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GUEST SPEAKER: Good morning.
Today we're going to start with a demo,
and we're going to show experimentally
some of the principles we've learned
about 4F systems in the past classes,
and actually we're going to see a comparison of an incoherent
versus a coherent 4F system when imaging resolution target.
So but since now we have all the tools to understand,
finally, the set up here, let me just describe it.
I've been showing some of these components
over and over the past demos, but I
guess now we can understand it, so
let me try to focus the camera.
So let me start with this is a green laser.
So the first component that we have here
is [? anti ?] filter that its job--
thank you-- so this component, its job
is to control the intensity of light
so we can attenuate it in this case for experiment.
This is an interesting component called a spatial filter
that is composed of a microscope objective and a pinhole,
and now that we know what Fourier transforms are,
the microscope objective is essentially
taking the Fourier transform of the laser beam, which
ideally is a plane wave, so it should be a delta function.
However, since this plane wave is not perfect,
it has high frequency noise.
So then the pinhole, what it does
is it does a low-pass spatial filtering.
So it basically removes all these high frequency signal,
and after these we have a nice spherical wave coming out.
This component here is an iris, control
the diameter of this spherical wave,
and this element here is my collimating lens
that will transform this spherical wave into a plane
wave.
In this case, this is just a beam splitter
that will allow me to couple from this side, where
I have a white light source.
In this white light source, we have a green filter,
so then we only have green light coming out.
We have an iris again to control the diameter.
Right now I'm blocking it, so it is a beam block.
And these are collimating lens.
So here it should be the planar equivalent
of the incoherent side.
This is the planar equivalent of the coherent side.
This mirror here bends the light 90 degrees to send it
to the side from here.
This is the one line of the incoherent line,
and then should go into the common path
of the coherent line.
Then they basically bend again 90 degrees,
and this is the 4F system section.
So here we have first an element, which is the object
that we want to image.
The object is called the resolution target,
and it's a clear substrate, so a piece of glass
that has patterned chrome lines of smaller and smaller sizes
that we're going to see.
Then we have our first lens of the 4F system.
This is a non-magnifying 4F system,
so these two lenses are exactly the same focal length.
So first lens.
Then this is an aperture in the Fourier plane
because we're going to start with low-pass filtering.
The second lens, and then we're imaging this
into our detector, which uses CCD seamless array.
So now we're going to switch to the camera
and start with a coherent case.
So the first case, it's like what we've seen in class.
We're going to illuminate this object with an inline plane
wave.
OK, let me just zoom in a little bit.
So this is the resolution target,
and it's in imaging condition, right?
So the object is in the front focal plane of the first lens,
and the detector is in the back focal plane of the second lens.
So what you can see here is we are actually
imaging all the way to groups two and three.
This resolution target has lines even larger,
which are one and two that we don't see here.
And the first thing we see is the following.
We see some fringes, which are produced
by the interference from the glass,
so it's basically the reflection of the plane waves interfering
with each other.
So this is very typical in coherent imaging.
We see other rings that are produced by dust particles
maybe along the way.
So all these fringes right now look destructive to the image,
and perhaps they are for certain applications.
However, for those of you doing digital holography,
you notice that these fringes contain a lot of information
that is encoded about the amplitude and the phase
of the signal that we can exploit.
So just for comparison, I'm going
to zoom in and see what is the maximum that we can see here.
So you can see that we can have a resolution now that we know
what the optimal resolution.
With good accuracy we see maybe down
to this side of the group number five.
All right, so in the homework we learned
about the low-pass filtering in Fourier domain.
So just to remind you, essentially
what we're going to do is that the first lens
is taking the Fourier transform, and it
happens in the Fourier plane.
So then in that plane we put an aperture,
centered, and then what we do is we're
going to start reducing the diameter of the aperture
to try to low-pass filter the signal more and more.
So what do we expect to see here?
Well, you did it in the p-set, so we
expect to see some blurring, progressive blurring.
So I'm going to do it, and hopefully you
can see it from there.
So right now it's all open, and I'm going to blur it.
So I'm going to do it again.
This is clear.
Blurry.
So I'm controlling with this iris.
And you can see how the signal that contains the higher
spatial frequency in this area goes first,
then this one and this one.
But all of them now--
in this blurry condition, if I try to zoom back again,
now we see how the signal is really blurry.
If I open the aperture, it becomes clear again.
So now let's do the comparison of the exact same thing.
What would you think it would happen if I switch illumination
now to incoherent light?
So I'm going to block the coherent light here, and then
the first thing that we notice is that we don't have
those fringes anymore, right?
So it looks like a very clean image.
So again, in this case, you can think about
that for microscopy, this is very nice.
However, from the digital holographic point of view,
we don't have that extra information
that we can use in encoding of the signal.
Now let me zoom in to compare the resolution in this case.
So you can see it's hard to tell if we don't
do a quantitative study to see exactly what is the smallest
line that we can see in each case,
but we can see the comparison again
versus the resolution in the coherent and incoherent cases.
And now let's do the low-pass filtering.
So again, this aperture right now is fully open,
and then I started closing it.
And I'm closing and closing and closing it.
Open again.
Close it.
So we see two things.
Remember that the incoherent system is basically regulated
by the OTF, as opposed to the ATF in the coherent system.
So now that you can think of it--
the easy terms is it's a [? combo ?] because the OTF is
a normalized cross correlation of the ATF, which
means that when I close this aperture,
you can see how the signal becomes weaker
and weaker because essentially I'm restricting the DC
component that happens to be in all the signal there,
but also gets blurry.
It's hard to see, but it's getting blurry.
It is getting blurry at a different rate
as the other system.
So now let's go to high-pass filter,
and I'm going to switch back to the coherent case.
So this is the coherent case.
So right now what I'm going to do
is that I'm going to remove the aperture in the Fourier plane,
and I just did a very improvised high-pass filter, which
basically is a microscope slide that has a little piece of tape
in the center just to block the basic component, or kind of,
but anyways, it works.
Somewhere around here-- let me crank up the light.
And we can see how basically we remove the DC term.
And you see the high-pass filtering effect,
which is essentially accentuating
all the edges of the signal.
Now I'm going to put the incoherent case
and the coherent superimposed just to show edge detection.
So this is a normal thing.
We can accentuate the edges.
So what is interesting also here is
that if you move the filter in different places,
you can, say, just accentuate the edges
in the horizontal direction or in the vertical direction, et
cetera, et cetera.
We see that the incoherent case, we
don't see the same effect, which is as expected.
Again, it's a normalized cross correlation.
But with a coherent case, we do see these drastic change.
GEORGE BARBASTATHIS: Thank you.
Any questions about the demo or about coherence, incoherence?
AUDIENCE: How do you tell whether something's
partially incoherent or completely incoherent?
GEORGE BARBASTATHIS: OK, so the test
is with one of these interferometers
that Se Baek showed last time.
You basically measure the fringe contrast.
For example, if you're worried about spatial coherence,
you set up a [INAUDIBLE] interferometer [INAUDIBLE],,
and the fringe visibility will tell you
the degree of coherence.
If it is full contrast, it is fully coherent.
And of course, you never see full contrast.
You never see-- it's always somewhere in between.
OK, so today we'll have a flashback
of geometrical optics.
I will go back to something we saw in geometrical optics,
and that's the single lens imaging system.
Oh, and before I forget, on Wednesday,
it is the project presentation for the graduate version
of the class, so I wanted to remind you.
I will also send an email from the website,
but if you haven't done so already,
you should have a meeting with your project mentors,
finalize your presentations, and then
on Wednesday, what we'll do is it'll be a conference style.
Each group will have about 15 minutes, approximately 12
for the presentation, approximately 3 for questions.
How many groups?
I think we have six or seven groups.
So typically, these things run over,
so we'll take probably the full two
hours with the presentations.
So for those of you in the 2.71, you don't have to attend,
but I think it would be very useful if you do because it's
sort of interesting.
It is more advanced material.
The graduate students will present some topics
from current literature, and you'll
get an idea not only over the basic optics that we'll do here
but also what kind of things happen in the research front.
For example, last week several of us--
Pat, Yuan, and Sean--
were at the conference in Vancouver,
so some of the topics that were represented on Wednesday
were actually big items in the conference.
The conference was actually called Advances
in Imaging, this spring congress of the optical society.
Anyway.
So going back to the lecture, this
is something that we've seen before,
but with geometrical optics.
Now we try to understand how the same system works
but with the wave optics.
And if I could have the tablet again, please.
So the way we analyze the system, of course, is we'll
basically have to propagate using Fresnel diffraction
to propagate the field from one element to the next
until we reach the end of the system.
So for example, at the input we have the input field.
I will write it in one [INAUDIBLE]
so we don't write too much.
The input field is simply the product
of the illumination field times the complex transmittivity
of the input transparency.
I will do everything for a spatially coherent
illumination, and we know now that if the illumination is
spatially incoherent, if we have already
solved the coherent case, we're in good shape
because we can simply take the modular square of the point
spread function and the autocorrelation
of the coherent amplitude transfer function,
and then we can compute the correspondent quantities
for the incoherent case.
So the coherent is the one that we'll always start with.
OK, so doing that now, we can calculate, for example,
the field immediately to the left of the lens.
So let's call it g sub L minus for left,
and I think I still use x double prime as my coordinate.
So this is now a Fresnel propagated version
of the input field, so we'll write out the big, bad Fresnel
integral.
e the i 2 pi z1 over lambda i lambda z1.
So this is the [? prequel, ?] and then we
write our big integral here, g sub
in of x, and then the quadratic.
And of course, two things happen at the lens plane.
First of all is the lens itself.
That will impose an additional quadratic phase delay
upon the signal, but now it's a little bit
different than the geometrical optics case.
I forgot to mention earlier that I
have added a thin transparency in front of the lens.
So this is again a pupil mask.
As we will see in a moment, it does
pretty much the same job as the pupil mask
did at the Fourier plane of the 4F system, but in this case,
we actually put it sort of in contact with the lens.
In fact, we don't have to.
We can put the mask anywhere we like,
but anyway, it makes the math a bit simpler in this case.
And in most implementations, for practical reasons,
they also do it this way because it is much better
to have everything stacked together,
as opposed to having different elements
sort of floating around in an optical system.
OK, so this means then that the field to the right hand
side of the lens, so the plus now means to the right,
still x double prime coordinates,
is going to equal to the field that was
on the left multiplied by 2.
Two functions now.
One is the thin transparency itself,
which I denote there as g sub pm.
pm, of course, stands for pupil mask,
like in the previous cases.
And then times another complex quadratic
exponential that corresponds to the lens itself.
So this would be e to the minus i pi x
double prime square upon lambda f,
where f is the focal length of the lens.
And finally, the field at the output.
Now the output coordinate is x prime.
It's going to equal another Fresnel propagation
from the pupil mask to the output plain,
so I have to write another propagation term here.
This would be z2 this time.
And what I will put now is gL plus because this is the field
to the right of the right of the composite lens and the pupil
mask, so that would be x double prime and then
another quadratic exponential.
OK, so this is it, and when you plug in all this math--
this is something we've done already,
so I will not do it in the class again,
but there is a supplement.
There is a scan of my handwritten notes
in the website where you will see a little bit more detail
on how I did the derivation here,
but I will skip some of the immediate steps,
and I will show the result here.
So the way you handle this kind of-- as you imagine,
when you combine all of these integrals,
you will get a sort of integral that involves all
of these functions, g sub in, g sub pm and so on,
but also you will have a bunch of quadratic exponentials.
So what you do is you expand the exponents
of these exponentials, and you rearrange terms,
and you write it out this way.
So these are the output coordinates.
They don't participate in the integration,
so we can knock them out of the integral.
And then of course, we cannot do much about this thing
because, well, we don't know what these functions are yet,
but we can certainly try to see what we can do with
the remainder.
So what is sort of glaring here is that in the exponent,
you have quadratic terms, and you also have linear terms.
The linear terms, we know from experience
that they lead into Fourier transform, so they're nice.
We'll know what to do with them.
The quadratic terms, they usually
correspond to nasty things like the focus, but in this case,
you can see that the quadratic term is multiplied
by a coefficient here, which allows
us to knock it out if we select the distance
as z1 and z2 judiciously, and of course that
is the result of the lens law.
This is the same equation that we
derived using geometrical optics,
and we'll call it the imaging condition.
So you can see here that when you satisfied this imaging
condition, then you essentially knock out the focus term
from the diffraction integral.
So that is gratifying because we actually
derived a result from geometrical optics,
but this time it is perhaps more rigorous because it
is based on electromagnetics.
What is the rest now?
Well, we'll deal with the rest in a moment,
but you see that there's another quadratic here that we cannot
quite so easily get rid of, and this extra quadratic that
popped up there is interesting because it is not predicted
by geometrical optics.
So this is something that--
well, it will happen only with spatially coherent
illumination, and geometrical optics cannot predict it
and actually cannot deal with it.
And in fact, not so long ago, about the time
I was born in the early '70s, this
was a topic of research, what to do about this term.
In fact, Professor Goodman, who wrote your textbook, one
of his early results as a young scientist,
he wrote the famous paper about this term,
what this term means.
And he wrote it--
I believe he wrote it in 1971, exactly the year I was born.
I'm not that old, so I guess that you can still
call it a relatively recent development.
So again, we're talking about this term over here,
what to do about this one.
And then of course, there's an entire section in Goodman
that talks about this term, and I would encourage
you to read the section.
I will show a summary here.
Something got messed up that with my animations here.
This was not supposed to be here, but anyway.
So Goodman actually goes over three methods
that you can use to eliminate this unwanted quadratic.
One of them is you lay out the input transparency as a sphere,
or in a spherical surface, instead
of the typical planar surface.
That may sound a little bit strange,
and it's probably very difficult to implement in practice,
but conceptually you can see what's
going on because now these points in the transparency
start with an original phase delay,
you can work out the curvature right here.
In fact, the curvature would have
to be exactly z1, the areas of curvature,
so that it cancels this unwanted quadratic.
So that's one way.
The second way is to actually stick
a lens in front of the transparency.
This is actually very often done in geometrical optics as well.
This is called the condenser lens.
So for example, we don't have this anymore,
but actually, we do have one in the classroom.
We used to use these projectors.
Basically, if you look at the top surface of this projector--
I don't know if I can take it out here--
but if you look at it, it's actually a lens.
After class, come over and look.
There is grooves on it, and basically,
sort of the grooves implement equivalent of a lens surface.
If you do the operation modulo 2 pi on the surface of the lens,
instead of having this surface that we typically
think of as a lens, if you do a modulo 2 pi operation,
you will get something like this, of course.
OK, so that's a way to make a thin lens.
That is called a condenser.
The reason we put it in a projector
is not to eliminate this quadratic factor,
it is simply because it makes the illumination uniform.
So when you project the slide on--
I don't know if you guys are maybe
too young to have ever seen anything
like this in operation, but the next time if I remember,
I'll bring an old fashioned transparency,
and I will show you how people of my age
used to give talks at conferences
when we didn't have PowerPoint.
But anyway, this is done actually very commonly,
to stick a lens in front of--
just attached to a thin object.
But there are some cases you cannot do that.
For example, in the microscope, it
is very difficult to imagine sticking
a lens behind the glass slide.
So in some cases, this is applicable,
and in some it is not.
Anyway, a big application of this type of lens
is of course in overhead projectors.
Does anybody know another application
where people use this kind of lenses,
or what might motivate you to make this kind of lens?
Can you think?
These lenses are commonly used--
actually, you know?
AUDIENCE: Do they sometimes stick them
on the backs of vans or cars to make it wide angle?
GEORGE BARBASTATHIS: That's right, yeah.
That's another application.
Exactly, to improve the rear view of the driver.
And that's a good one.
I wasn't thinking of that.
Yet another one is in very high power illuminators,
which are most commonly used in photography in shooting movies
because the glass, as you know, is not a very good conductor,
so it would hit up very badly if you illuminate
with very high power.
They're called, actually, babies,
these really big light bulbs that they
use when they shoot movies.
So they actually use these.
It is called a Fresnel lens, the same Fresnel
lens in Fresnel propagation.
So they use the Fresnel lenses in these so-called babies.
All right.
And lighthouses also, yeah.
So again, every time that you have a really huge light
source.
Of course, it's a matter of practicality also, right?
If you have a major lens, as you can imagine,
if the diameter is big, then also this size
would also be very big, right?
So in a lighthouse, you might have
a lens that is maybe 1 or 2 meters in diameter,
then it would have to bulge over another half meter or so.
That's not very practical.
It is heavy, fragile, blah blah blah.
So they make the Fresnel lenses to get around this problem.
A good thing to remember about those
is that they generally give very poor image quality.
So if you try to use this as an imaging lens,
you typically get a very bad blur and a very bad image
quality, but obviously in the lighthouse you don't care.
In an illuminator, you don't care.
And also the projector, it is not used as an imaging lens.
The imaging lens is a real lens, this one.
This is simply collimating the illumination
so you get a uniform image at the image plane.
So for jobs like this one where you basically
want to take a light bulb that is highly un-uniform.
It has filaments, all kinds of crap.
So this is a very nonuniform field.
This type of lens is very good to uniformize
the intensity, and then of course
if you need to image a sharp object,
then you need the real lens.
You cannot get around that.
OK, that's a bit about Fresnel lenses.
So that's a bit of a detour, but I
wanted to point out some practical ways
that you might be able to do this kind of thing
if you need to get rid of the lens.
And finally, what John Goodman proved back when he was young
is that there's a condition that you can
neglect this nasty quadratic.
Again, we're talking about the quadratic
that you see at the top of the slide.
So it turns out if the field of the object--
when we say the field, sometimes in geometrical optics
we mean the size of this.
So if the field is smaller than about a quarter
of the imaging lens, then it turns out again
that the effects of this term are negligible,
and therefore we can get rid of it in clear conscience.
So I will not go into the details.
There is discussion in the book and also
reference 303 from the book is an article
that Goodman wrote back then, and he
goes into this in a little bit more detail.
It's a pretty interesting article.
I'm going to leave a post on the website if you like.
It's pretty interesting to read.
So assuming then that will get rid of this extra quadratic--
let me go back one.
So we got rid of one quadratic from the lens law,
and I spent a lot of time arguing about this quadratic,
that maybe you can get rid of that
then as well, assuming you can get rid of that as well.
Well, anyway, even if we don't get rid
of that, what is left in here is actually Fourier transform.
I think I pointed that out before.
A Fourier transform-- why?
Because you have linear now, linear
exponents times some coefficient,
and then here you have the pupil mask.
So again you recognize something familiar.
The same thing happened in the case of the 4F system.
We again got a term like this one,
but it was slightly different.
Instead of z1 z2 in the denominators here,
it had f1 and f2, the two focal lengths.
Here we have the two distances, but it's pretty much the same,
and again the pupil mask appeared in the kernel here.
So this is a Fourier transform, and of course we
get rid of this term, and let me jump through these animations
now.
OK, so if we're doing that, then what we finally get
is this expression here, where, again, this
is the Fourier transform of the pupil mask.
I wrote it in a form that looks a bit more like a Fourier
transform, and of course u and v are now arbitrary arguments.
These are the spatial frequencies in the Fourier
transform, but what actually goes in there
is these parameters over here.
So basically we get again a term that looks like a convolution.
So we see that the image, the complex optical field
at the image is a convolution of the input field times something
that again we will call the point spread function.
And this point spread function, just as
in the case of the 4F system, again here the point spread
function is given as the Fourier transform over
the pupil of the pupil mask scaled
with appropriate coordinates.
OK, and there is various scaling forms and factors here.
For example, this one again ensures energy conservation,
which is kind of important, but very often
in optics we don't care so much about the absolute intensities,
which means that we don't really care about this term.
What we really care about is about the spatial distribution,
like the type we're solving, whether features make it
to the image or they get lost due to spatial filtering
and so on and so forth.
So very often these multiplicative constants,
we just drop them.
That's sort of convention.
Now, to get a little more insight out of this,
imagine for a moment that the point spread function
becomes extremely narrow.
It becomes a delta function.
So if you substitute the delta function
into the expression we had before,
then you get this expression, which
is of course again very gratifying
because now the output really looks like,
again, some multiplicative term, but then it really looks
like the input, except in the input
now, the coordinates have been scaled,
so therefore you can immediately see that this term is
the magnification.
In fact, the lateral magnification of the system
is given by this expression.
So we basically rederived the same result
that we got from geometrical optics--
namely, the lens law that is the imaging condition,
the magnification and so on and so forth.
Now we've got them back from our wave optics approach,
but we had to do some approximations.
For example, we had to pretend that the point spread
function is a delta function in order
to get the geometrical optics result. If it's not,
which is never the case, right?
Delta function is a very good approximation.
So if it's not a delta function, then the result
is not exactly like this, but it is a convolution.
That is, basically this will become blurry.
Like [? Pepe ?] showed earlier, you will get a blurred vision--
or in general, an especially filtered version
of the original input will actually
survive through to the output.
What I would like to point out now,
this is something that I believe Se Baek also showed
last time, the same slide.
What I would like to do is basically pay
a little bit of attention to the scaling factors here.
These scaling factors are actually very important.
So I mentioned that to drop multiplicative factors that
appear in the front here.
This, it is OK to drop, but these scaling factors
that go inside the argument, these
are very important because they determine
the amount of spatial filtering that goes on inside the system.
So what I will do actually for a while
is I will compare the 4F system with a single lens
similar system.
So the scale factors, again, they are different.
You see f1 is inside the transfer function, and z1
in the case of the single lens.
So what is the effect of that one?
Also in the incoherent case.
Of course, in the incoherent case,
the transfer function is the autocorrelation
of the coherent transfer function,
and since the coherent transfer function is basically
proportional to the pupil mask, then what
we call the optical transfer function, the incoherent case,
is an autocorrelation of the pupil mask itself.
This is again a very basic result, but again, be careful.
When we compute these autocorrelations,
we have to apply different scaling factors
in the argument.
OK.
So I will show in a second how this works, and I will show it
in two cases.
One is what we called before the Zernike phase mask.
Actually, this is not exactly accurate.
Zernike did not invent exactly this.
He invented another one that has pi over 2 phase
shift at the edges.
It's like a ring.
But anyway, the function is the same,
so everybody refers to all of these types of masks as Zernike
nowadays.
So this is familiar.
We saw a few examples in the past
when it was in the middle in the focal plane of the 4F system,
and this is the case when we put it as a pupil mask
in a single lens system.
It's the same mask.
It will do a very similar thing, but there's
a subtle difference, so this is what I want to point out.
So this is the subtle difference.
What you see here is, of course, the schematics,
so you see it is opaque outside some amplitude.
Then inside, you have this little extra phase
delay near the optical access in a small region.
And OK, this is the mathematical expression
that I don't want to dwell upon.
I'd rather focus on these plots here.
So these two plots are the magnitude
and the phase of the mask.
So the magnitude is the top.
The blue plot it goes from 0 to 1.
So it is 1 within the amplitude, so let's see if that's correct.
The amplitude has a size of 1 centimeter,
so indeed this is 1 between minus 0.5 and 0.5.
And the phase, well, the phase, nobody knows what the phase is.
Where the magnitude is zero, we can not
define the phase for a complex number that equals zero.
Nevertheless, let's set it equal to zero.
But what I want to emphasize is that within the past band
of the system, the phase is 0, except for a small chunk
of width of 0.2 centimeters where
the phase jumps to pi over 2.
Pi over 2 is of course i.
And this is the real and imaginary part.
These two are actually equivalent.
So the real part, it is 1, except at
the little extra phase protrusion there where
the real part goes to 0, and the imaginary part
becomes 1 because of course when the phase equals pi over 2,
then the actual complex amplitude equals i.
So that's what you see here.
So these two pairs describe both of the phase mask.
So the [INAUDIBLE] transfer functions of course
are the same.
I only plotted them here as a real and imaginary part,
but they have the same shape.
What is really different-- and unfortunately,
my plot is a little bit too small here.
You cannot see very well--
is that they actually have different sizes.
This one, if you do the scaling factors here for the numbers
that I used, I think I used--
so I used lambda equals 1 micron.
Lambda equals 1 micron, f1 equals 10 centimeters,
f2 equals 1 centimeter for the 4F system.
And then z1 equals 11 centimeters,
z2 equals 1.1 centimeters, and f1 equals still 10 centimeters
for the single lens.
And I worked those out so that in both cases,
you get the demagnification of a factor of 10.
That's why the numbers are a little bit strange there.
So in both cases, you get a demagnification factor of 10,
and I did that deliberately because the two systems give
you the same effect kind of, but you
can see in terms of geometrical optics, they are identical.
They both give you a demagnification
by a factor of 10, but you see here
that in terms of wave optics, they're
slightly different because one of them,
the 4F system is scaled by the focal length.
Its spatial frequency extends, goes
from minus 50 to 50 inverse millimeters.
This is what you see here.
Whereas in this case, it goes from approximately minus 48
to 48 inverse millimeters.
So the single lens actually does a slightly more severe
spatial filter than the case of the 4F system.
That is to be expected, of course, because the distance
z1--
you can see from here--
the distance z1 that you subtend from the object to the aperture
is longer, so therefore this system
is cutting off more angles, or more spatial
frequencies than this system.
We will see that in more detail in a second.
The other thing I want to point out to you
is that if you take that autocorrelation
of this function, which is the coherent transfer
function or the ATF, if you take its autocorrelation,
you obtain the OTF, which is, as Se Baek demonstrated last time,
it is the transfer function for spatially
incoherent illumination.
So again, this is a little bit of an exercise
that I will not do here, but it is in the notes.
I have posted another set of practice problems.
So it is in pages 16 and 17 of the last set
of practice problems.
I have gone ahead and derived this one,
and I would encourage you to go through.
It's a little bit of algebra, but you
can think of it as mental fitness right
because you might say, why do I need to do all this algebra?
When would I ever have to do so much algebra?
I can just plug it into Matlab, and I get it.
Well, I'm sure all of you do some kind of fitness.
You go to the gym, right?
When you do bench presses, what is the probability
that you need to do a bench press motion in real life?
It's actually very small.
So you do bench presses in order to keep fit.
So the reason we do these kind of calculations
is it's the equivalent of mental bench presses, right,
so I would encourage you to do it.
Because presumably you are at MIT because you have you
want to have mental fitness as well as physical fitness.
Let me talk a little bit more about these scaling factors.
So in this case, this is just a clear aperture that I put here,
and you can see that in the case of the clear aperture,
it's actually very easy to do sort of the ray diagrams,
and you can see also here that--
well, let me back up for a second.
So you remember from geometrical optics,
we had the definition of the numerical aperture.
So we defined the numerical aperture as the angle
that we subtend towards the optical system
if we place a point source on axis.
So therefore, the numerical aperture
is limited by one of the physical apertures that
are in the system.
So in the case of the 4F system, that would be the pupil mask.
At some point in any physical system,
the pupil mask cannot be infinite.
It has to be finite, so the physical diameter of the pupil
mask, assuming the lenses themselves are large enough,
the pupil mask will become the limiting factor.
And so we can compute the numerical aperture then.
It is the ratio of the radius of the mask
over the focal length, f1.
You can see it very easily from this triangle over here.
And of course this is an approximate expression.
In reality, the numerical apertures
would be the sine over the inverse tangent
of this quantity, but of course we're
doing [INAUDIBLE] approximations here,
so we can drop the trigonometric functions.
In the case of the single lens the numerical aperture is--
again, you can get it from this triangle over here,
but now the one of the orthogonal sides is z1, not f1.
So you can see that the numerical aperture is
r over z1, and of course if this system is supposed to produce
a real image, as opposed to a virtual image,
then z1 must be longer than f1.
You can see very clearly from the imaging condition,
1 over z1 plus 1 over z2 equals 1 over f.
If you want z2 to be positive--
so for z2 to be positive, it means I have a real image--
for this, it is required that z1 is bigger than f.
So because of that, you can see that if you have the same size
pupil mask, the single lens system would have a smaller
numerical aperture.
That is sort of an unfortunate fact of life.
So another thing I want to point out here
is that the numerical aperture actually
changes when you go to the second leg of the system
because, of course, the system has angular magnification.
So what started as a numerical aperture at the input
is of course an angle.
If you take the marginal ray, it is
propagating at a given angle with respect
to the optical axis.
By the time it comes out, this angle will have changed.
By how much?
By the angular magnification of the system.
So therefore, the numerical aperture at the output
is not the same.
It equals the numerical aperture at the input times the angular
magnification of the system.
So this is not a cause for confusion.
We can use either one.
We'll get actually the same conclusions whether we
use the numerical aperture or the input
or the numerical aperture at the output,
but we have to be a little bit careful
that we don't confuse things, as long
as we remember this simple relationship
that the two are connected by the angular magnification.
So the reason the numerical aperture is so important is
because if you consider a circular-- of course,
you cannot see the circle here.
This is just the projection.
But imagine that I have a circular pupil.
Then recall that the point spread function of the system
is actually the Fourier transform
of that circular pupil because this is a clear aperture now.
And we learned sometime ago--
I don't remember when--
but we learned that the Fourier transform
of a circular function is this crazy [? zinc ?]
so-called function that is given by a ratio of a Bessel function
to its argument, and it's probably better
to think of it as a plot like this one that
has a main lobe and then a smaller side lobe.
And actually, it has a lot of lobes.
It continues on forever, but the size of the lobes decays away.
How fast it decays away is 1 over the argument.
So what I want to point out here is that in both cases,
the point spread the function of the system
looks like this [? zinc ?] functions.
Sometimes it's also known as an Airy disk.
Not an Airy function, by the way.
Airy function is different.
This one is referred to as Airy disk.
If you're curious what Airy function, is
you can open the table of formulas.
Abramowitz and [INAUDIBLE],, and you'll see a monstrous thing.
That's called the Airy function, and this turns out
to be a special case of an Airy function.
But anyway, this is the Airy disk,
and what I want to emphasize is that in both cases,
you get the same shape of Airy disk, but different in size.
And you'll get different size because in the case of the 4F
system, the size of the aperture is the same,
but because the scaling factor is smaller--
it is f1-- then you actually get a bigger ATF.
The size of the disk in the frequency domain is bigger.
Therefore, it will give you a slightly narrower point spread
function.
If you work out these ratios over here,
and you work out these coefficients,
you will see that very clearly, but I
want you to get it sort of intuitively
by using the scaling theorem of Fourier transform.
In both cases, you start with the same physical aperture,
but what matters is not the physical size of the aperture,
but the numerical aperture.
So in one case, you have this numerical aperture.
This is a f1.
This is the physical size, and numerical aperture
is R over f1.
In the other case, you have again the same aperture,
but now you have a longer distance here.
This is still R. So now the numerical aperture
would be R over z1.
So the same physical aperture in the two cases
will actually give you a different size in the ATF.
This will have a bigger ATF.
So the size of the ATF, if you work it out,
it will be proportional to actually 1 over lambda f1,
and the size of the ATF here--
I'm exaggerating, of course--
will be proportional to 1 over lambda z1.
OK, so since we've got the smaller size of the ATF
in this case, it means that we'll get a broader PSF.
So which system is better?
Well, obviously this one because this one
will give you a narrower PSF.
Therefore, it will give you a smaller blur.
So of course there's a caveat.
I sort of took it for granted that the desirable
in an optical system is to minimize the blur, which
in most cases it's true.
If for some reason someone asked you deliberately
to produce a system that causes a lot of blur,
then of course you would go for this one, but in most cases,
we try to minimize blur.
So this means, given our resources--
that is, given our physical size of the numerical aperture--
we should try to maximize the numerical aperture,
and that is what the photo system is doing.
So there are a few comments about the resolution that
are in the rest of the slides.
I don't want to destroy the rest of your morning.
Because the PSF is finite, you can
imagine that if you have two point
sources that are at distance.
So now this is like an experiment.
I'm moving two point sources together.
There will come a point where the two point
sources will merge.
And if this is your image now, you
don't really know whether you had one point source or two.
Now, of course you say, well, you also
get twice the intensity.
That's true, but very often you don't know
the intensity you started with.
For example, if you're looking at the sky,
and you're looking at the bright dot
and you're trying to decide, is it one star or two stars?
Well, so far we cannot yet go to the stars and measure
their brightness, right?
We can only measure the total brightness
that we receive here.
So therefore, if you're looking at a telescope,
and you see this, you don't know if there's
one star or two stars that are too close to be resolved
by your telescope.
So actually, this sort of situation
was dealt with by a fellow called Rayleigh, and of course,
the size of the PSF, as I worked out
before, it depends on the numerical aperture.
So I will go over this in more detail
later, but this is just a preview
that the numerical aperture actually gives you an idea--
I actually believe also Professor [INAUDIBLE]
mentioned it in one of the past lectures.
The numerical aperture gives you an idea
about the capability of your system to resolve point images.
So imagine that each one of these lobes
here corresponds to the point spread function of one point
object.
So in this case, they were resolved because I spaced them
so that the diameter of the main lobe
falls exactly on one null of the other lobe.
So if you work it out, and you take into care
into where the zeros of the Bessel function
are located and so on, you come up
with an expression that looks like this.
The space in between the two sources
is 1.22 times the wavelength divided
by the numerical aperture.
And of course this would be at the input plane.
So this would be the spacing at the input required
so that the two sources can be resolved.
But when you look at the output, the distance of course
would be given by a similar expression,
but the numerical aperture at the output
would appear in this case.
And of course this is sort of the most common definition.
Some books define the resolution--
instead of the diameter, they use the radius,
so they come up with an expression that is exactly 1/2.
Which one is correct, it actually
depends on your requirements.
If you require a very crisp image,
or if you're dealing with a very noisy situation,
then you go for this one because as you
can imagine if you organize noise into this one,
then it becomes progressively more
difficult to resolve the two spots.
So in a very noisy environment, when
you have a very little light, then
you go for this definition.
If you have plenty of light, then you
can possibly resolve two sources that are very close like this,
right.
Here you see you have very poor contrast.
You basically have to rely--
in order to resolve the two sources,
you have to rely on this little dip in density.
So if your signal is noisy, this may be lost,
and then you cannot resolve anymore.
OK, so that's a preview of resolution.
We'll talk about this a little bit more.