In the last lecture we had seen that if for
a planar waveguide V is less than pi by 2,
then the number of modes is one. It supports
only one mode and when V lies between pi by
2 and pi there are 2 modes and if it is between
pi and 3 pi by 2, then there are 3 modes and
so on. If I keep on increasing the value of
V then the number of modes would increase,
but I also see that the propagation constants
can be approximated by certain expression,
and we can find out the propagation constants
even without solving the transcendental equation
for first few modes. So, let us see how.
So, if it is a highly multimode waveguide
where V is much larger than 1 typically V
is more than 10 or so then What I see if I
now again plot these transcendental equations
this is this is eta is equal to psi tan psi
and the blue one is eta is equal to minus
psi cot psi.
Now if I plot the sections of circles for
large values of V let us say V is equal to
10 or V is equal to 18 or 20, what do I see
there for the first few modes the points of
cross sections are somewhere here.
So, for first mode that is m is equal to 0
mode the point of intersection is close to
pi by 2. And for m is equal to 1 the point
of intersection is close to pi and so on.
So, and these are only for first few modes,
if you go to higher order modes than these
points of intersections are much further from
these values. So, what I can say that for
first few modes, the points of intersections
can be given psi m is equal to m plus 1 pi
by 2, where m is equal to 0 1 or 2. So, for
m is equal to 0 it is pi by 2 for m is equal
to 1 it is pi and so on.
So, in this case in this case I can find out
the beta in this way, I know psi m is equal
to kappa m d by 2. So, it is equal to m plus
1 pi by 2 and kappa m is k naught square n1
square minus beta m square is square root.
So, so this gives me beta m square is equal
to k naught square n1 square minus m plus
1 square pi square by d square. Now I can
increase the value of V by 2 ways I can have
large value of V by 2 ways - one is that if
I have very large value of d. If I have large
value of d then you can see there that is
beta m will approach to k naught n1 beta m
will approach to k naught n1. And it is obvious
it if I start if a start from a very thin
waveguide then there are discrete modes and
they have propagation constants which lie
between k naught n2 and k naught n1.
But when I increase this width of the waveguide
and the width is very large as compared to
lambda then it is as good as infinitely extended
medium. So, so the light will propagate with
propagation constant k naught n1 where n1
is the refractive index of the medium. So,
I will approach towards infinitely extended
medium if d is very large. Now if I obtain
the large value of V by having very large
index contrast, that is very large difference
between n1 and n2, then what happens is there
is a huge index contrast and this very large
index contrast will give very high reflection
a very high reflection ok.
So, it is it would be difficult for the wave
to penetrate into n2 region. So, it would
be something like it would be very close to
what is planar mirror waveguide and in that
case you can see the expression for these
beta closely resemble to the expression of
beta for planar mirror waveguide. Let us work
out some more examples here.
I consider a step index planar symmetric waveguide
with n1 is equal to 1.5, n2 is equal to 1.4
and d is equal to 4 micro meter. And let us
calculate the number of modes at 0.5 micrometer
and effective refractive indices or effective
index of modes of TE0 and TE1 modes.
So, again to find out the number of modes
it is the same I find out the value of V.
So, when I find out the value of V. It comes
out to be 13.53 at lambda naught is equal
to 0.5 micrometer if I divide this value of
V by pi by 2. This number comes out to be
8.61.
So, the number of modes supported would be
9. I can see this value of V is very large.
So, it would be now easier for me to find
out the propagation constants of TE0 and TE1
mode, even without solving the transcendental
equation.
Since it is much larger than 1. So, for the
first few modes the values of beta would be
given by this, and for TE0 mode m is equal
to 0 and for TE1 mode m is equal to 1 and
in this way if I find out the values of beta,
then they come out to be 18.83 and 18.78 correspondingly
the effective index of TE0 mode comes out
to be 1.498698, and for TE1 mode 1.494783.
You can see the very close to the refractive
index of the guiding film which is 1.5 ok.
Now, let us do extend all this analysis by
using normalized parameters. So, we had symmetric
modes antisymmetric modes.
And the transcendental equations were in terms
of psi and V. Now I already have one normalized
parameter which is normalized frequency V
which contains all the waveguide parameters
and wavelength. Let me also define normalized
propagation constant, that is let me normalized
beta and how do I normalized beta yet I take
beta over k naught square minus n2 square
divided by n1 square minus n2 square. If I
normalized beta in this way then for guided
modes the value of b lies between 0 and 1.
So, for guided modes the value of b lies between
0 and 1. So, this is now normalized propagation
constant. Now my task is to replace that psi
with b and V. So, how do I do this? So, this
is the transcendental equation for symmetric
mode this is the transcendental equation for
antisymmetric mode. Let me find out how can
I how can I get psi in terms of b and V. So,
psi square is equal to kappa square d square
by 4. So, this is nothing but d square by
4 and kappa square is k naught square n1 square
minus beta square. I do a little mathematical
manipulation I add and subtract k naught square
in 2 square here, and then I regroup these
2 terms and these 2 terms.
So, if I group these 2 terms and these and
also I take this term outside. Why I am doing?
So, because I can immediately identify that
if I club this term with this I get d square.
So, I do that. So, this becomes psi square
is equal to k naught square d square by 4
n1 square minus n2 square times one and from
here I take minus outside and I write it down
as beta over k naught square minus n2 square
divided by n1 square minus n2 squares. So,
this is nothing but b. So, this is b this
is V square.
So, what I have got now from here psi square
is equal to V square and this is 1 minus b.
So, this gives me psi is equal to V times
square root 1 minus b. So, as soon as I get
the psi in terms of V and b I put it in these
equations and I get the transcendental equation
in terms of b and V only.
So, for symmetric mode this transcendental
equation will transform to V times square
root 1 minus b tan V is square root of 1 minus
b is equal to V times square root of b ok.
Because from here V square minus psi square
V square minus psi square would be V times
square root of b. And for antisymmetric mode
this equation will become this. From here
I can find out the cut off condition of a
mode. A mode is said to be cut off when it
is propagation constant approaches k naught
n2. I know that if an effective for guided
modes I know for guided modes an effective
or beta over k naught beta over k naught beta
over k naught or an effective lies between
n2 and n1.
So, if this beta over k naught approaches
n2 then a particular mode is cut off. So,
mode is guided when the when the effective
index of the mode lies between these 2, but
as soon as the effective index goes below
this then it is no more guided. So, this value
of an effective will give me the cutoff. This
beta over k naught is equal to n2 gives me
b is equal to 0. So, this is the cut off condition
in terms of b. So, from here I can immediately
find out for what values of V for what values
of normalized frequency a mode would be cut
off. So, I simply put b is equal to 0 here
then the solutions of the equation that I
get will give me the cutoffs of the modes.
So, this will give me if I put V is equal
to 0 sorry b is equal to 0 and V is equal
to Vc which symbolizes the cutoff frequency
cutoff normalized frequency, then it gives
me Vc tan Vc is equal to 0 for symmetric modes.
And for antisymmetric modes it will be Vc
cot Vc is equal to 0. So, these 2 together
will give me the cutoff of mth mode is m pi
by 2 where m is equal to 0 1 2 and so on.
So, TE0 mode will have no cut off m is equal
to 0, TE1 mode will have a cutoff of pi by
2 TE2 mode will have cut off pi and so on.
So, this is how I get the cutoffs of the modes.
Now very beautiful thing about these normalized
parameters is that they are independent of
waveguide parameters. I get these 2 equations
and these 2 are purely mathematical equations
2, purely mathematical transcendental equations.
What I can do? I can solve these equations
for I can solve these equations for different
values of V. And I find out the roots of these
and plot these roots as a function of V.
So, I start with I start with a very small
value of V. When I do this as long as the
value of V is less than pi by 2, then what
I see that there is no root coming out from
this equation and there is only one root coming
out of this equation. I find out that root
for different values of V and plot it, it
go something like this. As soon as the value
of V crosses pi by 2 1 root starts one root
starts appearing from this equation. And the
value of V up to pi I will get one root from
here, one root from here I keep plotting them.
And then I further increase more and more
roots start appearing and I plot them ok.
The very first root which comes from here
which is closer to which is closer to 1 the
first root which is closer to 1 is TE0 mode.
The first root from here is TE1 mode the second
root from here is TE2 mode and so on. So,
I plot them, once I have plotted them then
these curves are these curves are universal
curves. So, once I have done this exercise
once I have done this exercise then it is
forever these are universal curves, now what
I can do using these curves?
If there is a given waveguide and wavelength
then I simply calculate the value of V this
should be V naught and there should not be
any 0 here. So, so I just calculate the value
of V, and then I go to these curves and read
the value of b corresponding to that value
of b for example, if I get this value of V
let us say V is equal to 1.4 then I draw a
vertical line corresponding to V is equal
to 1.4 and read the value of b corresponding
to this from b-V curves. Once I get the value
of b then I can find out the effective index
of the mode from here, this is coming out
from the very definition of the normalized
propagation constant.
If I am somewhere here, if I am somewhere
here if the value of V is let us say 3.5.
So, it will go like this I will have 3 points
of intersection and corresponding to that
I can read the value of b. So, I will get
the propagation constants of all the 3 modes.
So, in this way I can find out the propagation
constants for any symmetric planar waveguide.
Let us do some examples. So, if I have a waveguide
with n1 is equal to 1.5 n2 is equal to 1.48.
d is equal to 2 micrometer and I operate it
lambda naught is equal to 1.5 micrometer.
If I calculate the value of V from here it
comes out to be 1.0226. I go back to those
b-V curves and read the value of b from here
corresponding to this value of V it comes
out to be 0.4636. And then I find out the
value of an effective, which comes out to
be 1.4893053 ok.
Now, I change my waveguide parameters a little.
So, I now have n1 is equal to 1.5. n2 I have
change to 0.418. D is equal to 1 micrometer
and lambda naught is equal to 1.5. I have
artificially change these parameters for achieving
the same value of V actually. What I want
to demonstrate is that, if my waveguide is
different d one if my waveguide is different,
but it gives the same value of V then the
value of b is always the same. For same value
of V even if the waveguide is different the
value of b normalized propagation constant
is the same, but the effective index would
be different now, because n1 and n2 are different
n1 and n2 are different. So, in this case
the effective indexes this while in this case
the effective index would be this.
So, different waveguides can have same value
of V and hence the same value of b which means
that these b-V curves are universal and they
do not depend upon waveguide parameters. Let
us now look at modal fields, how do the modal
fields look like. Let us plot the field for
TE0 mode TE0 is symmetric mode.
So, the modal field will be given by Ey of
x is equal to A cosine kappa x in the region
mode x less than d by 2, and Ce to the power
minus gamma mod x for mod x greater than d
by 2. I know for TE0 mode psi would lie between
0 and pi by 2, because V would lie between
0 and pi by 2. So, the point of intersection
the value of psi for point of intersection
will always lie between 0 and pi by 2.
It will not exceed pi by 2. What is psi? Psi
is kappa d by 2. So, kappa d by 2 will lie
between 0 and pi by 2 now let us look at the
waveguide. And the modal filed in the region
mode x less than d by 2 the solution is cosine
function. So, I will have cosine kappa x what
would be the value at the boundary? What would
be the value at the boundary? At the boundary
I have kappa x is equal to kappa d by 2 and
minus kappa d by 2, but kappa d by 2 is always
less than pi by 2, if kappa d by 2 is always
less than pi by 2 then cosine kappa d by 2
will never cross any 0. The first 0 of cosine
function will occur at kappa x is equal to
pi by 2, but I have even at the boundary kappa
x is always less than pi by 2. So, I would
not have any 0 crossing.
So, Ey in the guiding film will be like this.
And then in the lower refractive index region
the exponential decaying part exponentially
decaying solution will take over and it will
go like this. So, I will have the modal field
which would look like this all right. What
I see that there is oscillatory solution here
and decaying solution here. The field extends
field extends to lower indexed region also.
How much does the how much does this extend?
Into this region I can quantify this, I find
that at x is equal to d by 2 plus 1 over gamma
because this field is going as e to the power
minus gamma times x. So, at x is equal to
plus x is equal to 1 over gamma this field
will decay to 1 over e of it is value at the
boundary.
So, if I go if I start from here and mover
distance 1 over gamma from the boundary then
the field will decay to 1 over e of it is
value at the boundary. Then this is how I
can quantify this distance up to which this
field extends into the lower index region.
So, it is defined by 1 over gamma, and it
is known as penetration depth into lower index
region. So, 1 over gamma is penetration depth
into lower indexed region let us plot the
field now for TE1 mode. TE1 mode is antisymmetric
mode and the solutions would be given by sin
function in the in the guiding film and exponential
decaying function again in the lower index
medium.
Now for TE1 mode, psi would lie between pi
by 2 and pi psi would lie between pi by 2
and pi, because this the green curve will
cut the blue curve only in this region. Now
if I which means that kappa d by 2 would lie
between pi by 2 and pi, and now in the same
way if I plot the field in the guiding film
first. So, there would be a sin function ok.
But at the boundary it has to be less than
pi. So, kappa d by 2 has to be less than pi.
So, it would not cross another 0 there would
only be 1 0 which is at x is equal to 0. So,
in the guiding film the solution would be
like this and in the lower index region the
exponential decaying term will take over and
this is how the modal field would look like.
So, I can see that if I go back then here
for TE0 mode there was no 0 in the modal field
there was no 0 crossing for TE1 mode there
is only one 0 crossing.
And if I go on then TE2 mode has two 0 crossings
TE3 mode has three 0 crossings which means
that if it is TEm mode then it will have m
number of 0s ok. It will have m 0 crossings,
this is one thing another thing that I see
is when I solve this, that with the number
of modes the propagation constant decreases
right, if you remember the b-V curves.
So, you have this is V which is normalized
frequency this is b which is normalized propagation
constant, and b always lies between 0 and
one for guided modes. I had seen that TE0
mode goes something like this. This is TE0
mode TE1 mode it goes something like this,
this is TE1; TE2 mode like this. So, if I
have a value of V somewhere here, then the
propagation constant of TE0 mode is here,
propagation constant of TE1 mode is here,
propagation constant of TE2 mode is here.
So, the propagation constant decreases as
the ode number increases.
I also see that the lower order modes are
more confined into guiding film. So, you can
see that TE0 mode it has more power here.
And as I and the and the tail into n2 region
is a smaller, but as I go to higher order
modes this tail extends more and more. So,
if the number of mode increases if the number
of mode increases then the corresponding confinement
in the mode decreases.
I can see the effect of this penetration depth
I can see this penetration depth. So, for
a given waveguide if beta decreases because
penetration depth is defined by 1 over gamma
and gamma is equal to beta square minus k
naught square n2 square it is a square root.
So, if beta decreases then, then if beta decreases
then gamma decreases which means penetration
depth increase and beta decreases with mode
number. So, that is why the penetration depth
increases. So, if I have this waveguide and
this wavelength and I see the mode profiles
of different modes. So, this is TE0 mode it
has penetration depth which is about 0.47
micrometer for TE1 mode this is 0.52 micrometer
and for TE2 mode it is 0.65 micrometer.
What is the effect of index contrast on penetration
depth? Well again, the penetration depth is
given by this, so for a given n1d and lambda
naught. Now if I change the value of n2 and
hence the index contrast, for example, if
I decrease n2 if I decrease n2 then the value
of V increases. The value of V increases because
n1 square minus n2 square increases. If V
increases the propagation constant if V increases
the propagation constant increases, beta increases
if beta increases gamma increases and penetration
depth decreases.
So, this is how it looks like. So, if I have
n2 is equal to 1.48. Then penetration depth
is 0.47 micrometer. For TE0 mode for the same
mode if I change the refractive index n2 from
one point 48 to 1.3 then you can see the penetration
depth decreases the field is pushed more towards
the guiding film. And if I change to one then
it is even more confined into the high index
region and penetration depth decreases to
0.1.
What is the effect of wavelength if for a
given waveguide if I decrease the wavelength?
If I decrease the wavelength then I am increasing
the value of V which means I am increasing
the value of propagation constant ok.
If I am increasing the value of b that is
beta square minus k naught square n2 square
increases gamma increases. So, penetration
depth decreases. So, this is the field of
a given waveguide when operated at 1.55 micrometer
you can see how much the field extends outside
and when I decrease the wavelength to 1.3
the field is pushed more towards n1 region
and when it is 0.7 micrometer than the penetration
depth is much smaller.
So, this is the; this is how this is how the
modal field would vary with different parameters
of the waveguide. And wavelength in the next
lecture we will have more inside into the
modes and we will understand what these modes
basically are. We will have physical understanding
of modes.
Thank you.
