.
Prof: I dropped a
bombshell end of last class telling you that all the
geometric optics I taught you has to be revised,
and before letting you go for lunch I said you've got to tell
me why I would do that. Student:
Experiments. .
Prof: Yes,
somebody said it's the experiments, and that's what I
want you to remember. That's the only way things
change in physics is experiments tell you something's wrong with
your theory, and in the end if your theory
doesn't agree with the experiment it's over.
And conversely,
if you're a newcomer unknown to anybody, make predictions and
they agree with experiment then you're a rock star.
Everything is based on
experiment. That's the only way we change
our mind. Now you might say,
"Why do you keep doing this to us?
We believe you.
We write everything down.
We do the problems and then you
say, 'Oh, these guys are wrong. Here's a better theory,' what's
going on? Are people really wrong?"
So I have to be very careful
when I say people are wrong because the news leaks to the
press and the media will say, "Physicists think they're
always wrong." In fact, I've gone on record
here saying, "We're always wrong."
What I mean by that is no
matter how many laws we find one day you will find some new
experiments which are not explained by these laws.
That's not really bad news.
That's what keeps us in
business because we want to find something that doesn't fit
anything we know. For example,
Newtonian mechanics in some sense is wrong because it
doesn't work when velocities approach the speed of light,
but it's not wrong in the sense that the predictions it made in
its proper domain do not work anymore.
It was supposed to work in a
limited range of experimental observations.
If you push the limit,
if you build accelerators that send particles at very high
speeds you may find they don't obey Newtonian mechanics.
Then you get Einstein's special
theory. Now if you have a special
theory like Einstein, or a new theory that overthrows
the old theory, and it explains new phenomena,
there is still one extra requirement.
Can you guess what that might
be with a new theory? One thing we demand of every
new theory, yes? Student:
That part of it coincides with the old theory?
.
Prof: Very good.
If you go back to the old
experiments, which are explained by the old theory,
well your new theory should continue to explain that.
In fact the new theory,
when it's a good one, will tell you why for hundreds
of years people fell for the old theory and then it'll also do
new stuff. If you do new stuff and it
doesn't fit the old stuff that's not good.
So relativity is just like that.
It works for all velocities up
to the speed of light, but if you let v/c go to
zero then you will finally get back Newtonian mechanics.
Similarly, quantum mechanics
works for very, very tiny objects,
the atomic scale, but when you apply it to big
things you'll find that the world begins to look Newtonian.
Similarly, the end of geometric
optics came from a series of experiments,
and I will tell you why we didn't realize there was
something wrong with it for a while.
So here's an experiment you
could do. Here a screen and there's some
light coming from here, and there's a hole,
and behind this screen is another screen.
So I'm not good at drawing,
so you should imagine this is coming out of the blackboard,
and the ray of light is coming in.
It'll form a shadow.
I happen to have in mind a
circular hole here. It'll form a disc shape exactly
like that. And another thing,
but you will find that as you shrink this, make it smaller and
smaller, suddenly that won't be true anymore.
You will find that the light
spreads out like that to bigger and bigger areas.
So it's no longer forming the
geometric image of this aperture.
It's fanning out,
so it's not going to come from geometrical optics.
That's one experiment where
something goes wrong. There's a more dramatic
experiment which really puts the nail on all kinds of ray
theories, and that's the experiment involving
interference. So you take a partition with
two holes in it, and you send light from here,
and you put a screen behind. Are you guys with me?
This is a cross section of a
plate coming out of the blackboard.
It's got only two holes in it.
Or here's a better way to think
about it. This is the top view of an
experiment. I'm looking down at my lab from
the top. There's a wall with two holes
in it, and there's the back wall, and the light is coming
from here. If I close this hole what we
will find typically--let me make them slightly bigger to make my
point. You'll find some light.
I didn't draw it properly.
Most of the brightest areas
will be in front of this hole. So let's see.
It'll tend to be big here and
kind of fall off. Now if you block the other one
and you shine light here you get something like that.
It's like two windows in the
room. You've got one window,
you get some sunlight. You've got another one you get
another sunlight. Now the question is if you open
both of them our expectation is that there is some energy coming
here, some energy coming here,
and if you open both you should get the sum of the two energies,
which I'm assuming this is the sum of these two graphs plotted
here. By the way, do you know how I'm
plotting this? I hope you understand that this
graph really means that distance is proportional to how bright
this point is. See, normally you're used to
drawing graphs like this as a function of distance,
but I've turned it around so that that is the graph of the
brightness at different points on the screen.
That's what you expect.
You open two windows you get
the light equal to the sum of the two.
For example,
if you're feeling warm it'll be warmer in this region because
you're getting light from two sources.
But this experiment also gives
you different results under certain conditions.
I'll tell you what the
conditions are. What you find then if one slid
open it looks like that. If the other slid open it looks
like that, but with both open it looks like this.
That's called interference.
The main point to notice is
there are some places, like right here,
where you get less light with one more slid open,
okay? And more light with only one
open, less light with two open. In fact, you can arrange it so
this is really zero. I mean, it can come all the way
down here. In other words that's the point
on the screen which used to be bright with one slit open.
You open the second one that
point becomes dark. That means if you're warm
there, and you tell somebody, "Hey, this is good,
open another window," suddenly you get nothing.
That happens.
So when did people realize this
is what happens, and why did people fall for the
other graph that looks like this before?
The answer is this.
In order to see this phenomenon
in which it instead of being a shadow like this you get a huge
spread, or instead of being simply the
sum of these two featureless graphs it starts oscillating,
and a certain condition has to be satisfied.
The condition is the following.
This light, if it has a
wavelength λ, the λ, if it is much,
much smaller than any distance in the problem then you get
geometrical optics. So in this example if this slit
is thousand times the wavelength of light then the shadow,
in fact, will be geometrical, but as you make the slit
smaller and smaller until it becomes comparable to the
wavelength then you'll find it begin to fan out.
Similarly here,
if this distance between the two slits,
d, is much, much bigger than the wavelength
and each slit is much, much bigger than the
wavelength, then you will find that you don't see any of this
interference effects. Because what happens is
interference effects are always there, but they oscillate like
mad on that screen. Suppose it oscillates like
this, and this could be 10,000 oscillations per centimeter,
and your eye is able to look at it only one millimeter at a
time, then you're averaging over a
region and you only find the average of this graph.
That looks like what you got
without interference. So in order to see interference
you need to make your apparatus probe the physics at a distance
scale comparable to the wavelength.
So that's the answer.
We will see all the details,
but it'll turnout that if you look at light using any
equipment, maybe hole in the wall,
two holes in the wall, whatever, whose dimensions are
much bigger than the wavelength you will not see the wave theory
of light. You'll be able to get away with
the geometric theory. But once you probe light at
length comparable to wavelength all these funny things will
happen. That's why people didn't
realize all these things about light.
For example,
the wavelength of light, typical light,
is 5,000 angstroms, or 500 nanometers,
that is 500�10^(-9), or 5�10^(-7) meters.
That means if you forget the 5
there are 10 million oscillations of the wave within
1 meter, so if you have an instrument
that can pick up that kind of fine features then you will see
the wave nature, but normally you won't.
So it's only when people probed
light with very, very tiny holes placed very
close to each other they saw it. The second important thing to
remember is the following. Even if you made the holes very
small, and if you shine white light on this,
you will not see any of these patterns.
Why is that?
Can you guess?
I'm going to tell you
everything, but can you guess why?
Yes?
You have a guess?
Student:
Because they're white light.
.
Prof: That's
right. White light is made of many
colors and many wavelengths, so each wavelength will form
its own pattern, but the maximum and minimum of
one guy won't coincide with that of the other.
So where this one wants a
maximum that'll want a minimum, it'll get washed out.
So in order to see this effect
you need monochromatic light sent by a source which is
sending it at a definite frequency and wavelength.
And these are all things you
have to get right before you will see these effects.
Okay, so now let's ask
ourselves, what is the nature of waves or what is the nature of
light that produces these funny effects?
Light, as you know,
obeys a wave equation. Let me call ψ
as whatever it is that's doing the oscillations,
generic name is ψ. For example,
ψ can be the height of this string that's vibrating.
If the string is doing this
from equilibrium ψ is the deviation from
equilibrium. ψ can be E or
B in the electromagnetic case.
ψ can be the height of
water in a lake above some reference level.
That's the reference level,
and the water does some oscillations,
ψ is the deviation from the normal.
In the case of a sound wave,
as you know sound wave is deviations in air pressure.
Again, this graph would stand
for the normal air pressure, ambient air pressure,
and when you talk you make some oscillations in the density or
in the pressure, and there's the deviation in
pressure. So I'm going to call ψ
as any one of these things. It's something that's
oscillating in the wave. And all these ψ's obey the
wave equation. The wave equation looks like
d^(2)ψ/dx^(2) minus 1 over the velocity
d^(2)ψ/dt^(2) = 0. The context is different but
the equation's the same. The velocity will vary from
problem to problem. So I'm going to give a short
name for this. I'm going to all it box ψ
= 0. Turns out it's not a crazy
notation. A lot of people actually use
that. Do you know why they draw the
box? This is for people who have
done somewhat more advanced mathematics.
There's something called del
squared which stands for the three spatial derivatives
squared. This has got the fourth one so
you use the box. Anyway, it just means I'm tired
of writing this. This stands exactly for this.
That's the main point.
Now when the wave obeys the
wave equation there's a very, very important property.
The important property is it's
a linear equation. It's a linear equation which
means if you multiply ψ by a number it also satisfies
the wave equation. Because if I multiply by number
3 I can take the 3 inside all the derivatives,
and I find 3 times ψ is also a solution.
Another important property is
that if someone gives you a solution called
ψ_1, another person gives you
another solution to the wave equation which is some other
function of x and t called ψ_2,
then you can check that if you take any number times
ψ_1, and any number times
ψ_2, and add this,
you will find box of aψ_1
bψ_2 is also 0. You understand why?
When the derivatives come it
doesn't care about a. It's a constant.
Box of ψ_1 will
vanish and the box of ψ_2 will vanish.
The whole thing will vanish.
But the important property is
you can take two solutions, ψ_1 and
ψ_2, multiply each one a number you
like, a constant, and add them,
that's also a solution. That's the principle of
superposition, and its origin is in the fact
that the underlying ψ obeys a linear equation.
It's very important that it's
linear. I've told you many times if you
have something like that plus ψ^(2) = 0 you can verify
that if ψ_1 is a solution and ψ_2
is a solution if you add them you will get
ψ_1^(2) ψ_2^(2),
but that is not the same as ψ_1 ψ_2
the whole thing squared. So for nonlinear equations you
cannot add the solutions. These are linear because they
involve the first part of ψ. Okay, now what that means is
first of all ψ can be positive or negative.
What if someone tried to prove
to you that ψ is always positive?
How will you shoot that person
down? Do you know?
Yes?
Student:
Well, if the constant is negative 1.
.
Prof: Very good.
If the constant is negative 1
then minus ψ is also a solution so no one
can ever convince you ψ should always be positive.
Well, if ψ
can be positive and negative, which means generally it can go
up and down in sign, it cannot stand for things
which are always positive such as the brightness of light.
The brightness of light is
always positive. The worst think you can have is
no light. You cannot have negative
brightness. Therefore, in the electric
field the brightness is not measured by the electric field,
the one that oscillates, but by something quadratic in
the field, and that's called the intensity.
The intensity is proportional
to square of whatever is oscillating.
It is the measure of energy
contained in the wave or brightness.
Intensity being positive
definite has some chance of being a description of something
that's positive definite. Not only does it have a chance
that happens to be the case, okay?
See, in all these problems it's
the square of the function that stands for something like
brightness, something like energy per unit second.
They're all given by the square.
So the important thing to
notice is that when you have one source producing some light,
and a second source producing some light,
and you turn them both on, together they'll produce a
field which is the sum of the two fields.
What you can add is the ψ,
which in this case happens to be E or B.
You cannot add the intensities.
No one told you the intensities
are additive. The correct process is if
ψ_1 has an intensity I_1,
let me call it ψ_1^(2),
and ψ_2 has an intensity I_2
which is ψ_2^(2),
then if you put the two waves together by producing them due
to the two sources, I_1 2 is
(ψ_1 ψ_2)^(2).
That, you see,
has got a ψ_1^(2) and a ψ_2^(2)
2ψ_1ψ_2 which is I_1
I_2 something which is indefinite in sign.
This can be positive.
This can be negative.
This can be 0.
So I_1 2 is
not equal to I_1 I_2.
It's got the extra stuff,
and that's going to be the origin of all the things we see
here. So it turns out for waves
there's two levels as which things happen.
There's a thing that actually
oscillates and then you have to square it to get things that you
normally associate with energy or brightness.
And the additive law applies to
the things that oscillates and not to its square.
This is the reason why if
you've got two sources each one of it produces a positive
intensity I_1. When you turn them both on the
answer is not I_1 I_2,
but this cross term, and the cross term can come and
neutralize these and make it vanish.
That's why the experiment when
you open a second hole some places can become darker than
they are. Now later on,
I mean, not later on, later this week when we start
quantum mechanics we'll deal with the ψ.
That's called the wave function
in quantum mechanics. Now that's a very bizarre
object. I don't want to even talk about
it now. Let me just say that it's
something which is intrinsically complex.
See, normally for
electromagnetic waves or harmonic oscillator if you've
got x = Acosωt we're
used to taking x = Ae^(iωt) then
taking the real part at the end. That everyone knows x is
real. You bring the complex
exponential because it makes it easier to solve certain
equations. But in quantum mechanics all of
ψ, real part and imaginary part together are needed.
It is, by nature, complex.
In fact the equations of
quantum mechanism, the analog of Newton's Laws,
or the analog of the wave equation will have an i
in them. There's an i in the
equation for ψ, so you cannot get rid of it.
So in those cases what you want
to think about the analog of intensity cannot be ψ^(2),
because ψ^(2), if you really mean by ψ^(2)
the real part of ψ i times the imaginary
part of ψ whole thing squared,
you see that is ψ_real^(2) -
ψ_imaginary^(2) 2iψ_real
ψ_ imaginary.
What's wrong with this?
If you want this to stand for
something positive definite this is definitely not it.
Can you see that?
In the simple case where ψ
is purely imaginary you see it's negative,
but neither positive, not even real.
So you must know what I should
do in the case where ψ is intrinsically complex to
define something positive. Any guess on what it is?
Yes?
Student:
Take the ________. .
Prof: Could take
the absolute value of ψ, or another way to say that is
you want to take ψ*ψ. But anyway, I'm doing some of
this today because I want you to be on top of this before
Wednesday. I've given you enough warning.
I ruined your spring break by
sending you notes on complex numbers so you can read them.
If you know it already it's
fine, but you should be able to understand the meaning of real
part, imaginary part,
absolute value squared, things like
E_Iθ, okay?
That's going to be used a lot.
By anyway, if I take
ψ*ψ that becomes ψ_real i
times ψ_imaginary times ψ_real -
i times ψ_imaginary.
And if you do the whole thing
you will find it's ψ_real^(2 )
ψ_imaginary^(2). That's, of course,
never negative. So when you do complex waves
they will also obey the superposition principle for
ψ, but the analog of intensity,
which is the positive definite number you extract from ψ,
will be that. You can also write this as
absolute value ψ^(2). When you do complex don't
forget absolute value. When you do real either you can
square the ψ or you can take the absolute
value. It doesn't make any difference.
For the real function the
simple squared and the absolute value squared are both the same.
All right, so now I'm going to
start with trying to understand interference of waves by doing
the following. Usually you get interference
when two people are sending you a signal.
One person is sending you a
signal ψ_1 and the signal may be traveling in
space, but I want you to sit at once
place and just register what's coming to where you are.
For example,
you could be in one part of a lake and someone is rocking the
boat there and sending out ripples,
the ripples come to where you are, and let's say it's
periodic, and at the location where you
are let it be of the form Acosωt.
So A is the amplitude
and ω is the angular frequency.
If you like,
it's 2Π times f where f is the normal
frequency. Then another wave is coming
towards you. It is
Acosωt, but it comes at a different
phase Φ. Now the phase Φ,
if you've got only a single wave you can choose Φ
to be 0. Because Φ
is--what does Φ do?
cosωt likes to
have a maximum of t = 0, cos(ωt
Φ) has a maximum somewhat earlier.
You can always reset your clock
so that whenever it hits a maximum you press your clock and
set it to 0 you can get rid of Φ.
But you cannot get rid of Φ
from two wave functions. You cannot choose them both to
have their maximum at the same time because they're not.
In fact this fellow goes like
this. This one has got an extra lead
on it, so it'll sort of start here.
So it'll go like this.
If you went backward it'll have
its maximum at some negative time.
So what we want to do is to add
these two signals to get ψ_1
ψ_2. Think of this as water waves.
That's the best example.
Okay, someone shaking the
water, ripples are coming out. They're coming to where you are.
With only the first vibrator
sending the waves you get this. With the second one you get
this. But now if both are doing it
then this is what you will get. And the height of the water
will actually be simply the sum of the two heights.
It's not obvious but true, okay?
If you agitate the water with
one source and it produces the height here,
and with the second one that produces another height,
if you do both it's not obvious, it's not a logical
necessity, but it's a fact that the height
of the water at any point will simply be the sum of these two.
So you've got to add these guys.
So you do that you get
Acosωt Acos(ωt
Φ). Now you've got to go back to your
trig identities and remember the following: cosine α
cosine β is twice cosine α
β over 2 times cosine α - β
over 2. If you've got a formula like
this, of which there are numerous ones,
and if you have any doubts you can check some special cases.
For example,
if α is equal to β I want to get cos α
cos α or twice cosine α.
On the right hand side I get
twice cosine α α over 2,
which is cosine α times cosine of 0 which is 1.
Now it doesn't prove it's the
right formula, but it will tell you that if it
is wrong in some way then you can tell.
Anyway, this is the right
formula applied to this problem. If I apply it to this problem,
I get twice cosine ψ over 2,
I'm sorry, 2A cosine ψ over 2 times cosine
ωt Φ/2. This is just plugging in.
It is an example.
This whole thing looks like a
pre-factor, doesn't depend on time and this is oscillating.
So this is a problem which
looks like some amplitude cosine ωt Φ/2.
The amplitude Ã
is 2A cosine Φ/2. And if I want the amplitude to
be always positive I will take it to mean the absolute value of
cosine Φ/2. So here's the answer, guys.
When you add these two
waves--if you draw the pictures it's kind of confusing,
but this is the simple answer. It says the total will be a new
wave whose amplitude is Ã.
à can be as big as
2A, or it can be as small as 0 depending on what the
cosine does. And then it will also not be in
step with this guy, not in step with this guy,
it'll be half way between them. Its phase will be Φ/2.
Now let's check the formula in
a couple of special cases. If Φ is equal to 0 that
means the two signals are identical.
We don't need any fancy stuff
to know the answer is 2Acosωt.
If Φ is equal to 0,
that's Acosωt,
cos 0 is 1 and you get the 2A.
The one other case you can do
very easily in your head is when Φ is equal to Π.
If Φ is equal to Π
cosine of ωt Π is just minus cosine
ωt. That means this wave starts
like this. Wherever this is going up this
is going, I'm sorry, it starts like this one.
So when that one is doing this,
this will do that and it'll always be opposite to it.
If you add them you'll get
identically 0. Anyway, they are the two
extreme cases where you can check your formula.
So this is the only formula you
will need, so you should get that into your head.
There is some natural frequency
ω and the phase is increasing at the rate ω
so it's ωt. The other guy has shifted by
phase Φ. Their sum has a new amplitude
and a new phase. I'm doing a special case where
the two waves have the same amplitude.
If they're different amplitudes
it becomes more messy, but that's not necessary.
So I'm going to do this once
more, but with complex numbers, because I want you guys to get
used to complex numbers. And I think the only way you're
going to get used to anything is if you use it a lot.
You can sit down and read it at
bedtime. It may help you go to sleep,
but it doesn't do any good. So I'm going to keep using the
complex numbers, because starting Wednesday
we're going to be doing a lot more.
But you don't need very fancy
complex numbers. It's at the level that I'm
doing now. So let us now do exactly this
addition using complex numbers. So let's take one single wave
first, ψ_1 = Acosωt.
We agree there's no complex
number here, but let's write it as the real part
Re(Ae^(iωt)). Now you remember the complex
number is z = mod ze^(iθ) looks like
this. It's got a length equal to mod
z and there's an angle equal to θ.
That's the complex number
z. If you want you can think of it
as that. It starts at the origin and
terminates at the point z.
Its real part is really this
one, this mod z cosine θ.
The imaginary part is mod
z sine θ, but I just want the cosine.
It's the real part of this guy.
So what does this look like?
Well, let me draw you a picture
here. This looks like this.
It's got length A and
the angle is ωt, and as time goes by it goes
round and round. So at any future time it'll be
on this circle. At this instant its real part
is just this one, A cosine
ωt. So if you like,
the complex number has fixed magnitude, goes round and round
in a circle. The real part of it has got a
maximum of A here, a 0 there, and -A,
and that'll be the cosine, because the horizontal part of
this is A cosine ωt.
So we can keep track of the
thing we're interested in by letting this vector rotate in a
circle by looking at a shadow on the x-axis,
okay? If you have some light shining
from here you ask, "What image does it
cast?" Well, it comes there.
So for a single complex number
it's not very useful, but now let me bring in a
second complex number. Let's bring in a second complex
number, so I'll draw a new figure for the two of them.
So you should understand this
is a dynamic situation. This is not a fixed vector.
As t increases this is
constantly rotating, okay?
And as it rotates if you follow
the shadow on the x-axis you get the real part,
which is our function. Now if you've got a second
function, so let me take a circle of radius A,
and here's the first guy who's angle ωt.
The second one is also on that
circle, but you've got an extra angle Φ.
You follow that?
So this ψ_1 this
is ψ_2, actual ψ
is actually the real part of these two guys,
but I'm going to work with the complex number first.
We all remember in the end to
take the real part. And I'm using the fact that the
real part of ψ_1 ψ_2 is the real
part of ψ_1 the real part of ψ_2.
Therefore if you want to add
two real waves you can add the complex waves and then take the
real part. You can take it before or you
can take it after. Our philosophy is going to be
to add the vectors first then take the real part.
I want you to think about,
what is the sum of these two guys going to do?
As usual I'll give you a few
seconds to think about it, okay?
By vector addition you do this
parallelogram. You draw one copy of this here
and one copy of that here, and you join this to that.
That line is ψ_1
ψ_2, so their sum would rotate on
that circle, but let's look at the sum.
What's this angle here?
You can tell from the
similarity of this parallelogram that that angle is Φ/2,
because the angle Φ between them is split evenly
between the diagonal of the parallelogram.
So you already tell that the
new wave, which is obtained by adding these two,
when this guy rotates we'll have a phase of Φ/2.
And what's its amplitude?
That we're asking,
what's the length of the diagonal of this parallelogram?
Then you use whatever theorem,
and to say Ã^(2) is the square of one guy,
square of the other guy 2 times length of A length of
A cosine of the angle between them.
I mean, cosine of this angle,
Φ. So this becomes 2A
squared times 1 cosine Φ. Now I remind you of another
trigonometry identity that 1 cosine Φ is 2 times cosine
squared θ/2. That is Ã^(2).
Then for à if you
take the square root of this it's 2A times cosine
Φ/2. So we just reproduced the
answer, but I want you to understand that the sum of these
two complex numbers is a number of length 2A cosine
Φ/2. It goes around at the same
ω, but it is shifted. Its phase is half way between
the phase of the two waves. That half comes because in a
parallelogram the diagonal will bisect the angle Φ.
So this answer is the same as
what we got from trigonometry. It's not going to give you
anything new, but it's helpful to think of
our actual ψ_1 ψ_2,
if you like, is the real part,
which is the projection on the x-axis,
but it's a lot easier to add these arrows than to add the
real parts. Okay, now I'm going to do the
last part of it one more way, so once more with feeling.
So let me do that,
ψ_1 ψ_2.
This is when I'm not going to
draw any pictures. Some people like to draw
pictures. Some people like to do the
algebra. So I'm one of these algebra
types because when they draw a picture I don't understand
anything. So for fellows like me
ψ_1 ψ_2 looks like
this, ψ_1 is
Ae^(Iωt), ψ_2 is
Ae^(Iωt )^(Φ),
which I can write like this. So let's write it as a common
Ae^(iωt) times 1 e^(i)^(Φ).
This much, I think,
is automatic. Your spinal cord can do this
calculation, but the next move requires a little bit of
planning and we do this. Ae^(iωt) I'm going
to pull half the phase of that e^(i)^(Φ)
^(/2) times e^(i)^(Φ)
^(/2) e^( i)^(Φ)
^(/2). You can see if I open all the
brackets this goes back to being 1.
This goes back to being
e^(i)^(Φ). The reason I do this is that I
want everybody to know this is a famous combination,
e^(i)^ (θ)
e^(-i)^(θ) is twice cosine θ,
so twice cosine Φ/2. So the sum of all these complex
numbers became 2A cosine Φ/2 times e^(iωt
)^(Φ) ^(/2).
You can see that this is the
amplitude. This is the length of the
vector, and this is angle by which it's rotating.
Can you see that?
That's like mod ψ_1
ψ_2, and there's the i times
whatever the phase of ψ_1
ψ_2 is. There are many ways to do this,
but having done it three ways you should at least remember the
answer. When you add two waves both at
the same frequency ω, with a relative phase between
them, their sum oscillates at the
same frequency with the phase half way between the two with an
amplitude which is 2A cosine Φ/2.
Now I'm going to start applying
this result to a variety of problems.
The problem I'm going to do is
the following. I'm going to take--let me start
fresh here. Now I'm going to do the
interference problem. For interference you take these
two slits and some wave come from here.
It hits the two slits.
Now here's some part of physics
that I'm just going to tell you without proving it.
I hate doing that,
but once in a while we have to do that.
I can only appeal to you sense
of logic. That is the assumption that if
you look at this wall which is completely covered except for
two tiny holes, let's say, then any light on
the back, on the other side,
will be in the form of two glowing objects,
and with the two holes in the wall will act like two light
sources. That's how it works,
for example. You open the window,
the window looks like a source of light.
It's coming from somewhere,
but to the people inside the room that a source of light and
that's a source of light. And if the things are very
close to each other, this hole, or very small it'll
look like a point source of light,
and for the point source of light the waves will emanate
like this with that as a center, and another set of waves with
this at the center, and the two will be in sync
because this wave front hits them both at the same time.
And I'm going to sit somewhere
here, and I'm going to measure what's going on.
So you imagine this is water
waves. This is a top view of the water
wave. This is a crest and there's a
trough in between, another crest,
another trough, another crest,
and I'm sitting here and these wave are going by,
so two waves will come to me. One will come from here,
and one will come from there. Let that distance be r.
Let that distance be r
δ. So I am sitting here.
So the water here will be
bobbing up and down by an amount equal to what this guy tells me
to do plus what that guy tells me to do.
So I'm only asking you,
"What is the amplitude of vibrations of the water here
with both of them open?" Well, you've done it many,
many times. If the first one sends a signal
Acos(kr - ωt),
by the way, you have seen waves in one dimension whether it's
cos or sine of kx - ωt.
That's the wave in which the
fronts look like this. When you advance the next the
lines of constant phase or maximum of the cosine are lines
like this. But this is not a plane wave
it's a circular wave. It spreads out in all
directions, and the points of constant phase are circles,
if you like. See, here if x is a
constant you get a certain phase, and as you vary x
at a given instant you'll get the various maxima here.
Here things depend on r.
At a given r the phase
is constant, so if the cosine is maximized at one value of r
it's maximized at all values of r.
So all the crests will lie on
that semicircle on this side, and all the troughs will be
half way in between. And I remind you that k
is related to the family of wavelength by this formula,
2Π/λ. You see that?
If this k was
2Π/λ, kr looks like
2Πr/λ. What that tells you is if to
any given r you add a λ it doesn't make a
difference because you're adding a 2Π to that trigonometric
function. That's the relation between
k and λ, it's inverted.
Okay, that is ψ_1.
ψ_2 has got the
same amplitude cosine kr kδ - ωt because
this path, I'm assuming is longer by an
amount δ, so the r measured from
this hole is going to be r δ.
So what happens when you add
them? It's the same trick we have
done many times. All you have to do is identify
kδ as this number Φ.
That's the phase difference
between the two. And that's the phase difference
because even though these apertures are acting as
synchronized sources of waves it takes this wave somewhat longer
to get to my point than this one,
so there is a delay and that's why it arrives with some shift
Φ. Yes?
So what's the amplitude here?
Well, we don't have to do this
over and over again. I already told you
à is 2A cosine of Φ/2 becomes cosine
of kδ/2. So when is the amplitude big
and when is the amplitude zero? That's what we're asking.
If you go along the
perpendicular bisector here you can see by symmetry that two
r's are equal and δ is 0 then amplitude is 2 times
A. So here the two waves are
completely in step and they give you a wave of double the height.
And double the height means
four times the intensity because intensity is proportional to
square of the amplitude. So this point will have an
amplitude which is 2A. I'm just going to draw that
part there, but now let's go to another place here where
kδ/2 is Π/2. Well, cosine Π/2 is 0.
At that point the amplitude
will vanish, and then you will get no vibration of water there.
Let's understand precisely why
that is true. So kδ/2,
k is 2Π/λ times δ/2 is Π/2
canceling the 2 with the 2 the Π with the Π
I find δ is λ/2.
That means this second wave has
to travel a distance half a wavelength more than the first
one. I think you can see what that
means. I mean, here's a wave going
along. If that guy hits there and
you're half a wavelength behind this guy will hit there.
When you go a half a wavelength
in a wave you're doing exactly the opposite of what you're
doing here. At every instant you can take
any wave you want, take a flash photograph,
go to some point and go half a wavelength you'll find you're
doing minus of that. Because half a wavelength adds
the phase of Π to the cosine or sine,
and sine when you add a Π goes to minus itself.
So that's why you get
destructive interference here because the signal from this one
is telling the water to go up. The signal from that one is
telling the water to go down so it doesn't do anything.
A little later the first one
says go down, second one says go up then also
nothing happens. So you can get two signals that
can actually cancel each other. That's the property of waves.
It's not the property of
intensity. You might naively think,
"Hey, this used to be bright without the second slit.
It used to be bright without
the first slit. Maybe with both it'll be twice
as bright," and that's wrong because what
adds is not intensity. What adds is the amplitude,
the wave is the function ψ, and ψ being positive or
negative has the possibility of getting canceled by a second
source. So one source can cancel
another source in ψ and therefore kill the
intensity. That's why if you now plot this
graph you'll get a minimum here, you'll get a maximum there,
and minimum here, and maximum there.
So, where are the places where
you get a maximum, and where are the places where
you get a minimum? So let's write that down.
You remember the
à was 2A cosine kδ/2.
Therefore, Ã is
equal to 2A whenever this cosine, I mean,
let's take the absolute value of this for the amplitude.
Whenever kδ/2 is
equal to Π, equal to 3Π plus or minus,
plus or minus 5Π and so on that means
2Πδ/λ is equal to Π--
let me see. I'm doing something wrong here.
I'm trying to ask,
when is it a maximum? The maximum is when this angle
is either 0 or Π, that's right,
so why am I getting half a wavelength is what I'm trying to
understand. Okay, delta over 2.
So this is 2Π/λ
times δ/2 = 0. Oh, I'm sorry.
I know what I did wrong.
This whole thing has to be 0.
That's good.
That means δ is 0.
What's the next time it'll be a
maximum? When this whole thing is equal
to Π, so let me see what I get for that.
These two cancel and I get
δ is equal to λ. Can you see that,
2 cancels 2, Π cancels Π,
δ equals λ, so you can have δ
equal to 0, λ plus or minus,
plus or minus. They're all going to be bright
if you're talking about light. They're going to be huge waves
if you're talking about water. On the other hand if this is
equal to Π/2, 3Π/2 etcetera,
plus or minus, then there's going to be dark,
that means that δ is equal to λ/2,
3λ/2 etcetera. So there will be bright and
dark, and bright and dark fringes on the screen.
If you go to the screen here
the midpoint will be bright, but it'll be oscillating like
that many, many times. This was the experiment that
was done by Young. Young did the experiment with
light and showed that it has an interference pattern.
He did not know what light was.
He did not produce
electromagnetic waves, but you don't need to know
that. You shine the light,
you get the dark and bright, and dark and bright,
and I will tell you in a minute how we can find the wavelength
of light from this experiment. You can probably guess,
but I'll do that in a minute. He got the wavelength of light,
but he didn't know what it is that was vibrating,
but he got the wavelength. So let's ask ourselves exactly
where on this screen I will get dark and bright,
and dark and bright. So for that you've got to do a
little more geometry now. So let's do that geometry here.
So here is the double slit.
I want to go in some direction.
I'm assuming the screen is so
far away that the rays can be treated as parallel.
And I want to go at an angle
theta so that these lengths are equal.
This is the extra length delta
here, and this is the distance between the two slits.
So you can see delta is equal
to dsinθ, because this angle θ
between the horizontal and the ray is the same as the
perpendicular of the horizontal and the perpendicular to the
ray. It's at that angle.
So dsinθ is
δ, and that had to be equal to
either 0 or an integer multiple of λ for a maximum,
or equal to plus or minus λ/2,
plus or minus 3λ over 2 half wavelengths for a
minimum. Can you imagine doing this
calculation? If someone says,
"Here is light of some wavelength,
5,000 angstroms or 500 nanometers, the slits are
separated by some microns or whatever,"
then you can say, "What's the angle θ
at which I will get a maximum or a minimum?
Well, you can already do this
calculation. If you wanted the first maximum
you said dsinθ equal
to wavelength. You solve for sine θ
and look at a calculator, it'll tell you what theta is.
Then you can go and see,
"Maybe I can get two lambda,"
then that can also give you another θ.
But remember,
it cannot go on forever because at some point sine theta,
which will be some multiple of λ/d where m is an
integer, if it's bigger than 1 no luck.
Sine theta cannot be bigger
than 1. So you'll get some number of
this maxima then it'll stop. That's because if you've got
two signals coming from here to here only if you go in this
direction will you get the maximum phase,
the maximum path difference if you go straight up,
and if that's not enough to be some multiple of λ
there's no use going any further because theta you can
get is 90 degrees, and the biggest path length you
can get is then the space between the slits d,
and your multiple wavelength should fit into that.
So I hope all of you understand
this very simple experiment, the double slit experiment.
So light comes from the left.
These two points act like
sources of light. They send out spherical or
circular waves. You go to a screen far away and
you add the contribution from the two.
If this and that differ in the
path length by an integer number of wavelengths it doesn't
matter. They will arrive in step.
I mean, the ninety-sixth
maximum will come when the ninety-fifth one from this one
comes, but we don't care. A maximum is a maximum.
If they differ by half a
wavelength or three halves wavelengths then they will
cancel, and what I said on the top also goes on the bottom.
Now sometimes there is a little
extra geometry you can do which is the following.
This may be some homework
problem. If you put a screen here at the
distance L and the patterns look like this,
this is the central maximum, and I ask you,
"Where is the central minimum measured from
here," well, you call that distance as
y. Then you see y/L is
tanθ and θ should be such
that dsinθ is λ/2.
Are you with me?
You find the directions of
minima. Once you find the found the
direction of a maximum or a minimum it is simple
trigonometry to find that if a screen is L meters away
how much do you have to move in the vertical direction.
That's very easy trig.
You can calculate and you can
predict. That's what Young did because
he knew the distance between the maximum and the minimum.
He knew how far the screens
were. He knew the slit separation.
All he did not know was the
wavelength and he got the wavelength without knowing what
the waves are made of. Now you can see why you got
fooled in the old days. If y/L is--so let's make an
estimation of how far apart these maxima and minima are in
this graph. Since they repeat let me just
take the first minimum, and that's a rough measure of
the oscillations. How far apart are these maxima
and minima? Well, the first one occurs when
dsinθ is equal to λ/2,
or sinθ, which is roughly going to be
θ for small angles is λ/2d.
Now if you pick any λ
you like, visible light is 5,000 angstroms.
That is 5000�10^(-8
)centimeters. And suppose your screen is 1
centimeter away? Then you get 5�10^(-3).
So it's roughly one-thousandth
of a radian is the difference between the maximum and the
minimum. The interference pattern,
even though I've shown it this way here, actually oscillates
very, very rapidly. Okay, the angle of difference
between the maximum and a minimum and a maximum and a
minimum is one-thousandth of a radian.
So even if you go a centimeter
away you will find that your probes, like your eyeball,
if your eyeball is this big, okay?
It's going to pick up so many
of these waves at one time. It cannot tell the maximum and
minimum. That's why we think there is no
interference because you only pick the average value.
That's why you have to do very
careful measurements. Okay, so that is double slit.
Okay, here's another thing
which will be very useful for people who are doing biology,
or maybe even chemistry. Let's call it diffraction
grading. In a diffraction grading you
take a piece of opaque glass covered with black soot then you
draw fine lines on it, evenly spaced lines so that the
effect is the following. This whole thing is dark except
you've made some tiny places where light can come through
like that, and the light's coming like this.
Are you with me?
This is my solid screen with a
few places through which light can come, and every one of them
becomes a source. It's like a double slit
experiment except you've got million of lines,
and you want to ask, "What will they do?"
In the forward direction if you
put an object, if you put a screen very far
away that's roughly infinity they all take the same path
distance. They all go to the same
distance. Therefore they will all be in
step. So in the forward direction you
will get a big maximum, because for every color the
path is the same in the forward direction.
But think of another direction
which is not the forward direction.
It's some other angle.
Then this guy and this guy are
out of step by an amount δ.
This guy and this guy are out
of step by an amount δ. This guy and this guy are out
of step by 2δ, so all the sources add up at
face difference of δ, 2δ, 3δ,
4δ and so on. And suppose you want them all
to be still coherent. What will you require?
What will ensure that all those
things are coherent in a certain direction?
Any guess on what δ
should be? Yep?
Student: It's equal to
the wavelength. Prof: If δ
is equal to the wavelength this guy and this guy are a
wavelength apart. We already know they are in
step from the double slit experiment.
This guy and this guy are a
wavelength apart. This guy and this guy are two
wavelengths apart. That is just fine.
Therefore if you pick a
direction in which that delta is either a wavelength or an
integer multiple of a wavelength all the things will add
coherently. Therefore the favorable
direction for a bright maximum satisfy,
again, dsinθ =
mλ where m can be 0 plus or minus
1 plus or minus 2. These are the maximum.
Except here in the forward
direction where θ is equal to 0--okay,
here's the point. Let us find the first maximum
away from the center. The center is not even counted,
center being white. All the colors will come to the
forward direction and you will get a big bright area here,
but look at what happens to the different colors.
This condition,
let's take m = 1, dsinθ =
λ. Whose λ
are we talking about, right?
Red light is one λ,
blue light's another λ. So each one will pick a
different λ, each one will pick a different
direction for this first order maximum.
So that means the white light,
when it comes in this direction, will be split into
many colors. It is just that they all shrink
to one line here in the center, but any other place the lines
will split. That's because λ
depends on the color; d is, of course,
the spacing between these little lines on the grating,
so sine θ is a function of λ.
So it acts like a prism.
It splits the colors.
So this is called the first
order maximum. Between this and that will be a
minimum. Then there'll be another
minimum. Then there's a maximum.
It's harder to calculate the
minimum, very easy to calculate the maximum.
So it's easy to calculate the
maximum. That's when everybody has to
add. Minimum is more complicated
because everybody has to cancel. You've got to add the little
arrows in different directions and ask when they cancel.
You can do it,
but it's a little more complicated.
But we can sort of imagine that
between the maxima would lie the minima, but the colors will get
split. That's important.
So this is how people find out.
So if you send white light it
will split into all the colors. If you look at white light
coming from the sun you'll find some colors are missing because
what happens is hydrogen in the surface of the sun will absorb
some of the colors because its atomic--
its atom likes to absorb light at a certain color.
That color is missing,
missing in the spectrum. So in this whole area you'll
have a little cavity. Somebody is missing and you can
easily tell from the wavelength that it's some emission line of
hydrogen. That tells you the sun contains
hydrogen. That's how people know what
elements are sitting on different planets.
How does anybody know what
they're made of? It was not at all clear in the
ancient days that stars and planets are made of the same
thing you are, right?
People thought moon was made of
cheese. Well, that's pretty much the
same thing that you are, but something else,
okay? But they didn't know.
They thought that heavenly
bodies and earthly bodies, but now we find it's the same
elements because you can find the spectra lines either by
their presence or by their absence.
So this is the kind of
spectroscopy you do. Okay, now I'm going to do one
other thing which is extremely important.
That is to explain why when
light comes through a hole of some size d instead of forming a
shadow like this it spreads out a lot into something much
broader. So let's understand that, okay?
Here is light coming in.
Now what you can do is this
aperture is going to be bright. When you see it from this side
it'll be glowing. So let's treat every point of
it as a source of light. In the forward direction
everybody's in step, but if you go off at some angle
they're not in step because you can see this guy is behind that
guy, and that guy's behind by that
amount. Each one is somewhat behind the
leader. Now my claim that if this
number here is λ, what do you think will happen?
In that direction if that
spacing is λ is that going to be a maximum
or a minimum? Anybody want to guess?
Pardon me?
Yes?
Student: Minimum.
Prof: And on what do you
base that? Student: Because the
λ/2 ________ the minimum, and...
Prof: But this thing is
λ. Usually λ
between two holes was considered good,
but now it's bad. The reason is that it's just
not the top guy and the bottom guy.
It's all these little guys in
between. So if this and that differ by
λ, by similar triangles, you see that's λ
over two? So this fellow will cancel that
fellow, and the second guy will be canceled by the second guy
from below. So when the whole phase shift
is λ, or the path difference is
λ, you can pair the points into
two at a time each differing by λ/2 and we know they will
separately cancel. So for a single slit the
fraction dsinθ =
λ is the first minimum, and after that some funny
things happen which are negligible,
but the light spreads out to that angle.
And this is the diffraction of
light by an opening. And let's look at it.
Sine θ
is λ/d. Suppose λ
is 10^(-4) and d is 1? Sine θ
= theta 10^(-4) that means the angular spread is 1 over
ten-thousandth of a radian. That's the extent to which
light won't go straight. It'll spread out,
but the spread is 1 over ten-thousandth of a radian.
It's very small.
But take a case where λ
and d become comparable, maybe d is equal to 2λ,
then sine θ is one-half. That's 30 degrees.
It's already beginning to
spread out. That's why I said earlier that
if you want to see the effect of wave optics you need to have the
length in the problem, which is the size of the hole,
and the wavelength in the problem comparable to each
other. If the wavelength is much
smaller than the dimensions of your experiment you can get away
with the ray optics, and you need wave optics only
when the condition is not met. So I remind you one more time
why people thought ray optics was good.
They were not able to produce
slits which are this narrow. Most slits you produce by
drilling a hole with your electric drill will be many,
many times the wavelength of light.
That means its angular spread
will be slightly non-zero, but instead of a shadow that
big you'll get a shadow that big.
You won't know.
Similarly the interference
pattern, the interference pattern for
microscopic separations between the slits will be so tight that
your eye cannot detect it even if you use light of one color,
and if you use light of many colors it'll wash out anyway
because each lambda likes to peak at a different angle,
so different colors would wipe it out.
So I wanted to let you know why
it is that people believed in geometric optics in the
beginning because they didn't do experiments.
This is why every theory will
fall. One day somebody has the
technology to produce slits which are very,
very near and to get light which is very pure of a definite
color then you will see something new.
Then you've got to figure out
what it is. But when Young did the
experiment and saw the interference they knew right
away it's waves because people had played with water waves,
and they knew with water waves that you can have interference.
Okay, here's another thing you
can do. Here is a bunch of atoms at the
surface of some metal. Metal's down here,
and I'm shining light. You know how reflection works?
The light hits the atom.
Atom absorbs the light and it
reemits the light. That's what reflection is.
You don't just bounce off.
You get absorbed and get
reemitted. Now the problem is if you
reemit you reemit in all directions.
You don't just reemit in one
direction. The question you can ask is,
why is it that i = r for reflection?
That's what I want to explain
to you. So this guy emits something.
There's a wave emanating from
this. This guy emits something,
wave emanating from that. So let's take two rays going
off in some direction. And let's just ask when these
two guys will be in step in terms of the emitted wave.
So you see the light signal is
coming from here. It hits this one first.
Are you with me?
So this emits right way.
Its signal has started taking
off in the final direction. This guy is delayed because it
hasn't hit this, but once they start moving
toward the eye this is closer than this one,
so you see this one and that one, if you want to make a real
comparison-- well, this is bad picture.
Let me draw you another picture
where there's a good competition between them.
So let's see.
Suppose I had a really bad idea
and I thought they reflect like this.
I'll show you that won't work.
That's all I want to do.
You see what you should do is
draw perpendicular here and that's the extra distance this
has to travel, right?
And you will draw a
perpendicular here and let's see, what's the extra distance
the other has to travel? It has to travel that extra
distance. So if you draw those two angles
you will find--let's see which angle I want to take.
If you call this alpha you'll
find d cosine α_1 is d
cosine α_2 or α_1 =
α_2. That means I = r.
You see that?
Maybe I should repeat the
picture here so everybody can see.
I just need two of them.
So one let me draw like this.
Let me draw the second at what
is definitely not i = r. So here I'm saying draw that
perpendicular and draw that perpendicular.
Starting from infinity these
two have the same length, those two have the same length.
This is an extra length
traveled by this guy. That's the extra length
traveled by that guy, okay?
So if you call that angle
alpha, you call that angle β,
and if this is d, you can see that d
cosine α must be d cosine β
or α should be β. That is a very interesting
result. It tells you why i = r
is true. See, if light is made up of
particles that bounce off a wall and ricochet with i = r
you can understand that from conservation momentum.
But light is a wave and when
the atoms absorb light they emit in all directions,
but only in this direction will everybody be in step because
once I made sure these two guys are in step you can repeat it
for these two and those two. You'll find that's the
direction in which every single guy is in step.
So i = r is the
direction in which the re-radiated light arrives in
step in only one direction, and that's the direction with
the angle of incidence is the angle of the reflection.
You understand why there is a
difference? Okay, one last thing I should
mention, because I've given a problem for you,
and I just want to tell you one trick you may not know.
And you can read up.
The book is very good in that.
If you've got some oil slick on
a rainy day, there's water here. This is an oil slick.
You see all kind of colors in
the oil slick? This is to explain why it is.
So white light comes or any
light comes, bounces off the top surface.
The light also bounces off the
bottom surface. And the sum of those two hit
your eye. Suppose you want to arrange it
so that blue light cancels out in the process?
What will it take to cancel?
I claim first of all the
wavelength in this medium is n times the wave.
It's λ/n.
If this is the wavelength here
the wavelength in a medium of refractive and n is
λ/n. If this whole thing is equal to
a full wavelength, this round trip,
I claim you'll get a dark spot. You may think that's wrong
because this guy's gone an extra full wavelength.
That should be additive,
but it turns out that when light bounces off in going from
a rare to a dense medium it changes by an extra Π.
At the other end when it goes
from dense to rare it doesn't. So you have to add an extra
Π to everything you thought. Then you will find out that
when this extra round trip is the full wavelength you get
destructive interference. So if that occurs for a
wavelength for blue light you take blue light out of white
light you get some yellowish greenish color.
That's what you see in the oil
slick. And if the oil slick thickness
varies different colors will be missing from white at different
point. That's why you see all kinds of
different colors in the oil slick.
So I agree I didn't do a good
job of this one, but this is the easiest thing
to me. The only thing to do,
when you do your homework, if you didn't read the section
don't forget there's an extra factor of Π
in the phase shift when you hit the hard medium.
