Let us continue our analysis on planar dielectric
optical waveguides. If you remember that we
were doing the analysis for confinement in
x direction and propagation in z direction.
And we had seen that if the confinement is
in x direction and the propagation is in z
direction. Then the electric and magnetic
fields associated with the light wave can
be given as E (x, y, z, t). In fact, here
you need not to take why because I am taking
the planar waveguide. So, it is simply E (x,
z, t) vector is equal to now it is E e 0 or
E of x e to the power i omega t minus beta
z.
Similarly, H is x z t is equal to H of x this
is H of x e to the power i omega t minus beta
z. So, they represent the modes they represent
the modes E of x and H of x and we know that
since this is this is propagating in z direction.
Then what we have is basically Ex, Ey, Ez
all these components will be there and Hx,
Hy and Hz all these components of E and H
would be there. This is a mode propagating
inside direction and you should remember that
these E and H are not constant. So, these
are not plane waves, these are not plane waves
that is why I will have all the components
here ok.
What I want to find out now is what are the
values of beta and what are the corresponding
E and H, they define the modes. And we were
doing the analysis of this; we were doing
the analysis for symmetric step index planar
waveguide where this n of x is defined like
this. And this waveguide is symmetric about
x is equal to 0, it has high refractive index
region n1 and low refractive index regions
n2 and the width of the high index region
is t.
So, what we did we categorize the modes of
this waveguide using the symmetry of the structure
into symmetric and antisymmetric modes and
for symmetric modes.
The field the electric field we were doing
the analysis of TE modes. So, for TE modes
the field is given by this, A cosine kappa
x for mod x less than d by 2 that is in the
guiding film. And Ce to the power minus gamma
mod x for mod x greater than d by 2, which
is in these 2 regions since it is TE mode.
So, the non-vanishing components of E and
H are Ey, Hx and Hz and Ey Hx and Hz are related
by these 3 equations. So, we apply the boundary
conditions at the interfaces x is equal to
plus minus d by 2, and the boundary conditions
are the tangential components of E and H should
be continuous at x is equal to plus minus
d by 2, and the tangential components at x
is equal to constant are Ey and Hz.
So, and since Hz is related to Ey if some
constant dEy over dx that is why Ey and dEy
over dx should be continuous at x is equal
to plus minus d by 2. So, when we applied
this then we got the transcendental equation
kappa d by 2 tan kappa d by 2 is equal to
gamma d by 2. Now our task is now our task
is to solve this equation, let us see how
we can solve this equation and get physical
insight into the modes of a waveguide.
So, to do that what we do when we define we
what we do we take this transcendental equation
kappa d by 2 tan kappa d by 2 is equal to
gamma d by 2. And here we see there are 2
terms appearing which contain kappa d by 2
and gamma d by 2 where kappa square is given
by this and gamma square is given by this.
So, if I add these 2 up kappa d by 2 square
plus gamma d by 2 square then I get k naught
square n1 square minus n2 square times d by
2 square. What does it have? It has all the
waveguide parameters and the wavelength because
k naught is 2 pi over lambda naught. So, so
by adding these 2 up I get something which
contains only the waveguide parameter waveguide
parameters, and the wavelength which are the
inputs basically.
So, let me define this since everywhere there
is a square. So, let me define this as V square
and where V is k naught d by 2 square root
of n1 square minus n2 square or 2 pi over
lambda naught times d by 2 square root of
n1 square minus n2 square. And this I call
normalized frequency, because it contains
lambda in the denominator. So, I call this
normalized frequency or generally I also call
it V parameter. So now, in order to in order
to solve the transcendental equation graphically
what I do I define this kappa d by 2 this
kappa d by 2 as some parameters psi, some
variable psi.
Because this is the transcendental equation
in beta kappa contains beta the only unknown
here is beta. So, let me define kappa d by
2 is equal to psi and therefore, if I put
this kappa d by 2 is equal to psi here then
gamma d by 2 automatically becomes a square
root of V square minus psi square. So, if
I put these 2 here in this equation the then
the transcendental equation becomes psi tan
psi is equal to V square minus psi squares
square root.
So, in this equation I have all the input
parameters in the form of V. So, if I have
all the waveguide parameters with me that
is n1, n2 and d. And I have the wavelength
at which I am operating my waveguide then
I can calculate V from here and for that value
of V. I can solve this transcendental equation
and get the values of psi and hence the values
of beta. Similarly, if I do the analysis for
antisymmetric modes in the same way then it
leads to the transcendental equation which
is given by minus psi cot psi is equal to
square root of V square minus psi square.
So, the guided modes of a planar waveguide
where a symmetric or antisymmetric there it
fall in the range of an effective line from
n2 to n1, and they are given by Ey of x is
equal to A cosine kappa x for mod x less than
d by 2 and Ce to the power minus gamma x for
mod x greater than d by 2 for symmetric modes
this is the symmetric solution. And the values
of beta which are allowed are given by psi
tan psi is equal to square root of V square
minus psi square.
So, these are symmetric modes, and similarly
the antisymmetric modes satisfy the equation
minus psi cot psi is equal to square root
of V square minus psi square and the model
field is given by in terms of sin function
sin kappa x in the guiding film and again
exponentially decaying function in the in
the n2 region or lower index surrounding.
So, by solving these equation first I need
to get the values of beta, only certain values
of beta or certain values of psi are allowed,
I first need to find out those values of beta
and from those values of beta I can find out
kappa and gamma those values I put here in
the field and then I get the model field that
is how I will get the complete solution for
the modes of planar dielectric waveguide.
So, let us solve it. So, I have again symmetric
modes antisymmetric modes in order to solve
these 2 equations what I do I equate both
the LHS and RHS of these equations to some
variable eta. So, in this way I will have
sets of 2 simultaneous equations for symmetric
modes as well as for antisymmetric modes,
and there are eta is equal to psi tan psi
from here and if I equate this to eta then
it gives me psi square plus eta square is
equal to V square. These 2 equations are for
symmetric modes and similarly for antisymmetric
modes I get the equations eta is equal to
minus psi cot psi and psi square plus eta
square is equal to V square.
Now, I can obtain the solutions of these equations
by graphical method. So, what I need to do
I just plot these 2 equations on psi eta plane
and look for the points of intersection. Those
point of intersection will give me the allowed
values of psi and hence the allowed values
of beta ok.
So, let us plot these. So, here the red line
shows the equation eta is equal to psi tan
psi, which corresponds to symmetric modes.
And these blue lines they show the plot for
eta is equal to minus psi cot psi they correspond
to antisymmetric modes. And these circles
for different values of V are basically psi
square plus eta square is equal to V square.
So, for a given V for a given value of V,
I take the circles. And see the points of
intersections of these circles with these
blue and red curves and they will give me
the allowed values of psi. So, let us examine
these. So, V is this. So, for a given waveguide
that is n1 n2 and d and given wavelength that
is lambda or lambda naught, V is known. If
this V lies between 0 and pi by 2 if this
V lies between 0 and pi by 2 we can see that
there would only be one point of intersection.
And that would be with the red curve that
is corresponding to symmetric modes.
So, this point of intersection will give me
the value of psi for this symmetric mode.
So, in this range if V lies in the range 0
to pi by 2, then there is only one mode, which
is possible and this mode is symmetric mode.
Because it is obtained from the intersection
of red curve with the circle and red curve
corresponds to symmetric modes. Now if V lies
between pi by 2 and pi for example, if V is
equal to 2.5. I see 2 points of intersection.
One is with red and one is with blue. So,
one is with symmetric and another is with
antisymmetric. So, I get 2 modes one symmetric
and one antisymmetric. Then if I look in the
range from pi to 3 pi by 2 if V lies from
pi to 3 pi by 2 for example, V is equal to
4 then I get 2 symmetric modes and one antisymmetric
mode ok.
So, in this wave for any given value of V.
I can find out the number of modes the number
of symmetric modes the number of antisymmetric
modes and their propagation constants.
So, in general the number of guided modes
I can find by looking at the value of V if
V lies between m pi and m plus half pi where
m is an integer 0, 1, 2 and so on, then I
will get m plus 1 symmetric modes and m antisymmetric
modes. On the other hand if V lies between
m plus half pi and m plus 1 pi then I will
get m plus 1 symmetric modes and m plus 1
antisymmetric modes.
In general, I can get the modes for any value
of V and I labeled these modes as TEm mode
where m is equal to 0 1 2 and so on. So, the
very first mode is T E 0 mode than there is
TE1, TE2 TE3 and so on. So, that is how the
modes are labelled. Now the number of modes
I can also find out by looking them at V line.
So, if I have V line and I divided into the
sections of pi by 2, then what I see that
from 0 to pi by 2 there would be TE0 mode
if V lies somewhere here than TE0 and TE1
if V lies somewhere here than TE0, TE1, TE2
and so on.
So, every pi by 2 I am adding one mode. So,
in general if I know the value of V then the
total number of modes can be found out by
dividing this V by pi by 2. So, I divide this
V by pi by 2 and get a number then the number
of modes would be an integer closest to, but
greater than that number closest to, but greater
than that number. So, if this V over pi by
2 comes out to be 2.1, then the number of
modes would be 3. Now what is the single mode
waveguide?
If V lies between 0 and pi by 2 that is V
is less than pi by 2, then there is only one
mode supported that is TE0 mode and this TE0
mode is the fundamental mode of the waveguide,
I give you an example, if n1 is equal to 1.5,
n2 is equal to 1.48 and d is equal to 3, then
this waveguide can be single moded or multi-moded
it depends upon the wavelength I am using
because V contains wavelength also apart from
these 3 parameters.
So, so by just looking at waveguide parameters
I cannot say whether the waveguide a single
moded or multi-moded I will have to say it
is single moded at this wavelength or it is
single moded in this range of wavelengths.
So, I can find out that range. I know that
V is equal to 2 pi over lambda naught times
t by 2 times square root of n1 square minus
n2 square and if V is less than pi by 2 then
the waveguide a single moded than this gives
me that for these parameters of waveguide
if lambda naught is greater than 1.46 micrometer
then the waveguide is single moded. That is
for all the wavelengths longer than 1.46 micrometer
the waveguide is single moded and for all
the wavelengths which are smaller than 1.46
micrometer the waveguide is multi-moded.
For all the wavelength shorter than 1.46 micrometer
the waveguide is multi-moded. Let us do few
more examples. So, let us consider a dielectric
step index symmetric planar waveguide, with
n1 is equal to 1.5 n2 is equal to 1.46 and
d is equal to 4 micrometer. For this waveguide
let us calculate the number of symmetric and
antisymmetric modes at wavelength lambda naught
is equal to 1 micrometer.
First part and second part the waveguide thickness
for waveguide to be single moded at lambda
naught is equal to 1 micrometer.
Well, so let us first solve the first part,
I know for the number of modes what I should
do? I should find out the value of V and divided
by pi by 2. So, I should calculate V over
pi by 2. V over pi by 2 comes out to be 2
d over lambda naught times square root of
n1 square minus n2 square, because we know
V is equal to V is equal to 2 pi over lambda
naught timesd by 2 times square root of n1
square minus n2 square. So now, I put the
value of d lambda naught n1 n2 here and I
find out that this V over pi by 2 comes out
to be 2.7527.
Which means that the number of modes should
be an integer closest to, but greater than
this and there it is why the number of modes
would be 3. If there are 3 modes what modes
are supported? Well we start from 0. So, TE0
then TE1 and TE2; TE0, TE1 and TE2 and we
see that all the even number modes are symmetric
modes and odd number modes are antisymmetric
modes that is how we got the solutions also.
So, so here we have 2 even number modes and
one odd number modes. So, so we have 2 symmetric
modes and one antisymmetric mode.
Let us do the second part, waveguide thickness
for waveguide to be single moded. We had seen
that that if the waveguide thickness is 4
micrometer, then at lambda is equal to 1 micrometer
the waveguide supported 3 modes. Now if I
still want to operate at lambda naught is
equal to 1 micrometer and want my waveguide
to be single moded, then what should be the
value of d? That is what I need to find out.
So, I know again for single mode waveguide
V should be less than pi by 2, which means
2 pi over lambda naught times d by 2 times
square root of n1 square minus n2 square should
be less than pi by 2 or d should be less than
lambda naught divided by 2 square root of
n1 square minus n2 square.
Now, I just put the values of lambda naught
and n1 n2 it gives me for all the values of
d is smaller than 1.5431 micrometer this waveguide
would be single moded at lambda naught is
equal to 1 micrometer. I can also find out
the number of modes with wavelength. So, if
I have a given waveguide let us say defined
by n1 is equal to 1.5 n2 is equal to 1.46
and d is equal to 4 micrometer.
Then, if I change the wavelength the number
of modes will change. If I start if I start
from a wavelength which is say 2 micrometer
then there would be only one mode supported,
let us say and that would be TE0 mode. If
I decrease the wavelength now gradually then
up to about little less than 1.4 micrometer
there would be only one mode supported, that
is TE0 mode.
As soon as I cross this then TE1 mode will
also come into the picture. So, here I will
have TE0 and TE1 up to this point let us this
is about 0.9 micrometer as I further decrease
the wavelength below 0.9 micrometer, then
TE2 mode will also come into picture now here
you will have TE0, TE1 and TE2 mode up to
about let us say it is little less than 0.7
micrometer. And then we I will have TE3 mode
and so on. So, so as I decrease the wavelength
as I decrease the wavelength of operation
in a given waveguide the number of modes will
change.
So, I go to shorter wavelengths I increase
the number of modes. So, this is how I can
find out how many modes would be supported
and how the number of modes will change with
wavelength, how the number of modes will change
with waveguide parameters. What is still remains
to find out is the values of beta, the values
of propagation constant, the values of the
model fields the expression for model field
the field plots. So, for that what I need
to do now I have already got the value of
beta, but approximate value of beta. This
approximate value of beta I can refine by
using numerical techniques. I can solve the
transcendental equation by various numerical
techniques and I can find out the much more
accurate value of beta and from those values
of beta I can find out the model field ok.
So, in the next lecture we would look into
the model fields and before that. In fact,
I would also like to define these transcendental
equations in terms of normalized parameters.
We have already defined one normalized parameter
which is normalized frequency V. I would like
to also define a normalized parameter with
respect to propagation constant. So, normalized
propagation constant and the idea of defining
these normalized parameters is to have certain
universal curves certain universal solutions
which do not depend upon waveguide parameters
or the wavelength. So, in the next lecture
we will do extend all this analysis by using
the normalized parameters and then we will
look into the model fields also.
Thank you.
