Now, in this lecture we will start the analysis
of optical waveguides using electromagnetic
theory. Let us first look at types of waveguides,
this is the bulk medium.
So, this is not a waveguide, when the dimensions
are very large as compared to wavelength.
We can have a waveguide like this where we
have a channel or a material in this geometry
and the dimensions are comparable to the wavelength
and it is surrounded by a lower index medium
like this. Then in this way I create a channel
for light to flow along this. This is a rectangular
channel waveguide. I can have this channel
in the form of a cylinder high index refractive
index cylinder which is surrounded by lower
refractive index medium. Again all are comparable
to wavelength the dimensions are comparable
to wavelength and this is optical fiber waveguide.
And then I can also have a configuration like
this we are this width of the channel can
be infinitely extended.
So, instead of having this kind of channel,
I can have a slab thin slab of high refractive
index. It is sandwiched between 2 other thick
slaps of lower refractive indices. Then this
is known as planner waveguide. In these 2,
I have confinement in 2 dimension and propagation
in this longitudinal direction. Here I have
confinement only in one direction and propagation
in this longitudinal direction. So, we are
going to now do the electromagnetic wave analysis
of such kind of media. We had seen that in
infinitely extended medium where there were
no such refractive index discontinuities,
the solutions of the wave equations were plane
waves.
If you remember that the electric field associated
with that wave in an infinitely extended medium
if it is infinitely extended medium.
I had seen that E was given by E0 e to the
power i omega t minus kz and H was H0 e to
the power i omega t minus kz and I had these
vectors.
So, they have components Ex, Ey, Ez, Hx, Hy,
Hz, this is Hz. And the point to be notice
here was that, these E0 and H0 they were constants.
S0, E0, H0 purely constants; what we could
see that in case of infinitely extended medium
our direction of propagation was z. So, I
had z part coming out like this and then time
part coming out like this. Here also what
I will do? I will take this direction which
is propagation direction is z and z part would
be like this itself. So, in order to now find
out the propagation of light waves in such
kind of medium where we have refractive index
discontinuity refractive index is not uniform,
when I consider Maxwell’s equations in homogeneous,
linear isotropic charge free current free
dielectric medium.
In the previous case it was homogeneous medium
now it is inhomogeneous medium. And again
I write down all the Maxwell’s equations
Maxwell’s equations go like this. But now
if I look at constitutive relations, when
D is equal to epsilon E this epsilon is now
in general a function of x, y and z, which
is nothing but epsilon naught times n square
of x, y and z. So, this is not a constant
scalar now, which was the case for infinitely
extended medium. And again I am talking about
dielectric. So, B is equal to mu H which is
closely equal to mu naught times H because
it is non-magnetic medium. So, what should
I do? Know again I should form the wave equation.
So, that I can find out what are the electric
and magnetic field solutions. So, the usual
procedure to find out the wave equation is
I take the curl of this equation and use other
Maxwell’s equations into it.
So, I get del of del dot E minus del square
E is equal to minus del t of mu times del
D over del t, thus in the same way as I had
done in the previous case. But now here I
should check that del dot E is not 0 I should
be careful. In the previous case del dot E
was 0, but now I have del dot D and D is equal
to epsilon E epsilon depends upon x, y and
z.
So, I cannot take that epsilon out of this
del operator. So, del dot E would not be 0
if del dot E is not 0 then how this wave equation
is going to change? So, let me know find out
what is del dot E for that I take del dot
D is equal to 0 and put D is equal to epsilon
not n square En square is a function of x,
y and z. So, I write it down as epsilon naught
del of n square dot E plus n square del dot
E and that should be equal to 0. This gives
me del dot E is equal to minus gradient of
n square over n square dot E. I put this back
into this equation and rearrange the terms
to get a wave equation in this particular
form.
What it is del square E plus del of 1 over
n square del n square dot E minus mu naught
epsilon naught n square del 2 E over del t
square. So now, you can notice an extra term
here, this is an extra term that we have got
it as compared to the case of infinitely extended
medium.
So, this is a wave equation in an inhomogeneous
medium. What is the implication of this term
now? What complicates it can introduce in
our analysis? Well if I expand this then I
write down this x, y and z components and
if expand this what I see that now it is not
possible for me to separate out x, y and z.
t can be separated out, that is not a problem.
But I cannot separate out x, y and z. x, y
and z solutions cannot be separated or which
I could do in case of infinitely extended
medium. So, what do I do now? Well similarly
if I do it for the magnetic field I will obtain
an equation in H something like this, again
there would be a middle terms which makes
it impossible to separate out x, y and z solutions.
So now let me consider a case where refractive
index where is only transfers direction which
is the case of optical waveguides and optical
fibers. So, so I take in general a case where
n square is a function of x, y which can be
a channel waveguide or optical fiber. So,
n square does not vary with z, in this case
I can separate out z and t parts. And if I
can separate out z and t parts then the solution
z and t solution can be written in the same
way as this. So, I write down z and t solution
like this. And x, y solution is still remains.
So, I put it with E0. So, the associated electric
field I can write as E(x, y, t) is equal to
E0 of x, y times e to the power i omega t
minus beta z Similarly H.
Now what I have got? I have got that the solutions
here have this form. So, this is a function
of x and y and this is the propagation in
z direction. So, as if some function of x
and y is propagating in z direction with some
propagation constant beta, similarly for H.
So, these are the modes of the system. And
as I will find out that there can be only
certain such functions possible which sustain
their shape and propagate with certain propagation
constant beta these are the modes of the system.
So now, my problem reduces to find out these
functions, E0 of x, y and H0 of x, y and their
corresponding propagation constants beta.
Let me start with doing a very simple problem
where I remove any index discontinuity even
in y direction. So, I take the simplest case
here the variation of refractive index is
only in x direction.
So, I have n square as a function of x only
and this is the case of say planner waveguide
something like this, where you have this is
x this is y and this is z. So, where y is
infinitely extended, z is infinitely extended
and you have index cn discontinuity only in
x direction. So, here you have different refractive
index here different and here different. So,
n square is the function of x only. Now if
it is a function of x only then I can separate
out y part also. And the solution I can write
as e to the power i omega t minus some gamma
y minus beta z. And x part will be now associated
with E0. So, E0 is not a constant is it is
a function of x some function of x. Similarly,
H is equal to H0x e to the power i omega t
minus gamma y minus beta z.
So, these are the form of solutions now. What
I can do? I can always choose my direction
of propagation if the light is propagating
in this direction I can label it as z or I
can label it as y it is up to me to choose
my access. So, what I do? I choose z axis
as the direction of propagation then without
loss of any generality I put gamma is equal
to 0. So, you can see that if this is infinitely
extended, this is infinitely extended index
discontinuity is only in x. So, you can launch
light into this in this direction and let
the light propagate along y or you can let
the light propagate along z you can launch
it from here. So, I choose to take the direction
of propagation as z. So, I put gamma is equal
to 0.
Then I can write the solutions as E0 of x
e to the power i omega t minus beta z and
x 0 of x e to the power i omega t minus beta
z. So, I have got similar form of solution,
but it is not the same. Here E0 and H0 are
constants; here E0 and H0 are the functions
of x and these functions now I want to find
out.
I need to find out how E varies with x and
how H varies with x, that is what I need to
know and they will give me the modes. So,
I have the solutions here and do not forget
that I have vector sins here which means this
E is nothing but Ex x cap plus Ey y cap plus
Ez z cap and similarly H. So, basically I
have 6 such equations 3 in E and 3 in H. So,
if I write down the components of this then
I can write them as Ej is equal to Ej x e
to the power i omega t minus beta z and similarly
Hz and similarly Hj where j can be x, y or
z.
Now, let me put these solutions into curl
equations. Why I am doing this? Ultimately
I want to find out how E varies with x and
how H varies with x. So, I need to form a
differential equation in E with respect to
x. In order to do that and I know from here
I will get del E over del x del H over del
x terms. So, that is why I put these into
these equations now. When I do this then this
will give me 3 equations, one corresponding
to Hx then Hy and Hz. And this will also give
me 3 equations, Ex, Ey an Ez x, y and z components
here, let me do it. The x component from here
we will come out to be if you expand this,
because you know that del cross E del cross
E you can write as x cap y cap z cap del del
x del del y del del z. And then you have Ex,
Ey Ez is equal to So, this is del cross E
minus mu naught del del t and then you have
3 components here Hx, Hy and Hz.
So, from here you can find out 6 equations,
the first one would be i beta Ey is equal
to minus i omega mu naught Hx. The x equation
from here would be i beta Hy is equal to i
omega epsilon naught n square of x times Ex.
The second one from here would be minus i
beta Ex minus de Ez over del x is equal to
minus i omega mu naught Hy. And here it would
be minus i beta Hx minus del Hz over del x
minus i omega epsilon naught n square of x
Ey.
Third one would be del E by over del x minus
i omega mu naught Hz and here it would be
del Hy over del x, i omega epsilon naught
n square of x Ez. So, I have got 6 equations
which relate the electric and magnetic field
components Ex, Ey and Ez and Hx, Hy and Hz.
What do I do with these equations? Well, what
I notice one thing that I can simplify the
situation to certain extent. And how can I
simplify the situation? Well if I have if
I have a waveguide and I launch light into
this when launching light into this I have
some control on light and that is I can launch
this polarization or this polarization.
This is x axis this is y axis. So, if I decide
to launch this polarization that is Ey is
non 0 and Ex is 0 then let me see which equations
do I invoke. Do invoke all the 6 equations
or I invoke only a few of them?
And what I find that the equations which have
E y non 0 and E x 0 are this one has E y non
0 E x is equal to 0. This one has E y non
0 and this one has E y non 0. So, so if I
if I launch y polarized wave then I invoke
these 3 equations and when I launch exploitation
I invoke these 3 equation. So, 3 equations
can be involved at a time. So, this gives
me this gives me a room to simplify the problem,
because I need to now consider only 3 equations
at a time. These 3 equations the blue ones
are these and what I see there is they have
only 3 non vanishing components of E and H
and they are Ey, Hx and Hz.
In these 3 I get that there is only one component
of E and that is transverse. Then these modes
are also known as TE modes or transverse electric
modes or transverse electric polarization.
While the other 3 have non vanishing components
of E and H as Hy Ex and Ez and I see that
there is only one component of magnetic field
and that is transverse component then they
are known as transverse magnetic modes or
TM polarization and this will correspond to
these 3 equations. So now, let me do the analysis
of TE modes first. So, what I want to do?
Again do not forget I want to find out how
E and H vary with x and I need to find out
a differential equation in E or H with respect
to x. So, for TE modes I write down these
3 equations.
And what I can do now since these 3 equations
relate Ey, Hx and Hz then if I know one of
them then I can find out the others. So, for
example, if I know Ey I can find out Hx from
here and Hz from here ok.
So, what I do let me find out Ey. So, I substitute
for Hx and Hz from these 2 equations into
the third equation. And when I do this I form
a differential equation in Ey. And this comes
out to be d2 Ey over dx square plus k naught
square n square of x minus beta square times
Ey is equal to 0. Where I have use the effect
that k naught is omega be c, which is also
2 pi over lambda naught where lambda naught
is free space wavelength. So now, I have got
now I have got a differential equation in
Ey for a given n square of x. So, if I know
my planner waveguide that is I know n square
of x then I can solve this equation for the
given n square of x and obtain Ey. And that
will give me the modes, that will tell me
how electromagnetic wave propagates in that
medium of n square x refractive index variation.
So, let me apply it to a very simple waveguide,
which I call planner mirror waveguide. What
is a planner mirror waveguide?
You take a very thin slab of refractive index
and let us say glass. It has got a width d
and refractive index n. And I polish and I
sorry not polish and I deposit metal here
and here. When I deposit metal on top and
bottom and if I launch any light then that
light will be reflected back and forth from
this mirror and from this mirror and should
be guided.
So, this is the simplest waveguide I can think
of let me do that. So, I deposit metal on
top and bottom if I look at the refractive
index profile. Then I find that in this region
between 0 and d. I have refractive index n
and here at the boundaries I have metal. When
it boundaries I have metal then the electric
field at the metal boundary should be 0 that
is what I know. So, what I do know I write
down the wave equation the equation which
I obtained in the previous slide. And I put
n square of x as n square and I write it down
in the region between 0 and d.
So, this would be the equation. So, I from
here I can find out how Ey varies in this
layer. And I know that the fields has to be
0 here. Now let me defined this k naught square
n square minus beta square as some kappa square
since I know that beta which is the propagation
constant of the wave in this region has to
be less than k naught n it cannot be greater
than k naught n, because propagation constant
cannot be greater than the propagation constant
of the medium itself k of infinitely extended
medium.
So, beta is less than k naught n. So, kappa
square is always greater than 0 which means
that the solution of this equation would be
Ey is equal to A sin kappa x plus B cosine
kappa x. Now my feel has to be 0 here and
here. So, I apply these boundary conditions
Ey is equal to 0 at x is equal to 0 and at
x is equal to d and this gives me B is equal
to 0 and kappa d is equal to m pi. So, since
B is equal to 0. So, this term goes off and
kappa is equal to m pi over d.
So, I put kappa is equal to m pi over d. So,
my solution now becomes E y is equal to a
sig m pi x over d where m can take integer
values now what are. So, I have got E y of
x what is left corresponding beta from where
beta are coming from here because I know kappa
d is equal to m pi and kappa square is equal
to k naught square n square minus beta square.
So, this gives me that there would be only
certain discrete values of beta defined by
beta m and given by beta m square is equal
to k naught square n square minus m pi over
d whole square.
So, I have got for a planer mirror waveguide
only certain functions only certain functions
which has certain propagation constants they
will be sustained. If I plot them then for
m is equal to one it will look like this for
m is equal to 2 it would look like this m
is equal to 3 like this. They are nothing
but if you if you look carefully they look
like as the modes of vibrations of a stretched
string modes of vibrations of a stretched
string. So, it is similar to that. So, in
the next lecture I am going to understand
what do they exactly represent. I know that
in a waveguide in a waveguide if I launch
if I launch ray like this then it will we
reflected back and forth or if I launcher
wave then wave would be reflected back and
forth, but how do these represent the guidance
in an a waveguide? So, let us understand it
in the next lecture.
Thank you.
