Optics (Lec - 7) - Electronics & Communication Engineering Video Lecture - Electronics and Communication Engineering (ECE)

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so today in some way is the reward for
what we've been suffering through for
the last few lectures because we'll
actually get to see some optical
instruments and how they work
so we'll basically put all these tools
at all the terminology and the matrices
and the formulas and all that
we'll put to work for the basic optical
systems namely microscope and telescope
and we'll also look at the sort of the
building block of a microscope which is
a magnifier land but before we do that I
would like to spend some time just
reviewing basically all the fundamental
material before we go into the
application systems I would like to
review the fundamental material so the
first step four or so four or five
slides I don't remember I think five
they're not really new they are just
brought them you know just copied them
from the previous lectures and the kind
of overview and I would like instead of
just describing it again I would like to
encourage you I will pause or need to
and I would like to encourage you to ask
any questions if you have or this
particular topic on something related so
the first topic is the Ray transfer
matrices and again very briefly to
remind you this comes about because
refraction at spherical dielectric
interface is very complex to deal with
not very complexity it is actually
pretty simple but algebraically
complicated three or I should say
trigonometrically very complicated so
because we don't want to deal with that
stuff and it also hides a lot a lot of
the intuition we derived this paraxial
approximation which is true for a pretty
limited typically ten to thirty degrees
actually thirty stop domestic maybe ten
to twenty degrees from the optical axis
we discovered that there is a linear
between them elevation and angle over a
that is arriving from the left to the
interface and the elevation and angle of
propagation of a ray that is departing
to the right of the interface the same
rate and and of course the elevation is
the same because of the paraxial
approximation we neglect the thickness
of the instrument so so basically no
matter whether a hits here or here or
here it is all the same so that helps
and the second is that the angle
basically the only satisfies Snell's law
with respect to the normal atmosphere at
this location over here but of course
because this orientation depends on the
relative position of the arrival of the
other a and the center of the sphere
then we simplified it using you know the
paraxial approximation and we discovered
this expression over here so the
relationship between the right and the
left angle it is it has to do with the
elevation so so if you look at this
equation it says that an right alpha
right equals and left alpha left that's
the first term and then the interesting
part is the second one that says equals
minus this quantity times the elevation
of the Ray so that tells you that the
more the higher the rival elevation of
the Ray the more it bends right and that
makes sense because the higher the
arrival the larger the angle that it
makes with respect to the to the normal
to the sphere and therefore if you wish
the more violent the Snell's law will be
and of course you can imagine as you
increase the angle eventually the angle
becomes so large that the paraxial
approximation breaks down but but this
is all too before the approximation
breaks down the progression is linear
okay it is linear in the position
and the quantity that appears here does
anybody remember how did you call this
quantity optical power correct
could you please push the bottom correct
it is called the optical power and it is
actually the first intense instance that
we saw this quantity it is inverse
distance units but we don't call it
inverse meters we call it how to call
the unit of this quantity diopters right
so professor have a quick question yeah
here we have quantity over radius R does
this describe hyperboloid or elliptical
shaped or a spherical lens and if so
that's a profession for example with the
spherical or elliptical lens that has
two different radiuses how do you
account for that alright so so this
derivation we actually did for a sphere
so the radius here is the radius of the
spherical interface and a very similar
relationship would hold if instead of a
sphere you had another avoid another
surface of revolution which would be
symmetric around the axis over here okay
so for example if I put a parabola
imagine the parabolic Bowl where this is
the cross-section of the parabola okay
but then if you spin it around this axis
it remains the same this is called the
paraboloid of revolution this is
actually I believe this was a homework
or actually we will do it a little bit
later today we'll discover that a very
similar actually the same equation holds
except in that case R is not a radius of
the parabola the parabola does not have
a radius but it does have a curvature
near the axis okay so the curvature of
the parabola actually the inverse of the
curvature is the radius of curvature is
what enters this equation
now we discussed an ellipsoid and the
uri to that the ellipsoid has two axis
but the ellipsoid that will solve the
two axes are oriented in a different way
so an ellipsoid the two saw we had react
so propagation Z then you have one of
the lateral axis X and we saw that the
perfect focus in element the perfect
focus in refractive element is an
ellipsoid that looks like this so of
course this is a major and minor axis
but you can see that the ellipsoid is
still symmetric with respect to the
optical axis so in other words it is
still an ellipsoid of revolution around
the z axis so I don't have to worry
about two radii of curvature now but it
is still a valid question what you are
asking what if I did away if what if I
do put a surface there with two radii of
curvature well it comes out you will
seldom this see this in books except by
basic but I don't think I've ever seen
it in a book but but you can reconstruct
what you should do right this is
slightly incomplete because I'm only
dealing with elevation of the Ray and
the angle in a cross-section of the
plane and that is fine here because I
you know it is a revolution force of
revolution but if it is not a surface of
revolution then I have to grow this to a
four vector with two angles and to two
elevations and then this would become at
four by four matrix and then I would
have a different power in each axis
depending on the Gaussian curvature of
the surface in the correspondent
location where their egg hits the axis
so you can imagine it will become a very
complicated kind of situation so
typically we don't deal with this I
don't even know if optical elements are
designed this way
I was using
can you press the button please oh
you're gonna hear I'm sorry I pressed it
but it didn't seem to do anything okay I
was you couldn't hear now yes I was just
I was just saying that I haven't come
across this for geometrical optics
either but but I have seen it for
diffraction optics in terms of the Vigna
distribution function and what I was
going to carry on to say though was for
the geometric optics case it might be
quite interesting because it would allow
you to look at the behavior of skew rays
right so it might actually give some
quite interesting information I think
the story and you could also look at
things like caustics right you know if
you make the surface non non
rotationally symmetric you get new
different caustics forgivers who got
into into a slightly esoteric
conversation here but that's a really
good question so so when I end up
writing a paper about this in a few
months
great any any other questions
isn't this problem separable in X&Y
actually before I answer that let me
bring up a trivial cache where you might
have this situation that's a cylindrical
lens right the analysis would did is
actually valid for two cases one is when
you have a surface of revolution like a
sphere that is if you have a cylinder in
which case everything is invariant in
the out-of-plane direction so those are
sort of a trivial case where you have a
radius of curvature in one plane and the
caricature is the reduced is infinite in
the other plane that is you have a plane
a planar surface that's a corner
cylindrical lens and of course the
cylindrical lens does not focus to a
point it focuses on a line right it
would it would well there is no focus in
here but you can imagine to focus on a
line out and so what will the parent
yeah the question was isn't the problem
separable in x and y so we could just
extend the matrix in y direction and yes
I think that's true either paraxial
approximation it is separable yeah yeah
but of course if you go off the paraxial
approximation things like screws and so
on then all bets are operated they
become highly coupled so for example
astigmatism is a highly coupled that you
cannot really trace it with them
astigmatism for those of you who are not
unfortunate enough to have it in your
own eye glass prescription astigmatism
is a situation where an optical system
focuses kind of difficult I'll go to it
in next week it's a form of aberration
where the optical optical system has a
different focusing properties for
tangential rays as opposed to the
sagittal days I don't want to confuse
you right now but it is a form of
aberration that has to do with two days
that are arriving sort of effects and
off plane later and it is actually very
common in people very often those of us
who wear does anybody know of you if you
have astigmatism behind you
that's right yes it means that his
glasses are not exactly spherical no
other questions
okay the next step to do so this we did
I believe it was two weeks ago or so
then the next thing that we did is we
took this formulation basically oh and
before I move on let me just remind you
very briefly why we did this this looks
kind of silly he tells you that if you
have a very proper propagating in free
space again you can relate the elevation
and unbe to the left of some chunk of
free space to the lay elevation and
angle to the right of that chunk of free
space once again that looks silly
because well we know the a propagates on
a straight line in free space or in
uniform space because of firma principle
and of course what will happen is that
the 11 I mean the angle of propagation
will remain the same that's what this
law says and also the elevation will
increase by the distance times the
tangent of the angle of propagation but
of course what is the paraxial
approximation so instead of the tangent
who put the angle itself so this is what
this equation says here if you multiply
this by this the indices of refraction
will drop out and you're left with D the
distance times alpha okay so it is
really the the tangent okay so this is
silly but there is we did it is so that
when with optical systems with a cascade
of uniformed space then have the
refractive interface and then more
uniformed space and so on then you can
basically write write those systems
simply by cascading these matrices so
this is why we did this particular part
but this really doesn't say anything
other than rays must propagate in a
straight line in uniform refractive
index because that is the shortest path
for them to follow and then what we did
after we developed this formulation is
we actually put it to practice in the
special case of two adjacent spherical
surfaces
so typically you have air outside and
between the two surfaces we have glass
and this kind of element is what we call
in everyday language the lens it does
not have to be glass for example in the
case of our eyes is not glass it is
actually tissue mostly mostly water in
fact and but anyway it is higher index
of refraction than the surrounding air
so it is acting is doing the same job as
a lens and what we did then if you
recall a will not do all of it again is
but what we did is we did the Cascade of
the matrices for the two interfaces and
based on this we derived the properties
of this element and what we discovered
is the basic property is that if you
have a right bundle that is arriving
from infinity therefore this rays are
parallel this type of optical element
will actually bring them to a focus so
the Rays after they will propagate a
certain distance after the element and
then they will focus to a point and will
also derive the other useful quantities
we derived the elevation of this point
we discover that it is proportional to
this focal distance times the angular
propagation and then we saw that these
different cases of focusing depending on
the nature of the lens namely its
optical power so positive lens will
focus the Rays at a finite distance and
negative lens will actually cause their
eyes to develop so the strictly speaking
do not focus but if you extend the
backwards then they create what is
called the virtual image and we will see
later today what exactly this virtual
image means and why it is an extremely
useful concept in optical design for now
it appears as an oddity what exactly is
that why did they move why did they
trace that is backwards but today
actually you will see why this is useful
the main point to take out of here is
that the lens is acting kind of like a
liver so if you have a if you imagine
that you are rotating the angle that the
Rays are arriving to the left of the
lens then this point is moving linearly
okay
is staying on this plane I deal with
lists within the paraxial approximation
it is staying on this plane but it is
moving so as you rotate this angle you
rotate the angle the elevation changes
and the constant of proportionality in
this change is the focal length of the
lens this is a very key point to
remember about lenses and how they
function and the other thing that you
said is that them it can also work the
other way around if you take this
picture the rule is that the Rays always
go from the left to the right it's not a
rule it is a convention there's no
physical law obviously that says that
but so conventionally so but
nevertheless there is a physical law
that says that if a light propagates in
one direction and then you flip the time
backwards that is you reverse the
propagation of the light then the
physical law actually says that the
propagation should remain invariant that
is if I knew that the light propagate in
this way focus's if I turn time
backwards like these old movies that you
see where they play the movie backwards
and you see a person you know moving
towards the back if I did that I would
have their eyes going like this okay so
this is against our convention so that's
why I flipped it here so now again we're
eyes are going from left to right but in
this case they're going through the
focus what is picture is telling us is
that if you have rise that originated at
the point one focal distance behind the
lens then this will form an image at
infinity that is the lens will column
eight this ray bundle and we'll send the
Rays out parallel at an angle that obeys
actually if you look at it is the same
relationship that was obeyed before with
an extra minus sign and why did the
minus sign come about
anything symmetric here except for this
annoying - and well kind of
but
so they're bothering go to some Destin
Java object distance is it maybe not the
distance we have distances and what else
here
the angle is negative the angle that's
right the only reason of this negative
sign appears there is the sign
convention this argue here we define it
as positive this angle here is the same
but that's not how we measure angles who
always measure angles this way okay so
therefore this angle is negative so to
account for this negative angle a
negative sign appears over there I have
another quick question let's say when
you have a lens where our left and our
right are the same which I believe is
pretty common in optics so from the lens
makers equation you would get that 1
over F is equal to 0 no or is that wrong
because you get 1 minus the quantity
right that's not true actually if the
edges to the left is equal to that well
depends how you are is it like this will
like this so like this like this yeah
sure ok let me give you a hint if it was
really like this we have 1 and 1 now it
is the same radius of curvature and that
would indeed have no power no optical
power the focal length would be infinity
or the optical power would be zero in
this case are the radii of curvature the
same or no also one's negative ones
positive is everything exactly they're
not the same they're opposite okay like
so they may be the same in absolute
value but they have opposite signs
therefore in this case you actually get
a finite focal length I'm sorry I don't
hear you could see what they do yeah so
so yeah ok
other questions
is a very good question I'm very glad
you asked that we have to be very really
careful when we apply these formulas to
use the proper sign so remember the
curvatures by again by the sign
convention which I really recommend that
you I don't have it available here but
they really recommend you go back and
studied the convention is that if the
curvature points like this it is convex
towards the left then it is called the
positive curvature so that in this case
you would like for example you would
write our equal sign or maybe it is ten
centimeters right in this case because
the surface now is concave then you
would say that R equals minus 10
centimeters so then the lens makes it
make it a question in this case would
say let's say N equals one point five
for example lens maker would say 1 over
F equals one point five minus one times
one over 10 minus 1 over minus ten and
this is whatever in I believe it is
three or something super fun but this is
or this is 150 so it is 0.5 times
one-fifth so it is 1 over 10 so it is
done
other questions
Oh
okay
so this was 204 what we'll call the
Finland's where it is thin because we
assumed that the distance between the
surfaces within the paraxial
approximation can be neglected if we
cannot do that if we cannot neglect the
the distance between the two surfaces
our product our formulation does not
abandon us who just get a more
complicated formula and there is really
complicated I'm sorry the reason the
formula becomes more complicated is
because in if we drop the proximation of
negligible spacing then the way reflects
twice before exiting it will refract at
the first interface and the second
interface okay we'll get a formula for
this the thick lengths I don't want to
go back into that formula it just comes
out of the algebra what I really want to
emphasize in this case is that even
though the way reflects EE we can
pretend that it only refracted once and
we can do that by extending the incoming
ray extending the outgoing ray and then
what the plane where they intersect that
will that is called the principal plane
actually that is called the second
principal plane if I do it for a
incoming ray bundle that is imaged at
infinity then it is the other way around
again I can extend the incoming ray the
outgoing array and again get the first
base principal plane and the reason we
did all this exercise were to discover
principal planes and so on is because
this allows us to do write tracing as
shown over there and no clue in history
MA so this microphone has a tendency to
fold okay you can still hear me yes okay
so so so the reason we did this exercise
of effective focal lengths and so on is
because you can play the following trick
I can give you an optical system with a
bunch of lenses I don't know whatever
very complicated and I'm asking you to
find the imaging condition here's an
object and I'm asking you where is the
image what is the magnification lateral
angular and so on this formulation of
principle planes it applies it applies
as we discussed not only on a thick
lines but in general in any compound
optical system that contains multiple
elements like this one and in order to
utilize it what you do is you throw out
so to speak all this entire system and
you replace it with its principal planes
so here is let's say that the principal
planes are like this here's the first
principal plane the second principal
plane and now I can do the imaging
business very simply here's my object if
I take an a that goes horizontal from
the object okay this way it is as if it
came from infinity right so therefore
this way will go to the second principal
plane it will Bend and will meet the
axis one effect we focal length to the
right of the second principal plane
right then I can take another ray which
goes through the first focal point that
goes one focal length to there I'm sorry
let's head to the left before I should
you said to there let me repeat so this
ray will go will hit the optical axis
one focal distance to the right of the
second principal plane I said
the wrongs in the court anyway to the
right okay now let me take a day that
goes through the focal point one focal
distance to the left of the sack of the
first principal plane okay we'll run out
of space here but that's okay
this way we'll have to hit the first
principal plane and then turn horizontal
because this is equivalent to a point
source one focal distance to the left of
the lens that would be image at infinity
so therefore this lay must come out
horizontal and where this layer is made
is actually where my image is located
this construction is very useful and
very basic and fundamental and that's
why the principal the principal planes
are important now let's try this any
questions about this since we since we
have principal planes for the system are
we going to substitute this system bias
by a matrix which is for single lens
like 1 minus 1 over F 0 but that's not
going to be possible because the single
lens doesn't change the angle but the
system can can change the position and
angle booth yeah the system is not
equivalent to a thin lens it's a little
bit weird is equivalent you can think of
it as having two thin lenses one of them
is operating for this way right this is
like a thin lens but for this way this
is like a thin lens but for this way
this will bit confusing to think of it
like that right so sure would rather not
yeah not go there yeah yeah but if you
look at the matrix of the system it is
not a matter within lunch of course not
yet yeah
you
okay that was actually my question so
he's asking what if the focal length is
negative how would you do that is what
you're asking right yeah so that was
actually by the next thing you're going
to do also here so let's take this out
of the way take this out of the way too
because the market lake the market lick
kill - okay
we're wasting too much paper okay so
let's do the same game now but now let's
say that the focal length is negative so
here is here's an object
and here the two principal planes that
let's say this is the first principal
plane
the second principle plane how did i
trace in this case where I have a
negative focal length the someone want
to help me trace it but you first
propagate to second principally in again
understanding the horizontal ray to the
second principal plane
now what come back and I in the sense
that you the the read diverges RJ from
the first plane but you left to go
well my eyes are not very good they say
that he yeah this is the formal point
there yeah so two level diverge in this
fashion right yeah yeah what about the
other one
does the mother where is the first focal
point they cannot hear you well probably
we should take rays from the object to
the first focal point there's a back
focal point so what is that that's on
the right side of the first principal
plane should be somewhere on here and at
the point where they intersect you make
it parallel so this is minus F this is
minus F right here
the do this is okay so then what do I do
the point there the rib meets the first
transfer plane we make it parallel to
the optical that's right
so what I have to do is first of all I
draw a dashed line that goes to this
point okay this tells me which way this
should be pointed this will have to go
all the way here the other side and then
where is the image
the image is actually where these rays
are supposed to meet right they never
quite meet but I have to extend one of
them backwards so they would actually
meet here and notice the significant
difference that here the image is erect
it is upside which is customary for
virtual images whereas in this case the
image if you recall the image quite
clearly was inverted
so it is it is really the same approach
haven't changed anything but the only
thing you have to do really is be
careful with the signs who is to the
left of whom they work who keep track of
that and we don't lose our head is by
faithfully following the sign
conventions so the definition is that
the positive focal length would be to
the right of the second principal plane
okay fine if I'm given a negative focal
length that means that I should go to
the left over the second principal plane
and so on and so forth it that's the
that's the principal what about this
case can add this
of course again in front of the homework
you will see a system that has that
property
the second principle plane is to the
left of the first principle plane
there's no rule or law against that what
we do nothing changed really can still
do my rate ration as before if I were to
do my rotation here well I would go up
to the second principle plane I would
shift the rate down let's say this is
one focal distance and then now let me
do the other one correctly so this is
the first principle plane so to go from
here something like that okay that
produce the ridiculously over here I did
something wrong here so they must ended
up very close to the principal plane but
this is how you would they trace the
system
any questions about the stuff I'm doing
here
so when you're given an optical system
it's a little bit of a judgement call
you have a happen some message flashed
something got disconnected are you still
there
yep okay so when you're an optical
system there is three ways you can
analyze it
one is you can just blindly write down
the Cascade of matrices figure out the
income in and out or in outgoing angles
and then try to solve it this way the
second way is you find the principle
planes and then you apply one of these
techniques that I just did on paper and
this is another way to analyze it
the third way which is what I think
Sabac they did the class and it is also
the supplement of your notes is where
you take each lens you apply the imaging
condition individually and then move on
I will not do it again it's in the
example in your notes but it's a little
bit of a judgment call and sometimes
it's a different a matter of preference
you know some people they are very good
with matrices so it is just better in
that case to just multiply the matrices
and get done with it the principle
planes they give you more intuition so
I'm personally not in favor of just
blindly multiplying matrices that's a
little bit mind-numbing and also very
prone to algebraic errors because we're
all human right but these methods of
principle planes they rebuild a little
bit less prone to algebraic errors
because with these methods you can
actually test your intuition you can see
easily whether what you're doing is
correct or not what if you do make an
algebraic mistake during the matrix
multiplications well it's very difficult
difficult to to know where it happened
and if I confess when I wrote the
supplement to the node which is I don't
know if you've had the chance to see it
but it is the same optical system solved
with the three different ways you know
with first of all with a cascade of
lenses then with matrices and then with
principle planes so I confess that I
first did it with a cascade of lenses so
I took the first lens i imaged it
and I used the image of the circle of
the first lens as object to the second
lens and Ament again and I found the
answer
then I took a plunge I got a big cup of
coffee I did all the algebra of the
matrices actually got the wrong result
it was not agreeing with the first
result so I went back and did it again
and finally I got the algebra correct
and I got the two results to agree but
for sure the first thing I did it with
the cascade of lenses was correct then
every time I got a different result with
the matrices I knew that I was making an
algebraic mistake until I finally got
the matrices to agree with it with the
actual result and I got it right so
there's no I'm telling you this because
I've taught this class for 10 years
believe it or not so every time when we
do the quiz number one your colleagues
in the past they all try to solve the
problem with matrices because it is easy
right you know maybe this is a
no-brainer you just multiply and get the
result about 70 percent however however
all these years have actually a
statistic about 70 percent they get
stuck into some of the black error and
then they cannot for solve the problem
because the nature of this is if you get
the rate racing wrong you cannot do
anything you cannot do magnification you
cannot do whatever so I would only
recommend the matrices if you are really
really sharp with algebra otherwise you
are better off and actually overall I
think you're better off if you learn the
physics which is not what the math
assist issue the matrices are convenient
for example who derived all these based
on matrices but after that we should let
our intuition take over and not rely
just on the math which is actually
pretty good principle in anything you do
in engineering but it is especially true
here any questions
hallo question so for the focal answer
as objective if you retrace the Ray
coming from the object say oh through
the center of the first
principal plane so this organization is
a center of the principle Glenn you
trash it to the focal to the focal point
and I mean Grammys so in this case for
the local folk audience is the positive
so it we have three say crouches
characteristics arrays of for the human
race ray so OC Rhea OC yes ray yeah so
for the focal focal lengths that is
negative can't draw that line so you you
will find that the intersection of those
rays would not meet the objects you find
oh it should it should it should if I if
I had drawn it properly it should if you
drew it this way if you drew this ray
and then you extend this very backwards
it should still go through my image
point know the student so if you're
right that they don't need because this
is a virtual image what does meet is
their extensions backwards yeah no but
when you draw it but so you see those
angle angle relationship so the less
because my is not very good if I had
done the drawing properly it should have
happened the outward anger is larger
than the larger that then the angle
coming going from the ah say
I mean if you if you take a look at the
Django relationship between this one
yeah
and one this one and this one and the
the one coming from the object to the
first focal point so this so this angle
is that larger than the rate coming
larger than than the one from the go to
the first focal point which means that
if you back this is my hand a there's my
handwriting oh I see I see
okay so this is how I first get in order
to make them meet I did not make a good
diagram I'm sorry okay let's see
they are not a very good rafter I'm
sorry the radically draftsman right
not a very good draftsman okay so I
don't want to dwell on this too much but
this is basically we already discussed
this idea that they're pretty much the
only case where you form aerial image
that is the Rays actually do meet at the
image point is when you have a positive
lens and you place the object to the
left of the front focal point of that
lens that will fall very any element all
the other combinations they actually
form virtual images which means that
what is coming out of the system is a
divergent ray and divergent ray bundle
therefore they don't really meet but
they do meet if you extend them
backwards and this backwards
intersection is what we'll call the
virtual image and you can see these
different cases here sometimes the
magnification is bigger than one
sometimes is less than one this actually
can never remember every time I have to
derive which is the case but anyway all
the possibilities can happen except for
false for this one where the
magnification is always negative it is
always inverted in the virtual image the
magnification also don't to say it is
always positive it is it is very much is
erect okay
any questions about this real and
virtual images
and the last thing we wanted to remind
you off is what we did there on Monday
actually the definitions of Apple tool
stops
superious windows chief and marginal
rays I really don't feel like going
through all this of it again they are
there this is probably very fresh in
your mind so are there any questions
from - lecture that I can answer about
this or the term relating applies to the
case when the numerical aperture changes
over the field of view that's right so
to exaggerate it suppose that you have
an optical system were on axis you get a
large acceptance angle but then somehow
and you go after off axis the acceptance
angle becomes very small right so now
you compare this with this clearly this
is smaller that is Benetton and that
would be would that be independent of
placing an aperture stop at the focal
plane of the lens what you can say about
vignetting is that it generally happens
when you place the Apple to stop in the
wrong place when a properly designed
optical system it it should not happen
here at least if there's a way to get
around it if you construct the optical
system in a way that you know you cannot
place the Apple to stop then that's a
badly designed optical system right so
in generally you try to avoid it by
placing your stops in in a strategic
location
there's no universal rule that says you
have to place the apples to stop there
in order to avoid the vignetting it kind
of depends you know if you have a multi
element optical system
if you always make sure that we place
their purchase top at the effective
focal plane then the vini ating doesn't
happen because there is coming at any
angle we'll that is for the case of a
telescope or in this way if it is true
for a general optical system yeah yeah
for a telescope where you if you place
it at the common focal point then yes
you have you minimize vignette engine
okay before we break let me go over what
we will do today so today whether two
items in the agenda basically one of
them is mirrors so far with that with
spherical refractive dielectric
interfaces we never said anything about
curved reflective surfaces like mirrors
so we'll do that very briefly it is
simply all we need is a modification of
the sign conventions and we're done so
that's a fairly painless topic there's a
little bit of terminology here who's
been Greek actually don't know if you've
been Greek make sense see but these are
relatively archaic words even in Greek
we don't use them anymore but a system
that uses mirrors is called kitab trick
from the greg from the ancient greek
word for mirror and then if it uses only
refractive elements you call it the
optic that's also where the term diopter
comes from and finally there is a class
of optical systems that use both mirrors
and refractive lenses they're very
popular in two areas in astronomy and
lithography in particularly fog rafi if
any of you have dealt with semiconductor
lithography semiconductor industry where
they make these huge machines that write
on the right patterns and silicon to
make chips and such for example she goes
for his project for what he does
mythography typically these machines
they contain the optics are probably
taller than me okay I'm not that tall
but anyway they're problems that the
basketball player and they contain
typically between 20 and 30 elements
some of those are mirrors some of those
are reflective lenses so this type of
system that contains both is called
catadioptric that's a mouthful that's
what is called so there we'll see some
examples of such systems and we'll see
them in the context of the basic imaging
systems that is a
the magnifier lens the eyepiece which is
basically a magnifier then the
microscope which is actually two
magnifiers the telescope which is a
slightly strange magnifier and finally
we will see different types of telescope
okay what is Leigh Leigh yeah this is
the answer to your question well you
guys took a break I do again a careful
diagram with array that you asked so now
you see so the red rays are the right
races for the case where the lens is
positive but the second principal plane
is to the left and the black lines are
there are the Rays that go to the that
intercept in cippolini on axis so you
have to be a bit careful here I got
tricked myself actually going the first
time edited that the when you depart
from here you have to meet the first
principal plane at the center of the
axis and then you start from the second
principal plane then you join to the
image point and you can see that the two
black lines they do look sort of
parallel don't they and of course the
images here actually
because they are similar it's a little
bit difficult to see here it is obvious
that they're similar if you do the case
of the single lens remember this diagram
right ok that's a really bad drawing but
but from this you can see that the
similar triangles right yeah and then of
course when you put the principle planes
it just is just dead space between the
two principal planes right but they're
geometrically relationships the similar
triangles and all that stuff they get
preserved
except you are dead space and in this
example it is even worse the dead space
is kind of reverse-thread because they
moved one principal plane on top of that
what did you know what a minute I think
if I attempt to express it I will
confuse issues okay let me talk a little
bit about mirrors now so for mirrors we
have a set of sign conventions that is a
little bit modified with respect to the
sign conventions for refraction and
there is an of course is that when you
have a reflective surface then the Rays
will indeed fold and they will start
going from the right to the left okay so
because of that fact then we really have
to modify our sign conventions and here
I put all the cases of positive
quantities so most of these are familiar
the positive curvature is the same as
before of these angles and directions
here the same as before the really
interesting ones are the ones that
happen upon reflection because upon
reflection the distance is now become
positive if they go to the right not to
the left and that kind of makes
intuitive sense right because now it is
as if actually the best way to make
sense of this is to simply unfold it
okay so if you want to take this the
part that happened after reflection and
then fold it so it could go
back in the forward direction then these
quantities would in vin would indeed
remain positive so this is the origin of
the sign conventions for the mirrors
okay so for example why is this angle
positive well because if I if I flip it
to them around it would indeed be the
same as a positive angle over here okay
whereas of course this angle if it had
happened in a refractive system that
would have been well it would have been
forgive forbidden to begin with because
that is going from right to left and
that's that's an acronym okay so having
said that I will come back to this but
having said that the next thing we would
like to do is derive the matrix
relationship for mirrors now a long time
ago I think it was in the first or
second lecture or so we derived the
ideal mirror we said that if you have
rays coming from infinity and you want
to focus them onto a single point out
here like so we said that the way you do
that is with the surface a reflective
surface and you remember probably the
water surface was it was a parabola so
we wrote it in questions here let's if
you call this elevation s you call this
axis X so we found the equation of the
parabola was s equals X square over 4 F
where F is the focal length of the lens
okay now so the parabola is ideal but in
some cases people make spherical mirrors
actually it is not as common as lenses
lenses are very commonly spherical
mirrors are very often actually
paraboloidal very close to ideal
especially mirrors used in in
concentrators things like like satellite
antennas solar concentrations that are
used in
in energy you know solar energy systems
and so on then because they want to
focus the light on a single point then
you actually go for the parabola it
turns out if you wanna form a sort of a
bigger image here then the parabola is
not very good so people actually make
spherical mirrors so for a spherical
mirror I did a little bit of a
derivation here which basically follows
the paraxial approximation to connect
the ideal parabolic shape and the focal
length that would be that we get from
the parabola to the equivalent sphere
and again the parabola and the sphere
they kind of look the same you know if
you look at the parabola maybe it goes
like this if you look at this fear and
you match the curvatures near the center
it might go like this okay so there so
what we're trying to find here what is
the radius of the sphere here that Max's
best the parabola there their parabola
so this is the DL parabola that is given
by this equation
and then here is the sphere that not as
very good looking sphere but anyway
let's call it a sphere and the question
is what is the radius of curvature of
this sphere that will match the parabola
ideally so this is what I did here if
you write the equation for the sphere
well we know actually let me leave this
up we know how to write the question for
the sphere in my chosen system of
coordinates this is X and this is the
positive axis Z so the question for the
sphere is X plus the radius square plus
Z square plus actually s square equals
the radius square so this is the
question for the sphere this is this
displacement here with respect to the to
the center of the coordinate and then
what you do this is already on the slide
but I will actually derive it for you
here so you can see it in a sort of
animated form then what you do is you
actually solve so you have an S
equals well s equals actually plus minus
R - what am i doing I'm sorry yes I did
didn't I yeah sure about it okay I think
I better do this over
so this is as for the parabola this is s
for the sphere and yes so of course it
should be s plus R square plus X square
equals R square so then this means that
s equals minus R plus minus square root
R square minus X square okay now which
signs should they keep the plus or the
minus when the minus sign you can see a
little bit by inspection the minus sign
is this part of the sphere which is
pretty fire phone from the paraxial
approximation clearly right so the only
way to deal with the paraxial
approximation is to keep a plus sign so
right then s equals minus R plus the
next times that we do to derived
paraxial approximation z' is you pull a
large quantity out of the square root so
the large quantity is the radius here
and you write it like this 1 minus X
square over R square then you apply a
Taylor formula that says that square
root of 1 plus small approximately
equals 1 plus the small / - okay
and now of course that applies also for
a for a negative sign so I can do like
this and this means that I s equals
minus R plus R 1 minus X square over 2 R
square and if you do it carefully here
you see that this kills the square this
kills this and you end up with s equals
minus I'm supposed to get over here
minus X square over 2 R and what is miss
now is if you compare with the question
of the DL parabola which was x squared
over 4 f it means that F equals minus R
over 2
okay the next question now is what does
this mean is the focal length negative
or positive well to answer this question
we have to first remember our sign
conventions is ourself as positive or
negative that is is our a negative or
positive a positive or negative quantity
so if you recall I will go back one
slide to the previous one this is a
positive curvature right if it is
pointing that way I will not use any
formal term because that may be
confusion if it's like that it's
positive right well this is not like
that it's the opposite right so
therefore here R is negative okay
so you have a negative sign that came
out of the math but R is a negative
quantity
so therefore F is still positive right
so this is still a positive length
okay
and it is clear that it is a positive
lens because it forms a real image for
an object infinity by construction we
demanded that if I have part of the
layers from infinity this thing should
focus them to a real point on axis so
therefore clearly this is what you would
call a positive lens and since the point
lens it actually has a matrix that is
very similar to the matrix of a lens all
you have to do is substitute the focal
length with a question that we just
derived for there for the radius of
curvature and this is the matrix that
describes a mirror a mirror concentrator
and of course it since it is a lens it
also forms images so notice here the way
I drew it these quantities are still all
positive of course s naught is positive
because I go from the object to the
right towards the instrument now notice
what happens from the instrument I go to
the left towards damage but that is
still a positive quantity because there
was a reflection that occurred that so
therefore s sub I is still a positive
quantity so I can write this a question
with a clear conscience so to speak that
s naught and si are all positive
quantities now can you imagine when a
lens might form a virtual image the
answer is a mirror so clearly the
miracles following is forming a real
image spray can you imagine how I could
construct a mirror form in a virtual
image
if the mirror diverges and you are
looking from the same side then that's
right that's one way if you make the
mirror like this now the mirror has a
positive R therefore it has a negative F
now this mirror has become the
equivalent of a negative lens therefore
it will form a virtual image
what's another possibility for forming a
virtual image with a mirror what about
the positive lens could put the positive
lens ever form a virtual image that's
right if you put the object in between
the focal point and the mirror itself
again it will form a virtual image it's
actually very funny to observe by
yourself you cannot quite do it with a
spoon because the spoon is really too
small you have to go really close and
you cannot focus your eyes in there but
those of you who go who've been at
hotels I'd never seen this in a home but
most hotels in the bathroom they have a
curved mirror and I don't want to
speculate why they put that curved
mirror there but anyway they have it and
this mirror is very lightly curved it
has a very long radius of curvature so
it is very easy actually to if you stand
in front of the mirror you can move back
and forth and you can switch your image
from real to virtual it is really
spectacular it is worth checking
yourself in the ritz-carlton in Boston
spend $400 for a night just to see that
actually it is it's a very very
instructive from the octave optics point
of view but the real it's nice because
as you go near the mirror will basically
when you get between the focal length
and the mirror then you see an erect
image of yourself that is virtual
because you see yourself behind the
mirror and is magnified actually the
magnification is so high that you
typically only see your eye
but anyway something like that if you
move back then you you see yourself
inverted as you see yourself in a spoon
by the way if you look at yourself in a
spoon you almost always see yourself
inverted because the spoon has a very
short focal length so you are always in
this scenario when you look at the spoon
any questions
so there's not much else to say about
Miller's really the same optics applies
with this little twist of the flipping
of the science opponent version so if
you face a situation like this all you
need to do is keep cool you know don't
let yourself panic or anything it's just
a matter of keeping track of the
positive things that happen to the right
or to the lower sorry the positive
things that happen after reflection and
typically with the kind of problems that
we have to solve by hand these are
pretty simple I mean at most you might
encounter one reflection now people who
make optical design software they have
to deal with arbitrary numbers of
reflections I wouldn't want to be the
programmer who has to solve that problem
but but but because you know if you have
a second reflection then the signs flip
again right as you can imagine if I
stick another mirror here so the lights
out sort of zigzag in every reflection
flips the signs in the Sun conveying
convention again so you can get a bit
tricky to keep track of all of this
flicking but you know we don't have to
worry about that yet our at you know at
this introductory level okay so let me
move on to them to the next topic which
is a few examples of optical systems hey
Jonas yes sorry to interrupt and then I
pass around these a parabolic reflector
so you can try that I look at it also
maybe you don't have to go to a hotel
okay Roger that what is the focal length
of this you know it's like 20
centimeters or so you can you can check
it out
right so yeah your tattoo I try to go at
40 and the panel whatever and you will
see the difference is actually quite
dramatic and very interesting okay so
the magnifier as the name suggests it is
a device that we use to aid our eyes
when we are looking at a very small
object
and this is a took a lot of time to make
this cartoon so I'm very proud of it so
let me start the animation here so if
you have a really small object of course
it will form a very small image on your
retina right and very often this image
is not big enough to observe with
sufficient detail so don't you think
that we would do if something looks
small to you and you'd like to make it
bigger you don't have to know any optics
what you do is you bring it near your
eyes right but from experience you know
that this doesn't work very well because
if you exceed the distance of
approximately 25 centimeters then you
can really not focus anymore and
actually for you guys who are young it
is about 25 centimeters as we grow older
this distance becomes longer and longer
and we basically lose your ability to
focus which is a very annoying annoying
thing actually it has just started to
happen to me as a as approach edge fault
we have started noticing that they
cannot focus as well as I do too
so it is a little bit annoying but
anyway it beats the alternative I guess
so so I'm very happy that they reach
that age with anyway and so the point is
that that there comes a point actually
for me not the only way to focus is to
take off my glasses so but even so there
comes a point where you cannot focus
anymore even you cannot focus ITIF you
will example to five or ten centimeters
you cannot focus the question is what
could you do
so what you do is you use a magnifier
the magnifier is something that you put
between the object and your eye and the
way it works is like this we can do some
simple very tracing here okay this is
the parameters of the magnifier let's
say it is a positive length they magnify
let's say the focal length is F and the
power is the inverse of that so this is
some ray tracing that shows how the
image is formed by this lens so all you
do is you take a horizontal ray this
will go somehow to the this guy will
bend it and let's pretend now that this
ray somehow I shoot things so that this
ray went through the center
of the ISO this is a weapon been trained
but you can see really clearly now
because what this lens did is it Bank
this array bundles so that it meets the
object over here the result is that
they've got a much magnified image on
the retina okay so that's about it
that's that's all there is to it the by
using this element you can get bigger
images now if I didn't have the eye here
this is actually the most instructive
thing I'm going to say today this is
really the meaning of the virtual image
if I give you this system by itself
obviously because this lens if you look
powerful it created a divergent ray
bundle what usually this object is
between the focal point of the lens ray
so therefore this is the situation that
went countered before and we said that
the system forms a virtual image well
what is the virtual image oh I have to
take this rays that the lens produced
and have to explain the backwards and
with a little bit of manipulation here
because this is a cartoon it is not a
real calculation but with some
manipulation to make sure that there is
Dmytro okay so there is me at the point
where there is me it is the virtual
image there is nothing there
okay nothing the object is here the only
physical entities here and you are i
right and the logical there's nothing
here but when your eye forms an image it
is as if you could bake this away you
could take away the combination of the
object and the magnifier and replace it
with a virtual image that the system
formed this virtual image is acting as
an object for your eye the eye images
this object and because this object has
been magnified a relative to the actual
object you end up absorbing observe in a
magnified image so this is the
significance of virtual images by
themselves that don't mean anything but
because at the end of an optical system
you always have the observers retina I
mean the observers eye lens and the
right
the virtual image and that become in a
real image hopefully on the retina if it
becomes a real image elsewhere you will
not be able to see it you will be out of
focus or even worse if you create a
virtual image and you send it to an
observer so it becomes a virtual image
on the retina again you don't see
anything there because you can never
observe a virtual image unless you can
image it using the optics of your eye so
this is a significance then and we will
see in a second that everything my
microscopes telescopes magnifiers they
all form virtual images which eventually
the observers eye lens converts to
elements on the retina ok the next thing
that we will do now is we put some life
around this topic and this is a little
bit unfortunate because the book is
generally very good the textbook that we
have hacked I do think that in this
particular topic the textbook is a
little bit confusing so I'll try to walk
you through the derivation in the
textbook and hopefully I will make it
less confusing but if when you go back
home and you read the textbook you get
confused again please bring your
questions on Monday so we can talk about
this again ok so I try to make it less
confusing but still I am using the
notation in the book so you don't in
addition to the confusion in the book
you don't have to overcome the to
overcome a notational confusion ok so
the definition of the magnifying power
whose cannot be argued against because
it is a definition is it is the ratio of
the antis appended by the tip of the
object before and after magnification
after the subscripts me here
un a u stands for unaided that is the
eye without the magnifier and a stands
for aided that is the eye with the aid
of the magnifier lunch and
now we need to place symbols to all of
these distances here and I think this is
why the book is a bit confusing because
it tries to do everything at once but
anyway so neither symbol for the
distance between the object of the lens
then another symbol from the lens to the
eye and in the symbol from the lens to
the virtual image which is of course si
and in this case you would expect the
site to turn out to be a negative number
because it is a vector element and
finally uses a symbol big L for the
distance between the virtual are the
virtual image and the eye so L use of
course simply the sum of these two
distances with a proper sign convention
okay and then the first thing to note is
that you don't really need to all these
notations because the magnifying power
is very simple to calculate as we did
with the past and all you have to do is
use similar triangles because you can
see that the magnifying power okay
the angry a sub a the ADA angle equals
the ratio of the size of the virtual
image over what look at the red line
okay from this a and B over the big
distance L right and of course there is
a tangent that should have a tangent
there but they drop the tangent because
of the paraxial approximation similarly
for the object for the unaided eye it
equals the ratio of the natural height
of the object to the distance between
the object of the lens okay now for the
unaided case the convention is to use
the near point of the eye that is the 25
centimeter distance beyond which the eye
cannot focus because presumably if you
go if you wanted to get the maximum
possible unaided magnification you
really have to bring the object within
these 25 centimeters right okay so
that's why we use these
zero that's a standard actually it is 25
centimeter and when you play around with
this you discover that it equals the
ratio of these two distances which is
easy to do and this is what we use to
call what did you call this quantity
when we did the emergent condition
button little magnification that's right
this is what we used to call the lateral
magnification it is the ratio of the
size of the image over the size of the
object so this is what I say here and of
course this is the lateral magnification
where the questions for it in terms of
the imaging distances and and we can
substitute these equations and we can
actually derive a number of equivalent
expressions for the magnifying power so
now none of this is particularly
intuitive and then the book goes ahead
and gives two possible uses of the
magnifying lens that are not
particularly intuitive either okay so I
will not even bother to deal with them
one of them is when you put the
magnifier one focal distance away you
never I don't know if it's necessary to
do that that is when you put the
magnifier you stick it to your eye which
is not a very good idea because there's
all kinds of germs around so you don't
want to stick magnifiers onto the money
the way we use magnifiers in general is
the number three description of the book
which says we place the object at one
focal point behind the magnifier so with
the magnifier focal length is five
centimeters
actually that's a pretty long but anyway
you place it five centimeters from the
object what do you accomplish this way
what happens if you place an object at
one focal distance in front of the lens
that's right
you form an image at infinity actually
in that case you form a real image of
infinity now is that good or bad
in the context here
you found a magic infinity and you still
have an observer with his or her eye
lens and retina okay let me draw this
now we'll draw the specific case
according to the book lingo this is case
number 3s I equals F so what does it
look like here's my magnifier here's my
eye my eye lands do this of course the
eye kind of looks like this and this is
the retina and here is the optical axis
and here is the object so let's look at
the tip of the object since it is one
focal distance away from the lens I will
get a parallel a bundle is that good or
bad it's a so equal self right yes thank
you
so he's a good man if we talk of a
variations it's good yeah let's not go
into aberration yeah our simple
geometrical optics is that situation
happy
are we happy that we did this but here's
a number of reasons yeah
yes because they could perfectly focus
at the point that's right first of all
this is a yeah this is assumed to be a
corrected eye so if it was me that would
be me where in my eyeglasses right if it
is me without my glasses have a problem
I will see in a second what problem but
if it is a you know a healthy eye side
where they I can focus perfectly then
this indeed will form an image on the
retina okay what is that good news about
the situation the fact that the image is
at infinity is that good news or bad
news your eyes remain your eye remains
unaccommodating that's right we learned
that the high is most relaxed when it
observes remote objects of infinity the
reason I cannot focus closer is because
I have to strain the length of my eye
and as I get older my ability to strain
it diminishes right so but this is true
for anybody is young or old when you try
to look at something close by you get
tired right if you read the book and
like this your parents probably told you
they told me you know don't read the
book like this you get the headache and
indeed you do it if you live for a long
time to get the headache from very clock
so the reason this is a happy situation
is because the I can be totally relaxed
as if it was observed in something very
far away and still see the object with
magnification okay and how much is the
magnification well that I will skip the
math the math is fairly simple here but
it turns out to be simply the product of
the near point to the 25 centimeters
times there
of the lens and the only penalty you pay
there isn't actually the book covers
this case over here is because this is
not the maximum magnification you can
get to the magnifier you can get
slightly higher if you stick the
magnifier next to your eye okay fine but
you never stick the magnifier to your I
actually unless your adventure also you
know if I did this I will damage my
glasses okay so the masses perhaps you
know important but they're really what
they are really hopeful to get you hope
that you get out of this is the
significance of the virtual image and
why the magnifier functions by forming a
virtual image that is erect and
magnified and this is also tells you why
you use a positive lens as a magnifier
if you recall it is actually one of the
previous slides a negative lens also
forms a virtual image but unfortunately
it is not magnified the magnification it
is the magnified the lateral
magnification is less than one so so we
would be out of luck in this case okay
the next element is actually the same it
is not different than the magnifier but
it comes under the name IPS because in
optical instruments such as by night
Euler's telescopes microscopes and the
like this is the last lens that you look
into the instrument with your naked eye
now I realize that this is all fashion
talk modern instruments you seldom look
at with a naked eye you actually have a
camera so you put a digital camera and
you register with image digitally
however even a digital camera functions
like a retina so in other words even a
digital camera would have an eyepiece
and then a condenser and then it would
it would go onto the chip right so so
really the situation here applies to
digital imaging as well as as human as
well as you know instruments meant to be
used by humans so there so the eyepiece
is basically the last element investment
just before you get to the to the actual
sort of observation stage and because
the eyepieces were traditionally
designed for humans
they typically designed like a magnifier
with the same principle that I described
before that forms an image at infinity
so therefore the observer when he or she
looks into the IPS they can use an under
kama dated eye and observe the magnified
image without having to strain their
eyes okay so this is the sort of
standard use of an eyepiece and there is
a second condition for the eyepiece
which which will become a little bit
more more apparent in when a in the next
slide when I talk about the microscope
it is this one over here
this looks a little cryptic it says
Santa the exit pupil I point where the
observers eye is placed at ten
millimeters from the estimate that
doesn't mean much but I already
described what this means in a second
when I talk about the microscope
notwithstanding this last point over
here any questions about the eyepiece or
the magnifier or anything of that sort
okay why is the eye relief ten
millimeter when we cannot see okay what
this really means is that ten
millimeters is the design distance that
you're supposed to place your eyes with
respect to the instrument when you look
at the microscope for example most
people don't do that but you're supposed
to be at about ten centimeters from the
from the last cell phone so that is
called the eye relief and what is it
really saying is that when you look at
this you are not supposed to see any
blockage of the image and you're not
supposed to see anything yet okay so
there's two things that can happen if
you place your eye at the wrong point as
you know if any of you have looked at
the microscope if you are not positioned
properly you cannot see the entire field
of view that's one problem you see a
black hole and you kind of have to move
around and you can never see anything
until you go to the proper place and
then finally you see everything right so
this is the eye relief this is by design
where you should place your life the
thing that can happen if you place your
eye at the wrong place or if the
microscope is not properly designed is
vignetting so this requirement really
assumes that there is no vignetting when
your eye is placed at eye relief
distance that is a ten millimeters from
ventolin and of course you don't have to
worry about focusing because the
microscope is supposed to have an
infinite conjugate red so you just use
with a can accommodate an eye and I
suspect I don't know if it's true but I
suspect there is many reasons why they
put over there but also it's kind of
convenient that when you look at the
microscope you're so close to the
cooked should be sent that you cannot
focus and you cannot say that right so
you can basically see what you're
supposed to observe but you don't see
the actual instrument because it is out
of focus first kind of convenient okay
partner said no this about the
microscope here is a microscope here's
what it looks like
it is deceptively simple because there
is of course a lot of engineering design
that goes into making all of these
elements but the principle is
surprisingly simple it consists of
basically two magnifiers one of them is
called objective that is called eyepiece
but the both function is magnifiers okay
so the job of the objective this is the
first lens it is called objective
because it is placed near the object and
it is job its job is basically to form
an intermediate image that is magnified
to begin with what is it magnified by
well here we simply have a actually this
is any element so we simply have a
positive lens for mineral image we can
compute the magnification it is equal
minus as I over as alright just like the
regular formula and then we have the
eyepiece or whom or for which the image
that the objective produced actually now
acts as the object so this intermediate
image is reimaged by the eyepiece but
the eyepiece is placed one focal
distance F sub II here is the focal
length of the eyepiece
it is one focal distance from that
intermediate image therefore the final
image is formed at infinity now the way
this is drawn for the one with a what is
prominent to this picture is of course
the own axis ray and there is no
magnification to talk about because this
is just an axis right nothing
interesting is happening what is really
interesting is if you look at the North
axis point which is at the bottom here
it's a little bit difficult to see it
looks better than the book actually so
if you place this off axis point then
you find that
final image is this ray bundle that is
coming off axes into your eye this
parallel rays that is the final image
that the microscope produced of this of
artists objects point so you can see it
is entering the eye at a very large army
therefore until you form a big image so
that's how the microscope magnifies
small objects so that you can observe so
then the angular the magnification of
the microscope very similar to the case
of the objective is defined as the ratio
of this angle the angle that the
microscope reduced for the off-axis
object for the same angle that your that
you know the I would observe unaided
right in a very similar case as the
magnifier so in order to compute this
now it's not very difficult to do we
need to two elements one is the lateral
magnification of the objective because
recall the objective actually founds a
real image so who have one here is the
object and here is its image right so we
need first of all to find out where the
intermediate image appears with respect
to the optical axis that's the lateral
magnification of the objective that is
whatever oh I will say in a second why
this 160 appears but the second the
second element that we need is the
angular magnification of the eyepiece
because this will allow us to say how
this distance over here translates it to
angle that goes into the eye final okay
so this one is actually a familiar
formula this is the same formula that we
had for the magnifier for go back 1 we
said that actually is over here and we
said that the magnifying power of an
eyepiece or a magnifier equals the near
distance for the eye which is about 25
and the meters times the power of the
lens over the positive lens that they
use so this is the same formula in
microscopes that don't use exactly 25
centimeters they use 25.4 centimeters so
this is the standard and how much is 13
inches is it then this is not eleven
twelve twelve inches then uses that will
be twenty two point although that's
pounds I'm sorry that's pounds oh yeah
it's the attending system very so this
is them the same formula that we had for
the magnifier what did this come from
well if you remember one of the ways to
write the lateral magnification is like
this one - si over F so you can write
this as one minus si over F equals F
minus si over F and now have to palaver
okay so minus F I answer it is f minus
si over F so f minus s I is actually the
distance between the between the
objective lens and there and they vary
with distance L that's what the time to
get so this is known as the tube length
of the microscope it is the distance
between the objective lens and the
location where you put the eyepiece of
course the eyepiece is this whole thing
right so that's where you stick the
eyepiece so this is a standard in
microscopes it is 160 millimeters and is
known as the tube length and this is
what appears in this formula over here
so basically you can see that if you
pick the focal lengths to be
sufficiently small then you can get very
high magnifications in the order of
hundreds maybe even a thousand
let me ask two questions now first of
all what has I said what would what do
you think might stop me from getting
there focal length that is let's say 1
micron I pick a focal lengths to be 1
micron then everyone these are all in
millimeters so 160 millimeters of a
micron that would be what there will be
about 160,000 and then another micron
here that I would get a magnification in
the orbit of a billion what stops me
from Greenland there's no one just to
get back to what is it from your
experience what is the typical
magnification that you get in
microscopes about a thousand right or
the mine but you know if your plan is
formula here you could easily get a
magnification of 1 billion by picking
the focal length would be one micron
what is wrong with my statement here is
something like other where people would
be doing it later is it because you
can't see past the wavelength of light
that one micron is the most you could
focus because that's the okay that is in
the correct direction yes
okay the other possible reason is the
value of the potential approximation but
okay
the production approximation may fail
all at once but they can always do
something more sophisticated to do non
paraxial calculation say okay maybe you
are saying that if people actually
approximation fails then this formula is
fatal as well right so then you don't
get as good magnification I think you
guys both captured that the two of you
captured the answer in its entirety you
have another I don't know if it's
another R to make such a high such a
short focal length lens will have to
either make a very big lens that so you
have to be very highly character it
remember the focal length of the lens is
proportional to the inverse to the
radius of curvature so you would have to
make a land with their edges of
curvature equal to 1 micron and that's
when you get into problems like the what
do you never get to wear in wear them I
think one side the question before D
yeah this is what the Dean yeah thanks
this is what Dean answered if you try to
make a curvature that is as sharp as one
micron then basically you get into the
range of very strong diffraction of
light so basically all these dramatical
optics that we do share they fail in
fact you discover if you did make such a
very small element that the light does
not really focus there's something
something that we will say what the dull
but there's no focus so reasonable
radial curvature
I mean reasonable focal lengths are in
the range of the centimeter a few
millimeters is a typical microscope
objective focal length in the eyepieces
are similar that's why you get
magnification of the order of a thousand
okay the next topic is the telescope but
I think I've exhausted that your
endurance