Prof: All right,
I'm hoping that this interlude
into optics will give you a
chance to catch up and be on top
of the class,
okay?
I take that to be a yes.
 
So if you have to give an
elevator presentation of
geometrical optics,
you're riding in the elevator
with some big shot,
you get what,
60 seconds and you have to tell
that person everything,
what would you say?
 
Any guess?
 
Student: 
>
Prof: Right,
Principle of Least Time covers
everything,
and I thought I would do that,
because I'm sure you guys,
most of you,
have seen this in one form or
another.
And I had to do something
slightly different,
and that was to show you that
it all came from Least Time.
Well, I haven't shown the whole
thing yet.
What I did show you is that
angle of incidence = angle of
reflection.
 
That came out.
 
Snell's law, that came out.
 
But I was about to do mirrors
and the case of mirrors,
I said if you want a parallel
set of rays from infinity,
to focus on one point,
then what do you need to do?
First of all,
focusing itself is a very
special case.
 
Focusing means many light rays
going in different directions
all end up in the same place.
 
That means they all must
satisfy the Principle of Least
Time.
 
How can all of them be the
least time?
Because they all take the same
time.
Therefore there's more than one
way for a light beam to go from
start to finish and that's when
you get focusing.
This is the first example of
that, and the way it works is,
you start with this guy here.
 
First of all,
the rays are coming from
infinity.
 
If you're going to keep track
of how long that takes,
you're going to get infinity
for everybody.
Maybe infinity 1 for one and
infinity 2 for another one.
And we know from the time we
were children that they're all
the same, so infinity's not a
good reference point.
So you draw the gate here,
a mental gate,
and you say,
"I'm going to attrack
these guys from this point on.
 
No one's got any head start.
 
Then I want to see how long any
of them takes."
Well, this guy hits that wall,
I mean, this mirror here.
The mirror is locally vertical,
so it will bounce right back,
and this focal point is where I
want everyone to end up.
So this ray does that,
loops around and comes back
here.
 
How much time does it have?
 
Well, it has the time to go
here, plus the time to go an
equal distance f on the other
side.
That means this light ray which
began here has the same amount
of time.
 
But for this mirror here,
it's got the time to go there.
But you want it to hit the
mirror and come here in the same
time.
 
That means it will happen,
only that distance and that
distance are equal.
 
And this must be true for any
ray parallel to the axis.
That means your mirror is a
curve with the property that
points on it have the same
distance from the focal point
and from this line here.
 
And there's a name for that
curve,
and that is the parabola,
and satisfies the equation,
y^(2) = 4xf,
where f is the focal
length.
 
I derived that for you.
 
Then the next question is,
it's not enough to have--then I
told you that if you want,
you can approximate this by a
spherical mirror.
 
You take a sphere or radius
R and you slice a portion
of it.
 
A sphere is not a parabola,
because sphere is going to--put
a sphere here,
it's going to start deviating
from the parabola.
 
But you can fool this beam of
light if you don't go too far
from the axis.
 
In that region,
it coincides with the parabola.
And you say, how big a parabola?
 
Well, it looks like a parabola
whose focal length is related to
the radius of that sphere in
this fashion,
where f is R/2.
 
That's called a spherical
mirror.
So whenever you do this,
you should be aware of all the
limitations.
 
A parabolic mirror will focus a
parallel beam of light,
no matter how wide it is,
measured from the axis,
because I've cooked this up to
work for any height.
If you approximate it with a
spherical mirror,
then you have to ask yourself,
"How long will this beam
of light think that the sphere
is a parabola?"
Well, you can see,
as you start moving off,
it's going to catch on,
so you cannot go too far off
the axis.
 
So that's an approximation.
 
And for a spherical mirror,
if you send a parallel beam of
light, if it's too thick in the
transverse direction,
it won't focus here.
 
It will focus for small
thickness, small deviation from
the axis.
 
But now our goal in making
mirrors is not only to form the
image of an object at infinity.
 
We want an object at a finite
distance.
We want object at finite
distance to form an image and
that's what we were talking
about.
So let's go back to that
problem.
Okay, I'm going to be writing
some equations today,
and I want to request people in
the last row,
if you cannot read something
you have to tell me if the font
is small.
 
Guys in the front row,
if you cannot read something
because handwriting is bad,
you've got to tell me.
People in the last row may not
know why they cannot read it.
Maybe the handwriting,
maybe the font.
But between the two of you,
you've got to keep me on the
lookout for that,
okay?
Because I did realize last
time, I was drawing more and
more and more lines in the same
figure.
It was hard to read.
 
All right, so I'm going to
write some equations.
I'm trying to plan my
blackboard strategy here,
but first I will tell you what
I'm going to do.
I mentioned that to you last
time.
We want to make sure that when
you follow the rules of ray
optics that you learned in high
school.
What do you learn in high
school?
You say, well,
take any object here.
You draw a bunch of rays and
you know where they're going to
intersect because the parallel
ray goes through the focal
point.
 
That's how it's defined.
 
Here's another ray that's a
little interesting.
Go through the focal point,
you'll come out parallel.
How do I know that?
 
I know that because if I run
this thing backwards,
it's a parallel ray going to
the focal point and ending up
here.
 
If this path was the path of
least time, the reversed one
should also be a path of least
time.
So these two meet here.
 
So we say that's where the
image is.
But that's not enough.
 
I told you two lines will
always meet.
You've got to draw more lines,
and here is one more that I
drew.
 
The one that hits the center
here, that's i = r,
very obvious incidence =
reflection, because the mirror
is vertical here.
 
I showed you that the third
line demands that
h_1/u is
h_2/v.
But that's already true from
these two rays.
Remember, the two rays
intersecting here gave the
condition 1/u 1/v is
1/f and they gave this
condition.
 
So the third ray which I drew
demands the same thing,
so it's consistent.
 
But if you really want to nail
this down,
you want to be able to prove
that no matter what height you
pick for the interception of the
mirror here,
you will take the same time.
 
Because I've only taken three
special values for the height
y.
 
One is y =
h_1,
this beam.
 
One is y =
h_2,
or if you like,
-h_2,
this beam.
 
And y = 0,
the third one,
and they all happen to agree.
 
I want to take an arbitrary
y and hope that the
answer will not depend on
y.
The time it takes to travel for
any y should be the same.
And that's where we will find
it is not exactly the same.
Yes?
 
Student:  What are
u and v?
Prof: u is the
distance of the object measured
from here,
and v is the distance of
the image measured from there.
 
I don't know what the book
calls them.
I don't know which book you
guys are using,
d_i and
d_o and
i and o.
 
There's a lot of different
symbols you can use.
So you're free to use any
symbol you want in your problem
set, as long as you explain to
the TA what you're using.
Unfortunately,
the only thing everyone agrees
on is that f stands for
focal length.
That's universal;
these are random.
All right, so do you understand
what I'm going to do?
Because it's going to take some
time and it's not worth doing it
unless you know why we are doing
it.
We're trying to verify that the
time it takes to go from here to
the mirror and over here is the
same for every conceivable ray
for any height y.
 
And I'm telling you,
so what would you like to show?
You would like to show the
following.
You take this here.
 
You want to hit that mirror at
some height y,
you make it bigger.
 
So these are all drawn as very
big,
so that you can follow them,
but the understanding is,
they are not really that big
compared to the dimensions of
the mirror.
 
So it goes there and it does
something, and it comes here.
So this is the point
(x,y) on the parabola
that it hits.
 
d_1 is the
distance--
I'm going to use Pythagoras'
Theorem to calculate it--
is that distance,
and d_2 is
that distance.
 
So d_1 depends
on y and
d_2 depends on
y and when I add them up,
that's got to be independent of
y.
That's got to be the same for
all y.
In particular,
it's got to equal the value
when y = 0,
= d_1(0)
d_2(0).
 
That's my reference signal.
 
That's the one that goes here,
hits the center of that and
comes here.
 
You guys with me now?
 
It should be independent of
y, therefore it should be
equal to the value at any
y I like.
I'm going to pick the y
= 0 as my reference.
You find the distance
d_1 d_2
when y is 0,
you move up and down,
calculate it,
and it shouldn't depend on
y.
 
So when I do the calculation,
what do I really hope will
happen?
 
If I plot--if I make a plot of
d_1 d_2
as a function of y,
I'm expecting the answer to be
flat,
independent of y.
And this is the reference
point, y = 0.
But when you do the
calculation, first of all,
if the graph of
d_1 d_2
looked like this,
that's no good,
because that means if you take
y positive,
you take a longer time than
y negative.
Obviously that y and that y are
not paths of least time,
because one is clearly taking
less time than the other one.
So you don't want anything
growing linearly in y,
so it's got to be quadratic,
at best.
But then even if it's
quadratic, you go far out,
you can see that the answer
depends on y.
So what I will show you is that
the graph,
if you plot it as a function of
y,
if you plot d_1
d_2 as the
function of y,
it's got some constant,
which is the time for y
= 0 some coefficient Ay
some By^(2) 
Cy^(3).
These are all possible things
that can happen.
If you want perfect focusing
for any value of y,
all of these guys have to
vanish, because it cannot depend
on y.
 
But we will find out it's the
first two guys that vanish and
there is no guarantee about the
others.
But of course,
from dimension analysis,
when you have a y and a
y^(2) and a y^(3),
for them to all have the same
dimensions,
they will be divided by some
numbers.
So this y will be
divided by let's say u,
this will be divided by
u^(2),
this will be divided by
u^(3).
It's an expansion in a small
parameter, and only when the
parameter is small will this
work.
So you have to notice two
things: in a parabolic mirror an
object at infinity will form a
perfect image,
no matter how wide the beam is.
 
But an object at a finite
distance will not form a perfect
image if the object is too tall.
 
If the height y becomes too
much, then you don't get a good
image.
 
But you get a pretty good
image, because the possible
dependence on y and
y^(2) will be 0,
but not to arbitrary powers of
y.
Do you understand?
 
If you have a graph that's a
power series in y, if it
begins with y^(2) it will
look like this.
If the first non-zero term is
y to the fourth,
it will look very flat,
but eventually it will move.
The flatter this is,
the further out you can pretend
nothing depends on y.
 
So we want to see how far we
can go.
That's the purpose.
 
It will also tell you the
limits of a real mirror for a
real object.
 
Any questions on calculation
I'm about to do?
Okay, before I do the
calculation, you're going to
need the following
approximation:
u^(2) θ^(2) to the
power ½.
You're going to get such
combinations all the time.
You can write that as u
times 1 θ^(2) over
u^(2) to the ½.
 
That's exact.
 
Yes?
 
Student: 
>
Prof: Oh,
R in the top?
R/2.
 
R stands for the radius
of the sphere from which you cut
out that piece.
 
Everybody understands that?
 
We had a parabola here,
and I'm going to approximate it
by taking a sphere of some
radius R and that sphere,
if you look at this portion
here, it looks like a parabola.
You can ask,
"What parabola does it
approximate?"
 
I showed you that near the
origin, it looks like a parabola
whose f = R/2.
 
So if you take a sphere of
radius R and you slice a
portion off it and you paint it
with silver so it becomes a
mirror,
its focal length will be
R/2.
 
But it won't be a perfect
parabolic mirror.
It's an approximate parabolic
mirror.
There are different levels of
approximation.
A sphere will not focus a
parallel beam even if the beam
is too wide.
 
A parabola will focus the
parallel beam no matter how wide
it is, provided the object is at
infinity.
If the object comes to a finite
distance, I'm saying even the
parabolic mirror will not give
you a perfect image.
But if it's very short,
short compared to u and
v and so on,
it will do a good job.
So let's see how that happens.
 
So I'm going to need this
approximation that looks like
u.
 
now this thing,
I'm going to write as u
times 1 θ^(2) over
2u squared ...
θ to the fourth,
u to the fourth,
etc.
 
I'm going to drop all this.
 
This is the approximate
expansion, (1 x)^(n) is 1
nx.
 
So that becomes = u
θ^(2) over 2u.
So let's bear that in mind.
 
This is just a mathematical
preliminary I'm getting out of
the way so we don't get stuck in
that.
Yes?
 
Student: 
>
Prof: It was squared
here, but when I multiplied by
this u,
it got un-squared.
So in this calculation,
I mentioned to you,
some numbers are big and some
numbers are small.
I said that near the end of the
class.
u, v, f are treated as
big numbers.
Compared to that,
the height h_1
of the object,
height h_2 of
the image and y, the
height at which the light hits
the mirror,
they're called small.
And the coordinate x on
that mirror where you hit it is
proportional to y^(2).
 
You remember that?
 
So it's proportional to small
squared.
Remember the parabola equation
is x = y^(2)/4f.
In fact, let me write it
exactly.
You can see that f is a
number that's called--you can
call these order one numbers,
these are small numbers,
these are smaller numbers
squared.
So in all these calculations,
I think you have no trouble
seeing why x is much
smaller than y, right?
The shape of the parabola is,
if you climb a lot of height,
you don't move a lot in the
x direction,
because it's quadratic.
 
And the square of a small
number is even smaller.
So if y is small,
x, which is
y^(2), is even smaller.
 
So I want to do this,
because no matter what field of
life you go into--
well, not every field,
but in most fields where you
have to do some kind of
calculation,
whether it's in economics or
biology or whatever,
you should know how to do
approximate calculations.
 
I'm trying to show you how to
do a systematic approximate
calculation.
 
Most of the calculations people
do are approximate.
In fact, every calculation ever
done by any physicist is
approximate, because we're
neglecting something.
You show me one result that you
swear is accurate,
it is not.
 
Everything has got corrections.
 
So we should learn how to find
the big numbers in the problem
and keep them and how to
organize our calculation,
if you like,
as a series in small and
smaller squared and smaller
cubed,
and you keep a few terms.
 
So let's calculate now the
distance, d_1
and d_2.
 
So we'll have to look at the
picture here.
Let me just write down what I
get.
d_1 d_2
is going to look like
h_1 - y whole
squared u - x squared.
That is d_1.
And d_2 will be
h_2 y squared
v - x squared.
Can you see in that figure in
the right that the right angle
triangle--maybe I'll bring it
down one more time to show you.
In fact, this is what we really
need, so let's keep it here.
This whole thing is
h_1. This
height is y,
so this side is
h_1 - y,
and this side here is u
- x,
because u brings you
from here to the mirror,
but this point is to the right
by an amount x.
 
That's this right triangle.
 
Then there's a second right
angled triangle here to find
d_2.
 
That height is h_2
y and that distance is again
v - x.
 
This is the point v and
this is the point u.
And I want that to be same as
when the beam goes to this
dotted line here.
 
That's the y = 0 answer.
 
That's very easy to do
Pythagoras on that one.
This dotted square is
h_1^(2) u^(2)
under root,
and this dotted square =
h_2^(2) v^(2)
under root.
So this whole thing has got to
= h_1^(2) u^(2)
h_2^(2) v^(2).
 
So the right hand side does not
depend on y.
Left hand side depends on
y.
In the best of all worlds,
somehow all the y
dependents will magically
cancel, but it's not going to
happen, I can tell you right
now.
What you will find is that the
term linear in y will
cancel, the term quadratic and
y can be made to cancel,
and that's all you can do.
 
So let's now look at these two
things.
Let me look at the right hand
side.
It's very easy to do this one.
 
I only paved the way for this.
 
Remember, h_1
is a small number,
u is the big number.
 
When you do this one,
you will get u
h_1^(2) over
2u.
You see that?
 
h^(2) is like
θ^(2) in my
formula.
 
And this will look like v
h_2^(2) over
2v.
 
So the formula I'm using is
u^(2 ) θ^(2) to
the½ = u θ^(2)
over 2u terms
involving θ
to the fourth and so on,
which I'm not keeping.
 
So the right hand side is very
easy to calculate.
It's that.
 
So let's look at the left hand
side now.
So let me open out everything
in the bracket here.
I get h_1^(2)
y^(2) - 2h_1y u^(2) -
2ux x^(2).
 
Now the second guy is
h_2^(2) y^(2 )
2h_2y v^(2) - 2vx
x^(2).
That's got to equal the u v
h_1^(2 )over 2u
h_2^(2) over
2v.
Remember, the right hand side
is not really--I've put a dot to
tell you it's an approximation.
 
This square root goes on
forever.
I've kept the leading,
first leading term.
The first leading term happens
to be quadratic in h.
Quadratic in the small
quantity, so to be consistent,
I should look at the left hand
side, also to the same accuracy.
So let's look at the left hand
side here.
This is quadratic in the small
guy, quadratic in the small guy,
small times small,
a big square,
big times small,
and small squared.
But you've got to be a little
careful.
x is not small;
x is small squared.
Remember that?
 
x is proportional to
y^(2).
This is proportional to
y to the fourth.
You cannot keep a term to
fourth order in the small
quantity on the left hand side
if you don't keep a similar term
on the right hand side.
 
So you systematically drop
things which are quartic in the
small.
 
So even though x makes
it, x^(2) doesn't make it
in this approximation.
 
Yes?
 
Student: 
>
Prof: Yeah,
it can be anywhere it wants.
You may think that because it's
under square root,
when you take the root of it,
it may become x.
Is that your worry?
 
Is that what you're saying?
 
Student: 
>
Prof: The point you're
making is, I see an x^(2)
like that, right?
 
But it's really not
x^(2);
it's really x.
 
That's certainly true.
 
But you're adding a whole bunch
of numbers and then taking the
square root of that.
 
In that sum,
these numbers are going to be
big compared to this one,
whether you take the square
root r or not.
 
So it is dominated by the terms
under the same square root.
That's why you drop it.
 
Okay, so this fellow is out.
 
Here's what I want to do now.
 
I'm going to write the first
thing.
So let's decide what to call
what.
This is my u^(2).
 
The rest of these guys are my
θ^(2).
You understand that?
 
Unfortunately,
they're not next to each other,
but all of that is
θ^(2),
this is u^(2).
 
Then the formula tells me,
this = u h_1^(2)
y^(2) - 2h_1y - 2ux
over u v
h_2^(2) y^(2)
2h_2y 2vx divided
by v = u v
h_1^(2) over 2u
h_2^(2) over
2v.
This is what I want to this
order.
Now I have an x
everywhere, but x is not
independent.
 
x = y^(2)/4f.
 
So what is -2ux.
 
-2ux will be
−y^(2) over
2u over f.
 
Yes?
 
Student:  Should the
denominator for those be
2u and 2v?
 
Prof: Yes, thank you.
 
Very good.
 
Okay, therefore--so let's
compare everything.
The big numbers have to cancel
the big numbers,
otherwise you've got big
problems.
You guys have a question?
 
Student: 
>
Prof: It was to
the½,
but I've taken this
approximation that when you take
it to the power ½.
 
Look at the top right.
 
u^(2) θ^(2) to the
power ½
is u θ^(2) over
2u.
Okay?
 
Yes, so I've done this a lot of
times, but I want you to follow
this, because this is telling
you how to organize the
calculation.
 
This is how we organize.
 
We identify quantities which
are quantities of importance,
and then smaller and smaller
quantities and you balance the
equation,
starting with the big numbers,
then go to the smaller numbers.
 
So the big numbers better match.
 
So that tells you u v
matches u v.
That's fine.
 
Now take the terms proportional
to y.
They have to separately match.
 
When you have a function of
y, then if it functions
equal on the two sides,
every power of y should
match.
 
Of course, we only have two
powers here and they should both
match.
 
So take the term linear in
y from both sides.
Oh, let's get it to one more
stop.
Do you see this
h_1
^(2)/2u canceled this
one?
And this h_2
^(2)/2v cancels that
one.
 
So I've got nothing left.
 
So everything on the left hand
side should add up to 0,
because there's nothing on the
right hand side.
So what do I have to linear
order and y?
This guy is linear and
y, this guy is linear and
y and that's all there
is.
So the coefficient of y
looks like y times
-2h_1 over
2u 2h_2
over 2v and that's got to
vanish.
And that simply says
h_1/u is
h_2/v.
 
So this is what you get by
usual geometric optics
arguments.
 
I showed you last time with
similar triangles,
but it comes from the Principle
of Least Time,
by demanding that the function,
the time doesn't depend on y,
to first order and y.
 
Then I have to do the
y^(2) term,
understand?
 
You understand if you've got
numbers on the two sides,
big numbers,
square of small numbers,
linear and y,
quadratic and y,
everything has to match,
order by order.
So this is a linear and
y term.
What is the quadratic and
y term?
Quadratic and y,
look at the coefficients.
y^(2) is coefficient
½u.
You see that?
 
And this one has got 1 over
2v, coefficient of
y^(2).
 
So I've got 1 over 2u 1
over 2v,
but then I've got -2ux
and -2vx.
So what is -2ux,
is −y^(2) u over
2f divided by 2u -
2 y squared v by
2f divided by 2v =
0.
Student: 
>
Prof: Yes, I'm sorry.
 
I'm not being consistent.
 
So let me just call this the
coefficient of y squared.
Do you agree?
 
It's the coefficient of the
y^(2).
If you want,
it's in front of everything,
all of this is multiplying
y^(2).
So this has to vanish.
 
Okay, so cancel the 2's
everywhere, you get 1/u
1/v.
 
This cancels.
 
Let me see.
 
There is no 2 here,
because 2ux = y^(2)
u over 2f.
 
This 2 is not there, sorry.
 
This is all you have.
 
You get -1 over 2f -
another 1 over 2f = 0.
Look guys, we know where we are
going, right?
So it's going to happen.
 
It's happened for the last 300
years and I'm not going to
change it, so it will happen.
 
So the 2's will all work out.
 
What does this tell you?
 
It tells you 1/u
1/v is 1/f.
So you can ask,
what's the point of doing this
exercise?
 
This tells you that because of
the Principle of Least Time,
the time it takes to go not
only for these two privileged
rays,
but any ray you can think of
drawing from here to the mirror
and back here.
They all take the same time,
in the approximation in which
the height over u,
for example,
to the fourth power,
is neglected.
Basically, anything that looks
like small over big to the
fourth power and higher powers
are neglected.
So it's not going to form a
perfect image for everything.
It will form a perfect image in
a world in which you are able to
keep the term to leading order
in u,
to the order
h^(2)/u^(2),
and stop.
 
This is an approximate thing.
 
So for it to successfully work,
you've got to keep your object
and the image all to be much
smaller than u and
v and f.
 
That's the rule.
 
If you take any parabolic
mirror and you have an object
which is as tall as the focal
length, it won't form a good
image.
 
What it will mean,
what do you think it means not
to form a good image?
 
Rays will cross at different
points and you will get a
blurred image.
 
If you put a screen,
if you put a candle here,
if that's a candle,
you can actually put a real
screen here and there'll be a
glow on the screen,
image of the candle upside down.
 
It will be a very sharp image,
if the candle's not too tall.
Once it gets taller,
image will get blurred,
because the rays will not meet
at the same point.
All right, I want to do one
other thing, then I'm going to
stop with the Principle of Least
Time.
That has to do with lenses.
 
You saw how a mirror does a job.
 
Let's take a lens.
 
Let's say we have the following
lens.
And I have an object which is
infinitesimally tall to begin
with.
 
And I wanted to form an image
somewhere on the other side.
So you can ask yourself,
how can the light rays leaving
this in different directions
arrive here and all take the
same time?
 
It's clear that this guy is the
shortest path.
These are taking longer,
and yet they have to be
competitive with the one in the
middle.
So how do you think that
happens?
How do you think they will take
the same time?
Yes.
 
Student: 
>
Prof: That's correct.
 
The answer was,
it is true that this path is
shorter than one that goes to
the top of the lens and goes
down here, but this is through
glass.
It's like saying I have a lake
here shaped like this,
and I tell you to run so that
you go from here to here in the
least time.
 
It's not at all clear that
heading straight for the lake is
the best thing to do,
because you've got to swim.
This person doesn't have to
swim at all, but has to go
further.
 
You can imagine somehow cooking
things up so that they all take
the same time.
 
So I'm going to do it only for
the two extreme rays.
Since I've done a lot of
algebra, I don't want to keep
doing it.
 
So let us say this height is
h, height of the lens,
and this distance is u,
this distance is v.
And the refractive index is n
of the lens.
What that means is,
1 meter in air--I'm sorry.
1 meter in glass is like
n meters in the air,
you understand,
in terms of travel time.
Everybody with me?
 
That's what it means to say it
goes slower.
So a meter of lens is worth
n meters of air.
That's the consequence of the
velocity of light.
We're trying to find equal
times, so when it travels here,
it travels with a velocity
c/n,
therefore the time it takes,
when you divide the distance by
time,
will be n times as big.
Okay, then there's one other
result I'm going to need,
so I don't want to interrupt
you later,
which is the following:
I'll take this glass here,
this face of the lens,
and let's say it is coming from
a sphere of radius R.
 
So that's the center of the
sphere.
And draw a normal here and call
that distance delta and call
this height h.
 
This is all exaggerated.
 
In reality, delta's a very
small number.
I'm going to need an
approximate formula for delta in
terms of h and R.
 
So since it's a sphere,
that distance is also R.
Right?
 
This is a sphere of radius
R from which I've cut out
that portion,
which is this face of the lens.
The lens ends here,
but the mental sphere can keep
going like that.
 
So what can you say about this
distance delta?
You can say the following?
 
delta = R - this side.
 
So let's write the formula this
side, which I don't know,
let's say is l,
is R - l.
But we also know that l^(2)
h^(2 )= R^(2).
Therefore l^(2) = R^(2)
− h^(2),
or l is square root of
R^(2) −
h^(2),
and by the same trick I did,
it is R −
h^(2) over 2R.
The same θ^(2) formula.
 
So l is shorter than R
by an amount h^(2)/2R,
so this delta =
h^(2)/2R.
That's the formula I'm going to
invoke.
You will see why I bother to
prove that in a minute.
So you should now think about
what equation you will write
down, okay?
 
I'm going to give you a few
seconds to think about what two
things should be equal,
for the condition that
everything takes the same amount
of time to go from start to
finish.
 
So first I'm going to find the
time it takes to go on that
straight line.
 
By the way, since I'm going to
divide by velocity everywhere to
get time,
distance is all you need,
except here you've got to
remember 1 inch of glass is
n inches of air.
 
If you remember that,
you'll be fine.
So let's see how long is that
path.
That path = square root of
u^(2) h^(2) square root
of v^(2) h^(2).
 
By right angle,
this right angle triangle,
that right angle triangle,
very easy to get that.
And that's going to = the
effective length going straight
through.
 
So let's look at the effective
length in two portions.
That's a segment here of size
delta and a segment there of
size delta in glass.
 
So let's look at the distance
in the air first.
Distance in the air is u
- delta on the left v -
delta on the right.
 
That's the air distance.
 
Now I've got to add to that 2
delta times n.
That's what you have to
remember.
That's the only non trivial
part of this,
is that this distance,
2 delta, is should be
multiplied by n,
because you're traveling at a
speed reduced by n in
terms of time.
It's the distance factor
increase by n.
So this becomes u v 2
delta times n - 1.
That's the right hand side.
 
The left hand side,
it's u h^(2) over
2u v h^(2) over
2v.
Again, in the same
approximation,
in the quantity
h_2/u
_2, or
h_2/v
_2.
 
There are higher powers of
h/v which I'm not keeping
in either side.
 
Even here, there are higher
powers of h/R that I'm
not keeping.
 
So get used to the notion, okay?
 
u, v, f, R, big.
 
h, y, etc.
 
are small.
 
So you cancel the u and
the v.
Write the formula for delta.
 
Delta = h^(2)/2r.
 
I'm sorry, did I make a mistake
here?
I guess not.
 
Okay.
 
So you compare the two sides,
you find 1/u 1/v.
In other words,
drop these 2's and cancel the
h^(2) everywhere.
 
And 1/u 1/v = 2
times n - 1 over
R, which I'm free to
write as 1/f if I like.
In fact, that will be
1/f.
How do I know that it's
1/f, can you tell me?
Why should I call this quantity
that I get as 1/f?
It is what I got,
but my question is,
what allows me to identify
that?
Student: 
>
Prof: Remember,
v is not the focal
point.
 
v is the location of the
image and u is the
location of the object.
 
But what do you want the focal
point to be?
What's the definition of the
focal point?
How is it defined?
 
If you had a lens and I said,
"What's the focal
point?"
 
how will you find it?
 
Haven't you guys played with
ants when you were kid with a
magnifying glass?
 
Come on, what's the focal point?
 
Where a parallel set of rays
will converge on the ant.
So if you send rays from
infinity, if u =
infinity, v should be the
focal point.
If u is infinity,
you drop this guy,
then v,
the image should be at the
focal point, therefore
1/v is 1/f.
That's what allows you to
identify then this quantity as
1/f.
 
Because in the limit,
u goes to infinity,
1/v should equal
1/f, and therefore this
is what it is.
 
But this is a calculation of
the focal length.
See there are two things.
 
One can define focal length
operationally by saying take a
lens, send a parallel beam,
see where it all focuses.
That's the focal length.
 
But you can also calculate it
from first principles by saying,
if I made a lens whose faces
are cut out of spheres,
the two phases,
of radius of curvature
R,
and the refractive index of the
glass is n,
then I can actually calculate
the focal length in terms of the
radius of curvature and
n.
 
So one is a derivation of the
focal length in terms of other
given parameters,
whereas parallel rays focusing
at f is the definition of
the focal point.
It doesn't tell you how to
calculate it for a lens.
But this will tell you.
 
So if you've got a focal length
that is too big or too small,
you can vary it by varying
either n or by varying
R.
 
You can take different
materials or you can take
different radii of curvature.
 
Question, yes?
 
Student:  In that
drawing of the lens,
what is the f?
 
Prof: Which one?
 
Student:  In the drawing
of the lens, what is the little
f?
 
Prof: This little
f?
Student: 
>
Prof: You want to know
what f means?
Student:  Yes.
 
Prof: It means take this
lens and send a beam from
infinity.
 
That means it comes in parallel
rays.
That means u = infinity
and 1/v then becomes
1/f.
 
That means v = f.
 
v is the image point,
so that f will be here.
So that's what it means.
 
Okay, so I've gotten these
formulas, spent a lot of time
deriving them.
 
I'm assuming you have used them
a lot in your younger days.
I just want to give you a few
examples,
but I don't want to do all of
them,
because you can do endless
calculations involving
1/u 1/v is
1/f,
just punching in numbers.
 
I will just tell you a couple
of things to bear in mind.
Then I want to sort of wrap it
up.
So let's take some examples.
 
Let's take this focusing mirror
now.
This is f and this is
2f.
So let's see what happens when
I start with an object here.
You should be able to sort of
draw the ray diagrams,
and I'm just drawing you a few
of them.
So from now on,
we quit Fermat's Principle of
Least Time.
 
We just go back to the usual
diagrammatics,
having convinced ourselves it
all follows from the Principle
of Least Time.
 
So you don't want to go back to
that every time.
So what will be the situation
here?
You need only two rays to draw.
 
So one ray, you draw a
horizontal ray,
and you know that will go
through the focal point.
That's the definition.
 
Then you draw a ray through the
focal point.
That will come out horizontal,
and again, why is that true?
That's true because if you sent
a parallel ray at this point,
it will go to the focal point,
and if it ends up here,
then the reverse is also true.
 
Any ray that's possible,
then the time reversed version
of that's also possible,
because if one takes the least
time to go from A to B,
other takes the least time to
go from B to A.
 
That's how you draw this and
they intersect somewhere here.
In fact, one can show that if
you're to the right of
2f,
the image will be to the left
of 2f.  If you're
exactly at 2f,
you can show the image will
also be at 2f.
Let's see why.
 
Now 1/u 1/v is
1/f.
I'm saying if u is at
2f and I believe v
is at 2f,
I get them to add up to
1/f, so that's correct.
 
So 2f is the point where
the image and the object cross.
See, objects at infinity focus
on f.
As you move the object to find
a distance, the image moves away
and they cross when you put it
at 2f.
Then you can of course go
closer than 2f and you
can ask, "Now what image
will I get?"
So maybe I'll draw one more
picture for that one.
So this is closer than
2f.
Closer than 2f but more
distant than f,
something here.
 
So draw that ray,
draw it parallel,
draw that ray,
draw that and they will meet
like this.
 
And finally,
if the object is at f,
the image will be at infinity,
because that's just the reverse
of the other picture.
 
If all the rays from infinity
focus here,
if you put an object at the
focal point and run the lines
backwards,
they'll go in parallel,
so they'll never meet.
 
They're trying to meet at
infinity.
But of course,
you can also take an object,
even closer to the mirror than
the distance f.
That's what I want to talk
about now.
What if you go even closer than
f, so let's try that.
So here it is, same guy.
 
This is f.
 
I want to take an object here.
 
So the first ray I can draw is
the parallel ray going through
that.
 
First of all,
when you do the algebra,
you can see some problems.
 
1/u 1/v is
1/f, therefore 1/v
is 1/f −
1/u.
But if u is smaller than
f, 1/u is larger
than f,
and 1/f −
1/u will be negative.
 
That means v itself will
be negative.
So v is the location of
the image,
and it's negative,
which means on the other side
of the mirror,
you can ask,
how can light go to the other
side?
It doesn't go to the other
side, but you will see that it
looks like something is behind
the mirror on the other side.
So let's see why that happens.
 
You will have some difficulty
drawing the second ray.
Remember, the second ray you
draw through the focal point,
and you say it comes out
parallel.
If I draw a ray through the
focal point, now I've got to go
backwards.
 
Previously, you drew an object
and you drew a line through the
focal point heading towards the
mirror, where the focal point is
behind you.
 
So what's the fate of this?
 
Here is what you do.
 
Continue it backwards and I
claim it will come out like
this.
 
Why do I know that is correct?
 
Because I know the reverse of
that is correct.
In other words,
a parallel ray,
going like this will end up at
the focal point,
therefore this is to guide the
eye.
If you want to know what
happens with one particular ray,
join it to the focal point,
hit the mirror and come out
horizontal.
 
That's how you do the ray
diagram.
So this is a little delicate
thing, where if you really went
to the focal point,
you'd never hit the mirror.
You just do that to get the
direction, but you continue the
other way and you come out
parallel.
And the logic is,
if I draw it backwards,
I know that's correct.
 
But now look what's happening.
 
These rays are not focusing
anywhere.
In fact, this ray seems to come
from somewhere in that
direction.
 
That ray seems to come out in
that direction and the two of
them meet here.
 
That means to a person looking
from here,
from this side with the eye,
the two rays,
when you extend them backwards,
seem to emanate from this
point,
so it will look like there is a
candle behind the mirror,
emitting.
So that's a virtual image.
 
So virtual image is when there
is no real image,
but everything seems to come
from a point behind that.
Here's another variation they
probably exposed you to in high
school.
 
That is, what if I have a
mirror like this and I sent some
parallel rays into that guy.
 
You agreed this was not a
focusing lens.
It's not a focusing mirror,
because you can already see,
this light ray will come here,
take off like that.
That ray will go right back.
 
This will come here,
take off like this.
So it is not a converging lens.
 
But what is true is that if you
extrapolated all these lines
backwards, they seem to emanate
from this point.
You understand?
 
If you look from the other
side, if you have a sun,
and the sun is forming--sending
parallel rays,
the image of the sun will be
here.
It will be behind the mirror.
 
It will all seem to be coming
from here.
So instead of the lines
actually meeting at one point,
they'll seem to come from one
point.
So that's what happens.
 
But how do you do the
calculations in this problem?
Does anybody know?
 
What's the mirror equation with
this problem?
Pardon me?
 
Student: 
>
Prof: Yes,
you say f is negative.
It will turn out that
everything will work out,
1/u 1/v is still
1/f,
but if you want,
you can write this 1 over mod
f,
because this will be a problem
where the focal length,
if that is 5 inches,
f should be treated as -5
inches,
then everything will work out.
 
Similarly, a lens which looks
like this is a diverging lens.
If you send a parallel beam,
it will spread out like that.
But to a person on the other
side, everything will seem to
come from a point here.
 
You can do that lens again by
using focal length to be
negative.
 
So u and v and
f are positive for a
certain standard Mickey Mouse
problem where you have a
focusing lens and u
bigger than f,
then v is positive and
everything is positive.
But you can then move around in
this world of parameters where
v becomes negative.
 
That means it's a virtual image.
 
u is never negative,
because you put the object
where you like,
and you can also have the focal
length being negative if the
lens,
instead of being a focusing
one, is a convex one that causes
divergence,
or a lens which is causing
divergence also is described by
f negative.
Okay, there's one interesting
point.
You can take it or leave it,
so relax.
Don't kill yourself over this,
but I thought I should mention
it to you, which is the
following.
Suppose I have a lens like
this, and people tell you it's
going to all seem to come from
this point.
How do we know that's true?
 
You cannot use the Principle of
Least Time to argue.
Principle of Least Time says if
you have same starting point and
the same ending point,
then there can be many ways to
reach.
 
Then you have a focusing effect.
 
But here, there is no end point
where they all come.
They don't ever come to the
same point.
They seem to be coming from
here, but the rays never meet.
So starting from infinity,
you cannot na�vely use the
Principle of Least Time.
 
But I'll tell you why this is
still going to be the correct
answer.
 
Take any mirror surface you
like, some part of a mirror.
That's the normal to that
mirror.
Imagine that a light ray does
that.
This is the silver side of the
mirror.
And I ask you,
what if I silver this side and
shine light from the other side,
what will happen?
I claim the answer is that if
you continue this ray here and
continue that ray there,
then that's what light rays
coming from the other side will
do.
Why is that?
 
Because this light ray had i
= r.
You know that.
 
But the normal to this surface
is the same normal on the other
side, and by opposite angles
being equal, this is i
and this is r.
 
Therefore in any mirror coated
one way, if you continue the
lines to the other side,
they'll still obey i =
r.
 
Consequently,
take the following problem:
take the focusing mirror coated
like this.
We know these lines will come
and join here.
Then use the result I showed
you.
That will then show you that
the mirror image of that on the
other side will also obey i =
r,
because if you draw the normal
here,
if these two angles are equal,
those two angles will be equal.
Therefore I know this will lead
to that and that will match this
one.
 
Therefore rays coming this way
will go there,
but this one continued,
we know goes to f.
That's the reason why,
if you know the fate of one
kind of mirror,
you can find the fate if you do
it on the other face.
 
And this is the construction
that tells you why,
if you shine light from the
right, they will all seem to
come from this point.
 
That's something optional.
 
You can take it or leave it,
but I wanted to tell you where
that result comes from.
 
If you just do the algebra,
everything will work out if
f is negative,
but you need to know why that's
a valid picture.
 
Okay, so there's one final
gadget I want to talk about,
and that's the human eye.
 
So the human eye is pretty
impressive when you consider the
following situation.
 
So here you've got--look,
I don't want to even try this.
I'm just going to focus on the
lens in the human eye.
Here are the eyebrows and here
is my cornea over there.
And you put some object here.
 
It goes and forms an inverted
image somewhere here.
You must know that the image of
everything you see is actually
physically upside down in your
brain.
Did you know that?
 
Or you did not know that.
 
That's a very operationally
well defined thing.
If I can look into your cornea,
I look at the candles upright
here, the candle flame is here,
the flame of the candle will be
here like that.
 
But it doesn't seem to bother
us, because where do you really
see something?
 
It's not very clear.
 
Do we see it in the cornea or
do you see it in the brain?
There's a 1:1 correlation
between what's registered in the
cornea and what I run into in my
real life.
I'm walking around,
I bump into the upside down
table.
 
After the while,
the fact that it's upside down
in the cornea is not relevant.
 
I know if I take two steps,
I'm going to hit this object.
The brain has learned how to
translate that into what you
will encounter.
 
In fact, I'm told in some
bizarre experiment,
they put glasses on people so
that everything got inverted one
more time.
 
And after a few days,
those guys were just fine.
So the eye has got a lot of
software.
It's not purely hardware,
but software.
I learned it the hard way when
I had an eye operation and the
doctor pulled out the stuff and
said, "Ta-da!"
and I couldn't see anything.
 
So I was in a panic.
 
My doctor didn't seem
particularly worried,
and I said, "Are you
worried?"
He said, "No,
no, you'll be fine in a few
days."
 
Now I've heard that before so I
didn't believe that lie.
But slowly, I got better and
better.
So it's not that my eye was
operated any more.
He already fixed it,
but the brain began to
reprogram it with respect to the
new parameters.
You change the parameters of
the eye and he reprogrammed it
so the brain had to correlate
everything that it saw with
everything that's registered in
the brain.
So you just need to 1:1 map.
 
You can flip it any way you
like.
You can flip it once,
or twice or any number of
times.
 
After a while,
you will learn how to use it.
But that's not the only
impressive thing.
If you have an object here,
you go to the Lens equation.
You want the image to be at
that distance,
and that's not negotiable.
 
The distance from your lens to
your retina is fixed.
So no matter what you have,
as u varies,
you want v to be fixed
to obey this equation,
so how do you think that's
going to happen?
Pardon me?
 
You've got to change f.
 
You've got to change the focal
length of the lens,
and that's the amazing thing
about the human lens,
is that it's not made out of
glass.
It's made out of some jelly
like stuff, and there are some
muscles pulling it.
 
And if the muscles pull it,
it will become longer and
thinner, it will have one focal
length.
If they relax,
it will have another focal
length.
 
So as I look at an object which
is very far, that's the normal
relaxed state of the eye.
 
As the object comes closer,
it takes a certain effort to
focus on the object,
to keep reading whatever is
coming near you.
 
And that's the process in which
your eye is trying to change its
focal length.
 
But it cannot do more than some
amount.
The eye cannot focus anything
that is closer than a certain
distance called the near point.
 
And the near point is 25
centimeters, well,
approximately.
 
This is not a law of nature,
but this number has crept into
all the textbooks.
 
So 25 centimeters is all you
can do.
See, if you didn't have the
limitation, you don't need
microscopes.
 
You want to see a piece of
bacteria, just pull the sucker
really close to your eye.
 
Why doesn't it work?
 
Because what we really see with
the eye is the following:
if this is your eye,
what the eye cares about is the
angle subtended by the object
with respect to your eye.
We all know that.
 
So if you have the little
bacteria guy here,
you can see it just as well if
you can bring it that closely.
That will look as big as a
pencil 1 meter away,
except the eye cannot
accommodate that.
That's why you need a
magnifying lens,
and that's the last thing I'm
going to tell you,
is the principle of the
magnifying lens.
So here is the deal:
the eye really cares about the
angle subtended by some object,
but there is a limit to how
close you can bring it.
 
So if you have an object of
height h--you don't have
any lens.
 
You just have your eye and
you're looking at it--
then if that's the height
h and that's the angle
θ,
the best you can do,
put θ_0,
is
tanθ_0 =
h over this 25
centimeters.
 
If you can come to 15
centimeters,
θ_0 will
be even bigger,
but you cannot.
 
So for an object of given
height h,
the smallest distance is 25.
 
The biggest angle is
tanθ = h/25.
And for small angles,
we forget the tan and we say
θ_0 is
h/25.
That's the best you can do
without a magnifier.
So what does a magnifier do?
 
A magnifier,
this is the focal length here.
You put the object here,
somewhere there,
and you try to form the image
for that one,
so the parallel ray goes
through the focal point on the
other side.
 
Then, let's see,
a parallel ray going to the
focal point here,
that's where you're going to
have to get a virtual image now.
 
Understand?
 
Another ray you can draw is one
through the center that goes
straight through.
 
If you continue these guys
here, you will find there's an
image here, a virtual image.
 
In other words,
again what I'm telling you is
the rays will not actually meet,
but they will seem to be coming
from a point here.
 
So if this is
h_2 and this is
v, the θ you
are getting is
h_2/v,
which is h/u.
This is your actual height and
this is u.
So basically what you've done
is the following:
you managed to bring the object
closer to your eye than the near
point, if you want.
 
That's because the actual image
seems to be much further where
you can see it.
 
So the magnification,
the angle of opening you get
with the magnifier is this,
so the magnification for angle
is θ over
θ_0 and
that equals h over u divided by
h over 25.
That's 25 times 1/u.
 
25 is in centimeters.
 
That we're going to write as
1/f −
1/v.
 
Now you've got to be careful
here.
v is a negative number
because you're forming a virtual
image.
 
So I'm going to rewrite this as
the absolute value of v.
Therefore the gain you get from
using the microscope can be
written as 25 over f  25
over the absolute value of
v.
 
v is where the image is.
 
Now where do you want this
image to be?
Of course, the smaller the
v, better your
magnification is,
but v cannot be any
smaller than 25.
 
The best you can do with
v is 25,
in which case you will get 1 25
over f as your
magnification.
 
But if you're a jeweler working
with jewels all the time and you
want to see your piece of
diamond,
you don't want the image to be
at the near point of your eye,
because you're really
struggling to see it,
because that's the best your
eye can accommodate.
It's okay for a few seconds,
but it hurts.
Ideally you want to see the
object at infinity.
Therefore typically,
you pick v to be
infinity that means you put the
thing you're looking at right at
the focal point.
 
If you put it at the focal
point, image will go to
infinity.
 
Now you might say,
"Look, I've never seen an
object at infinity.
 
What are you talking
about?"
It turns out that infinity
could be 100 meters,
or 10 meters is good enough for
a lens.
It's far enough so that it
looks to be infinite.
How can you see something at
infinity?
The answer is,
as it goes farther out,
it also gets taller,
so you don't lose anything by
moving it farther out,
as long as it grows
proportionately.
 
It is just that distant objects
are when the eye muscles have to
do no work.
 
In other words,
the human eye is designed to
focus on objects at infinity.
 
It can manage if you bring it
closer by those muscles working,
but if you don't want the
muscles to be tired,
you want to have the image at
infinity.
So you take this little thing
you're looking at,
and you use the lens to create
an image at infinity,
and again, forgetting the extra
1, most people don't bother with
the extra 1.
 
So the point to remember is the
angle of magnification is really
given by 25 centimeters divided
by the focal length of your
lens.
 
So if you have a lens,
a focal length 2.5 centimeters,
you get an angle or
magnification of 10.
Most of the time,
the magnification you get is 3
or 4 or 5.
 
Any of these things you buy in
a supermarket to read
something--
you don't know what I'm talking
about,
but you will in a few
years--then that will give you
magnification of 3 or 4.
All right, so this is the end
of geometrical optics,
and next time I'm going to tell
you why you want to change the
rules of the game.
 
But before you go,
I won't let you go until you
tell me why I would want to
change anything?
Why would anybody change this?
 
The minute you answer me,
you go to lunch.
What's the reason I would
change all this?
Yes?
 
Student: 
>
Prof: Yes, he saved you.
 
There are some experiments that
contradict this.
That's the only reason to
change anything,
and I'll tell you what they are
next time.
 
 
