Optics (Lec - 19) - Electronics & Communication Engineering Video Lecture - Electronics and Communication Engineering (ECE)

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okay so for the ones that just arrived
I'm going to be giving the lecture today
but George is here so if you've got
questions you can ask ask me that I
don't mind defendant erupted at any time
but also if you want George to answer
something he's sitting here as well so
he can also contribute so last time
George was looking at the Fourier
transform in prophecy of a lens and he'd
looked at they derived the 1d case where
you've just got a function of X rather
than a function of X Y you've got this
transparency a distance then in front of
the lens and and then you've got a
screen placed a distance F behind the
lens and so the question is what do you
get on that screen and what I'm not
going to go through all the algebra of
that again because George did it last
time
this was the calculus this was the
answer what you find is that you what
you get in in this plane here is
actually the Fourier transform of the
thin transparency right so G of X is
this transparency here you get the
Fourier transform of that and multiplied
by a parabolic phase factor which
depends on the position of this screen
what the of the transparency what is the
value of this Z you'll notice that you
get this factor here 1 minus LED over s
so if Z equals F this is going to be 0
and this phase facts is going to
disappear so for the special case where
Z equals F all you see here is just the
Fourier transform of this very simple
very simple answer and people actually
have known this for quite a long time
and at one time it did look really
attractive that you
actually imagine using this as a real
way of doing Fourier transforms
especially as as we're now going to do
you can immediately generalize that to
the 2d case so you can do a 2d Fourier
transform and when you think that you
know the light propagates through this
at the speed of light you can do a 2d
Fourier transform at the speed of light
and when people first realized this
which was back in the 1950s 50s or early
60s you know computers were very slow
then so doing a 2d Fourier transform was
not trivial so a lot of effort was put
into trying to design instruments to do
that yes do you know who was the first
one to realize this was abandoned looked
or the question is who was the first to
recognize that yeah Vanderloo it's more
valued on books was that he was at
Michigan wasn't he and and interestingly
the holography guys were also in the
same lab what's the guy's name the one
that mmm at leaf and interestingly
Emmett leaf and Rand alert both
published virtually the same paper in
the same year it's a very very
interesting historical thing they're
they're both from the same lab and they
both invented almost the same idea as
far as I can see and neither of them
refer to the other guy very peculiar
anyway I don't know who really came up
with this idea first but but actually I
think the I suspect that actually even
an stabber really realized this you know
he and stabber was the scientist who
designed the Zeiss microscopes 120
hundred and thirty years ago and and and
he came up with a theory for image
formation and a microscope which is
actually what we're going to be getting
onto pretty well the next thing the
imaging with a 4f system he he
hood that 130 years ago and I don't
think he's often given this much credit
is you should get give anyway you can
use this as a way of doing a Fourier
transform and even the 2d Fourier
transforms so you see going from this
stage to this stage is trivial really
all you've always done here is assuming
that it's a separable function of x and
y so you get exactly the same sort of
integral for for Y as you go get for X
so it now becomes a double integral this
is a to the two dimensional Fourier
transform of G of G of XY and and you
can see that you get a gain the phase
factor here because it's separable
you've got this e to the I PI x squared
and you've got an e to the I PI Y
squared and of course x squared plus y
squared
very nicely equals R squared so this
becomes a very cool type wave so we've
still got this one - then over F so if
Zed equals F then again this this phase
factor vanishes and we end up with just
the the pure Fourier transform yeah as I
say so a lot of people thought that this
was quite a neat way of doing Fourier
transforms there was a lot of work on
optical computers all the way through
the 60s and 70s but the trouble is that
as fast as they could
develop these systems the computer
people developed their computers even
faster so they were always one step
behind nothing so I think in the end
people realized it was a waste of time
and that it was never really going to
become a you know a practical way of
doing real calculations but in the early
days it was and Emmett Leith we were
just mentioning he first developed these
systems for actually analyzing the
the images they took from synthetic
aperture they you know an aeroplane
they'd fly an aeroplane over North Korea
taking pictures and then the idea is
that the the plane takes many pictures
as its traveling along and they're
coherent with respect to each other so
you can actually combine them digitally
afterwards and get an image which has
got amazing resolution and you can see
the guy in North Korea reading his
newspaper or whatever so they really did
that sort of thing way back then right
so that's so that's the equation here
then and yeah so this is just going
through how that thing was derived so
we've got this we've got the screen
we've got the screen here the G of X by
the time we get here there's a is
convolved with some face parabolic face
function and then after it goes through
here it's multiplied by another
parabolic phase function and then as it
propagates here
it's convolved with another parabolic
face function and if what you have that
magic condition that this is there and
this is there most of these parabolic
phases cancel out and we're left with
this very simple final answer so that's
basically what's happening and George
went through you know the the various
stages of doing this of course you end
up with a couple of integrals of a
products of three phase functions here
so the algebra was quite complicated
we're not greed complicated but just
lengthy and but but everything in the
end cancels if you haven't got any cause
to F though then you can see that you've
got you have got this phase factor so
you don't actually get exactly the
Fourier transform in this plane in that
case you get the Fourier transform
multiplied by this
factor and that can be important in some
cases that's something that you need to
take into account there's another
special case actually the that's the
case when said it cause no the case
winds n equals nor is if this screen
coincides with the lens immediately
close-ups of the lens and I think
Goodman does like that one as well
doesn't he
and so if Z equals no this this goes and
so we're still left with this phase
factor each of the I PI X Double Dash
squared over lambda R right so and also
Goodman does the case of the
transparency after the lens who did not
doing the notes right okay yeah so
Goodman yeah
Goodman does this case of the
transparency placed in here that this is
the negativeness yeah yeah well it's all
it's very similar sort of answers okay
so yeah so this is just writing down
finally then we're defining then the
Fourier transform is called the Fourier
transform of little G is big G right so
the final answer is that you've got this
parabolic face function times the
Fourier transform of G
okay so this is pointing out that we've
now looked at two different ways you can
do a Fourier transform one way is to
just propagate the light a very large
distance you remember if you have Fram
off for diffraction if your screen is a
very long way away from the from the
diffracting screen from the diffracting
screw object whatever it was called
transparency then again you get this
Fourier transform relationship but of
course here is put a proportional side
there's some things out the front layer
and in particular there's going to be
again a parabolic phase term isn't there
because in this case here what you'd
expect actually is that this Fourier
transform is going to be actually on the
surface of some curved surface not of a
plane surface right so if you put the
screen if the screen is plane then it's
going to be multiplied by some parabolic
phase term and now we've just looked at
this other case where we don't have to
have a very large lab in order to do
this experiment we can actually make the
experiment quite small so that it fits
on top of the table just by using a lens
and you know really what you can see
that that is actually doing it according
to your geometrical optics ideas now you
can see what this is doing this is
producing an image of this at infinity
so you can see that these two things are
actually equivalent
what it's doing is bringing what would
have been an infinity to some finite
plane isn't it right so that's that's
all it's doing there but again we've got
this this thing here this proportional
sign which means there's some some other
factors constant factors out there
including in general this parabolic
phase factor which for this particular
case when
you got F and F it disappears the
parabolic phase function disappears when
you've got this special very very
special case with the two F's there
okay now this one
what is this really trying to show is
that really there's two ways of looking
at this problem you can what these two
diagrams here I'm really trying to show
this that is trying to show the same
case but trying to get you thinking
about a different way of viewing the
problem so in this case here you can
think about any point on the screen on
the on the the transparency it's going
to act like a point source so it's going
to produce a spherical expanding wave
and then this is a distance F so that
spherical wave is going to be converted
into a plane wave the angle of
propagation of this plane wave depends
on the distance at this point from the
origin from the origin of course right
and so you can think then of what you
see on the screen is being made up of an
angular spectrum what's called an
angular spectrum of plane waves where
the strengths of each of those angular
spectrum components comes about from the
value of G at that point right so so
that's one way of thinking about it and
this way is that it the other way this
way you're thinking of the object is
going to differ if you imagine your
transparency as being like a grate it
produces diffraction it producing like
grating orders example if this was a
very simple case of a of a sinusoidal
grating you'd expect it to produce some
grating orders and these grazing orders
chorus and
so plane waves traveling in particular
angles so this shows the angular
spectrum of the diffraction of the light
given by the transparency and then their
lens collects these the the different
these different plane wave components
all of them are going to be converted
into a spherical converging wave that
converges a distance F away it doesn't
say but this must be a and different the
plane waves moving at different angles
are going to converge to a point which
of course is proportional to the the the
sine of the angle of the propagation
right so here you at any point on the
screen you're going to sum up these
again these angular spectra components
each of which is going to give us a very
cool way so you can see looking at that
diagram again they're like sort of
inverses of each other
I think this is another example where
very often you can think of a reciprocal
type behavior you know you could yeah
you could you can just flip this over
and it would look the same the other way
wouldn't it okay so so this way here
then we're thinking this this here we're
thinking of this as as just like the
Huygens sense that we spoke about when
we were first doing diffraction they
okay so this is giving some nice
pictures I'm gonna I'm gonna steal this
presentation afterwards because there's
some really good
displays in this thing that I'm going to
use in my own Nexus in the future so so
this is giving some good ideas they so
we start off then this is a rectangle oh
I'm going to say something in a minute
though they're written so this is a
distance F this is a distance there so
this lens then as we've just described
it's going to produce a Fourier
transform of that so the Fourier
transform of a rect is a is a sync so
you'd expect to get this a sync in X
syncing x times of thinking Y now the
thing I was going to say though George
this looks to me that this is longer
than that this is longer this way than
that way so this should be longer that
way than this way I think you know
that's the aspect ratio of this Kleenex
conversation bringing home I think a
very important point that you've we've
mentioned before in the course that if
you if you make a function smaller it's
Fourier transform is wider if you if you
make the function wider is Fourier
transform is smaller right so and you
notice then this product of these two
sinks gives a bright spot in the center
a series of bright spots along the y
axis
there are also some bright spots in this
region here but they're so weak that you
can't actually see them you really you
can only see the ones of only AXYZ all
right so that's the first example then
this one's showing how if we make the if
we make the transparency smaller the
rectangles smaller so this pattern gets
bigger now actually we can just about
manage to maybe see someone's lot other
than along the axis
and then this one is an interesting one
so now we've got not just one aperture
but three apertures and so so we're now
how can we work out what what the the
Fourier transform of three apertures
would be anyone got some ideas
I'm sure shall in eleven idea anyone any
way how would anyone think of doing this
sort of thing
have you come across a thing called the
convolution theorem and probably you
should have done somewhere I would have
thought so there's the thing in Fourier
transforms called the convolution
theorem so this is actually it this is
actually like three three spikes
convolved with a square which really
really means all it means is we place a
square on each of the three spikes and
the convolution theorem says that the
Fourier transform of a convolution is
the product of the Fourier transforms so
what you'd expect is that the Fourier
transform that you're going to see here
is the Fourier transform of the
rectangle which is this multiplied by
something else which is the Fourier
transform of the three spikes and the
three spikes is like a delta function
and a plus a minus Delta function that
the plus and minus delta function
together e to the I something plus e to
the minus I something gives cause and
the the the one in the center just gives
a constant so the Fourier transform of
these three Delta functions is what is
one plus a cosine or something like that
and so that's what we expect to see
there this thing this pattern that we
saw before is going to be multiplied by
a cosine function which comes about
because of the separation of this and of
course because this this distance is
actually much bigger than the width of
the rectangle the spacing of the of the
cosine is much finer than the Fourier
transform of the rectangle and so that's
why these fringes you see these fringes
but very fine fringes become they become
a result of the fact that these
components here are quite well separated
so that's another very nice example and
so all these theorems that you learn in
aurilla transforms can all be directly
taken over into optics and sometimes
they can save a lot of work because you
can just use the standard forms for
things that you could look up in a book
at tables and would saves you a lot of
algebra or even worse arithmetic right
so this is another example so now we're
back to the the bigger rectangle or
something close to the bigger rectangle
the Fourier transform with the rectangle
we saw before but now this one you see
this is multiplied by some fringes so
this is the light and the opposite of
this one really so this is a this is a
convolution this is a product so we used
the same convolution theorem again the
the fringes of course correspond just
like we said before the Fourier
transformed the three Delta functions
with the fringes so in the Fourier
transform of the fringes is that it's
like three Delta functions so what we'd
expect this is a product of the fringes
and the rectangle so we expect here to
see the convolution of the Fourier
transform of the fringes and the Fourier
transform of the rectangle and that's it
that's what we see so we see three of
these diffraction patterns situated on
these three spots corresponding to the
Fourier transform of that grating right
so very nice display that one I like
that one okay so you know so this is now
introducing the idea of the what's
called the for F system so we're going
to do like this not just once but do it
twice so this is shows what we've had so
far if we've got some particular
transparency we get some particular
diffraction pattern in the back focal
plane
and then in the 4f system we do the same
again we now have another lens which is
going to do exactly the same it's going
so look at this pattern and produce its
Fourier transform so if remember if the
if this is F and this is F and this is F
and this is that we won't have to worry
anything about these parabolic phases
it's just straight Fourier transforms
with nothing else and so you can see
what you get is very simple you start
off with this object so here you get is
Fourier transform and if you take the
Fourier transform of the Fourier
transform what do you get if you do a
Fourier transform of a Fourier transform
yeah it's not quite is it you remember
you inverted that's right yeah so
someone got that the if you remember if
you if you do a Fourier transform of a
Fourier transform you get the original
function inverted you have to take
Fourier transform four times and then
you get back to where you start and so
that's what we can say anyway so this is
the inverted well of course inverted
it's just what we expect isn't it
because we know that when we have an
image imaging system we get an inverted
image so it's everything fits together
it gives exactly the same results that
we expect from our geometrical optics
and so in this case of course this is a
rather symmetrical object object so the
fact that this is being inverted I guess
you can't really even see
okay so that so this is saying then that
you can make a four F system which
produces then an inverted version of the
original object you might wonder what
the point of that is I mean I guess in a
way there's no point in making an image
which is much the same as the object
really but one reason why you might want
well there's two reasons one is that
it's actually of course mood it's
produced this image in a different plane
so it's actually it's like transported
the this distribution of light here from
this plane for this plane so that's one
type of application where you might want
to do something like this
and the second thing of course is that
although this is always called a four F
system the we obviously have to have
this and this F cool for
this to work we have to add this and
this F to be equal for it to work but
this F doesn't have to be the same as
that air so you can actually have a
system like this where you actually
produce a magnified image or at the
magnified image and now that of course
is getting a lot more useful so so
that's how you can produce an image of
an object in another plane now the next
thing to note is of course somewhere so
far we have assumed that all this light
gets through this optical system and
gets here so if that was the case you
get a perfect image if all the
information from the objects was getting
to the image then you wouldn't lose
anything but in practice of course
that's never going to be true and well
one one way it's not going to be true of
course is because in practice these
lenses are never going to be infinitely
large right so but also we could if we
wanted to actually what
some sort of mask in this plane here to
actually stop some of that like getting
through to the focus of the image plane
and so now we're going to look at what
happens if we do that this is much the
same overall effects as George spoke
about in previous lectures actually not
imaging case but he spoke about you
remember the results with the building
with the windows that you could actually
you could actually but quite just by
doing processing digital processing on
the Fourier transforms you can
manipulate the strength of the grading
components in the final image so the
same is true here except now we're doing
it optically rather than digitally right
so what you'd expect then is if we put
something in here so here this is
showing a passband so let's say you're
put in here some a pape screen with a
small hole in the hole is this passband
and the screen is the block bang so it
only lets throw the region in the center
according to how big that passband is
you'll get different answers in the in
the image plane yeah of course it says
the transparency may be a grayscale it
doesn't have to be a binary type thing
black and white or it could even be a
phase mask it could be that that this is
changing the phase by let's say PI by
two relative to this or PI relative to
this or whatever and you'll get you'll
get different answers according so what
you choose are obviously for the
geometry of this what we call spatial
filter
okay so yeah so this is going back to
this idea of these two ways of thinking
about the same problem so this is the
first way where we think of our objects
as being made up of lots of point
scatterers each of these scatterers
gives way gives rise to like Huygens
spherical wavelets these are each of
these is collected by lens the lens
converts that to a plane wave the plane
wave reaches the second lens and because
it's a plane wave it's converted back to
a spherical wave that converges the
distance F away from that and now we've
actually introduced the fact that these
X can be different so f1 f2 f1 F sorry
f1 f2 this has to be equal to this and
this has to be equal to this otherwise
the the phases won't cancel out properly
and so you can see that what you'd
expect then is that because in this
diagram here f1 is bigger than f2 this
image is actually smaller than the
original object and a note now you can
also see directly the the geometry
showing the inversion that the point
above the axis has come to an image
below the axis here right so that's the
one way of thinking about it and and
then the other way is to think in terms
of spatial frequencies we think of this
object as being like a grating which is
going to produce grating orders which
travel in different directions
each of these grating orders is like a
plane wave and when it reaches the lens
it's going to produce a converging
spherical wave which comes to a focus a
distance F behind the lens and then
after that of course it doesn't stop but
it carries on diverges again reaches the
second lens and then the second
because this is a spherical wave coming
from a distance that away is converted
into a plane wave again alright so again
you've got this this nice symmetry
between the the what goes in and what
comes out so there are these two
different with really different ways of
thinking about how this is all going on
end of show right then right okay so so
all that was that was just a warm-up
that was last week's lecture last
Monday's lecture we've done so far
right oh yeah there look lots of
interesting things that's coming up
later okay
so we're going to carry on straight away
with this more of this four F system and
spatial filtering so we're just getting
starting on this and so there's lots of
very interesting things you can do with
these so this is back to the same thing
again red showing again but bit more
detailed now because we're actually
going to really look in more detail of
where these orders are formed and what
you can do with them so we have a
grating object let's imagine this is
just a one plus a cosine which means
then it produces a zero order and then a
plus and minus one order so three plane
waves and as we've said each of those
plane waves is going to be converted to
a a converging spherical wave and so in
this back focal plane you'll see three
spots corresponding to the three orders
and and then carrying on these plane
waves then converted back into a
spherical wave and you'll get an image
which is back to the grating again right
so at the moment the optical system has
transmitted all the information and so
what we would expect is that the image
is exactly the same as the object apart
from the fact that it's inverted and it
might be magnified or or the opposite so
but the the resolution you'd expect
would be the same all the detail in here
you'd expect to see in there right now
so the next thing to look at is what
happens if we now play something in this
plane that changes the relative strength
of these different orders and they're
lots of things that we could try the
first one the simplest one to look at is
what happens if we just put in something
a mask here
but let's throw the zero order and not
the others the two that's plus 1 and the
minus 1 order are abstracted by this
screen so all we let throw is this
single order which is going to be
converted to a plane wave and so on your
screen here all you'll see is the plane
wave right so the image that you see
yeah well I call it an image but it's
not really much of an image it's
completely featureless it's just like a
well grain spot a gray region there's no
detail there so we've lost we've lost
that structure the structure actually
because we've put this mask in here the
structure is below is beyond the
resolution limit of the optical system
and so we can't see it ok so this is
what we might call a low-pass filter so
if you think of these as being like
these grating orders basically represent
different spatial frequencies so just in
analogy with what you have in
electricity where you've got frequencies
and time in space we think of we've got
distance and we've got spatial
frequencies so this is why we talk about
low-pass filtering we're filtering the
spatial frequencies and just like in
time you remember when you when you've
got your your hi-fi system if you turn
up turn down the treble that's acting
like a low-pass filter it's it's cutting
out the high frequency sound and netting
through the low frequency sound only
isn't it so that's what we call it a
low-pass filter and this is exactly the
same we've put in the mask it's a bit
different in a way from the electronic
case because in optics we've got a lot
more freedom as to what we can do we can
actually put in something here which has
got real
sharp edges to it in electronics if you
try to make a filter there's there's
always a limit to how much you can make
the roll-off of the filter all right so
you can't get something that suddenly
cuts off like that very easily okay so
this is going through the mass of this
our original transparency then is a one
plus cosine and the one plus cosine we
Fourier transform that to find the what
we call the object spectrum and so in
this case finding the Fourier transform
of cosine is very straightforward
because we just expand cosine as e to
the I something plus e to the minus I
something over two and each of those
each of the I type terms will correspond
to a single frequency right so we can
end up now with three terms at different
frequencies the one Fourier transforms
to a delta function of U which is a
spike where u equals zero and the two
components of the cosine correspond to
another to another two Delta functions
each situated this one's at U equals
some value use u naught and this one is
some value u equals minus you got right
so you've got these three three spikes
that we've been talking about from the
last few slides now so this is the this
is this the object spectrum it's the
angular spectrum that we've got here so
actually you can see that you here it
basically represents the angle that
these these plane waves are traveling in
and there's three different angles that
you live the bike and travel in and so
yeah so that will propagate then and so
what as we've said before you can see
that these these rays are moving at an
angle and so after you transfer as you
go along a distance there you get a
shift in the x position which is given
by of course the this the slope of this
the more of the slope the more it
deflects by the time you travel the
distance F so this then gives the shifts
of these two things so that's before the
screen then then sorry people before
them up mask things have happened a
sorry
nothing happening Oh some of them and
then
but what have been second
Oh
ah okay right so yeah so this is showing
then so these rays then as they come
down here if we put in this pupil mask
which is going to get rid of these two
and it just leaves the other one so we
just left them with our single beam
which as we said before gives an image
which is not really an image right yeah
so after the pupil mask we multiplied by
the mass get rid of those so just leaves
the other one and then finally then the
final image you'll see is going to be
the Fourier transform of that and the
Fourier transform of the Delta function
is a constant and so you see something
which has got no no detail in it okay so
this gives another example then here
we're looking at some binary amplitude
grating so this is a particular example
another example so that last time we
looked at the cosine sight thing
so this one is now a binary grating
which means that the object is something
which is either black or white and we
shine light onto this so it's like a
square wave grating and so unlike the
previous case where cosine of course
just produced those three three grating
components in this case there's going to
be a whole load of grating components
they're going to go on forever and but
they get weaker and weaker and so you
can see here it shows a few of them the
first the second there'll be the third
the fourth and so on and then minus ones
are honest on and each of these is
moving at a different angle so when when
the plane wave gets to the pupil mask
it's sorry when the the plane wave is
focused to the pupil mart it's going to
reach there at a point the distance from
the axis of course
depend on the angle of this plain wave
component the grating component right so
so if you're not here you're going to
see all these different grating orders
all arrayed in this position here and so
by putting some sort of face all sort of
masks here we can monitor and change the
relative strength or phase or whatever
we like of those different components
and then they'll be added together and
produce some nice effects that we might
want to look at so let's see what we're
going to get in this case so in this
particular case in the picture the size
of this aperture is enough to let
through the center of three components
the zero order and the plus or minus one
order but it's stopped all of the other
ones right so what you'll see here is
the Fourier transform of this if we map
it's before is the three it's these
three spikes again the Fourier transform
of the three spikes is just a 1 plus a
cosine the relative strength of these
might change according to the properties
of these of the transmission of the mask
but anyway so what you see
finally is an object which is like a you
see the 1 plus cosine type variation in
amplitude and so you notice that this
thing here started with these very
square edges we no longer see the very
square edges so this is an example of
how we've lost resolution the optical
system has got a certain finite
resolution
there's detail in this original object
that we can't actually see in the final
image and this is because not all the
light has got through the system
okay so we're going to look at a
particular case now this is the bit I
was saying I wasn't very keen on no no
it's all right but anyway there's a load
of maths here which I might actually
gloss over a bit and let you look at it
yourself later but basically then we're
looking at our particular example the
it's a binary grating with perfect
contrast every equals 1 ie this goes
from black to white that's what that
means it's got a period of 10 microns
right the spacing that the period is the
distance between the repeat on this this
grating and it's got a duty cycle of 1/3
which means that though the the white
bit is 1/3 of the black pit is that what
that means 1/3 of the whole yeah 1/3 of
the period ok and and then we're also
told that the for F system consists of
two identical lens's of focal length F
equals 20 centimeters so both F 1 and F
2 or 20 centimeters and we put a pupil
mask of diameter 3 centimeters here at
the the forea plane what is the
intensity observed at the output image
plane so that's the question
okay the secrets to solve this kind of
problem what you've got to do is to work
out what you're going to get in this
play which is basically the Fourier
transform of this and then you're going
to have to multiply by the mask and see
which of those orders get through right
so the relative scaling of this
diffraction pattern and the math is
obviously all important for that if you
don't get that right then you'll get the
wrong answer obviously because that
you'll you all you might get organiz
getting through that shouldn't or the
other way round and so once we've done
that we know now what the grating orders
the the the Fourier components in the
final image are going to be and we then
do another Fourier transform and we've
got the image amplitude
there we are okay we've got to do
scaling to get it to the right size
normally of course a very aim we would
be interested in intensity not hydrogen
as well okay so this is our example
low-pass filtering a binary amplitude
grating so this is this grating as we've
chosen and so you can see this is the
dimensions in microns so this is quite a
fine grating so you can see that it's
not symmetrical by the way the the duty
duty cycle is not 1/2 right so the the
width of the bright regions is smaller
than the width of the dark regions and
that's important when it comes to
looking at the Fourier transform and
therefore the spectrum of that and then
the pupil mask that we're going to look
at looks something like that we let
through the frequencies near the center
and we keep our rejects the frequencies
which are further from the center
okay so we we've got to the Fourier
transform of that this is a repetitive
function so the Fourier transform just
becomes a Fourier series and this is an
expression for the Fourier series of
this it basically consists of two parts
that there's it's kind of sort of
envelope which is the sync basically
this is going to consist of a infinite
number of discrete frequencies isn't it
it's going to consist of a constant term
which is the zero that it's going to
consist of a as first harmonic and in a
second harmonic and the third harmonic
and so on if the if the beauty cycle had
been chosen to be a half I it was a
square wave then by symmetry it turns
out there's a second harmonic and the
fourth harmonic and all those all vanish
so you'd only end up with the zero the
first the third fifth and so on but in
general there will be all the orders
there and the relative strength of those
orders here you can you can see is
decided is it depends on the value of
this sinc function actually that's
that's a very easy way of seeing about
the even orders Cantley because what
happens is if the duty prepared the duty
factor is a half it turns out that the
even the even orders when Kira calls or
even number coincide with the zeroes of
the sinc function and that's why they
vanish right so so where this comes from
is that so that we're saying that this
function is a maybe it's easy if I go to
the next line of me which is the Fourier
transform yeah the Fourier transform the
so the the Fourier transform the
strength of these components is what
we're going to find out and so this
function Fourier transforms to something
which is a series of Delta functions
corresponding to the different orders
different values of Q and the strength
of the different components is given by
the value of the sync right so you can
see again if if there are some values of
alpha Q which can be 0 and therefore
would mean that those components will be
there ok so so when it gets to the to
the pupil mask all that's happening
maybe you were to go back to the this
diagram so you can see that the what we
do what we need to know is this this is
coming to a focus at this point here but
we need to know what that distance if
the ear is but we need to know that's
relative scaling of where where that
diffraction order appears
along
that was what I cleared there
okay so it's this is this lambda ref
that comes in here because it's going
through a distance F so that gives the
the scaling of this distribution you get
there and so so that's where you now is
replaced by this X double dashed over
lambda ref right the scaling of that
size of that the diffraction pattern and
and then this is putting the answer in
is that you get this oh yeah we told
that alpha is the duty cycle so that's
the third so that becomes alpha is a
third in here and where do we get the
one centimeter from
yeah I think that saw something like
this on the crib sheet which you've all
had by the way so we've got one right
here
we've got lambda equals North Point 5
micro F equals 20 centimeters lambda
period is equal to 10 microns so that
gives that lambda F over R into lambda
is no point 5 times 20 over 10 that's
notthat's this is this is microns oh
yeah that's lead these are in microns
these are in microns these are in
centimeters so this is in centimeter and
this is equal to 1 centimeter right okay
so that's the what's going on there then
that's the 1 centimeter
okay that's so that's just plotting then
these different orders and so you can
see here the zero order dad plus one
plus two plus three so on but the screen
blocks out all the ones except for these
central ones and so this thing that
these relative strengths must come from
well there's a the fact of third day at
the front is this one and then this is
multiplied by the value of sync of a
third which must give this this the rest
of this thing here so I have posted in
the website a supplement to this lecture
so in the supplement I have worked out
all these calculations
how you compute the sink of one over
third sink of two over third and so on
so you can see some more details in
there in the in the website yeah you
remember I think you've defined this
sine of pi X over PI X look for that so
that's the that's the way that people
are using to define a sink of X equals
sine of pi X over pius all right so it's
it's so defined in that way because then
then lots of nice and nasty pies and
things cancels out in everything else
okay so we're kind of then multiply that
by the the size of the mask so as it
shows here the mask going from minus 1.5
centimeters to one point that's 1.5
centimeters which means that this
grating component which is at 2
centimeters doesn't get wrote right so
so that's what that what gets through
the aperture and then we Fourier
transform that again and so the Delta
function transforms to a exponential
shifted Delta function and corresponds
to exponential and therefore our output
of course this this F was the same in
this case the two F's of the two
elective lenses were the same so we get
have 1 over lambda ref again there yeah
what will happen if we make the width of
the this mass minus 2 2 plus 2 will
there be any deflection due to this
second-order
it was exactly - yes then I guess if it
really was exactly there you'd expect
half of it to get through I mean
normally we we would a serious either
less it through or it doesn't but so if
you let if you made the the size of the
aperture 2 / 2.5 then of course this
next order would get true wouldn't it
yeah I do it is exit will that means do
you have to consider a deflection of no
I think they're basically you know I
think that effectively yeah I think
you'd have to say 1/2 of a go through
actually that's correct because in
actuality the grating is finite so these
are not Delta functions they're sync
patterns so if you place the great in it
exactly - then you allow 1/2 of this
thing to go through right so there we
are
1/2 is yeah yeah that's
good way of thinking about here we
started off by saying that this is a
continuous repetitive grating and so we
can think of it in terms of a Fourier
series but of course it's not actually
infinitely big I guess it's not going to
be more than a few light years across
so this is multiplied by another rect
function really right so the Fourier
transform of this whole thing multiplied
by rect is something convolved with a
sync so each of these these things is
not really a delta function but it's
really a very narrow sync which depends
on the size of our great e so so the
point George is making is if you think
of it as a sync then you it really would
be letting it only through half of it
all right
I think mathematically adaptive be like
that as well you get used to these when
you do you must remember from solving
partial differential equations when you
do the boundary conditions you always
have to do that sort of thing it seems a
bit of a cheat but okay so this is
putting the numbers in then so this is
the the three components these are these
root 3 over 2 PI's we had there and and
then finally these two combine together
to make a cosine and that final
amplitude of our image is going to be a
constant plus of cosine you notice that
it's not the the amplitude that this is
not the same as the amplitude of that of
course because the fact that it was this
rectangular crazy ya know finally though
as I just mentioned briefly a little
while ago what you actually see is not
the amplitude what you see is the
intensity which is the modulus square of
the amplitude right so you remember we
said that what we normally are going to
call
the intensity just the modular square of
the amplitude there should be some
halves and funny things out there if we
were really doing according to the
pointing vector but normally we don't
bother with that we just say is the much
of the square so the modular square of
this we're going to get this squared and
twice the products of these two and then
this one squared so we get this terming
cos squared and then you have to
remember your formula that you learn in
math cos 2 theta equals 1 minus 2 cosine
squared theta which allows us to write
cos squared theta is 1 plus cos theta
over 2 so this cos squared we can now
replace by a constant plus a cosine of
twice the angle and so that this is the
constant bit and this is the cosine of
twice the angle all right so that like
gives a another harmonic in the image so
that's something that people are a bit
confused by sometimes first of all
because we've gone through a lot of talk
about how it's only the first time on it
there's got throw but actually now when
we look at the intensity we've now got a
second harmonic that's that's appearing
in the final insensitive so note the
second harmonic term intensity due to
the magnitude square operation this term
explains the ringing incoherent low-pass
filtering systems alright so yeah so
it's the ringing is the the Wiggles you
get an honest on the sharp edge alright
and that's a sketch of what it looks
like so it doesn't look quite like a
cosine grating anymore because when you
square it the the low values get
depressed more than the high values
right so it become makes it the more
peaky
like that
okay so perhaps we oughta start again I
was just waiting for one of our people
to come back our were there we are so
just in time okay so yeah before we
carry on then I there was a question
just before the break so we're gonna
have a question okay actually a little
question that's the first question in
the regarding the physical meaning of
the low-pass filter just now in a
previous slide
yeah okay so the question is about the
low-pass filter so you remember in you
know in fact I'll draw something from
electronics so you know you remember
from your high amplifier so this is
frequency so these these are high
frequencies and these are low
frequencies so so this corresponds to
and this corresponds and if we if we
multiply this by some function that
looks like that this is what we call a
low-pass filter because we should we let
throw the low low frequencies and we
stop the high frequency alright so the
same is true here so when we say low low
pass we we mean that the we let through
the the the spots which are close to the
axis and stop the ones that are far from
the axis so that's the first question
then there's another one the second one
is that regarding that our Bayes rule
because I've been through tell us if we
can create two orders out we've we can
form an image but just now we know that
even with the three orders I mean the
wave being still get very strange we've
compared to the original one so when the
check was a reason right okay yeah so so
this is going back so adverse rule which
I don't know I'm that I've never really
heard what he said actually but have you
got that in a book or something
it's probably in Chinese
but anyway the Chinese Abba came up with
this rule that you must have at least
well we gave an example just now didn't
we of one where you only had one b1
order got through and we didn't get an
image so yeah because you're not
obviously it's not gonna get any
interference if you've only got one beam
so abbath according to Albert you've got
to have two beams interfering in order
to get some variation in amplitude in
the image and so he said he says at
least oh I guess so it could be you know
if you think of these it could be you
could let throw the last time we had the
zero and the plus 1 and the minus one
which is three beams and the point that
was made in the question was that
although we got an image it was put in a
very good image so I think that's
probably right that that you know I
guess abbis criterion is Abbas rule is
all to do with resolution of an optical
system the smallest detail you can see
but it doesn't necessarily mean that
you're going to get a good image and
changing the relative strengths we're
going to see this in lots of examples
now actually I guess the whole of the
rest of this lecture will be about these
more examples of stopping beams and so
on and changing beings and seeing what
it does to the pattern and so very often
what you'll find then is you get
something which may be not a good image
of the original object but it it's an
image which is better than nothing
okay so all right that can't answer the
question okay so we're now looking at a
bandpass filter so going back to what I
was saying here the bandpass filter of
course is is something that looks like
this it's a bandpass filter stops the
low frequencies and also the high
frequency
and so you can see here the the optical
equivalent of this it's now stopping the
zero order but allowing the first orders
to get through and then the later ones
would also not get through
oh yeah that was another point that
George mentioned to me during the break
that I hadn't appreciated fully and that
is that this particular example which
had this duty cycle of a third actually
it turns out that the third order of
this one vanishes so I guess the third
the sixth the ninth all these and this
is just like we said that for a square
wave grating all the even orders cancel
so it's all to do with the zeros of that
sinc function we had sinc of alpha cube
right so where alpha is the duty cycle
so for any duty cycle there will be
particular values of the particular
integers for the order which would
vanish in the Fourier transform right
okay so we're here we're letting through
the plus 1 and the minus 1 what did
something do then ah right yeah okay so
okay so the amplitude is 1 I'm going to
be a 1 plus cosine and intensity we're
going to get a second harmonic in it
just like we had with the previous case
okay and then we're never going to go
through this in more detail so it's the
same grating as before so we haven't got
to work out all that stuff again the you
remember it was one centimeter was where
the first order was so we've got this
series of these different grating
components and these ones are 0 as I
just described the next one wouldn't be
0 that
and we're putting this mask in which is
just going to let through these two and
so we mathematically we can write the
mask as being the sum of two rectangles
rectangle and another rectangle so then
it lets through those two components and
stops the others so we've got these two
two Delta functions and then we just got
to do a Fourier transform again each of
the Delta functions produces a plane
wave and it's a complex exponential we
then gonna add those together what we've
got I've got to change the inter
distance of course and then combine
these two complex Exponential's to make
a cosine and then finally we've got to
square it again in order to get the
intensity out and so now notice what we
get now you see we get the constant term
C here is just cosine there's no
constant term anymore because the
constant term is gone so it can't the
cause when we square it we get cos
squared the cos squared we go into
double angles again we get a constant
plus a cosine and so you see that
actually we get an image where the first
time on it doesn't appear at all but the
second harmonic does so this comes back
to your question in a way you end up
we're getting an image which is actually
really not like the objects because you
see that this has got twice the
frequency of this alright so although
Abba says that you can get an image he
doesn't say that you can get a correct
image and in this case you
be very confused by this couldn't you
you could look at in your microscope at
something and you could see this and and
you'd be completely mistaken about the
object that was actually producing this
because it's actually twice as big as
what you expect so the right question
yeah
if you grab just the zero order and say
the first order but not the minus one
then you get the same period as the
original grading yeah I guess so with
that right if you take this one and this
one so you've got and then you square it
you're going to get one plus and eat in
the eye you're gonna I like yeah he's
working it out he's working it out we
have a race Shelley you're going to have
a one plus e to the oh I something
around and then this modular squared so
and then you can take out on each of the
I something over to easy you've got the
answer
yeah something like that
so this is cosine so you're going to end
up the cosine squared it's the same as
this isn't it would just be the phase
term no it can't be because it's up as
far apart yeah you get the first time
with it yeah because it's the square of
half of it yeah okay so it's it's it's
double half of it so you do get the
first time on it yeah you're quite
correct so but you know I mean it might
work for this object but I guess there
be other objects which it would also
produce funny results and and of course
as we know while some of us know most of
you won't know but for example Shalem
will know that if you have a
a mask which is a symmetric like that
it's gonna have some quite interesting
properties so I don't know whether you
gonzo these at some later point maybe in
the exam ok so yeah everyone SWAT up on
a symmetric gratings yeah yes single
sideband modulation yeah and what's the
other word for it what's this
well when they measure the flow of the
air in aerodynamics clearer and clearer
it's basically clearer okay
and now we're looking at tilted
illumination so same sort of system as
the last one but we're now not
illuminating it with a plane wave along
the ax axis we're going to tilt this and
so what we find what do we find if we if
we got a grating and we illuminate it
with a plane wave what do we find
happens if we tilt the the plane wave
you come across this it was you did this
earlier on did you I don't think I was
there for this one but actually all that
happens is it just tilts the diffraction
pattern very simple so you can see now
the zero order is still in this that
this direction and and all the other
orders just move follow these they all
rotate through the same angle this is
actually only true if you can think of
this as being a thin grating and you
know if you if you're go into a theory
of volume Holograms and things like that
then you'll find that things become much
more complicated the relative strengths
of the gratings can change and so on but
if this is a thin grading so that you
can use this property that you just
multiplied by the transmitter
transmission of the grating then that
will be the case all right so you can
see what that's going to do all the
gratings move up
which means that they all get moved up
there as well so the you see the zero
order grating which was appearing down
at the origin is now our tilted upwards
and appears at this point here so with
this band pass filter that we've chosen
we now get the zero order throw so we're
getting the zero order throw through
this one and we're getting the - - order
through this one right so it's we
illuminate it with a we eliminate the
grating with an off-axis plane wave at
an angle of two point eight six five
degrees which is obviously very
carefully designed so that this this
grating this a grating order moves here
so what happens okay so yeah so it's
just describing what I just said
yeah this is just describing why that
happens then in terms of multiplying the
the the see this tilted wave is just
going to have a complex a complex
exponential phase variation like this so
you multiply that by the grating and
then the the year is just as this result
of rotating the thing
okay so so there there are the grating
orders the zero order one is this one
and this is the plus of minus one and
the plus and minus 2 and so we get let
through those so so this is a bit like
so we're now letting through the the
we're letting through the zero order and
the - - order so it's a bit like the one
in the question but but now it's the -
tool they're not the plus either not not
though - one order so this is going to
be another example where it doesn't give
a very good image I think from that
result okay so yeah that's just really
by here we are so it's it's 1
centimeters the the the magic figure
that magic angle was designed so that
they're all shifted by one centimeter so
after passing through the mask we just
get those two components and so these
are the two components oh yes so ah but
they said no it isn't yeah that's right
because they're still so it's bit more
complicated than where I was eyes tricky
this so not only of course that they're
electro the the zero order and the - -
order but those orders have got this
complex phase shift is that right
yeah exactly exactly because of the face
because of the yeah okay so that's what
I'm saying so although this is a zero
order and a minus to order they're
actually shifted and so when you do the
Fourier transform they're they're just
as though they were the plus 1 and the
minus 1 order and that's neat very
interesting but with different
amplitudes given by these two strengths
here and then so yeah so then we have to
put the values of u in terms of the X
and then we combine the two
Exponential's
well can't do that anymore yes okay yeah
exactly
so have you now you actually done that
right so you've got to do the the the
modulus square of this so you're going
to have this squared plus this squared
oh yeah okay so that's the approach you
are so to touch then twice the product
of these so and okay no sorry that's not
right yeah that's right yeah
okay and this is what it looks like
there are so okay so the field is the
first time on it the intensity is the
second harmonic because of the squaring
and the contrast is not 0.7 right so
again we end up with a bad image as it
shifted because it's double frequency
and also the contrast is reduced in this
case because the strength of the two
beams was not the same this is
calculating here the contrast this is
the normal definition of contrast of
fringes the difference over the sum and
it comes to a point 706 to the previous
of case of course it was the contrast
was one right another example a new
pupil mask consisting of two holes each
of diameter one centimeter and centered
further away from the axis plus or minus
two centimeters from the optical axis
respectively what is the intensity
observed at the output image plane right
so at least it's told us the two
centimeters we know where that is we
don't have to worry about working out
what 0.347 nine degrees is we know that
the the first one first order was at one
centimeter so this this is the second
order is where the aperture is so the
second order is going to get through the
plus 2 and the minus so and did that
something that appear then up there
right okay so the second harmonic is
going to get through and when you do the
square of that you're going to get the
fourth harmonic so this is going through
the mass of that thing so we've got
these just these two components so these
have now got two centimeters in rather
than one centimeter so that means that
you get a two
multiplied in here there wasn't there
before so that's the where the second
harmonic comes in and then when you do
the square you know again you're only
going to get the cosine and no constant
term and then you do cos squared and
change that into double angles or full
angles actually and now we're going to
see what it looks like and now it's it's
even worse than image of what we wanted
so this I guess is I mainly telling us
what we shouldn't do other than what we
should do
yeah another question is this is still
assuming f1 equals f2 and what happens
if F 2 is smaller than f1 if you change
the relative values of F and F 1 and F 0
all it would do is is is change the size
of the final image right if you if you
think about this system let's go back to
one where it shows there we are you can
see that it's only the relative sizes of
the the rest relative distances from the
axis only depend on the this part of the
optical system and then after the after
the the pupil mask we've already decided
then what orders get into the final
image and then the final part all that's
going to do is just change the relative
size of the final image you see
right now we're going to do another a
bit more complicated one a phase pupil
mask and so what is what this is shown
as as you can see it's got an opaque
part of the screen here and then in the
central part it's it's transparent but
it's got some some sort of structure
there some some diffractive dielectric
element which changes the relative phase
of this order compared with the other
orders so the question is what's that
going to do to your your your image so
the rest of the system is completely the
same so this is this is what we are
considering putting in there it doesn't
matter what the thickness of this is
because all that's going to do is change
the relative phase of all of these so it
won't affect how they interfere with
each other who just changed the phase of
the final image which you don't see
anyway but what is important is this bit
here the thickness of this s equals 0.25
microns and the wavelength was half 0.5
microns so that's lambda by 2 so that
means it changes the phase by 180
degrees all right so any one I minus one
oh yeah n minus 1 so it's not going to
do that take in fact okay it's PI by
doing right okay so okay this it's known
as a phase pupil mask or a pupil face
mask or a mask pupil phase yeah yeah
that's right so that must be in the next
lecture because it's not in this one
so that is getting on to the real
application okay so George George said
that I would have just failed
exam because I forgot the minus one so
of course this light this light goes
through the glass but this light goes
through the glass and then the air right
so to work out the relative phases of
these you have to look at the relative
change in the refractive index times
their optical thickness right so this is
0.25 microns and this is at 0.25 microns
divided by the wavelength and it's gotta
be multiplied by 1 point 5 minus 1 so
this is assuming it's glass so this
means that the phase change is PI over 2
all right so not naught PI by PI over 2
ok so exactly the same grating these are
all the orders and so you know you see
first of all we get rid of these outer
ones we let through just these 3 but of
those we're also going to change the
relative phase of this one relative to
the other 2 that doesn't matter of
course if you in your son whether you
change the the phase of this or you
change the phase of the other two it's
only the relative phase which is the
important thing okay so
okay so these are the three conditions
then but the ones that are not don't get
through at all the strength of those is
zero then he's here we're taking this
these ones to be a strength one and this
one in the center to be a strength of e
to the I Phi and we know that e to the
we know that Phi is PI over two so this
is e to the I PI over 2 which is just I
is that right well I've got this I minus
one so one way to write this pupil
function is as an act that is as big as
their opening yeah yeah okay sorry yeah
so yeah so you can think of this as
being the whole thing the - well this
this this on top of the whole rectangle
right otherwise you'd have to do this
little bit of rectangle and this bit of
rectangle and this bit of rectangle
there's three things so you couldn't do
it either of those ways I guess but the
way that George has said done it here
it's the rectangle of the whole lot it's
all of this - the central one plus the
central one with the phase change right
so that's the - the one that's not there
plus the one that is there with a phase
change and yeah so now we well of course
we still get all these outer functions
but now we've got some eyes coming in
sorry just one oh I yeah and and then
these two combine to produce cosine step
I guess if you change the face of one
relative to the other it'd be more
complicated but it needs to produce a
cosine so we've got an eye a plus a
cosine and then we do the modulus square
of that and the modulus square of that
of course is
you're not going to get across product
term are you because these two are in
phase quadrature so when you do the
modular square
you're only gonna is that right yeah
you're gonna get this squared plus this
squared and this squared you can write
as these two terms when you do the the
double angle and I guess the next one
we're going to see what it looks like
that our this is what it looks like
first time on it infield
second harmonic in intensity because of
the squaring and again the contrast is
now much reduced actually in this case
down to 0.25 so I guess this hasn't
turned out to be a very strategy this
this lecture seems to be mainly showing
you what you shouldn't do you make your
microscope okay right so all right so
now we're going to go on to think of
this in terms of what we call a point
spread function of the low-pass filter
so do you remember that going back to
our analogies with electronics we can
think of in electronics we think of you
can either think of you think of your
hi-fi amplifier as being like a black
box don't you and you put in something
and you get out something and you can
either think of what happened is inside
there as being like H of T an impulse
response or you can think of it as being
a frequency response right so there are
two what two different ways of thinking
about this black box the same is true in
this case so up until now we've been
thinking of this in terms of the
frequency response what frequencies get
into the image and now we're going to
think of what that means in terms of the
point spread function we've come across
the point spread function before of
course so so we can the point spread fun
is basically the image of a point object
and our point objects we make by having
a screen with a very small hole in
alright so this thing is going to expand
reach the lens is going to be collimated
because we've assumed this to be on the
axis this collimated beam excuse me
it's going to arrive at our pupil mask
part of this collimated beam is going to
go through the aperture it's going to
get to a lens it's then going to be
focused down and produce a converging
wave to form a spot on the screen all
right so this is what we think of as the
final spot is going to be what we call
the point spread function H of X Y ok so
how do we how do we work out this point
spread function so what we do is we can
again work through this system in terms
of Fourier transforms this is a delta
function so the Fourier transform of a
delta function is just a constant we
multiply the constant by the pupil mask
and and then we've got another system
here which does another Fourier
transform so what what we're going to
get in the in the final plane the image
of a point object is going to be the
Fourier transform of the pupil mask and
yeah so it makes the point here that we
have to be careful of course how we
scale things and of course the scaling
is going to depend on the relative sizes
of these f1 and f2 as well of course so
example if we got the low-pass filter
right so zeros in the center it lets
through the low frequencies stops the
outer frequencies
so this is the this is the pupil mask
what does the point spread function of
that give and so we're assuming that the
input the input transparency is Delta X
are Y on the axis so this is illuminated
on axis the field after the after the
pupil mask is just the same as the as
the pupil mask and then we Fourier
transform that and it's going to give a
sink
the Fourier transform of rectus sink and
and we're here we've got the the
relative sizes of this as well so that
we know how big it is and so the
amplitude point spread function then is
is the is this thing this Fourier
transform rescaled with this X dashed
over lambda F we replace the U by the X
left oh Alam direct F and this finally
then is a answer and that
which three which three times its own
there this this three is that one that's
right yeah yeah okay yeah of course the
energy has to be conserved by this thing
so remember there's this theorem in
Fourier transforms again that says the
area under the function squared is equal
to the area under the Fourier transform
squared right so it's conservation of
energy and so this is the amplitude
point spread function it's sink so it
goes negative sink sink is sine PI x
over PI X so it goes negative so that's
why it's plotted here this is zero okay
it Wiggles up and down above zero the
zero like and so what you'll actually
see of course if you looked at the image
of a point object you see the modulus
square of this one here right just like
when we looked at our examples of the
gratings we finally look at the actual
intensity rather than the amplitude the
intensity of course is always positive
so you'd actually be looking at the
square of that but if you're trying to
think of this thing as being you know
what we're going to carry on to say of
course is in Fourier space Fourier
domain the spatial frequencies of the
object are multiplied by the mask in in
in terms of
this real space you're going to get a
convolution the product becomes a
convolution so we're going to convolve
our objects with an amplitude point
spread function right so but we have to
do it with the amplitude point spread
function find the find the total
amplitude in the image and then doing
the modular square to find the final
intensity
okay and this is an example of the point
spread function of this phase filter we
looked at
the the pupil mask this is the same as
we had before so it's this is its
modulus this is the phase and if you
calculate the point spread function of
this it comes to this this is the
modulus this is the face so you can see
that that is why we got a pretty
horrible image it's because it's got a
very nasty point spread function its
point spread function for a start
there's got these very large side lobes
which normally we don't like normally
like to have our point spread function
to be nicely narrow and smooth and
without big side lobes and then more
than that it's also got these horrible
face jugs which I think very likely to
produce pretty awful image which is
probably what we saw all right so this
goes through the this the some I'm
actually doing that so this is the same
as we had before then the big rectangle
minus the little rectangle with the
phase and then we've got to do the
Fourier transform of that and so the way
we've done it here then we just got this
the sum of two rectangles each of them
becomes a sync but you've got a you've
got this eye here you have to be careful
of and then finally we work out the the
modular square of the out of the of the
point spread function and the phase of
the point spread function right so this
is the modular square of the point
spread function so these side lobes are
actually quite huge quite quite big
okay yeah that's just looking at the
difference between the two then so if we
have our low-pass filter
so this is what I'm saying about these
being huge you see the the low-pass
filter these side lobes are pretty small
by comparison I guess you could you
could use some sort of shaded aperture
or maybe an aperture with some structure
that would allow you to make those those
side lobes even smaller this is this
process that sometimes called
apodization apodization means to cut off
the feet I think if George was giving
the talk he'd say something about the
Greek four feet being something to do
with pod right there we are so we over
time really do what you ought to stop
now yeah maybe that's a good point I
think that's good you've got the
concepts of the point spread function
now you've got a week to think about the
point spread function and and then next
time it'll all be very clear okay so any
questions from either side