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GEORGE BARBASTATHIS: The key
point of the previous lecture
was to simplify the
equations for refraction
from a spherical surface,
such that we can write them
in a simple form.
OK, what I just
said was redundancy.
But we want to simplify them.
I should have said we
want to simplify them
so that we can write them
in a linearlized form.
And the linearlized form
is actually very useful
because with this form,
as I'm about to show,
we're going to use a very
familiar form of math, namely
linear algebra matrices.
And we can retrace
through almost arbitrarily
complex optical systems.
Of course, there's a downside
that if write trace this way,
we do not really get
very accurate results.
Our results, as we
discussed last time,
are only accurate within
the paraxial approximation.
On the other hand, if we do this
analysis even approximately,
we can get an idea of how
the optical system behaves.
And then if we get this
approximate design,
this approximate idea,
then we can plug the system
into a more sophisticated
numerical tool,
for example, Zemax
or CODE V. There's
a number of different
optical design
and software suites available.
And then we can actually
get a more accurate analysis
of how the optical
systems would behave.
So the other reason we
write these matrices
is that they give
us actually quite
a bit of information
about the basic properties
of optical systems.
So it is really worthwhile to
do this first, every time we
face an optics
problem, before we jump
into the numerical analysis.
And for those of you who really
have a real-life problems
to solve and you really
need to go to software,
we will tell you
later during the class
where we'll give you some
information about what
types of software are available
and what you can do with those.
But generally, this idea
of using paraxial optics
is very powerful in order to
understand optical system.
So the summary then of what
paraxial optics standard is
summarized.
These are shown on
this slide over here.
The most important
is the results
on the right that allows you
to write the angle of departure
of a ray to the right-hand
side of the card interface
and the elevation of
the ray, with respect
to the right-hand
side interface.
It allows you to write
them as functions
of the corresponding quantities
on the left-hand side
of the interface.
And as you can see
there, it actually
looks like it two by two
set of vectors, which are
related by a two by two matrix.
And the matrix are
not very interesting.
As you can see, the most
relevant, the most interesting
element is the element on
the first row, second column,
which equals the
index of refraction
to the right of the interface,
minus the index of refraction
to the left, divided by
the radius of curvature
of the spherical interface.
And, of course, the
radius of curvature
is simply the radius
of the sphere.
We're talking about the sphere.
And then the other result
that we put here and looks
like a propagation
through free space, it
looks kind of too much trouble,
really, for what it's worth.
But you will see in a second
that this is actually a very
useful tool to have in mind.
One of the thing that says is
that the ray, as it propagates
from the left over a
distance d, if it propagates
through free space,
then what happens
is the elevation
of the ray changes,
but the angle of
propagation does not change.
So this verifies what we
know from Fermat's principle
that a ray propagating
in uniform space,
it minimizes its path.
Therefore, it must
propagate in a straight line
because the straight line is
the minimum path between two
points.
Now, you might be wondering.
We talked last time about
ellipsoidal surfaces,
hyperboloid surfaces,
and spherical surfaces.
You might wonder, why
did we do this analysis
in the context of the sphere?
Well, that says it doesn't
matter very much because all
of these three
possibilities, ellipsoids,
hyperboloids, and
spheres, near the axis,
they actually all
look kind of the same.
If you compare a sphere--
so I should specify now
what I mean by axis.
So in the case of a sphere, as
Professor Sheppard pointed out
last time, in the
case of a sphere,
really, any lines going through
the center of the sphere
acts as the optical axis.
But for the hyperboloid and
the ellipsoid, that's not true.
There's a very well-defined
major axis and minor axis
for the surfaces.
And it turns out you can match
the curvature of these surfaces
with the curvature of a
sphere so that [INAUDIBLE]
a common axis, they
all look the same,
as I've done in this diagram.
It's a little bit
of an algebraic mess
to actually work
out how to do this.
So I didn't do it in this slide.
The idea is basically to do
a Taylor series expansion
on the explanations that
describe all those surfaces.
And then when you do
that, each one of those
is described by a different
set of coefficients.
So what you can
do then is you can
match the coefficients in
order to match the curvature.
So I will let the graduate
version of the class, those
of you who are taking
2710, as a homework,
I will let you work
out one of these,
a simple case for matching--
I forget.
I think it is matching
an ellipsoid to a sphere.
So let you do that.
What I really want you to
live with out of this lecture,
for now, is that the
paraxial analysis remains
valid for all three surfaces.
OK?
As long as we
match the curvature
of the sphere, ellipse,
and hyperboloid,
and we constrain ourselves
to the paraxial regime,
then our results remain
approximately correct.
So this is kind of
another useful property
of the paraxial
approximation that it is not
only valid for a
sphere, but it is also
valid for other
surfaces of revolution,
provided the curvature near
the center of the surfaces
matches the curvature of
the corresponding sphere.
OK.
Yeah?
Push the button, please.
Yeah, so that's a good question.
AUDIENCE: Could you
repeat the question?
We didn't hear the question.
GEORGE BARBASTATHIS:
The question
was if there's an
angle that we'll
consider as a cutoff
for the breakdown
of the paraxial approximation.
So the accurate answer is zero.
Any angle other
than zero violates
the paraxial approximation.
In actuality, you can calculate,
with the help from this Taylor
series, actually, that we
saw in the previous slide,
you can calculate the
accuracy of the approximation.
So treat it as a first order
of a Taylor series expansion.
And then you can calculate what
is the next order accuracy.
So this is actually
algebraically quite complex.
And typically, we
resort to software
to do it for us,
numerical software.
As a rule of thumb, anything
in the range below 30 degrees,
you can kind of trust the
paraxial approximation.
But that's a rule of thumb.
In some cases, it's not true.
In some cases, you
need more accuracy.
So the guideline is
something like that.
But, for example,
for sure, if you
have a ray propagating
at 80 degrees,
you know for sure that
approximation breaks.
If it goes to 1 to 3 degrees,
then most likely, you're OK.
And then there's a
gray area that you
have to be a little
bit more careful.
If the accuracy demanded
by the application
is much higher, than you
have to be more careful.
There's also surfaces that are
not very well-described by this
at all, for example, a cubic.
If you make a
refractive interface
that looks like a cubic surface,
would look kind of like this.
And then this totally
breaks down, right?
So you have to go to then to
non-paraxial [? methods. ?]
Another thing to discuss is
in one the previous equations,
I wrote down ray
elevations and ray angles.
But I was not very
careful about defining
the signs of these
angles, namely,
whether they go up, or down,
or left, or right, and so on.
I was deliberately left
this little bit vague.
Again, this may appear like
too much trouble for nothing.
But you will see, as we go
later on with the lecture
today, you will see that this
is a very useful and highly
nontrivial aspect of
geometrical optics,
namely, the conventions
of the signs,
for when the quantities
that we're dealing with
are positive or negative.
So bear with me
for a second, and I
will show examples of
how this works later.
So I want to read this because
I don't know of a better way
to do it.
I will read it.
And then as I read, please
look down at the slide.
And you will see on
the left-hand side
I have examples of all these
qualities being positive.
And on the right-hand
side, we have
example of all of these
quantities being negative.
OK, so the convention number
one is that the light always
travels from left to right.
So you know immediately when
someone draws an optical system
or attempts to draw
an optical system,
you know immediately if this
person knows their optics
or not by the way they
pick the light propagation.
If they show light rays
going from right to left,
it means they don't know optics.
So if they learned optics, they
learned it very deficiently.
AUDIENCE: [CHUCKLES]
GEORGE BARBASTATHIS: So
always light goes from left
to right in this diagram.
There's one
exception, of course.
Can anybody guess?
One exception that I'm
not dealing with here.
Mirror.
Push the button.
AUDIENCE: Mirror.
GEORGE BARBASTATHIS:
Mirrors, yes.
So in this diagram over here,
I do not consider mirrors yet.
We will deal with mirrors
later, in about a week.
And then we'll see a revised
set of sign convention.
But as long as there's no
mirrors in your optical system,
if you have what is called a
purely dioptric optical system,
then the light is always
going from left to right.
Taking this as a
given, then we'll
define the positive and
negative radii of curvature
as solved over here.
So if the surface is
convex towards the left,
then we'll call it positive, OK?
Then this sounds a
little bit weird.
It says that
longitudinal distances
are positive if
pointing to the right.
I'll show you an
example a little bit
later where distance can
become negative by pointing
to the left.
But for now, take this
definition for granted.
The same is for
lateral distances
that this measures perpendicular
to the optical axis.
If they point up, then
they're called positive.
If they point down,
they're called negative.
And finally, angles
are positive if they
are acute in the
counterclockwise sense,
with respect to
the optical axis.
So clockwise, we have to be
very careful when we define it,
so we don't get
any mirror effects.
So hope that the cameras
that we're using here
do not make any mirrors.
But counter-clockwise is
basically like this, OK?
If in this direction
you get an acute angle,
then the angle is positive.
If you have to go all the
way around to get your angle,
or another way to put
it is if you can make
it acute by going
clockwise, then you
call the angle negative.
OK.
So this set of conventions is
self-contained and consistent.
To add to the confusion,
set and optical text books,
they use the opposite
sign of conventions.
It is also possible, by the way.
The set of
conventions I use here
is consistent with
a Hecht text book
and the majority of text books.
There's a minority
of textbooks that use
the exact opposite conventions.
But we will not deal with those.
OK, using these
set of conventions,
we can define the power
of an optical system
based on whether rays
that propagate outwards--
by outwards, I mean a
ray bundle that expands--
whether a surface,
a spherical surface,
causes the expansion to
become slower or faster.
If the surface causes the
expansion to become slower,
as in the top diagram
over here, then
the surface is said to
have positive power.
If the surface causes the
expansion to become faster--
that is, it takes
the expanding bundle
and spreads it
further outwards--
then the surface is said
to have negative power.
So now, let me proceed to define
a specific optical system.
And then we'll see more examples
of this thing happening.
So the system I would like
to consider is a lens.
And this is actually the
first example of a lens
that we see in the class.
And I will take a very
special case with a lens
that we refer to as thin.
What's a lens?
A lens is simply
a piece of glass
that has been shaped to have
spherical surfaces on one
or both sides.
Typically both, but
we'll see examples
of lenses that are actually
flat on one surface
and curved on the other surface.
So when we call the lens
thin, well, we call it thin.
Consider an expanding
spherical wave that
propagates through the lens.
And consider the
maximum angle that can
be subtended through the lens.
In other words, angles
larger than this one
would actually miss the lens.
So therefore, this alpha maximum
is sort of the biggest angle
that we have to worry about.
Well, what does the
paraxial approximation say?
It says that this angle
has to be relatively small.
So assume that
this is true, then.
The lens also has a thickness.
So this thickness, I denoted
as T in this diagram over here.
So provided this
thickness, if we
do a little bit of
an analysis here,
we can calculate this thickness
relative to the radius
of curvature of the sphere.
And the angle phi that
is defined as shown,
from the center of the
sphere relative to the edge
of the ray, that subtends the
maximum angle alpha maximum.
OK?
So we can see the definition
phi for this angle
and the radius r.
So if, given this
angle of arrival
is alpha maximum, if it is
true that the angle phi is also
very small, then we can
consider the lens to be thin.
So we can basically pretend
that the space between these two
points on the lens, the
space between the entrance
to the optical axis and the
exit of the optical axis
through the lens, we
can neglect the space.
This is what the thin
lens approximation says.
That's very convenient because
we can now analyze the lens.
As you can see, if I
neglect this space,
then the ray elevation
before and after
the lens remains the same.
And also can
neglect a refraction
that might happen
within the lens.
So basically, we
have the system where
the ray arrives then is bent
in one step and then propagates
outwards at a slightly
different angle.
So in order to analyze the
system, now, what I will do
is I will break it
into two components.
One of them is the curved
surface on the left-hand side.
And the second is the curved
surface on the right-hand side.
OK.
And now, for each
one of those, I
can define an angle on the right
and then angle on the left,
as I have done in the
simple [INAUDIBLE]
of the refractive surface.
Now, each one of those is
a simple refractive surface
interface that we
know how to deal
with from the previous lecture.
And I can add equations.
Remember the equation
for the refraction
from the spherical surface.
I will add it here on my
pad, just to remember.
So it goes like this.
Hopefully, someone will notice
and they will project it.
So n to the right, alpha to the
right, x to the right equals--
can anybody see?
OK, we're having technical
difficulties here.
Oh, there we go.
AUDIENCE: Can you lower the
piece of paper a little?
Yeah, got it.
Yeah.
GEORGE BARBASTATHIS:
So this equals 1, 0, 1.
This is easy.
And then we have apply our
mnemonic, minus n to the right,
minus n to the left, over
the radius of curvature,
times the same
property for the left.
And left alpha left, x left.
OK.
Now, what we're have
to do is we have
to apply this rule successively
for the two interfaces.
So here's the question that we
get for the second interface.
So this case, n
to the right is 1
because to the right of
this interface, I have air.
And n to the left is n because
to the left of this interface,
I have glass.
OK.
Now, what I will do is I
will apply the same rule.
But I will now apply it
on the first interface.
And if I do to the first
interface, I get this equation.
Now, notice that
in the numerator
of the element in the
first row, second column,
the quantities are reversed
because now n to the right
is the index of refraction of
glass, whereas n to the left
is the index of refraction
in air, that is 1.
So therefore, I get this matrix.
Now, what it can do, I can
cascade the two matrices
because I can substitute
the vector n over 1 x1 that
is on the top equation.
I can substitute it from
the second equation.
And if I do that, of
course, I will get
a cascade of the two matrices.
And notice that
in the cascade, it
is very important to
remember that in the cascade,
the matrix of the last element
to the right appears left.
And that is very important
to remember and not
confuse because as you know,
matrix multiplication is not
commutative.
If you mess up the
order of these matrices,
you will get the
wrong result. So you
have to remember that
you start from the right.
That is, you start from the
end of the optical system
where the rays
depart, and you start
cascading the matrices going
backwards towards the beginning
of the optical system.
So this is what I've done here.
I've put the matrix
corresponding
to the second interface.
This matrix appears first.
And the matrix corresponding
to the first interface
appears second.
And then I cascade them.
I do the matrix
multiplication, which
is a relatively
straightforward task to do.
And I get the second,
the last equation
that appears on the slide.
OK.
Now, let's go ahead and
interpret this equation.
It looks very nice.
But let's see what
it really means.
Also, before I move on, let
me notice one more thing.
Notice that in the
vectors of the ray, angles
of the ray elevations, the index
of refraction has disappeared.
It has not really disappeared.
It is simply because the angles
alpha right and alpha left
are in air.
And therefore, the index
of refraction is 1.
That's why there's no index of
refraction in that location.
OK.
Let's look at this
equation again.
The top quantity that appears
relatively complicated,
it has a name.
It is called 1 over
the focal length.
Now, you might
wonder, why did they
call this the focal length?
I will justify that in a second.
But if you believe me that
the focal length follows
from this equation,
then this equation
is known as the lens
maker's equation.
And it is very useful
because it gives you
the focal length of the thin
lens as a function of the three
quantities that define the thin
lens, that radii of curvature
to the left and to the right
and the index of refraction
of the lens.
And what I want to
emphasize, again,
is that this equation
is valid if the lens is
surrounded by air.
If the lens is surrounded by
another material, for example,
water, or oil or
something with index
of refraction different
than 1, then this equation
does not hold anymore.
And as a bonus, in the next
homework, which has already
been posted, you will actually
derive the corrected equation
if the lens is immersed
in a material that
is different than air.
But for now, this
equation is correct
because we put our lens in air.
So therefore,
everything is fine.
And this is a lens
maker formula.
So then let me justify that
this quantity that appeared over
there and has units 1 over
distance, why on over distance?
Because this quantity has 1
over the radius of curvature
in the denominator.
So therefore, it
is 1 over distance.
So let me justify why
this quantity is actually
the focal length of the lens.
So to do that, let me
consider a ray that
is arriving from infinity
horizontally at an elevation x1
with respect to
the optical axis.
So this ray, I can retrace
through the system.
Basically to retrace
through the system means
that I have to propagate
this ray by a distance z.
And then I have to find out
what is the ray angle alpha
2 with respect to the optical
axis and the ray elevation x2,
also with respect
to the optical axis,
as function of this
propagation distance z.
So now you can see why it
was convenient to define
the matrix for propagation
through free space.
Because now, in
order to figure out
where this ray lands, what is
its elevation and propagation
angle, all I have
to do is cascade
the matrix that
describes the lens
with a matrix that describes
propagation through free space.
And we have seen this matrix of
propagation through free space.
It is simply 1, 0
in the first row.
And then the distance
and 1 in the second row.
And there's actually, if you
look back at your equation,
it is distance divided by
the index of refraction.
But, again, because we are
propagating in air here,
the index of refraction is 1.
So we don't have to worry
about this at the moment.
And because the free space
is following the lens,
the matrix corresponding
to free space
will actually appear first
before the matrix corresponding
to the lens.
So this is the cascade
[INAUDIBLE] correctly here.
And we can solve this.
And we could find that is given
by this equation over here.
So what we observe now is
that if we set the propagation
distance to-- first of all, I
did this in the general case.
So as the ray entrance angle, I
left a general symbol alpha 1.
But I said already that the
ray is arriving horizontal.
That means alpha 1 equals 0.
So therefore, the
elevation of the ray,
with respect to
the optical axis,
is given by this equation.
It is simply x2 equals
x1, 1 minus z over f, OK?
And we can see very easily over
here that if z equals f then
x2 will equal to 0.
So this means that if
I let the ray propagate
by a quantity equal to
this magical amount that
appeared in the lens
maker's formula, then
the ray will hit the axis.
So the ray will land
on the optical axis.
And moreover, we can
very easily verify
that this happens independently
of x1, because as you can see,
x1 is outside the
parenthesis here.
If I set this quantity
to equal 0, then
independent of what x1
was, we will actually
get all of the rays to
go through the same point
of the optical axis.
So this is a central focus.
You can see that the ray
bundle that arrived parallel
from infinity actually
now comes to focus
at this location of the
optical axis because
of this simple equations
that we just described.
And this property over
here justifies the name
focal length.
It is the length at which
the rays come to focus.
OK.
The inverse of the focal
length, just on its own,
as a sort of standalone
quantity, has a name, also.
It is called the optical power.
And that is a little bit
confusing because in our minds,
usually, power is
measured in what?
Well, this power is
not the usual power
that we measure as
energy per time.
It is a different power.
And for historical reasons, we
use the same name in optics,
but it is measured
in inverse meters.
And the inverse
meters, they're also
known as diopters in optics.
And they define
the optical power
as the inverse of
the focal length.
So, for example,
if the focal length
is 1 meter, which is a
pretty long focal length,
you call it one diopter.
If the focal length
is 10 centimeters,
it is actually 10 diopters.
If it is 1 centimeter,
it is 100 diopters,
and so on and so forth.
Very often, people like
myself who are myopic--
is anybody in the class,
is any of your myopic?
Anybody use corrective lenses
because you have myopia?
OK.
Do you know what is the
power of your prescription?
AUDIENCE: 0.75.
GEORGE BARBASTATHIS: 0.75,
you're pretty likely.
You have very small correction.
AUDIENCE: [INAUDIBLE]
GEORGE BARBASTATHIS: That's
a pretty small correction.
Can you say it in
the microphone?
AUDIENCE: 0.75 in the one
eye and in the other, 1.25.
GEORGE BARBASTATHIS: OK, 0.25.
Is it 0.25 or minus 0.25.
AUDIENCE: Minus 1.25.
GEORGE BARBASTATHIS:
Minus 1.25, OK.
So that is also in diopters.
AUDIENCE: Yeah.
GEORGE BARBASTATHIS: It
means that the focal length
of your lenses is approximately
80 centimeters, right?
AUDIENCE: Right, mhm.
GEORGE BARBASTATHIS: Yeah.
Do you have a--?
Yeah.
AUDIENCE: Mhm.
GEORGE BARBASTATHIS: OK.
AUDIENCE: Yes.
GEORGE BARBASTATHIS:
So we will see later
why then correction for
myopic people is negative.
I'm also myopic.
And my correction, I actually
have pretty high correction.
Mine minus 7 and
minus 5 diopters
in my left and right
eye, respectively.
So it means, again, that the
focal length of my glasses
is approximately minus
1 over 7, which is--
I don't know, what--
17 centimeters or something.
OK.
So that is the power.
Now, why we use the
term optical power?
Actually, it is best
explained if you
look at the final animation
on this slide, which
is what happens if the ray
bundle comes from infinity
again, but it is not horizontal.
It is coming at
an angle alpha 1.
So now, the angle alpha
1 is not 0 anymore.
But I let it have a
finite non-zero value.
And if you do that, then
you find relatively easily
from the equation over here.
Let me write the equation.
We have that x2 equals alpha
1 z plus x1, 1 minus z over f.
OK.
So if I set z equals f
to the above equation,
then I get x2 because
alpha 1 times f.
This term will
disappear, of course.
This will go to 0.
And z will become f.
So get x2 equals
alpha 1 times f.
So this is the equation
for the elevation
of the focus for a
parallel ray bundle
that is arriving
non-horizontal from infinity
at some angle alpha 1.
So you can also
write this equation--
it is not on the slide--
I will just add it
on the white board.
x2 equals alpha 1 over the
power, the optical power
of the system.
So the power is kind
of like a lever.
It tells you as you
change the angle alpha 1,
it tells you how does the
elevation of the focus change,
with respect to
the optical axis.
So it is basically,
again, if your thinking
is if you order transduction
amplification or something
like that, it tells
you as you change
the angle that the rays arrive
into the optical system,
how does the focus move with
respect to the optical axis.
OK.
Now, basically, these
equations that we just
described, let's see some
different cases of what
would happen to the rays.
So what I've done here is I put
the different types of lenses
together with a simple
spherical refractor.
And the different
types of lenses
are plain or convex where
you have one convex surface
followed by a
planar surface, Then
biconvex where you have
two planar surfaces,
then planar concave
and biconcave.
So you can see from these that
if you have a ray bundle that
is arriving horizontal, now,
horizontal from infinity,
you can see that what
will happen at the output
is actually different, depending
on the type of refraction
that you get in the
different surfaces.
So I will let you work out,
based on the equation, the lens
maker's equation, I will
let you work out why
these cases are all different.
But what I would like to
do is actually work out
the location of the
image, depending
on the different
cases, and also,
see how to ask with respect
to the sign convention.
So starting, for example,
with a biconvex lens,
we can see that
the biconvex will
focus the parallel bundle to
the right-hand side of the lens.
This is the case that we
have implicitly done so far.
In my drawing so far, I have
assumed that this is the case.
However, you can see that
if I have a biconcave lens,
like the one shown at
the bottom of the slide,
you can see that each
one of these rays
will actually expand
outwards after it
passes through the lens.
So in a sense that
we described so far,
this lens does not
really focus the rays.
It creates a
divergent ray bundle.
You can get a sense of focus if
you extend the divergent rays,
if you extend them backwards
towards the optical axis.
If you do that,
you will actually
see that the rays meet.
They do meet.
But they meat on the
left-hand side of the lens.
So now, this is an example
of focusing that happens not
on the right-hand
side of the lens,
as our equations have shown,
but on the left-hand side.
And we can now justify
this fact if you actually
define z to be negative
and f to be negative.
So this is what we
call a negative lens,
or a lens with
negative focal length,
or a lens with negative power.
And if you do that
now, then you will
see that the focus will
appear at the left-hand side,
consistent with our
sign conventions.
And this is what we call a
virtual image, as opposed
to the case of the biconvex lens
that we'll call a real image.
So this is an example of a sign
conventions, not convections,
the sign conventions in action.
Are there any
questions about this?
This is a point when I often
get a lot of questions.
Let me move on to
a different case.
And if you think of a question.
Please interrupt me,
or say something aloud.
Push on your microphone
button first.
Let me do a slightly
different case, which
is what happens if you try to--
so in the previous
example, we saw
how a lens can take
an object at infinity
and focus it to
a finite distance
equal to the focal length.
What I will do now
is I will argue
that the positive lens actually
can take a point object located
at one focal distance to the
left and image it at infinity.
Now, why is that true?
I will do this in a second.
But you can very easily
justify it to yourself
if you simply reverse the
orientation of the rays
in the previous case.
What we did before is we
had the ray bundle that
was starting from infinity.
And the lens was focusing it.
We can very easily reverse
the radiation of the rays.
And if we do that,
then we can see
that if I have a point
source over here,
it will actually get
imaged at infinity.
Now, of course, according
to my sign conventions,
this is not the proper
way to write it.
We have to make sure
that the light propagates
from left to right.
So therefore, the
correct to write,
it is not what is showing over
here, simply by convention,
not because this is
physically wrong.
But it does not
obey the convention.
So we never actually
draw something like this.
What we draw is what is shown
on the slide where the rays are
propagating from left to right.
They have a focus
that is a point
object on the left-hand
side of the lens.
And then the lens, of course,
will collimate these rays.
They will convert them from
a divergence spherical wave
to a parallel plane wave.
And therefore, we
limit at infinity.
The same thing can be said
about the negative lens,
except in this case, the ray
bundle arriving at the lens
has to be convergent
and has to come
to a focus on the
right-hand side of the lens.
So therefore, in the same
way that this type of lens
formed the virtual image in the
previous case, in this case,
we talk about the virtual
object for this type of lens.
OK.
So let me now try to put
this all together so it
can make some sort of sense.
Again, recall the
first sign convention,
which sees that the light
propagates from left to right.
If it propagates from
the left to right,
it means that the object should
be to the left of anything that
is of interest.
And the image should be
to the right of anything
that is of interest
to the optical system.
So that is why if an
object is to the left
of the optical element, as
it is in this case, then
we say that the distance from
the object to the element
is positive.
If, on the other hand,
an object happens
to be on the right of
an optical element,
as it happens in this
case with a negative lens,
then we say that the
object is virtual.
And the image from the
object to the element
is negative, because it has
to be on the right-hand side,
whereas in the proper case,
it had to be on the left.
That's why we put
a negative sign.
Now, similar things can be
can be said about an image.
Except, now, the image
is properly positioned
or the right-hand side
of the optical element.
So everything is really
on the right-hand side,
as it happens with a
positive biconvex lens.
Then the distance from
the lens to the image
will be referred to as positive,
whereas on the other hand,
if it happens to be
on the left, as it
is for the case of
the negative lens,
then this image will be
referred to as virtual,
and the distance will be
referred to as negative.
Finally, and I will
conclude with this,
and I will let Pepe do
his demo, is how this
applies to off-axis objects.
So remember we
derived this equation,
x2 equals alpha 1 times f, for
an object that is at infinity.
Now, for a positive lens,
because the focal length
is positive, if
the angle alpha 1
is also positive,
as shown here, it
means that x2 must
also be positive.
And this is indeed
what we can verify.
If we apply Snell's
law repeatedly
for each one of these
rays, we will see
that indeed this is the case.
On the other hand, if
I use a negative lens,
where the quantity f, the
focal distance, is negative,
but the angle of arrival
is still positive,
then I will get x2 is negative.
That is, x2 will also appear
below the optical axis, OK?
So this virtual image
not only appears
on the left of
the negative lens,
but it is also below
the optical axis.
OK?
Now, what about an
off-axis image at infinity?
I will skip this derivation.
It is very similar
to the derivation
that we did before for the
case of an object at infinity.
Basically, what I've done
here is I have reversed it.
I have used an
object at a distance
f to the left of the lens.
I have cascaded the propagation
matrix for this free space
propagation to the propagation
method corresponding
to the lens.
And I have proceeded to solve.
If I do this, I will
let you do it at home
and then come back
tomorrow if something
is unclear about this.
But what I've done
is I have solved it.
And when I find an equation
that looks like this,
I find that the
angle of propagation
of the ray bundle to the
right-hand side of the lens
actually equals minus the
elevation of the object divided
by the focal length.
Or another way to
write this equation is
alpha 2 equals minus
x1 times the power
of the optical
element, according
to the definition of power
that I remind you by definition
equals 1 over the focal length.
OK.
So there's a minus
sign now, which
means that if we apply this
equation to our positive lens,
the ray bundle that will
emerge from the lens
will actually now
propagate downwards, OK?
Because in this
case, x1 is positive
and the focal
length is positive,
alpha 2 has to be negative.
And negative alpha 2
means that ray bundle
has to propagate downwards.
OK?
If I do the same exercise
for the negative lens,
then we can see, again,
very easily that because x1,
If I pick it to be
positive, f is negative
and I have a spare negative
sign from the equation.
Then I can see that the
propagation angle, alpha 2,
will be positive.
That is, the image
at infinity will
be on the positive side,
above the optical axis, OK?
I hope the positive and negative
signs are not confusing.
Let me restate.
When I learned
optics, I found this
to be also horribly confusing.
So I realize and I sympathize
how at the beginning,
this can be a little
bit confusing.
I've done my best
here to present them
better than the professional who
taught them to me solve them.
So I hope, because I sort
of used my own confusion
as a guide, I hope
that I managed
to make it a little
bit clearer for you
if it is the first
time you see it.
But, of course, it
will become even more
clearer as you practice, OK?
So the homeworks and the
examples that we see later,
they will make it
progressively clearer.
For now, what I want
you to do between now
and tomorrow's lecture is
to go back and make sure
that these pictures that I've
drawn, that they make sense
with Snell's law.
In other words,
what you do is you
go to the spherical surface.
You imagine a normal to
the surface at each point.
You apply Snell's law.
And you convince yourselves that
the rays bend the correct way.
And this is consistent with the
way they image is formed, OK?
And then you apply all the
methods, all the equations,
and so on and so forth.
I broke my promise and I did not
leave Pepe with his 5 minutes.
But that's because
it took us 5 minutes
to fix the figures, the
display, or whatever
was happening at the beginning.
So hopefully, you guys can
stay for a little bit longer,
so we can do the demo.
Is that OK?
PROFESSOR: Sure.
I'll try to do it.
We'll try to do it fast.
GEORGE BARBASTATHIS:
Also, and look at the--
PROFESSOR: So now--
GEORGE BARBASTATHIS: Yeah.
PROFESSOR: We can switch
to their cameras, please.
OK.
So here we go.
OK so we tried to
set up these webcams
because last time it was a
bit hard to see the demo.
So this demo has
actually two parts.
But today, we're just going
to show one part, namely,
the refraction.
And we're going to basically
witness how Fermat's principle
that we've been learning in
the past couple of classes,
the Snell's law, applies
in these systems,
like lenses or prisms that
we've been learning so far.
So before actually
changing the exposure,
let me just show the set up.
So we are going to be focusing
on this half of the set up.
So hopefully, you
can see it fine here.
Let me just move this window.
So to avoid any problems
about chromatic dispersion
or wavelength-dependent
behavior,
we're using a laser,
a green laser.
So this component
here is the laser.
Then this is a new
component that we're
going to call-- it's
called spatial filter.
This, for now, it has a
very interesting property
that we'll learn in the
second half of the course
when we learn about
Fourier optics.
For this first part, just think
about this produces a very nice
point source object.
So therefore, after
this lens, there's
going to be a spherical
wave emanating, propagating.
So what we want to
do after that is
to convert that spherical
wave into a plane wave, right?
So this is what this
lens here is doing.
And that's what we
call collimation.
So after this lens,
as you can see-- well,
you'll see it in a second.
OK.
So now, I reduce the
exposure of the camera,
and I can see the rays.
Now, I see parallel rays.
Hopefully, you can
see them clearly.
Could you zoom in a little bit?
OK.
And we have basically
parallel rays coming out
of the collimated lens.
So now, this brings
another interesting.
As you can see,
our eyes, and we'll
see it next lecture,
how they are really
robust to focusing in
different conditions,
to change illumination.
And more particularly,
they're adapted
to high-contrast in light,
like very high intensity
in the light and low intensity.
Your eyes can see it
fine, whereas this camera,
we just need it to
change exposure,
so you could be
able to see that.
So this is a problem
of dynamic range
that occurs in optical systems.
So, all right, let's look
at the first component.
It's this lens here.
I'm going to put
cylindrical lens.
It's a plano curve lens
here that it's cylindrical.
Basically, it means
that the surface only
depends on the x-coordinate.
So we'll put it here.
And [INAUDIBLE] wavelength.
GEORGE BARBASTATHIS: If
I may interject here.
Pepe, if I may
interject something?
PROFESSOR: Sure, sure, sure.
GEORGE BARBASTATHIS:
That equations
that we derived,
strictly speaking,
they're for cylindrical
lenses, right,
because we only use
x in our equations.
PROFESSOR: Yeah.
GEORGE BARBASTATHIS:
If you have an x and y,
then you get the
proper spherical lens.
So strictly speaking, this is--
PROFESSOR: Yeah.
GEORGE BARBASTATHIS:
--what that equations--
PROFESSOR: Exactly.
GEORGE BARBASTATHIS: OK.
PROFESSOR: So now, you can see
now the ray tracing, but now
done optically.
You can see how the rays started
focusing in to this point here.
Well, in this case, now, since
this is cylindrical lens,
we see that it's a line.
And I can put this card here
and we see that we actually
form a line instead of a point.
Other nice things to
see about this lens
is that so far, we've
been ignoring reflections
that occur with this lens.
But let me just show you
some of the reflections
in this cardboard here.
So we'll see in the
second part of the course
that these reflections
are occurring,
transitioning from the air and
glass interface and back again.
So in these interfaces are
very sensitive to the angle
of incidence and to
the refractive indexes
of both the lens and
the surrounding media.
So in this case, you can see
that quite a lot of power--
so this is the incidence wave.
And you can see
quite a lot of power
is actually going to this side
because this lens might not
be AR-coated, or coated with
an anti-reflection coating
as a better lens would be.
So now, let's witness Snell's
law with this prism here.
So you see the little rays
here that are going straight.
So now, I introduce
this prism shown here.
It looks very dark.
But believe me,
there's a prism here.
And then as soon as I started
putting it into the system,
you see how the light starts
bending to one side more.
And then you can see how
basically, the tilted surface
is bending the light, and I
can just change the angle.
And let's now see
the other phenomena
that we learned in the
previous class, which is TIR.
So I still rotate this more.
I make the angle to be larger
than the critical angle,
and boom.
All the light gets TIRed
and we can actually see, now
that it's getting to this side.
But in this side of the
prism, there is a diffuser.
And that's why you see
the diffused light here
very bright.
So I'm going to do
it one more time.
Without TIR.
All the light goes, escapes
out straight through.
And now, I start
rotating this more.
And then you see TIR here.
There are any questions?
So I also brought a parabolic
reflector for you guys,
like the people here
in MIT if you guys
want to see the reflection,
how you see your face.
Look at your face with
this parabolic reflector.
It's actually quite
funny to see the virtual
and the real images.
And in this case,
you'll see, and we'll
learn in next class, how
the virtual image generated
by such a reflector, it's
actually not inverted.
It's erected, as it's called.
So I'll leave it here, so
you can come after class.
