most destructive way.
And you are actually
evaluating what
will be the resulting
amplitude by looking
at the vector sum of the
first and second field.
So I hope that this will give
you a some more intuition
about what we have been doing.
Any questions so far?
So now, we are actually
in a very good position
once we understand
this superposition
of the two electric fields
and the interference.
Basically, the size of
the resulting intensity
will be highly dependent
on the phase difference
between the two fields.
Then we are in a
very good position
to discuss the phenomenon
which we just see in the demo.
So before I actually
perform the calculation
and give you the
explanation, I would like
to take a vote, as of usual.
So the question we are
asking is, in addition
to what we see in the demo--
we see a colorful bubble--
how thick is the soap film
such that you can see color
from the reflected light?
The first option is maybe
it's like 1 millimeter, which
is possible.
And that is about the
size of the head of a pin.
Or it can be 100 micron, so
that's actually about the size,
the thickness is about the
size of the human hair.
Or 100 nanometer, which
is the size of the virus.
How many of you
think the thickness
is roughly 1 millimeter?
Raise your hand.
Nobody thinks so.
Really?
Actually nobody think
that's the case.
How about 100 micron?
How many of you think so?
How about 100 nanometer?
Me How many of you?
So that is actually the vote.
And we are going to know
the result very soon.
And how about the rest?
Cool.
So now we are going
to solve the puzzle.
So just a quick
reminder about what
we have learned from
the last lecture.
So there is a reason why
we have the lecture first
on the reflection of an
electromagnetic wave before we
discuss the color of the bubble.
So from the last
lecture, suppose
I have two materials, which form
an interface between material
number 1, with
refractive index n1,
and the second material
has a refractive index n2.
If I have an incident
wave, incident
electromagnetic plane wave,
and the incident angle
is actually, in this case, 0,
that means this incident plane
wave is actually propagating in
a direction which is actually
hitting the surface directly.
So if the initial
amplitude is A,
what we have learned
from last time
is that there will be a
reflective wave, which
is actually R times
A. R is actually
reflective coefficient.
And finally, you have
also the transmitted wave,
which I call T times A, where
is the transmission coefficient.
From the exercise, which
we actually already
done last time, R is equal to n1
minus n2 divided by n1 plus n2.
And the transmission
coefficient is actually
T equal to 2n1
divided by n1 plus n2.
So basically, what I am
actually talking about here
is a conclusion
from the exercise
we have done in the last
lecture, just a quick reminder.
So I would like to discuss with
you various situation related
to R value.
So the n1 and the n2 are
related to the property
of the first medium
and the second medium.
So it could be that n1 is
actually greater than n2.
So if n1 is greater than n2
in the experimental setup,
that means that R will
be greater than 0.
Because R is
actually n1 minus n2
divided by the sum of n1 and n2.
Therefore, what I'm going to
get is something like this.
So basically, I'm going
to have an incident
wave like this, where, say,
I use the notation pointing
upwards.
Once they got reflected, it
is actually still like this,
pointing upward, because the
R is actually greater than 0.
There's no changing
sign in the amplitude.
Therefore, there's an
no flip in amplitude
if n1 is actually
greater than n2.
On the other hand,
the transmitted wave,
if you look at the functional
form of the transmitted wave,
and transmission
coefficient, T is actually
equal to 2n1 divided
by n1 plus n2.
It's always positive.
Therefore, will there be any
possibility to flip the sign?
No.
You are absolutely right.
So therefore, what
is going to happen
is that I will use this
little arrow to keep
track of the sign change.
Basically, you'll see that after
it pass through the boundary,
there will be no change in
sign in amplitude no matter
what happens.
On the other hand, if I have the
situation n1 smaller than n2,
what is going to happen?
If you calculate the R value,
it will be negative, right?
In this case, R will
be smaller than 0.
So what is going to happen is
that, initially, the incident
wave has positive amplitude.
And I keep track of the
sign of this amplitude
by this arrow pointing up.
Because the R is
actually smaller than 0,
therefore, there is flip
in sign in the amplitude.
So what is going to happen?
So the reflective wave
will look like this.
And I use this arrow to keep
track of the flip in amplitude.
And finally, as I
mentioned before,
the transmitted wave, the
T, is always positive.
Therefore, there will be no
change in sign in amplitude.
Finally, the third
example is, if I
have somehow two
different materials,
but they have the same
refractive index, what
is going to happen is that
there will be no reflection,
and everything goes through.
Even if you have two different
kinds of material, but if they
have the same refractive index,
then what is going to happen
is that everything
will pass through.
And what you are going
to get is that you
will have no reflected light.
Meaning R is
actually equal to 0.
Any questions so far?
I would like to make
sure that everybody
understands the consequence
of this calculation.
So if I introduce no
flip in amplitude,
this means that this
contribution will
introduce a filter equal to 0.
So basically, there will
be no change in the phase,
because there's no
flip in amplitude.
On the other hand, if there's
a changing sign in amplitude,
what would be the
resulting filter value?
Can somebody actually tell me?
AUDIENCE: Pi.
YEN-JIE LEE: It would be pi.
Very good.
So that means you
are getting hit
by a phase difference of pi.
Therefore, the amplitude
changes by a factor
of cosine pi, which is minus 1.
So that is actually
something pretty important
when we have the discussion
of the soap bubble reflection.
So let me give you
a quick example
about why is actually the amount
of the reflected light and also
what is the amount
of transmitted light.
Let me give you a
concrete example.
For example, if I have n1
equal to 1, which is actually
the refractive index of the
air, and n2 equal to 1.5
If that happens what is
the resulting intensity?
Just a quick reminder, average
I, the average intensity
will be equal to c times
n times epsilon 0 A
squared divided by 2,
where A is the amplitude
of the electric field.
Just a quick reminder.
And this 1/2 is coming from the
time average, just a reminder.
So now I can go ahead and use
these two formula, R and T,
to calculate the
reflection coefficient
and this transmission
coefficient.
So R will equal to 1 minus
1.5 divided by 1 plus 1.5.
So basically, you get
minus 0.5 divided by 2.5.
And that is actually going
to give you minus 0.2.
Of course, I can
also calculate what
will be the T, which
would be a 2 divided 2.5.
So basically what you
are getting is 0.8.
So I can now calculate what
will be the resulting intensity
of the reflected light.
Everybody's following?
So what would be the intensity
of the reflected light?
This will be equal to
minus 0.2 squared, right?
Because the average intensity
is proportional to A squared.
A is actually the amplitude
of the electric field.
R should tell you
what is actually
the relative amplitude
between the reflected light
and the incident light.
Therefore, you are getting
hit by 0.2 squared multiplied
by the initial intensity.
Basically what you
are going to get
is 0.04 I, initial intensity.
So basically 4% of the
light is reflected.
That may surprise
you a bit, right?
Because when you see, for
example, the soap bubble,
you see that it is still
pretty bright, right?
But in reality, only 4% of the
light or 4% of the intensity
got reflected.
That is because your eyes is
actually having nonlinear.
Your eye responds to the--
or, say, receiving or
interpreting the intensity
is really highly nonlinear.
So basically, you
get 4% reflected.
And the rest actually
goes through.
And just to convince you that
the total intensity is 100%,
we can calculate what
would be the intensity
of the transmitted light.
This will be equal to 1.5--
this is actually related to
n2, because the intensity
is proportional to c n
epsilon 0 A squared over 2--
times T squared.
So basically you
have a 0.8 squared
and the I initial intensity.
And if I calculate this
value, basically you
are going to get 96% of
the initial intensity.
So 96% of the initial
intensity actually passes
through the boundary and
continues and propagates
in the second medium, which,
actually, in this case,
is the soap.
So the picture is the following.
When 100% of light
intensity going
towards the boundary,
what is going to happen
is that 4% of the
light got reflected.
4% of intensity got reflected.
And also, because n1
is smaller than n2,
therefore, there is a flip
in sign in the amplitude.
And the rest
continues, 96% of them.
And there is no flipping
sign in the amplitude.
Any questions so far?
We're really pretty close.
So now we are in a position to
discuss what is actually really
happening to this soap bubble.
So I'm going to keep
this result here.
And I will now discuss
a situation in which
you have two interfaces.
So suppose I zoom in, zoom,
and zoom in this soap bubble
and put it on the board.
So this is actually
the soap film.
And I have now an incident
wave, which is actually
going into this bubble.
So now I have 100%,
which is actually
going toward this film.
So after this light,
this plane wave
hits the film, what
is going to happen?
The first thing which
happens is that there will be
4% of the light got reflected.
n2 is equal to 1.5.
It's the same setup,
just a reminder,
just to make sure everybody
is on the same page.
So 4% of the light
got reflected.
Of course, the sign changed.
96% of the intensity
actually continue.
And what is actually
happening is
that there will be no
change in amplitude in sign.
And this is actually not
the end of the story, right?
Because the light will continue
and continue to propagate.
What is going to
happen is that it
will reach another boundary,
where the incident light is
you're traveling, from n2
refractive index material,
to n1, which is
actually the air.
Now I have a situation
where the light is actually
going through the boundary
and going out of the air.
That means the light is
actually going into the bubble.
So this is actually
inside the bubble.
So what is going to
happen is the following.
Basically, the calculation
is the same, except that now
the R is actually 0.2
instead of minus 0.2, right?
Because now n2 minus
n2 is actually 0.5.
0.5 divided by 2.5
is positive 0.2.
So basically, what
you are going to get
is a reflected light, which
actually doesn't change.
It doesn't change the
sign of the amplitude.
And what is actually
the intensity?
The intensity will be 96%
times 4%, because only 4%
of the light got reflected.
And of course, a large
fraction of the light
actually pass through the
bubble, 96% times 96%.
And this would be,
again, pointing upward,
because T is always positive.
Any questions so far.
You can see that this
actually really interesting,
because most of
the light actually
pass through the bubble.
So that's actually
already one thing
we've learned from
this exercise.
Now, what is going to happen
to this light if I continue
and increase the time?
What is going to happen
is that this reflected
light, from the second
surface or second boundary,
will go backward and pass
through the first boundary
again.
What is going to happen
is the following.
So basically, we're going to
get, again, transmitted light
and the reflected light.
What will be the sign
of the reflected light?
Will the arrow be
pointing up or down?
AUDIENCE: Up.
YEN-JIE LEE: Up,
yeah, very good.
Right now, if you
are bored, then that
means I am very successful.
So that means I'm getting
4% times 96% times 4%.
What would be the sign
for the transmitted light?
Pointing up or down?
AUDIENCE: Up.
YEN-JIE LEE: Up.
Very good, so everybody gets it.
And 96% pass through, 96%
times 96% times 4% will pass.
And of course, I can now
continue and continue.
What is going to happen is
that now you have learned 8.03.
You will see that this
is a crazy phenomenon.
What is going to
happen is that there
will be a tiny fraction of the
light which is trapped forever
between the two surfaces.
They are going to be
bouncing back and forth,
boo, boo, boo, boo, boo,
boo, boo, boo, forever.
Of course, the fraction
of the intensity
is really, really small.
Because every time you've
got the reflection actually
happening, you take a hit of 4%.
But since we are talking
about theoretical physics,
so, theoretically, that
would continue forever.
That's actually
pretty interesting.
And going back to
practical situation,
basically, I can safely
ignore any further reflection,
because they are
hitting so hard,
because every time I
get a 4% hit, right?
Therefore, I can ignore
all the other contribution.
And what we are
actually seeing is what?
Our eye is here.
We see the contribution
of the first pass,
which is actually reflected
from the first surface.
The second pass, OK, it pass
through the first surface,
got reflected from
the second boundary,
and pass through the first
boundary, in the second round,
and then, also,
reaching your eye.
So what are we looking at ?
We are looking at
the superposition
of two electromagnetic
waves coming from one,
which is like this, and two,
which is actually like this.
The question now is what is
the thickness of the film?
Now I can define the
thickness or, say,
the width of this film to be d.
Now the question we
are actually asking
is, what would be
the thickness, d,
which is needed such that I can
have constructive interference?
Now the question
becomes really clear.
And we can actually
calculate that
by evaluating the phase
difference between the path
number one and the
path number two.
So now, in order to have
constructive interference,
I need a specific
phase difference.
But before that, I need to
calculate the phase difference
first between path number
one and path number two.
What will be the
phase difference?
The phase difference delta will
be equal to, of course, pi.
This pi contribution is coming
from the flip of the amplitude.
That will actually give
you an pi phase difference.
The second phase difference
is coming from the difference
in the optical path length.
You can see that the first path,
it doesn't go into the film.
It got reflected, directly.
And the second path,
which is path number two,
it takes more
effort or more time
for the light to go
back and reach your eye.
How big is the path
length difference?
The size of the path length
difference is 2 times d, right?
Of course, I need to
actually translate that back
to the phase.
So first, I need to actually
calculate how many period.
So the length divided by
lambda will be the period.
So lambda is actually
the wavelength
of the incident light.
But I am missing a factor here.
And can somebody help me?
Because this lambda is actually
inside the material, right?
So which factor, I'm missing?
AUDIENCE: n2.
YEN-JIE LEE: n2, right?
Yeah, thank you very much.
So basically,
inside the material,
since the speed of light
is 1.5 times smaller
than the speed of
light in vacuum,
therefore, the wavelength is
actually lambda divided by n2.
And this is actually
the number of period.
And now, I need to translate
that to phase difference.
Therefore, I
multiply this by 2pi.
So you can see that now I
have successfully evaluated
or quantified the phase
difference between path number
one and two.
That is there are
two contributions.
The first one is pi.
It's related to the
flip in amplitude.
The second contribution,
the blue one,
is actually coming from the
optical path length difference.
And of course, we can evaluate
that really precisely.
Therefore, we can
now quickly conclude
that, in order to have
constructive interference,
I need to have filter equal to
2N pi, where N is an integer.
And in order to have
destructive interference,
I need to have
filter equal to 2N
plus 1, pi, which is
actually the result
of the calculation which we
have done, I think, before.
Yeah, there.
So this is actually
based on the calculation
we have done in the beginning.
So we are really close.
So now we have this result,
delta is equal to pi plus 2d,
times 2 pi divided by
lambda divided by n2, right,
so this complicated formula?
Now we are in the
position to evaluate what
would be the phase difference.
So the first thing which
I would like to discuss
is that, when d goes to 0, what
does is actually the limit?
The limit is when
the width of the film
is really, really
small, it goes to 0.
What is going to happen?
You are going to have
destructive interference.
Why is that?
That is because, even when
you have d equal to 0,
the filter is pi because of
the flip in sign in path number
one.
The second thing
is that now I can
calculate what would be the
constructive interference
width.
So this will happen when d is
equal to 2N minus 1 lambda,
divided by 4 n2.
So basically, you can use that
formula there and solve d.
Then basically that's the
formula we are going to get.
And I will not go
into detail with this.
Any questions so far?
So now, the third
conclusion is that, if I
fix d and the change in
lambda, that is actually
the more practical situation.
Because I have the soap bubble.
And it have a well-defined
width, which is d.
And what is happening
is that I am trying
to shine this soap
bubble with light
with different
wavelengths, right?
So that is actually
the third situation.
If I fix the width of the film,
and the change the wavelength,
lambda, what I am
going to get is
that the lambda max, which
is the wavelength needed
to have constructive
interference,
will be equal to 4d n2
divided by 2N minus 1.
So basically, I can solve the
lambda if I am given a d value.
So actually, we
already get the answer
we are asking in the beginning.
The first question is,
why do we see color?
The second question
is, when I see color,
what is actually the
width of the soap film?
We are going to know
the result in a moment.
So now, I have this
formula in hand.
If I have d roughly equal
to 100 nanometer, which
is the third option
we were discussing,
that is going to give you lambda
maxima equal to 4 times 100
nanometer times 1.5--
n2 is 1.5-- divided
by 2N minus 1.
So that is actually
600 nanometer
divided by 2N minus 1.
Suppose I have N equal
to 1, basically I
am getting 600 nanometer.
Suppose I have N equal to 2, 2N
minus 1 is actually 4 minus 1
is 3.
Therefore, you get 200
nanometer and 120 nanometer,
et cetera, et cetera,
which are the required
wavelengths in order to have
constructive interference
between path number one
and path number two.
Everybody is following?
If I plot the spectra
of this lambda max,
assuming d is 100 nanometer,
what I am getting is like this.
So this is a situation
of very thin film.
So this is the lambda.
What I am getting is that there
will be a maxima here, which
is actually 600 nanometer.
Red color is actually
roughly 650 nanometer.
This is red light.
And this is actually
roughly the range
of the visible light, which
is actually between lambda
equal to lambda violet--
violet is equal
to 400 nanometer.
So you can see that the
first maxima, lambda maxima,
where you have
constructive interference
is at 600 nanometer.
So that means you are going
to see what kind of color
in your soap bubble?
AUDIENCE: Red.
YEN-JIE LEE: You are
going to see red, right?
And then the next
wavelength which
you can have constructive
interference is 200 nanometer.
That is actually shorter than
the wavelength of the violet
light.
It's out of the range
of the visible light.
What is going to happen?
Your eye will not see it.
So the next one would be here,
whatever, blah, blah, blah,
blah, which I don't
care, because they
are so short in wavelength.
And you cannot see them.
So you can see that,
if I have a width which
is roughly 100 nanometer,
very same situation, what
is going to happen?
What is going to
happen is that you
are going to get only one maxima
in the visible light range.
And therefore,
you can see color.
Any questions so far?
Now, what I'm going to do is
take the same formula here,
but now I would like
to change this d.
So now I would like
to change the d
to consider a situation where
you have a very thick layer.
So now I would like to change
the situation to a very thick
layer, so maybe I need to
erase this part of the board
to make some space.
So now if I have d equal
to 100 micron, what
is going to happen?
So I can now still
use this formula
to calculate what would
be the lambda maxima.
So lambda maxima will be
equal to 600 micron, which
is when you have N equal to 1.
But this wavelength is way,
way larger, much, much larger
than the wavelength
of the visible light.
So it's not going to work.
Therefore, you
have to be patient.
You have to increase the N
value until N is equal to 500.
So I am calculation 1,
2, 3, 4, 5, 6, until 500.
Ahh, we are in the visible
light range, right?
Now, 5000 will give
you 600.6 nanometer.
Phew.
Suddenly, your eye can see it.
Very good.
That's very nice, right?
So I can now put
it in my diagram.
This is actually the
wavelength, again.
And, ah, I get one line here.
How about the next
one, N equal to 501?
I'm going to get
599.4 nanometer.
It's pretty close to this one.
And the next one would be 598.2
nanometer if N is equal to 502.
And what you are
getting is that you
can see that, no, things
are not going very well.
They are full spectra,
all very, very narrow.
Very, very large
number of wavelengths
can give you constructive
interference.
So what is going to happen?
What is going to happen is that
your eye will see reflection
with all kinds of
different wavelengths.
And what color is that?
AUDIENCE: White light.
YEN-JIE LEE: White!
You are going to see
something which is white.
So that is actually the
answer to our question.
So what would be the
required thickness?
The required thickness is
something like 100 nanometer.
So we can see how thin
is the water bubble.
That may surprise
some of you, right?
Most of you actually
didn't think
that's actually that thin.
Secondly, if d
equal to 0, you are
going to have
destructive interference.
That means there will
be no reflected light.
Everything is going
to pass through.
And the bubble is
like transparent.
Finally, when the
bubble is really thick,
you are going to see white.
So let me finish this lecture
with a demonstration here.
First, before I
turn off the light,
I would like to turn this on.
So what I have here is a
very complicated machine.
It's not that
complicated, actually.
So basically, I have
a light source here,
which emits light with all
kinds of different wavelengths.
And I have this
little device here.
There's soap solution inside.
And I can actually
rotate from outside.
You see how sweet is this setup.
And I can actually create
a soap film out of this.
You can see that I am rotating
and trying to actually project
the result on the wall.
You can see that, initially,
there's nothing really
striking in the beginning.
You can see that the
light is which color?
AUDIENCE: White.
YEN-JIE LEE: It's white, right?
Remember, because of
the optical setup,
this image is
actually upside down.
So the upper edge of
that image is actually
the lower edge of
my setup, which is
the lower edge of my soap film.
We have gravity, right,
so that I can walk around.
Due to gravity,
you can see that it
will form a thicker
and thicker layer
in the bottom of my
experimental setup
or in the upper
edge of the image.
On the other hand,
due to gravity,
the upper edge or the
lower edge of the image
will become thinner and
thinner as a function of time.
At some point, the color
will start to show up.
As you can see now, since
we wait long enough,
the soap film becomes
thinner and thinner.
And you can see that there
are colors popping up.
It's like a rainbow.
Why is that?
Because I am varying the
thickness of the film
as a function of the
vertical distance.
So therefore, you can see that
this is actually showing you
that different d value will
give you very different colors.
And if we wait long
enough, basically what
we are going to get is
that the whole film will
become more and more colorful.
And I am sure that
after this class,
you can walk out
of the classroom
and explain to your friend why
the soap bubble is colorful.
Thank you very much.
And if you have any
questions, I will be around.
And of course, if you want
to make your own soap bubble,
you can actually go ahead
and play the demo here.
So hello, everybody.
So we are going to show
you a demonstration, which
we can see colorful interference
pattern from a soap film.
So basically, the experimental
setup up is like this.
Basically, we have
light, which is
trying to shine this
thin layer of soap film.
And the result is actually
projected on the screen.
And you can see, at
first, you don't really
see a lot of colorful
pattern, because the thickness
of the film is still
rather large, rather thick.
Therefore, you don't really
see a lot of pattern.
But as a function
of time, you can
see that this pattern
is actually changing.
Because of
gravitational force, you
will be able to see that
the lower part of the film
becomes thicker and thicker.
And the upper part of the film,
which you see that upside down
on the screen, is actually
becoming thinner and thinner.
As we actually discussed during
the class, when soap film is
thin enough, there will be
only one or only a few maximas
in the interference pattern
as a function of wavelengths,
which happened to be inside
the visible light range.
And you can see it now,
already, this colorful pattern
really develops.
It's really beautiful.
And you can see that, in the
lower part of the experiment,
really have authentic
color, because there
are multiple maximas in
the visible light range.
On the other hand, in the
upper part of the film,
you typically have very
little number of maximas
or only have one maxima,
in the visible light range,
as a function of wavelength.
Therefore, you see
really, really dramatic
and very, very colorful pattern
develop from this experiment.
