We will continue with a course on Optical
Engineering. In the last class; we looked
at how we can start designing systems and
we took material with a certain refractive
index and said what should it shaped be such
that we get a perfectly focused pointer image
given a point object.
So, we said in order to do that in order to
achieve that; we need in this image we need
every ray of light that is leaving this source
S over here to get imaged at this point P
over here. And the equation that satisfies
or should be satisfied in order for that to
happen is what we have written over here.
This is nothing but the optical path length
for every ray traveling through the system.
So, we have written it for two rays over here
one is the on axis ray. So, that is the ray
traveling the distances S naught in medium
n 1 and S i in medium n 2. And another way
from the same point that is this point. It
will travel through some point on the interface
and then it too should come to the same point
P over here. However, it has travelled a distance
l naught in medium n 1 and a distance l i
in medium n 2.
In other words we are saying this should be
a constant. And we went on further to say
that; there is such an interface there is
such a shape that will satisfy this equation
that shape is the Cartesian over ok. Now I
can see a number of you wear glasses I am
wearing reading glasses. Do you think, the
curvature of those glasses are the Cartesian
ovals?
No.
No. So, somebody say no that is right they
are not Cartesian arm. Do you know what shape
most lenses have?
Convex.
Convex concave is the nature of. So, convex
I will say this is convex this is concave.
But if I had to describe this shape, is it
a Cartesian oval, is it cylindrical, is it
spherical does it have some other shape. So,
most lenses its not correct to say all lenses,
but most lenses will be spherical ok. So,
we have just said; the best shape the shape
that ensures any ray of light coming from
a point object point will get image to the
same image point is a Cartesian oval. And
now I tell you nearly every lens that we use
does not have a Cartesian oval shape, but
has a spherical shape. Why do you think that
is?
.
Sorry.
.
Absolutely, it is easier to build, it is easier
to manufacture spherical then it is any other
shape. Today we have very advanced technology
where you can input a complicated shape and
3 D print almost any shape but the costs in
mass production 3D printing is not very good
for mass production. So, the cost even today
to mass produce something other than spherical
would be very high and of course, lenses and
mirrors have been used in optical systems
for centuries now.
And we are talking about manufacturing requirements
of the ability to manufacture these shapes
hundreds of years ago spherical has always
been relatively easy to manufacture. So, that
might have been the reason historical people
started with spherical shape, but even today
is for mass production spherical shape would
give you a cheaper to mass produce and therefore,
spherical shape is used.
Now, you have to think about this. Because,
we are saying the shape that gives me the
best optical performance is something and
because it is not easy to manufacture that
I am doing something else. Clearly the optical
performance we get from this other shape has
to be adequate or better than adequate. We
are not sacrificing quality for ease of fabrication.
So, we are going to look at under what conditions
does the spherical shape satisfy this equation
over here ok.
Again I am not yet starting with the lens
I am just taking a single interface ok. So,
let us say I am right now going to start off
by looking at refraction through a single
surface ok. So, that is my single surface
that is the optical axis. The difference is;
I am now specifying that this curvature is
of a sphere ok. So, if I were to continue
this I would actually draw circle over here
right ok.
This is medium n 1, this is medium n 2. The
object I want to image is a point on the axis
let us say it is at point s and of course,
there will be a ray that travels along the
axis there may be millions of rays. I am going
to look at 2 rays the one that travels along
the axis and the one that travels at some
other angle. We will call this distance l
naught and this distance S naught; S naught
is from S to the vertex here.
Let us say the image is at point P. So, this
ray goes to point P sorry and this distance
is S i. Because, this is a spherical surface;
there is a center of curvature c and this
line can be considered the normal to the surface.
Because, this is nothing, but the radius of
this surface right. So, in yesterday’s class;
we said we define angles of incidence as the
angle between the incident ray and the normal
to the surface.
So, this is now the angle of incidence theta
i this is the angle of refraction theta t.
And we are going to need this you will see
shortly why I define this angle psi ok. So,
let me see do I have everything I need this
point let us call this point on the interface
as A. So, the optical path length is n 1 l
naught plus n 2 this is l i, but I can write
l naught in terms of this triangle right.
So, I will have n 1 l naught is R squared
plus S naught plus R, this distance is again
R. This whole squared minus 2 R S naught plus
R cos of psi yes. Again, just using some trigonometry
here, plus the second part of this equation
is n 2 and now I am interested in this triangle
right.
So, that is n 2 R squared plus S i minus R
the whole squared plus 2 R S i minus R cos
psi to the power half. We are going to use
Fermat’s principle; Fermat’s principle
states that the path that light takes is the
shortest path. So, I am going to minimize
this path length, shortest path here we are
deciding how this ray travels and reaches
this point. If I were to change how it reach
this point it would basically be moving along
this surface in other words this angle psi
would be changing.
So, the variable I am minimizing it with is
d psi ok. They using Fermat’s principle
ok. I am minimizing optical path length here.
So, I keep the product n into l unlike yesterday’s
example where we were minimizing the time.
Then I remove the refractive index ok. So,
if I carry out this differentiation; I am
going to have n 1 by 2 minus 2 R sorry S naught
plus R sine it will be minus sign.
So, this actually a plus in the denominator
this entire square bracket is going to now
go into the denominator remember that square
bracket was nothing but the distance l naught
right. So, I will just put l naught into the
denominator. Plus n 2 here I will have 2 R
S i minus R sine psi. But, I will have a minus
sign over here 
and again I have a 2 in the denominator and
I have l i in the denominator.
We are minimizing this; according to Fermat’s
principle. So, we are saying this has to be
equal to 0 ok. So, if I cancel out this, this,
this, this, this, this with respect to this
0. What I am going to be left with is n 1
by l naught S naught plus R is equal to n
2 by l i S i minus and I can rewrite this.
So, that I get n 1 by l naught plus n 2 by
l i is equal to 1 over R n 2 s i by l i minus
n 1 S naught by l naught.
The equation in this form is not; you are
not going to look at this and say that really
makes a lot of sense. You have applied Fermat’s
principle; you got an equation. This equation
is telling you something, but it is not at
this moment telling you anything very useful,
why? Because, the goal was to say every time
a ray leaves the point S. I want an expression
that tells me how that ray gets to p.
And if I look back at this figure; I can see
well before I go back to the figure, if you
look at this equation. This equation tells
me; every time I change the ray i trace. Now,
when I change the ray i trace what is happening.
I am changing basically if I change l naught
here. If I change the ray i trace; I am changing
l naught which means l i is changing. But,
that means; that equation the right hand side
is also going to change.
In order for it to be equal. So, if I trace
another ray, if I put another ray, so I have
a new l naught right. S naught an S i have
not changed, because those are the points
that S naught is the point I want to image.
So, it should not change, but if I look at
this equation if I change l naught and l i.
The only way this equation is going to stay
constant is if I change S naught and S i.
So, it does not seem useful at all to me and
we should not be too surprised. Because, we
have said the spherical interface is not the
interface that satisfies the equation.
So, it should not be surprising, but I started
out this derivation by saying under what conditions
can we use a spherical interface in order
to give us imaging. So, in its present form
it is not carrying out the imaging that we
want. Is that some condition we can apply
that will make this equation work for us.
And it turns out if you go back up to this
part of the equation you see you have a cos
psi here.
Now, what if I said I can make cos psi or
let us take the condition when cos psi is
equal to 1. When would that be what constraint
am I now putting on the optical system if
I say cos psi is equal to 1?
.
I am now saying; let us assume, only rays
making very small angle. So, strictly speaking
if i say cos i equal to 1; that means, I say
only on axis rays, but I will be a little
generous I would not say only on axis rays.
I will say rays making a very small angle
with the axis. Let us look at what happens
to this equation; if we only look at rays
that make a very small angle with axis does
that help us.
Now, the moment I say cos psi is equal to
1. What will happen to my definition of l
naught l naught was R squared plus S naught
plus R the whole squared minus 2 R S naught
plus R to the power half. Now I have put I
have put it ok. So, I should actually do this;
because I have set cos psi is equal to 1.
But, the moment I do this I can see let us
let us expand this. So, I have an R squared
plus S naught squared.
So, R squared plus R squared minus 2 R squared
this and this cancel. And finally, my l naught
is approximately equal to S naught. And of
course, the same will happen for l i as well
it will be approximately equal to S i. If
I now apply this condition to this equation
I have. So, this is for the condition oh sorry
cos psi is almost equal to 1 ok.
Then I have n 1 by S naught plus n 2 by S
i is equal to 1 over R n 2 minus n 1. In other
words; I have removed the l naught and l i
and I am saying any ray coming from the point
S will reach the point P as long as the rays
make a very small angle with the optical axis
ok. We call such rays paraxial rays. This
is also sometimes called first order optics
first order approximation, because, we have
made cos psi approximately equal to 1 and
not to the 1 minus the next term right. You
can also do third order higher order approximations
right.
This is why the spherical surface is used,
yes it is easy to manufacture, but it also
gives us good optical images under certain
conditions ok. What is the condition as long
as the rays are paraxial? This is now a very
useful equation. And the right hand side of
this is sometimes termed as relating to the
focal length. So, n 2 minus n 1 divided by
R. So, remember your surface had radius of
curvature R and it had the medium n 1 on one
side and then 2 on the other.
And we are now saying; this surface has in
fact, this 1 over f is the power of the surface
right. Power and units always given in meter
inverse. The power of this surface is now
determined by the refractive indices on either
side of the surface and the curvature of the
surface.
So, if I want to change the power of the surface;
these are the variables I have I can change
refractive index or I can change curvature.
Now in a lot of cases of course, one medium
will probably be air and you do not have an
infinite basket of optical materials. So,
you can say I want a material with this refractive
index and I will go and pick any refractive
index. You have a limited set of optical materials.
So, you are there not that many refractive
indices you can just pick up. So, very often
the variable you will be playing with is the
radius of curvature. We want to change the
power of the focal length of an optical lens
or mirror you will do that by playing around
with its curvature.
Now, this is still not that useful. Why? Because,
my a lens it does not consist of air on one
side and then glass all the way right. It
has one surface, it has a second medium and
then that medium ends in another interface
and then I go back into air or I go back into
some other medium right.
So, I now use this formula as a base. The
image created by this interface will become
the object for the next interface; that interface
will see a certain radius of curvature. And
it will have a medium of some other refractive
index after it. And it will create its image
and I can now do this forever the image it
creates can become the image for the next
surface and so on ok.
So, let us do that now and arrive at an equation
for a lens. Keeping in mind all the time that
we are talking about paraxial optics; that
assumption is not going to be explicitly mentioned
again and again, but all of this works on
the basis that you are saying cos psi is almost
equal to 1.
So, in order one more point we talked about
conjugate points also and here we are saying
S naught and S i these are the conjugate points
of this system. That means, in this particular
example I said let us assume; we want to get
an image of the source at S and we arrive
at the image at P I could interchange I could
put the source at P and I will get the image
at S ok. So, these are the conjugate points
of this system.
So, that was the equation for a thin for a
single refracting surface. Let us get the
equation of a thin lens. As usual I start
with an optical axis; it is a lens there are
two interfaces. So, I am going to say this
is one interface remember its part of a sphere
or circle and this is another. So, the this
is the lens this is all that we are going
to use ok. One side has radius of curvature
R 1, the other has radius of curvature R 2.
So, I am not drawn this properly. Please these
circles should be going the optical axis is
going through the center of the circles. This
I call C 2 the center of curvature of this
circle and this is C 1 center of curvature
of the first interface light will see. This
has rad R 2 this has radius of curvature R
1. And let us say our object is sitting somewhere
over here, so this is point S.
So, the distance of S from the first interface;
we shall call it S o 1 it is the first object
point right. The thickness of this lens let
us call it d right and finally, let us say
that the image is forming at P, but before
it forms at P; there is an image formed by
the first interface. And I am going to put
that image over here and I will explain why
I am putting that image over here ok.
So, light traveling from point S hits the
first interface; if the second interface did
not exist in the case that I am setting up
now its image point would actually be at this
point over here ok. And I will explain to
you why I have done that ok. So, if this is
the image point, whenever we are dealing with
lenses; we always measure distances from the
vertex of the lens ok. So, this is my first
image, so I call it S i 1. This image acts
as the object for the second interface.
So, the second interface; this is now object
distance 2 and this is image distance 2. P
is where the final image is formed. Remember,
we have just arrived at this expression right.
For a single interface, this was refraction
at a single interface. Now, what happens in
this expression; the left hand side is constant
for a single interface. You are not changing
the refractive indices, you are not changing
the radius of curvature it is a constant.
Now, let us say I go on decreasing the distance
S naught. What does that mean? I go on bringing
the object I want an image of closer and closer
to the interface, because the left hand side
is a constant. What does that mean for S i?
If I go on decreasing S naught; it means,
S i is going to go on continuously increasing.
At some point S naught will equal n 2 minus
n 1 by R, in other words it is equal to this
or sorry S naught is equal to f naught this
at some point I can make s naught equal to
f. What does that mean for S i? Where does
the image happen if S naught is equal to f?
That means, S i is at infinity what happens
if S naught decreases further? What happens
to S i?
How will this equation get satisfied? It means
S i is negative or as he said it will be on
the same side as the object. And that is the
reason why in this drawing I have put the
first image formed on the same side as the
object. Because, if you think about it by
doing that; what am I saying. I have my object
distance and let us say S naught is my object
distance.
I have the focal length f which is of the
combined lens ok. Same combined lens I mean
both the interfaces, but I have f 1 which
would be the focal length; if there was only
one interface. And why we do this is this
condition is almost always going to be satisfied.
Two interfaces together have more power than
a single interface right.
So, this condition is what you would expect
or is not surprising that this condition arises
when you compare what is happening at one
interface to what happens when you have two
interfaces ok. So, that is why, the first
image I have kept it on the same side as the
object ok. Is that clear that point?. So,
now, I want to use the same equation that
we arrived at for a single interface, but
apply it iteratively; first for the first
interface and then the image that is formed
will be the object for the second interface
and finally, arrive at an equation for the
full lengths.
So, let us do that. So, I hope you all have
this figure; because, I am going to be using
the variables from this figure. For the first
surface right. Keep the equation in mind right.
So, I am going to have and let us let me define
this as the refractive index surrounding the
lens as n subscript m. So, that is true here
as well that is the medium surrounding the
lens and the medium of the lens I will call
n l.
In the first case; this is equal to n 1 by
my earlier definition and n l is n 2 ok, for
the first interface ok. So, I have n m by
S naught 1 plus n l by S i 1 is equal to n
l minus n m and the first radius of curvature
Cs is R 1. That is the equation for the first
surface.
And the equation for the second surface n
l by S naught 2 plus n m by S i 2 is equal
to n m minus n l by R 2 ok. At this point;
I have to be very careful, because I need
to know what signs I am going to attribute
or assign to each of these variables. There
is a sign convention that we need to take
into account ok.
And the sign convention and let me just go
to this other. So, the sign convention I have
given it to you and I will share this with
you on moodle. So, do not worry about it too
much now, but the first column is the sign
convention that we are using with lenses.
So, concentrate on this for the moment. What
are the signs we associate with different
parameters, what are those parameters.
So, the parameters of interest are the focal
length of the system; the object distance,
the image distance and the magnification.
And you can see that for a convex or a converging
lens we will use the focal length as positive
for a concave lens it will be negative. And
in fact, we will keep coming back to this
because I add a few riders as we go along,
but I think at this point it is enough to
just give you this information.
The object distance and this is what you need
to note because we need to use this information
now. If the object is towards the left of
the lens; we say that the distance is positive.
So, typically, historically, optics has always
been designed by tracing light traveling from
left to right. And that is why; here if the
object is towards the left or is left of the
lens we say the distance is positive if the
object lies to the right of the lens we say
its distance is negative.
The reverse is true for the image; if the
image forms to the right of the lens it is
positive. If the image forms to the left of
the lens it is negative ok. That is really
what is important for you right now. We will
come back to this, but keep this convention
in mind. So, I have to be now careful with
these distances I have written out two equations,
but this equation this term S naught 2. It
actually is a sum of this term and this term.
And I need to take the convention the sign
convention into account when I add up these
two terms. So, S naught 2 is actually going
to be S i 1 plus d. And what was the convention
for object if an object was to the left of
the lens it was positive. So, S naught 2 can
remain like this, but S i 2 is an image distance
and in this particular case it also lies to
the left of the lens. So, by the sign convention
this is going to be negative.
So, I will just write that here maybe 
object distance to the left of the lens is
positive image distance to the left of the
lens is negative. So, this is what I am going
to get, following the sign convention. There
are other conventions and there is no problem
with you using any convention.
You just have to ensure that; if you are solving
or working out a problem you stick to the
same convention from beginning to end and
it is also good practice to state; what is
the convention you are using especially, if
it is going to be different from this one
just. So, that it is clear ok. So, if I now
take these two equations and I add them up
keeping in mind this sign convention let us
see what we get.
So, let us take all the n m terms 1 over S
not 1 plus 1 over S by 2 equal to and on this
side I have an l minus n m it is 1 term plus
I have n l 1 over S i 1 minus one over S i
1 plus all I have done is add up these two
the equations for each of the interfaces.
Now to make this again a little more useful
immediately; let us make an assumption that
the lens we are dealing with is a very thin
lens. So, for a thin lens in other words the
thickness d is 0.
Why is it valid to make this even in the morning
class I talked about geometric optics as being
the regime where we consider lambda to be
0 and that is a valid assumption to make,
because we are talking about imaging objects
much bigger and dimension to lambda.
So, here again; if I say thin lens how far
away is the object from the lens if we are
talking about distances much larger than the
thickness of the lens with respect t to the
thickness I can consider that distance so
much larger the thickness can be considered
negligible ok, but the advantage of doing
this is. If I look at the equation above now,
if d goes to 0, this term here will disappear.
So, with that condition; I can now say and
let us make some more assumptions let us say
the medium surrounding the whole system, medium
n m is equal to 1. So, let us say it is air
I will have this equation. I have an equation
which is in terms of the object distance in
terms of the final image distance the object
distance S naught 1 is the location of the
object on the optical axis from the vertex
of the first surface.
The image is the location of the image point
on the optical axis from the vertex of the
second surface right. And the medium surrounding
this lens is air. So, I am not bothered about
its refractive index and the medium off the
lenses n l that appears in the equation both
the radii of curvature appear in the question.
It is not very different from the equation
we got for the interface that also the power
was a function of radii of curvature radius
of curvature.
Here it is a function of both the radii of
curvature again this is called the power of
the system and often you will see it written
in this form where u is used as object distance
and v is used as image distance. This, if
you write this in terms of u and v it is also
called the lens makers formula. Sometimes
this is called the Gaussian lens formula ok.
Pretty straightforward.
So, let us just run through some possible
scenarios here, but what happens now, when
the object is located at different places
ok. This is again high school optics, but
now you should have a little better understanding
of where it is coming from right. If you go
back to your equation; 1 over object distance
plus 1 over image distance is equal to 1 over
the focal length.
If the object is at infinity this term is
disappearing the image is happening at the
focal length itself right. And you can move
your object from infinity to a position to
f away from the lens to something in between
two f to less than f and you can see how the
image moves accordingly ok.
And that is one of the first exercises you
will do in the lab, the simulation lab is
to simulate by changing object position where
the image is forming and also looking at the
nature of the image, because mostly you may
get a real image, but at some point the image
will appear to be on the same side of the
object and then you get a virtual image ok.
So, I will just scroll through these, but
this is actually what you need one of the
exercises you need to do in the lab today
is actually send light through a lens and
change the object position and then monitor
where the image is forming.
And you can see this is the last case; when
the object moves to a distance less than focal
length the rays now they are still of course,
going through the lens, but they are diverging
here. They are not going to converge to a
point. They appear to come from a point over
here and that is why we say it is a virtual
image.
It is virtual because, we would see it with
our eyes as if it was coming here forming
here when actually the rays are diverging
over here you could not place a camera or
screen and capture, you could not place a
screen here and capture that image you need
extra optics to capture that image.
