So, in this lecture, let me continue with
the previous lecture where I was doing the
analysis of planar mirror waveguide.
So, what I was doing was to find out the modes
of a planar mirror waveguide and what I did
that I took the planar mirror waveguide which
has refractive index n and metal deposited
at x is equal to 0 and x is equal to d, and
the modes of this waveguide I found out as
Ey is equal to A sinm pi x over d and corresponding
beta are given by beta m square is equal to
k naught square n square minus m pi over d
square.
So, what I got that I got this kind of variation
of electric field Ey with respect to x or
this or this and these fields propagate in
the waveguide and sustain their shape as they
propagate. They also propagate with certain
propagation constants beta1, beta2 beta3 and
so on. Let me understand what they are.
So, the field is Am sin kappa mx where kappa
m is equal to m pi over d if I write the complete
solution this is only Ey m of x, x part of
the solution remember that there is t and
z parts also. So, if I write down the complete
solution there it is some Am sin kappa mx,
e to the power i omega t minus beta mz. Let
me write this sin kappa mx in the form of
e to the power plus minus I kappa mx.
So, if I do that I get this and then let me
put it in this form. So, what I have got from
here, I have got it e to the power i omega
t minus b time z minus kappa mx and then e
to the power i omega t minus b time z plus
kappa mx what they are and remember you see
now outside I have a constant it is not a
function of x, the x dependence I have included
in the exponential now, the x dependence which
was earlier here I have included in the exponential
and when I see this now this is nothing, but
the plane wave because outside is a constant.
So, and this is the plane wave which is moving
in this direction, moving in x-z plane making
certain angle from z axis. And this is a plane
wave which is moving in plus z minus x plus
z minus x direction k. So, what I have got
that this mode, this mode is nothing but the
superposition of 2 plane waves, this plane
wave and this plane wave with some phases.
So, it is the superposition of 2 plane waves
1 propagating in this direction another propagating
in this direction. If you look at the waveguide
if you look at the waveguide this is x this
is z. So, this is x-z plane this is x this
is z. So, one wave is moving like this another
wave is moving like this and this is what
you have, that if you launch wave like this
then it will get reflected then reflected
from here, reflected from here.
So, all the time you will have 2 waves one
going in this direction and another going
in this direction. Let me put them together
here in x-z plane.
So, one is moving an angle theta m with z
axis and another one is moving minus theta
m angle with z axis and this has to be the
propagation constant in the medium of a plane
wave these are now plane waves plane waves
we will have propagation constant k naught
n. So, this is the component of propagation
constant in x direction this is the component
of propagation constant in z direction. So,
if I now do kappa m square plus beta m square
is square root then it comes out to be k naught
n and it comes out to be k naught n because
you remember that kappa m square is equal
to k naught square n square minus beta m square
which means that this is indeed k naught n
this is indeed k naught n.
So, the propagation constant in z direction
is k naught n cos theta m. Let us understand
it. If I take the components in x direction
I have 1 wave going in positive x another
wave going in negative x, which means that
in x direction I have 2 counter propagating
waves and when I have 2 counter propagating
waves they give you a standing wave solution
they give you standing wave. So, in x direction
I have a standing wave. In z direction if
I see this also gives me propagation in positive
z.
So, I have a standing wave in x which propagates
in z. The energy does not flow out in x direction
the wave stands in x direction, but that standing
wave pattern flows in z direction this is
the mode. So, these modes are nothing, but
the standing wave patterns and they are made
out of two waves one going in this direction
another going in this direction.
I can now find out what is the effective refractive
index of a given mode or effective index of
the mode by beta m over k naught which is
nothing, but kn cos theta m. So, for every
mode I have different beta m and therefore,
the angles that the constituent plane waves
make with z axis would be different and these
angles can be found out from here. For example,
for n is equal to 1.5 d is equal to 1 micrometer
and lambda naught is equal to 0.633 micrometer
helium neon wavelength. I find out that there
are these modes four modes. Let me look at
them closely.
The first one has effective index 1.4662 and
the angles of constituent plane waves with
z axis is z axis r plus minus 12.18 degrees.
So, there are 2 plane waves one making plus
12 degrees and other making minus 12 degree
with z axis and this is the a standing wave
pattern in x direction corresponding to those
waves.
The second one is with n effective is equal
to 1.3599 with angles plus minus approximately
25 degrees. Yet another one is this with the
angles plus minus 39 degrees and this one
plus minus 57.565 degrees.
So, these are the modes how many modes are
there for a given waveguide. Now let me look
at this picture again and from here I know
that kappa m d is equal to m pi. So, this
m would be kappa m d over pi and kappa m is
nothing, but k naught n sin theta m and the
maximum value of this would be k naught n
because the maximum value of sin theta is
1.
So, the maximum value of k m is k naught n
and this will give me what is the maximum
value of m. So, I put maximum value of k m
here which is k naught n. So, m max would
be k naught n by pi and k naught is equal
to 2 pi over lambda naught. So, this would
be 2 nd over lambda naught. So, the maximum
number of modes that can propagate in a planar
mirror waveguide would be a number and integer
which is closest to, but less than this, it
is closest to but less than this. For example,
if I have n is equal to 1.5, d is equal to
1 micrometer and lambda is equal to 0.633
micrometer there should be micrometer here
then 2nd over lambda naught comes out to be
4.739 and this gives me the maximum number
of modes as 4. So, there would be 4 mode supported.
So, this is the analysis of planar mirror
waveguide which gives me insight into the
modes of a waveguide which structure is quite
intuitive and light guidance and data structure
is also quite intuitive, but this is a structure
planner mirror waveguide is of very little
practical use for 1 it is not feasible to
make this waveguide that you have one micrometer
film of certain material and you deposit metal
on top and bottom of that and use it. Second
since it involves metal coatings and metal
is highly absorbing material. So, as these
modes propagate they will attenuate very quickly.
So, they have very high loss. So, this kind
of waveguide is of little practical use although
it gives a good understanding of mode propagation.
So, now what we are going to do is to look
at more practical waveguide which is asymmetric
step index planar waveguide dielectric waveguide
planar dielectric waveguide which is obtained
by sandwiching high index lab between 2 lower
refractive index labs. So, I have n1 n2, n2
here and n1 is greater than n2. Again if you
look there is refractive index discontinuity
only in x direction, y direction it is infinitely
extended in z direction also there is no indexed
discontinuity and it is infinitely extended.
If you look at the refractive index profile
of this waveguide then it looks like this
and let me put my x is equal to 0 access in
the middle of the waveguide so that I can
make use of the symmetry of the problem let
it have of the width d. So, this is x is equal
to d by 2 this is x is equal to minus d by
2. I can write down the refractive index variation
with respect to x in this fashion.
So, this region is mod x less than d by 2
while these 2 reasons can be represented by
mod x greater than d by 2. So, in mod x less
than the d by 2 I have refractive index n1,
in mod x greater than d by 2 I have refractive
index n2 and now my problem is to find out
Ey and corresponding beta for this given n
of x. So, how do I do? This again go back
to the wave equation corresponding to t e
votes. Let me first solve the problem for
TE-modes.
So, this is the wave equation for TE-modes
and this is the n(x). So, how do I solve it?
Well I find that in this region and in this
region and in this region in all the 3 regions
the refractive index is constant although
when I go from here to here and here to here
I encounter index discontinuity, but on this
side and on that side the refractive index
remains constant.
So, I make use of this and write down the
wave equation in this region and in this region
that is in this region and in these 2 regions.
So, for mod x less than d by 2 I write it
down as d2Ey over dx square plus k naught
square n1 square because nx is equal to n1
in this region minus beta square times Ey
is equal to 0. And I also keep in mind that
my effective refractive index which is given
by beta over k naught. Effective refractive
index n effective which is given by beta over
k naught this should lie between n2 and n1
for guided modes for guided modes n effective
would lie between n2 and n1 which means that
beta would lie between k naught n2 and k naught
n1. So, beta would be greater than k naught
n2 while it would be smaller than k naught
n1.
So, when I write this equation now in these
2 regions which are represented by mod x greater
than d by 2 I write it as d2Ey over dx square
minus beta square minus k naught square n2
square times Ey because n x is equal n2 here.
And why I have written it in this fashion.
So, that this is positive here and this is
also positive here because I am solving it
going to solve it for this condition which
is the condition for guided modes for which
the energy would be confined into this region.
So, I represent this as kappa square and I
represent this as gamma square and this kappa
square and gamma square are positive for guided
modes as I have explained. So, let me write
these equations down again here.
So, for mod x less than d by 2 I have a got
a got an equation d2E by over dx square plus
kappa square Ey is equal to 0 and for mod
x greater than d by 2 it is d2Ey over dx square
minus gamma square Ey is equal to 0 where
kappa square as this and gamma square is this.
What are the solutions? I know the solutions
of these differential equations very well,
this equation gives me oscillatory solutions
e to the power I kappa x or e to the power
minus i kappa x or sin kappa x cosine kappa
x while this equation gives me exponentially
amplifying and decaying solutions. So, if
I write the solutions in the regions mod x
less than d by 2 and mod x greater than d
by 2 I find that in this region it is oscillatory
solution A cosine kappa x plus B sin kappa
x and in the region mod x greater than d by
2 I have exponential and decaying solutions,
but out of these 2 solutions I retain only
exponentially decaying solution while exponentially
decaying solution and not exponentially amplifying
solution because I want a solution which gives
me guided modes and for guided modes the energy
should be confined into the high index region
and it should decay down as you go away from
the high index region.
As you go towards infinity x is equal to plus
minus infinity the energy should decay down.
So, I cannot take exponentially employee find
solutions that is y for x greater than d by
2 I have chosen the form of solutions C e
to the power minus gamma x and for x less
than minus d by 2 I have chosen the form D
e to the power gamma x this A, B, C, D are
some constants which can be determined by
the boundary conditions. What are the boundaries?
Boundaries are x is equal to plus d by 2 and
x is equal to minus d by 2, but here I can
make some more simplification what simplification
I can make is to utilize the symmetry of the
problem how can I utilize the symmetry of
the problem let us see.
I have the wave equation for TE-modes like
this and this is the n of x. I see that this
is symmetric about x is equal to 0 which means
that n square of minus x is equal to n square
of x.
Now, in this equation if you replace x by
minus x if you replace x by minus x, then
you get d2Ey of minus x divided by d of minus
x square that is x square plus k naught square
n square of minus x minus beta square Ey of
minus x is equal to 0 and this is nothing,
but n square of x because of the symmetry
of the profile which means that which means
that I now have 2 possibilities which are
Ey of minus x is equal to Ey of x if I put
Ey of minus x is equal to Ey of x I get back
the same equation or if I put Ey of minus
x is equal to minus Ey of x then also I get
back the same equation.
So, there are now 2 possibilities one is this
another is this and they are known as symmetric
modes because this is the property of a symmetric
function and this is the property of antisymmetric
function. So, these modes are known as symmetric
modes and these modes are known as antisymmetric
modes. The simplification it introduces is
that out of those 2 functions cosine and sin
I can pick one here and one here since cosine
is the symmetric function and sin is the antisymmetric
function.
So, symmetric modes would now be represented
by Ey of x is equal to a cosine kappa x for
mod x less than d by 2 and C e to the power
minus gamma mod x for mod x greater than d
by 2 for antisymmetric modes I will have Ey
of x is equal to b sin kappa x for x less
than for mod x less than d by 2 and then I
have D e to the power minus gamma mod x for
mod x less than d by 2, but I shall have to
take care of sin when I go from left side
or right side. What do I mean to say is this
if this is x and this is d by 2 this is minus
d by 2 this is x is equal to 0.
Now, if you plot a sin function in the region
mod x less than d by 2. So, it goes something
like this something like this and now in the
n2 region for positive x values you will for
x greater than d by 2 you will approach from
here and then it decays down while for x less
than minus d by 2 it you will approach from
here from negative side. So, here you start
from positive side and decay down here you
start from negative side and then decay down.
So, to take care of that sin I will have to
put x over mod x here.
So, these are the symmetric modes and remember
that I am doing the analysis of TE-modes have
non vanishing components of E and H has Ey,
Hx and Hz. What is left now? I need to find
out the relationship between A and C I need
to find out how these solutions are connected
at the boundaries for that I will use boundary
conditions. And I have learnt that the boundary
conditions are when you encounter an interface
between 2 dielectric media the boundary conditions
are tangential components of E and H should
be continuous.
Tangential components of E and H should be
continuous what are the tangential components
to the boundaries. This is the waveguide this
is x this is x is equal to plus d by 2 this
is x is equal to minus d by 2, this is y this
is z. So, to this interface which is x is
equal to plus d by 2 or x is equal to minus
d by 2 here the tangential components are
y and z y and z, x is a normal component and
here the nonvanishing components are Ey, Hx
and Hz. So, the tangential components are
Ey and Hz. So, these Ey and Hz should be continuous
and if you look back to the 3 equations which
relate these 3 components of E and H then
I find that Hz is nothing, but some constant
times dEy over dx. So, Ey and dEy over dx
should be continuous at x is equal to plus
minus d by 2. So, let me apply these boundary
conditions, let me do it for x is equal to
plus d by 2 the same would be obtained for
x is equal to minus d by 2.
So, I put Ey should be continuous at x is
equal to plus d by 2 which means a cosine
kappa d by 2 should be equal to C e to the
power minus gamma d by 2 and then for derivative
dEy over dx. So, I will get minus a kappa
sin kappa d by 2 is equal to minus gamma C
e to the power minus gamma d by 2. If I divide
this by this I get kappa tan kappa d by 2
is equal to gamma and remember that these
kappa have dimensions of 1 over meter. So,
I can make them dimensionless, so I can multiply
both sides by d by 2 and get an equation like
this which is kappa d by 2 tan kappa d by
2 is equal to gamma d by 2.
What is kappa and what is gamma? So, you remember
that kappa a square is equal to k naught square
m1 square minus beta square and gamma square
is beta square minus k naught square n2 square
k naught is equal to 2 pi over lambda naught.
So, for a given waveguide and wavelength for
a given waveguide and wavelength the only
unknown in this in these kappa and gamma is
beta. So, this is nothing, but a transcendental
equation in beta. So, beta satisfy this equation.
So, there are only certain values of beta
which are possible and those values of theta
satisfied this equation only those values
of beta are possible. So, this makes this
makes the modes discreet.
So, from here I find out what are the possible
values of beta and for those beta I can find
out the fields. So, that is how I can get
the modes of a symmetric planar waveguides
and these are symmetric modes.
In the next lecture I will continue with this
to solve to find the solutions for antisymmetric
modes.
Thank you.
