Till now we have carried out the analysis
of symmetric planar wave guides where the
high index region is sandwiched between two
lower index regions of same refractive index.
Now in this lecture we will extend the analysis
to asymmetric planar waveguide asymmetric
planar waveguides are more practical wave
guides. So, we will extend this analysis to
asymmetric planar waveguide and see how the
waveguide hence is done in these kind of waveguides.
So, what is an asymmetric planar waveguide
you for example, take a glass of substrate
of refractive index ns, and you deposit a
film of refractive index nf of thickness d,
this film can be for example, of a polymer
material, the refractive index nf is higher
than the refractive index ns and then it is
surrounded by air. So, air works as cover;
also you can put a cover of polymer material
itself of refractive index nc, where nc is
smaller than ns and ns is smaller than nf.
So, this kind of waveguide we can see that
is has asymmetric structure, because ns is
not equal to nc here. If I look at the refractive
index profile of this waveguide, then this
is the substrate an example is glass, this
is the guiding film the example is polymer,
and this is the cover region which can be
air or another polymer of lower refractive
index. Here at x = 0, I have made the interface
between the film and the cover, and at x = -d
represents the interface between the substrate
and the film.
Since the refractive indices of the individual
layers are uniform. So, this waveguide is
step index waveguide. I can also have a waveguide
in which the refractive index in the guiding
film is not uniform, but it varies with x.
So, how I can make this waveguide well I take
a substrate of material glass, and then I
diffuse certain ions into this pay over a
certain distance. So, I will have a diffusion
profile for example, I can diffuse silver
ions into glass and that will change the refractive
index of the material near the surface.
So, now this nf would be a function of x now,
and then I can have cover as air or cover
of a polymer layer. So, now, this nc is smaller
than ns, and then the refractive index of
the film at the surface of the film nf at
f = 0. So, ns is smaller than that. If I look
at the refractive index profile of this structure
now, so, I have a substrate and this is the
guiding film. This guiding film now has refractive
index with varies which varies with x and
ultimately it merges into the refractive index
of glass.
So, this is the surface index nf(0), and this
nf(0)is larger than ns and which is larger
than nc.
So, this is a typical graded index ion exchange
waveguide. Let us do the modal analysis of
this step index waveguide, the step index
planar asymmetric waveguide. So, this is the
refractive index profile and I can write down
the refractive index in various regions as
nx = nc, which is defined by the region x
> 0 which is the cover, then in the region
minus –d < x <0 the refractive index is
nf which is the film region, and for x < -d
which represents the substrate region, the
refractive index is ns.
Now, the modes of this waveguide are here
guided modes, whose refractive index is lie
between ns and nf. So, these are some representative
guided modes, and for, the field would radiate
out in the substrate region and forthe field
would radiate out in the cover region as well.
So, in this region you will have radiation
modes.
We are interested in the guided modes of the
structure, which are defined in this range,
and let us first do the TE mode analysis of
the waveguide, and as I can see that the refractive
index variation is in x-direction, and I am
considering propagation in z-direction. So,
the non- vanishing components corresponding
to TE modes are Ey, Hx and Hz.
I know from my previous analysis that the
wave equation satisfied by TE modes of such
a waveguide is given by . So, the procedure
is again the same I write down this equation
in all the three regions, I write down the
solutions of these three equations and then
apply boundary conditions to match the solutions.
So, this is the equation in cover region x
= 0, which 
is , because in this region nx = nc, and I
have taken minus sign outside and represented
it in this form because , so that this quantity
is positive for guided modes.
In the film region which is defined by minus
-d < x < 0, the equation becomes, again for
guided modes this quantity in the square brackets
is positive. In the substrate region defined
by x < -d, the equation is 
and again for guided modes this quantity is
positive.
So, now I can represent this as , this as
and this as and then find out what are the
solutions of these three equations.
So, these are the three equations now, where
, , and .
So, what are the solutions? The solution of
this differential equation is. And as you
know that we are interested in guided modes.
So, we cannot have exponentially amplifying
solution. So, I am only considering the exponentially
decaying solutions in cover regions and substrate
region.
Now, in the film this equation will give you
the oscillatory solutions, which I write as
, 
and in the substrate region it is because
x < -d. So, it is negative. So, this again
gives you exponentially decaying solutions.
Now a b c and d are constants and these constants
are determined by the boundary conditions.
So, let us apply boundary conditions and obtain
relationships between A, B, C and D obtain
the Eigen value equation or characteristic
equation which is satisfied by the propagation
constants of the guided modes.
So, these are the solutions in different regions
in substrate in the film and in the cover
and what are the boundary conditions? Boundary
conditions are again the tangential components
of E and H are continuous, and the tangential
components here are Ey and Hz. So, they give
me that Ey andbecause Hz is related to Ey
as. So, they are continuous at the interfaces
x = 0 and at x = -t.
So, let us apply these boundary conditions.
So, if I look at the continuity ofat. So,
I apply it here. So, if I approach from the
film and if I approach from the cover towards
x = 0 is equal to 0. So, I will get B + C
= A. Now continuity of at x = -d which is
the substrate film interface, this gives me
.
Now, let 
us see the continuity of the derivatives at
x = 0 and at x = d. So, at x = 0 gives meand
continuity ofat x = -d interface gives me
.
So, I 
have now four equations relating these four
constants A, B, C and D and by mathematical
manipulation of these four equations. I can
do two things one is obtain B, C and D in
terms of A, and then eliminating A, B, C,
D all together and form a transcendental equation
inwhich is also known as Eigen value equation
or characteristic equation. So, that equation
comes out to be:.
So, this is the equation which is satisfied
bybecause the only unknown here is, appears
in and . So, if I solve this equation and
find out the roots of this equation then I
can know what are the propagation constants
of the modes, what are the modes, which are
supported by this asymmetric step index planar
waveguide.
As we have done earlier also that it is always
a good idea to obtain everything in normalized
parameters. So, that we are more or less independent
of waveguide parameters, here we cannot have
complete independence from waveguide parameters,
but to certain extent it is still we can draw
some universal curve. So, how do we define
normalized parameters here? First is normalized
frequency V.
If you remember that in symmetric planar waveguide
we had only 2 refractive indicesand , . So,
there we had defined it in terms of .
Now, now in place of, I have, but low refractive
indices are now up toand which one we should
take in order to define V. So, it is very
simple to choose becausedefines the cut-off
of guided modes. So, I should choosebecause
as soon as the effective index of a mode falls
belowit is no more guided. So, I chooseto
define the normalized frequency v here, and
similarly for defining the normalized propagation
constant b I choose.
So, b is defined as . There is another parameter
here in this structure, which tells you how
asymmetric this waveguide is. The asymmetry
in the structure is introduced by the different
values ofand. So, what is the difference between
and? So, let us define by asymmetry parameter
a, which is given as.
So, now I have here three normalized parameters
in place, and now I can represent my transcendental
equation or Eigen value equation in terms
of these normalized parameters. If you remember
that in the transcendental equation the three
terms appear which are, and , which contain.
So, now I will have to represent these , and
in terms of V, b and a. And we can see that
. 
and .
So, keeping these in mind and looking at the
expressions here I would now try to obtain
, and in terms of V, b and a. So, let us first
try for, and I can see that if I do 1- b then
I can obtain this in the numerator of this.
So, b is given by this. So, if I do 1- b it
becomes, and .
So, if I multiple here by d /2 and then I
can havehere or I havedirectly here and this
is nothing, but V2d/2. So, this becomes . So,
this gives me.
If I look atgamma s square, . So, it is directly
appearing herethis thing. So, this will immediately
give me . What about now. So, I can see from
here that if I do b + a then I can only obtainin
the numerator. So, I do b + a. So, it becomesdivided
by this. So, which isdivided byand this gives
me .
So, in this way I have now obtained, and in
terms of V, b and a. So, I can now represent
my eigen value equation in normalized parameters.
So, this equation simply becomes: .
And of course, since and for cut offguided
mode,lies between ns and nf. So, b would lie
as usual between 0 and 1. So, I solve this
equation for a given value of V, V is if you
remember. So, this V contains all the waveguide
parameters and the wavelength. So, for a given
waveguide and wavelength I have V and for
that value of V I can solve this equation
to obtain the normalized propagation constant
b.
What are the cut offs? Cut offs are defined
byor b = 0. So, if I put it there then the
cut off equation becomes or cut off for TE
mode mth TE mode is now given by . You can
see here that the cut off of even TE0 mode
is finite.
So, in a particular range of V which is defined
by V even the TE0 mode is not guided.
This is the difference as compared to the
symmetric planar waveguide. In symmetric planar
waveguide TE0 mode has 0 cut off. So, TE0
mode was always guided, but here it is not
the case. Now let us solve this equation the
characteristic equation or Eigen value equation
for different values of V, and plot the roots
as obtained as a function of V. So, here I
have plotted the roots for both the cases
a = 0 which represents the symmetric planar
waveguide, and then for a very high value
of a which is 40 for asymmetric planar waveguide.
These solid lines are corresponding to symmetric
planar waveguide a = 0, and these dash lines
are corresponding to asymmetric planar waveguide
And what I can see that TE0 mode has cut off
here, TE1 mode has cut off here, TE2 mode
has cut off here. In case of symmetric planar
waveguide this was the cut off was the corresponding
cut off. Now all the cut offs have been shown
shifted by this much amount which is . So,
this is one thing another thing that I see
here is that if I take a particular value
of V then the propagation constant of the
mode is now smaller than the propagation constant
of the corresponding symmetric waveguide mode.
And it is it is understandable that because
asymmetric is introduced by introducing the
cover region. So, if this is the symmetric
waveguide, this is the symmetric waveguide.
So, this is ns this is nf this is again the
level ns, but now in asymmetric planar waveguide
I have nc. So, I am reducing the refractive
index in the cover region. So, the effect
is to pull down the effective indices of the
modes towards lower side so that we can see
from here itself that the propagation constants
of asymmetric planar waveguides are smaller
than those corresponding to symmetric planar
waveguide.
Let us look at modal fields. So, these are
the modal fields of TE0, TE1 and TE2 modes
of a typical asymmetric planar waveguide.
So, I can see that the modal fields are asymmetric
and you can see that the penetration gap is
more in the substrate and less in the cover
which is understandable, because here at this
interface the index contrast is smaller as
compared to the index contrast at this interface
ok.
So, the field extends more into the substrate
region as compared to in the cover region.
How many modes are supported? If you go back
to symmetric planar waveguide the number of
modes are the integer which is closest to,
but greater than , but now all the cut offs
are shifted by this much amount. So, the number
of modes would now be. So, you obtain this
number and find out the integer closest to,
but greater than this number.
What are the cut off wavelengths of various
modes and if I change the wavelength how the
number of modes would change. So, I can see
from here that the cut off wavelength for
mth TE mode would be given bywhich comes directly
from the definition of normalized frequency
V. So, I can write it down also which we will
quite often use.
. So, from here I get these cut off wavelengths,
now if I find out the cut off wavelengths
corresponding to various modes then I see
that for TE0 mode the cut off wavelength is
1.7187 for these parameters of waveguide,
and for TE1 mode it is 0.4914 and for TE2
mode it is 0.2867.
So, if I start from a longer wavelength let
us say 2, then until I cross this TE0 mode
is not guided. So, from here to here there
is no mode guided by the structure, and as
soon as I cross this go below this wavelength,
then TE0 mode starts appearing and TE0 mode
would be guided now for all the wavelengths
is smaller than this.
If I further reduce the wavelength and as
soon as I go below 0.49TE1 mode starts appearing
and below these this wavelength I will have
both TE0 and TE1 and so on. So, this is how
the waveguide would guide different modes
if I change the wavelength. How the thickness
affects the guided modes. So, from here I
can find out the cut off thickness of mth
TE mode from here itself. So, if I find out
d in terms of V now and put the cut off of
various modes in terms of V, then I can find
out the cut off of thickness ok.
So, if I have 0 thickness it means no waveguide
no mode. If I start increasing the thickness
then up to 0.58 there is no mode guided because
TE0 mode has finite cut off and as soon as
I cross this then TE0 mode starts appearing,
and when I cross 2.03then TE1 mode starts
appearing and so on. So, as I increase the
waveguide thickness the number of modes start
increasing I should have the label here which
is d in. So, the label here is t in micron.
Let us work out few examples. So, I take a
dielectric step index asymmetric planar waveguide.
Defined by nf =1.5 and ns =1.48 and nc = 1,
and d = 4. Now the first thing is to calculate
the number of modes at lambda naught is equal
to 0.5. So, I first calculate the value of
asymmetric parameter a which comes out to
be about 20, then I find out what is the value
of V at = 0.5, and this comes out to be about
6.13.
Then I find out this number. So, this comes
out to be 3.4. So, the number of modes are
4.
The second is wavelength range in which the
waveguide does not support anymore. So, I
know there would not be any more support it
if , which if I put the value V here the expression
for V here. Then this gives me the condition
in terms ofasshould be greater than this,
and if I plug in all these numbers I find
out for > 4.5427 the waveguide would not support
anymore.
Third is what 
is the cut off wavelength of TE2 mode? So,
I find out what is the V value for cut off
of TE2 mode. So, which is given by this with
m/2. So, if I put m /2 then Vc is equal to
this, and the cut off Vc for cut off V for
TE2 mode is this if I translate it to the
wavelength it comes out to be 0.8038.
What is the range of 
d so, that only TE0 and TE1 modes are guided
at.
So, I know that for mth mode the cut off 
thickness is this, and I want TE0 and TE1
both the modes guided. So, I find out this
dc for both the modes, and I also know thisfor
mth mode is given by this. So, I find outfor
d0 mode, Vc is this for TE1 mode Vc is 
this correspondingly if I now find out dc
for these 2 modes, then for TE0 mode cut off
thickness is 0.88, and for this is 2.93. So,
if TE0 and TE1 modes are 
to be 
guided 
then thickness should be 
between 
this 
and this.
So, this is all in this lecture in the next
lecture, we will extend the analysis to TM
modes.
Thank you.
