We are continuing with this course on Optical
Engineering. So, there is a quick recap of
what we did in yesterday’s class. We looked
at basically the importance of studying the
subject irrespective of what background or
field you may be working in and we started
by looking at the postulates of geometric
optics.
So, we said light travels in a straight line
in a homogeneous medium. When will that not
happen? If the medium is not homogeneous and
that brought us to the next postulate which
was every dielectric medium has a refractive
index associated with it. This in turn brought
us to the next postulate we said in geometric
optics what is really important is not just
the physical distance that the light is traveling,
but the product of the refractive index into
the physical distance, a parameter called
the optical path length.
We came then to the final postulate which
was Fermat’s theorem.
And, Fermat’s theorem is basically that
light will travel a Fermat’s theorem a Fermat’s
principle is that light will travel on the
shortest path or the path of least time. We
then used this principle to arrive at a very
well known law Snell’s law.
Now, we also came to a very important observation
when we use Fermat’s principle to figure
out the way light is traveling. We are minimizing
the path and I can always minimize the path
with respect to the optical path length itself
or I could minimize the path with respect
to the time. If however, you are taking it
with respect to the optical path length then
you would have d optical path length with
respect to some parameter let me call that
parameter phi for example psi for example.
Now, I can look at this law that we all know.
You have studied Snell’s law all the way
from high school right; you should be very
familiar with it. Fermat’s principle still
does not really explain why Snell’s law
is true. It stated a theorem, we used that
theorem and using that theorem we arrived
at a well known law. So, that might give you
the confidence that Snell’s law is correct,
still does not tell you why it is correct
and really if you want to understand why Fermat’s
principle is correct you have to go back to
the wave theory. You cannot escape Maxwell’s
equations, you have to go back to the wave
theory and you are all supposed to have done
basic course in Electromagnetics or Wave Propagation.
You might have derived Snell’s law in that
course. Do you remember what you used there
to arrive at Snell’s law? You did not use
Fermat’s principle to arrive at Snell’s
law, but you would have arrived at Snell’s
law in some way. How, forget Snell’s law.
In electromagnetics when you talked about
a wave hitting an interface and then you used
the Maxwell’s equations to figure out what
happened to that wave when it hit the interface
what did you use to allow you to calculate
that?
We imagine a boundary conditions.
So, that is right, you use the boundary conditions
and when you use the boundary conditions a
boundary conditions in electromagnetics allow
you to calculate how much light gets transmitted
into the next medium or how much light gets
reflected and there you see already a big
difference from geometric optics because we
said this yesterday geometric optics allowed
people to trace the path that rays took.
So, they could understand concepts or phenomena
like reflection and refraction, but it did
not tell them how much light got reflected
and refracted. Whereas, when you go to wave
optics you can you not only can you do quantify
exactly and accurately not only the directions
in which light will take, but also the amount
of light that goes in a certain direction.
And, you can use the boundary conditions to
arrive at Snell’s law and there you will
understand why.
Now, we talked about Snell’s law and generally
if you say Snell’s law you are referring
to this equation n i sin theta i is equal
to n t sin theta t, where n i is the refractive
index of the medium of incidence theta is
the angle of incidence and n t and theta t
are the similar parameters in the second medium.
This is actually the law of refraction.
Of course, we all know the law of reflection
and again this can be proved both using Fermat’s
principle as well as wave theory that theta
i is equal to theta r and a little later on
today we will look specifically at how we
define these angles because that is important.
Any questions till now? No. So, I want to
go to the next part.
And so, we are looking at using reflection
or refraction to enable us to design optical
systems right and before we go down that path
I want to ask you a simple question. It is
a simple question maybe it is not a simple
question, but it is a question that we should
be asking our self because it is something
that we are doing every day. How do we see
things? Why do we see things? You are seeing
things all the time, you are looking at me
now, you are looking at the board, you are
writing notes; how do you see things?
When light hits the things so, that can be
reflected so that we can.
So, he is saying that we see things because
of reflection and that is right. So, I if
I am looking at a light source I could make
a distinction I could make a further distinction.
I could say if the object I am looking at
has it is own light a luminous object that
light travels from the object reaches my eye
and that is how I see it, but most of the
objects around this table is not luminous,
it does not have a natural light source. How
we see it is that light from some source falls
on the object reflects and comes to our eyes.
Now, when we talk about reflection, I can
talk about very clean and neat reflection.
The reflection that follows that nice that
seems all reflection will follow Snell’s
law, but from specular reflection you can
easily see that Snell’s law is being followed.
Specular reflection is a reflection you see
from a mirror you have a nice optically flat
smooth surface. However, most objects are
not going to be mirror like and what you would
get from most objects is diffused reflection
what is shown in this lower picture here and
yet whether the imaging system is our eye
or a camera if the goal is to create an image
we must work with this kind of reflection
and get a good quality image.
So, if I look at maybe before we go there
I will just let us run through again the angles,
the definitions. So, if I had a nice flat
surface so, as shown in this green surface
here let us say that is our interface and
light is say reflecting off this interface.
So, from the incident light there is a reflected
ray and both the incident and reflected ray
are going to lie in the same plane. If this
interface is not 100 percent reflective then
there is also some light that is traveling
through and that is going to continue to lie
on the same plane.
So, I have a plane of incidence which contains
the incident ray, the reflected ray, the transmitted
or refracted ray and the normal that lies
in this surface that is it is the normal to
the interface between the two media. So, if
I have a diffused surface I can think of having
many planes of incidence because a normal
will lie depending upon the particular angle
or curvature of the surface.
So, for every ray in a particular plane of
incidence it will reflect in that plane and
it will transmit in that plane and you can
see this messy picture that you get down here
ok. Again, this is just been shown as a 2-dimensional
picture, but it could be reflecting in two
different planes.
So, this is a reflection of an aluminum foil,
it is fairly shiny and reflective, but you
can see it is not mirror quality by any means.
You cannot really make out is the person standing
and the image has been taken. If it was a
mirror you would I would not have to tell
you here is a person standing, you would see
that right. So, this is not specular reflection
and yet we see the world around us every surface
that is reflecting light is we are not hopefully
you are not seeing images like this right;
that means, our eye is able to take diffuse
reflection and make sense of it make proper
images out of it right.
So, very large part of this course is going
to look at how do I design optical systems
such that I take light that is reflected maybe
as specular reflection maybe as diffused reflection
or take light that has been transmitted and
even when it is transmitted if the surface
of the interface is not smooth, but has a
lot of curvature and variation when it is
transmitted it can the plane of incidence
may be changing and therefore, the light is
being refracted in many different ways.
For example, you can imagine you are using
an optical system to image just below the
skin. Light is traveling through tissue reflecting
off traveling back through tissue. So, there
is a lot of different phenomena that is going
on there is some reflection refraction and
there is a lot of scattering and yet the doctor
has to see a clearer image to make a proper
diagnosis.
So, this course will look a lot in the almost
the first 6 weeks we will look at how do we
design optical systems so that you capture
this light in a proper way that gives you
a proper image. Just as a recap in yesterday’s
class we said what all can we do with light
and we talked about three things that you
can do with light you can.
Ma’am.
Yes.
How is diffused reflection is different from
scattering?
So, his question is how is diffused reflection
different from scattering. So, when I talk
about reflection I am talking about what is
happening on the surface of an interface ok.
Scattering is more a phenomena you talk about
when light travels through a medium. For example,
I gave a tissue light is traveling through
this medium now the scattering happens because
the medium through which light is traveling
has particles in it and those particles have
certain sizes depending on the next dimension
with respect to the wavelength you get different
types of scattering but, in reflection we
are saying light hits an interface.
So, an interface means you had a medium with
a certain refractive index, you have now a
medium with a different refractive index or
you have a medium where you which is the refractive
index is a function of position. So, there
is an interface right. So, like I have shown
over here let us say this is n i incidence
and below this interface I have n r; n r may
be a function of x, y, z, but at this interface
what happens? The reflection that happens
here may be diffused if this interface is
not a smooth polished interface, then I talk
about diffused reflectance.
If this medium has particles in it of varying
size of different sizes then as the light
travels it is getting scattered by this particle.
So, that is that is the basic difference and
it is not that I get information from that
light that is scattered it is not that I have
lost that information, but I have to capture
that light in some different way and our analyze
it in a different way.
So, our goal then is to design optics so that
we can collect light in a manner that makes
it usable to us and I said I was going to
write out what are the different applications
main applications I guess the very broad applications
anything with optics in it can be thrown into
one of these buckets. So, if I think the first
thing I you would answer if I asked you what
is an application of light, well, it is illumination
the very first application of light is illumination.
So, you would have illumination.
Yesterday when I asked for applications some
of the first applications people said were
camera, microscope, telescope all image right.
So, the whole host of applications that come
under imaging and of course, now there is
a lot of use of light in biomedical applications
in medical diagnostics. So, it is again imaging,
but you are imaging in order to help with
some diagnostic technique. And, the third
applicant third broad area what would it be
communication or let us put it let us stick
with the I letters words information.
So, the projector here is displaying an image
for you, it is giving you information now.
Fiber optics which is used in the for internet
we are using optics to transmit information
and that data travels on the fiber in the
form of light and that carries information
right. So, there are million examples, but
almost any application would fit into one
of these 3 and imaging here is imaging as
well as information. So, it is not that you
can completely separate these three, but these
are the three broad areas.
So, a very large part of this course is going
to focus on the optics required for this for
imaging.
That the wave model is the one that really
explains why things happen the way they do,
the ray model does not explain why. It allows
us to analyze and understand certain phenomena
without going into the why without doing a
lot of detailed maths. It works and that is
why we use it and we use it for a lot of applications.
You are going to spend 6 weeks using the ray
optics model.
When we talk about imaging what exactly do
I mean? So, let us say I wanted to take an
image of this apple right. So, what does that
mean? That means, every point on this apple
say I want to image this point; now, if it
is not a luminous object, so that means, light
has to be incident on it from some external
source. Let us say I have some external source
light is incident on it is incident on the
whole apple.
Let us just concentrate on this one point.
There are a large number of rays that are
being sent out from this point. All the rays
from this point has to be captured by my optical
system and focused to one point and this is
true for every point on this apple. Every
point and you can see how messy this looks,
but if I want to create a good image I must
capture all the rays from one point and my
optical system must focus that down to a single
point and that is the challenge.
Now, it turns out I can never capture all
of the rays, it is impossible. I have a huge
cone of rays that is being reflected of every
point. So, my optical system at best will
capture some cone of rays and we will see
the implications or the results of that. All
the rays coming from one point on the object
must be focused down to one point in the image
and this must happen for every point on the
object.
You can see it gets complicated when I have
a system where in this case I have water here.
So, not only are all these I here I have not
shown the incident rays at all right. This
fish is not a luminous object. So, obviously,
light is incident on the fish and it is reflecting
off the fish all the rays from this one point
here and now they are going through an interface
so, they are bending.
So, there is a lot more happening and yet
my optical system whatever it is must take
all these rays and focus it to one point over
here. And, we must do that for every point
on the object that is what my optical system
must do and that is what you are going to
learn how to do in the next few weeks.
So, if my object is luminous it is clear how
we see, if it is not luminous we need an external
light source, but how do we see light itself
ok, just a small a site to get you thinking.
How do we see light itself?. So, how many
of you watch science fiction? Star wars? You
see images like this um? What is wrong with
this picture? Star wars is famous for its
light lightsaber there is something very wrong
with the lightsaber too, but will not go into
the lightsaber.
What is wrong with this picture? This is a
starship out in space and here is this laser
beam. We will try to use laser beams for non-violent
things, but here’s a laser beam presumably
as a weapon being shown somewhere what is
wrong with this picture? You are saying something?
.
So, how do we see light? Actually we see light
through scattering right. It scatters off
particles in the atmosphere. So, say in this
room right now it is fairly well lit, but
say the room was dark and it is brightly lit
outside and there is a gap in the curtain
and some sunlight comes through you often
actually see the ray of sunlight or we live
in a beautiful campus. You go out for a slightly
early morning walk, you will see the sunlight
filtering through the trees.
Now, you could see it in different ways you
could see the patch of light on the ground.
That way how are you seeing that it is this
reflection and that reflected light is reaching
your eyes, but you often actually see the
passage of light. Now, when you go for a walk
in the campus or if light was filtering through
this room and you saw the passage of light
you would see it because it is scattering
off may be dust or particles in the air. It
is because of scattering. Scattering is an
important phenomenon. It is not always bad
I would not see things if there was no scattering,
sometimes.
But, in this picture this is out in space,
there is no atmosphere. So, while you the
light is leaving this place and maybe there
is some scattering or reflection just as it
leaves so, at the edge you may see some reflected
light and the light may reach your eye because
of that. So, you may know that there is light
leaving this point, but you should not actually
see this path of light because there is nothing
for it to scatter off.
So, as I said we are going to concentrate
on geometric optics for image. So, any optical
system we start off by saying every optical
system has what we call an optical axis ok.
This is a straight line that runs through
the system and I will be more clear and say
runs through the center of the system ok.
Sorry, what is the optical axis? That is the
line through the system that if you sent a
ray the ray would travel undeviated through
the system.
So, if you consider a lens and if I had a
ray that was traveling like this you know
the lens is going to bend the ray right. If
I have a mirror and I had a ray traveling
like this you know that it is going to travel
in some axis angle. The optical axis is that
line which travels through the center of the
system. I have not really drawn it through
the center, but I hope you understand what
I mean right. It is the path that a ray would
travel undeviated through the system. I can
use a mirror to change the direction of the
optical axis ok.
In most optical systems or I should say in
many optical systems the optical axis is also
the axis of symmetry. This is because a lot
of our optical elements are circular in nature.
So, this does not have to be true I can have
non-circular lenses, I can have cylindrical
lenses for example,. I can have optics that
is not circular in nature, but a lot of our
optics is circular in nature take microscopes,
cameras, the lenses will be circular in nature
and therefore, more often than not the optical
axis is also going to be the axis of symmetry.
We can even think of it as the axis of rotation
if you.
Any optical system has a pair of conjugate
points. So, if I think of an the optical axis,
this is my optical system. I am drawing a
lens that lens is representative of an optical
system, it does not have to be a single lens
could be a number of lens. So, if it is an
imaging system; that means, if I have a point
source S and it is getting it imaged at a
point P that the conjugate points are S and
P this means that if I were to put the source
here the image would form here.
So, the optical path is reversible. So, this
is the reversibility principal. Now, while
the path is reversible; that means, source
is at S, image is at P or source is at P image
is at S the quality of that image need not
be exactly the same. The locations will be
the same, but you could design the optics
such that it gives you a better image when
light travels in one direction as compared
to another direction.
So, this only tells us where the locations
will be and that would not change and that
is a pair of conjugate points, but do not
think that means, that the optical system
works exactly the same way irrespective of
the direction . Now, what is the goal of any
optical imaging system? You want to collect
light and create an image and you want to
create as best quality an image as possible
that that is your goal.
Ideally, I want to collect all light ok. So,
this is the ideal goal. This is ideally what
I wanted to, but I can never ever collect
all light my. If you take this example over
here, let us say I have an object over here
I have light traveling. So, there is light
incident and then there is reflected light
and it is reflecting in a very large cone.
Some fraction of that gets through this optical
system.
And, you learn a few classes on that I you
might think I have drawn this set of rays
this cone of rays where the upper ray hits
the top of the lens and the lower ray hits
the bottom and this cone is the cone that
will now travel all the way through the system.
But, it could be there is some element later
on in the system that blocks this part and
this part. So, so actually could be some much
smaller cone that is making it through the
system ok. So, we will look at that later
on.
But, the point to take is never do you collect
all of the light reflected of the object you
only collect a part of the light and the most
that you can collect with the particular diffractive
sorry, with the particular optical system
we call that a diffracted limited image. So,
that is the best you can do alright.
The amount that you collect is limited and
we would not go into the details of that process
in this course, but it is limited by a phenomenon
called diffraction and that is why the best
image you can get. If there is no nothing
else wrong with the optical system, if your
lenses are perfect, your mirrors are perfect,
everything is in exact place where it should
be, then you have a diffraction limited system
so that is the best you can do.
This also relates to the resolution with which
you can image. Clearly I want my imaging system
to have a high resolution. I want to capture
the fine details in the object that I am imaging.
Optical systems are limited by the wavelength
of the light that you are using. So, the rule
of thumb you can say roughly lambda by 2 where
lambda is the wavelength; lambda by 2 is the
best resolution that you can get with an optical
system.
But, optical systems are not the only way
you can image right. Do you know of any other
ways of imaging, non-optical? You can use
electron microscopy right. Again, the resolution
is determined by the wavelength, but it is
a wavelength associated with the electron
and that is the orders of magnitude higher
better right. So, in optical microscopy your
best resolution with the very good optical
microscopy microscope would be in hundreds
of nanometers; with an electron microscope
you can go down to tens of nanometers ok.
You can use ion beam microscopy, again tens
of nanometers; you can use atomic force microscope,
again very high resolution right.
So, you have many different ways of imaging,
but in optics it is diffraction limited and
you are stuck with around lambda by 2. Even
here I have to say that in the last 10 years
or so, people have taken optical images and
then used computational methods 
to get what is called super resolution ok.
So, then you can computation you will capture
optical images with the best resolution you
can and then you can improve the resolution
using some computational techniques. So, this
is a big area of research right now.
Funnily enough in 2014 a Nobel Prize was awarded
to somebody called Stefan Hell from Germany
for a technique called STED microscopy. It
was actually awarded I think in chemistry
the Nobel Prize right and he used diffraction.
So, we say optical systems are limited by
diffraction and therefore, you cannot get
better images, but he actually used diffraction
and improved resolution ok. So, it is a very
clever and once the technique is explained
a very simple technique and you think you
know, anyone could have thought of this, but
of course, Stefan Hell did.
So, maybe if we have time later on in this
course we will look at this technique in some
detail, but in this case diffraction is actually
used to develop some element that could be
used to improve resolution. So, the main point
is in an optical system wavelength is what
will limit your resolution. There may be other
ways of getting there are other ways of getting
much better resolution, but in an optical
system this is what your limit really.
One way to think of this is if I had since
the resolution is limited to lambda by 2 you
can think of let us go to lower and lower
wavelengths right and people do that some
of the more sophisticated tools use ultra
violet – UV, so, you are going to lower
and lower wavelengths. But, of course, you
cannot always use lower wavelengths right.
How do you think of geometric optics then?
We think of geometric optics as and even when
I say lower wavelengths you are saying you
are reducing the wavelength, but reducing
it with respect to what right?
So, it is reducing the wavelength the wavelength
you are using for imaging with respect to
the size of the object that is being imaged.
So, the size of the object in relation to
the wavelength that is what will determine
your resolution. So, in geometric optics we
say because you geometric optics you are talking
and imaging of large with respect to wavelength
you can consider lambda is equal to 0 here
ok. That is a assumption we are making because
the wavelength is so small compared to the
objects that we are typically imaging ok.
So, that is assumption that we are making.
So, if I am imaging objects in the order of
nanometers I cannot make this approximation
and geometric optics will not work there,
I have to go to wave optics to explain what
I am seeing right. You would have seen if
you take a CD or a DVD right if you show hold
it against the light you see it splits up
into colors right. Their structures on that
are in the order of hundreds of nanometers
right it is not a geometric optics phenomenon
that is splitting up that light. So, I cannot
use geometric optics to explain.
So, only where that condition is satisfied
can I use geometric optics and that is what
we are going to do in the first part of this
course. So, if I am talking about an imaging
system we have already said that clearly any
imaging system has to capture as much light
as it can from any point from every point
on the object and all the light from that
point of the object has to form one point
in the image plane right. If one point in
the object plane forms ten points in the image
plane I have a blurred image right.
So, of course, what is very important in any
imaging system is the focus. So, what element
can I use that will provide the best focus
um? So, let us take first thing I will do
is always draw my optical axis then I am going
to take just one interface and I would say
I have a point source here, light travels
from this point source to this interface.
So, it is an interface; that means, I have
refractive index n 1 here and right now I
am not limiting it right. So, it is just an
interface. So, after this interface it is
completely another refractive index n 2 we
are talking about homogeneous medium media.
So, it is n 1 on one side and n 2 on the other
side.
This travels to some point over here. So,
this point I will call S this is P and this
distance to the vertex of this interface let
us call it S naught again from vertex to here
we will call it S let us call it S i this
is l naught this distance let us call this
O and this is l i. The optical path length
of the ray traveling along the optical axis
right will be n 1. So, this is optical path
length n 1 the refractive index into the physical
distance S naught plus n 2 into the physical
distance in this medium which is S i.
Now, our goal is every ray coming from this
point S should arrive at point P. In other
words, for the second ray that I have drawn
this should be equal the path length should
be equal. So, n 1 into l sorry l naught plus
n 2 into l i and this should be true for any
ray that I draw. If I draw drew another ray
like this sorry from S and all the way to
P. You agree with this? If every ray traveling
from S reaches P and this condition is satisfied
I am saying the optical path lengths are equal,
right then I would say this interface that
I have chosen it is not a flat interface,
it is an interface with some kind of curvature.
This interface, that I have chosen is doing
the job of imaging.
Now does such an interface exist? It turns
out it does right and the interface that satisfies
this condition is called a Cartesian oval.
You can think of how it works if you break
it down you know look at this 
if you have. If you have rays traveling along
the optical axis and a ray traveling at some
distance, this ray hits the interface before
this ray does right. So, the ray along the
act is enters the interface and starts traveling
slower.
While this is traveling slower here this ray
is still traveling faster in air. By the time
it reaches this interface this ray has already
traveled some distance within the interface
right oh sorry, what is exactly happening?
You can see that the wave that is hitting
this interface is going to bend because this
has traveled to this point this is still traveling
at a higher velocity this is slowing down
over here and so, there is the bending of
the wave and that is what that interface is
doing. That is what any optics does right
it changes the shape of the beam that is incident
on it.
It changes a wave that is diverging into a
converging, it is creating that focus ok so,
that that is the primary goal of what we are
trying to do. So, I think we have run out
of time. So, we will continue in the afternoon
class at 2’o clock. Keep this image in mind
we will continue from here right, but the
goal is to shape the incident beam either
to collect it and focus it to a point or shape
it to give us some particular application.
