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 in these kind of
waveguides
so what is an asymmetric planar waveguide
you for example take a glass of substrate
of refractive index n s and you deposit a
film of refractive index n f of thickness
d this film can be for example of a polymer
material the refractive index n f is higher
than the refractive index n s 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 n c where n c is
smaller than n s and n s is smaller than n
f
so this kind of waveguide we can see that
is has asymmetric structure because n s is
not equal to n c 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 is equal to zero i have made
the interface between between the film and
the and the cover and at x is equal to minus
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 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 of the material ah near the surface
so now this n f would be a function of x now
and then i can have cover as air or cover
of a polymer layer so now this n c is smaller
than n s and then the refractive index of
the film at the surface of the film is n f
at f is equal to zero so n s 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 ah with x and ultimately
it merges into into the refractive index of
glass
so this is the surface index n f zero and
this n f zero is larger than n s and which
is larger than n c 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 n x is equal to n c which is defined
by the region x greater than zero which is
the cover then in the region minus d less
than x less than zero the refractive index
is n f which is the film region and for x
smaller than minus d which represents the
substrate region the refractive index is n
s
now the modes of this waveguide are here guided
modes whose refractive index is lie between
n s and n f so these are some representative
guided modes and for beta over k dot smaller
than n s the the field would radiate out in
in the substrate region and for beta over
k naught is smaller than n c the field would
radiate out in the cover region as well so
so in 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 t e 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 t e modes are e y h x and h z
i know from my previous analysis that that
the wave equation satisfied by t e modes of
such a waveguide is given by d two e y over
d x square plus k naught square n square of
x minus beta square e y is equal to zero 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
greater than zero which is d two e y over
d x square minus beta square minus k naught
square n c square e y is equal to zero because
in this region n x is equal to n c and i have
taken minus sign outside and represented it
in this form because because beta over k naught
is greater than n c so that this quantity
is positive for guided modes
in the film region which is defined by minus
d is less than x is less than zero the equation
becomes d two e y over d x square plus k naught
square n f square minus beta square e y is
equal to zero again for guided modes this
quantity in the square brackets is positive
in the substrate region defined by x less
than minus d the equation is d two e y over
d x square minus beta square minus k naught
square n s square e y is equal to zero and
again for guided modes this quantity is positive
so now i can represent this as gamma c square
this as kappa f square and this as gamma s
square and then find out what are the solutions
of of these three equations so these are the
three equations now where gamma c square is
equal to beta square minus k naught square
n c square kappa f square is k naught square
n f square minus beta square and gamma s square
is beta square minus k naught square n s square
so what are the solutions the solution of
this differential equation is a e to the power
minus gamma c x and e to the power plus gamma
c x and as you know that we are 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
b e to the power i kappa f x plus c e to the
power minus i kappa f x and in the substrate
region it is d e to the power gamma s x because
x is a smaller than minus d so it is negative
so this again gives you exponentially decaying
solutions now now a b c and d are constants
and these constants are determined by the
boundary conditions
so let us ah apply boundary conditions and
obtain relationships between a b c d and 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 of
e and h are continuous and the tangential
components here are e y and h z so they give
me that e y and d e y over d x because h z
is related to e y as d e y over d x so they
are continuous at the interfaces x is equal
to zero and at x is equal to minus t
so let us apply these boundary conditions
so if i look at the continuity of e y at x
is equal to zero so i i apply it here so if
i approach from the film and if i approach
from from the cover towards x is equal to
zero so i will get b plus c is equal to a
now continuity of e y at x is equal to minus
d which is the substrate film interface this
gives me b e to the power x is equal to minus
d so b e to the power minus i kappa f d plus
c e to the power i kappa f d is equal to d
d e to the power minus gamma s d
now let us see the continuity of the derivatives
at x is equal to zero and at x is equal to
minus t so e y prime at x is equal to zero
gives me i kappa f b minus i kappa f c is
equal to minus gamma c a and and continuity
of e y prime at x is equal to minus d interface
gives me i kappa f b e to the power minus
i kappa f d minus i kappa f c e to the power
i kappa f d is equal to gamma s d e to the
power minus gamma s d
so i have now four equations relating these
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 in beta
which is also known as eigen value equation
or characteristic equation so that equation
comes out to be tan kappa f d is equal to
gamma s over kappa f plus gamma c over kappa
f divided by one minus gamma c over kappa
f times gamma c over kappa f
so this is the equation which is satisfied
by beta because the only unknown here is beta
beta appears in kappa f gamma s and gamma
c so if i solve this equation 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 to obtain everything in normalized
parameters so so that we are ah 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 ok so how
do we define normalized parameters here first
is normalized frequency v if you remember
that in symmetric planar waveguide we had
only two refractive indices n one and n two
n one is greater than n two so so there we
had defined it in terms of n one square minus
n two square
now now in place of n one i have n f but low
refractive indices are now up to n c and n
s which one we should take in order to define
v so it is very simple to choose because because
n s defines the cutoff of guided modes so
i should choose n s because as as soon as
the effective index of a mode falls below
n s it is no more guided so i choose n s to
define the normalized frequency v here and
similarly for defining the normalized propagation
constant b i choose n s
so b is defined as beta over k naught square
minus n s square over n f square minus n s
square there is another parameter here in
this structure which which tells you how asymmetric
this waveguide is the asymmetry in the structure
is introduced by the different values of n
s and n c so what is the difference between
n s and n c so let us define by ah asymmetry
parameter a which is given as n s square minus
n c square divided by n f square minus n s
square
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 the
three three terms appear which are kappa f
gamma s and gamma c which contain beta so
now i will have to represent these kappa f
gamma s and gamma c in terms of v b and a
and we can see that kappa f square is given
by k naught square n f square minus beta square
gamma s square is given by beta square minus
k naught square and s square and gamma c square
is given by beta square minus k naught square
and c square
so keeping these in mind and looking at the
expressions here i i would now try to obtain
kappa f gamma c and gamma s in terms of v
b and a so let us first try for kappa f kappa
f has k naught square n f square minus beta
square and i can see that if i do one minus
b then i can obtain this k naught square n
f square minus beta square in the in the numerator
of this so so b is given by this so if i do
one minus b it becomes k naught square n f
square minus beta square divided by k naught
square n f square minus n s square and two
pi over lambda naught is k naught
so if i multiple here by d by two and d by
two then i can have kappa f square here or
or i have kappa f square directly here and
this is nothing but v square divided by d
by two square so this becomes kappa f d by
two whole square divided by v square so this
gives me kappa f d by two kappa f d by two
is equal to v times square root of one minus
b
if i look at gamma s square gamma s square
is b dash square minus k naught square n s
square so it is directly appearing here beta
square minus k naught square n s square divided
by k naught square this thing so this will
immediately give me gamma as d by two is equal
to v times square root of b what about now
gamma c square which is defined as beta square
minus k naught square n c square so i can
see from here that if i do b plus a then i
can only obtain gamma c square in the numerator
so i do b plus a so it becomes beta square
minus k naught square n c square divided by
this so which is gamma c square over v square
divided by d by two whole square and this
gives me gamma c d by two is equal to v times
square root of b plus a
so in this way i have now obtained kappa f
gamma c and gamma s in terms of v b and a
so i can now represent my eigen value equation
in normalized parameters so this equation
simply becomes tan two v square root of one
minus b is equal to square root of b divided
by one minus b which is coming from gamma
s term plus b plus a divided by one minus
b square root which is coming from the cover
term gamma c term divided by one minus square
root of b divided by one minus b and the square
root of b plus a divided by one minus b
and of course and of course since since b
is equal to b dash square over k naught square
minus n s square divided by n f square minus
n s square and for cut off beta over k naught
is equal to n s or for guided volts beta over
k naught lies between n s and n f so b would
lie as usual between zero and one so i solve
this equation for a given value of v v is
if if you remember v is defined as two pi
over lambda naught times d by two times square
root of n f square minus n s square 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
by beta is equal to k naught n s or b is equal
to zero so if i put it there then the cut
off equation becomes tan two v c is equal
to square root of a or or cut off for t e
mode m th t e mode is now given by v c m is
equal to m pi by two plus half tan inverse
a square root of a you can see here that the
cut off of even t e zero mode is finite so
in a particular range of v which is defined
by v less than half tan inverse square root
of a even the t e zero mode is not guided
this is the difference ah as compared to the
the symmetric planar waveguide in symmetric
planar waveguide t zero mode has zero cut
off so t zero 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 the roots as obtained as a function of
v so here i have plotted the roots for for
both the cases a is equal to zero which represents
the the symmetric planar waveguide and then
for a very high value of a which is forty
for asymmetric planar waveguide these solid
lines are corresponding to symmetric planar
waveguide a is equal to zero and these dash
lines are corresponding to asymmetric planar
waveguide
and what i can see that t zero mode has cut
off here t one mode has cut off here t two
mode has cut off here in case of symmetric
planar waveguide this was the cut off v c
m is equal to m pi by two was the corresponding
cut off now all the cut offs have been shown
shifted by by this much amount which is half
tan inverse of square root of a so this is
one thing another thing that i see here is
that if i take a particular value of v then
then the propagation constant of the propagation
constant of the mode is now smaller than the
propagation constant of the corresponding
symmetric waveguide mode
and and it is it is understandable that because
asymmetric is introduced by introducing the
cover region so so this is the if this is
the symmetric waveguide ok this is the symmetric
waveguide so this is n s this is n f this
is again the level n s but now in asymmetric
planar waveguide i have n c so i am i am reducing
i am reducing the the refractive index in
the cover region so the effect is effect is
to pull down the effective indices of the
modes towards lower side so that we can see
from here itself that 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 t e zero t e one and t e two
modes of a typical asymmetric planar waveguide
so i can see that 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 so the 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
the number of modes are are the integer which
is closest to but greater than v over v over
pi by two but now all the cut offs are shifted
by this much amount so the number of modes
would now be v minus half tan inverse square
root of a divided by pi by two 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
m th t e mode would be given by lamda c m
is equal to two pi over v c m d by two square
root of n f square minus n s square which
comes directly from from the definition of
normalized frequency v so i can write it down
also which we will quite often use
v is equal to two pi over lambda naught times
d by two times square root of n f square minus
n s square so 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 t e zero mode the cut
off wavelength is one point seven one eight
seven for these parameters of waveguide and
for t e one mode it is zero point four nine
one four and for t e two mode it is zero point
two eight six seven
so if i start from a longer wavelength let
us say two micro meter then until i cross
this t zero mode is not guided ok so from
here to here there is no mode guided by the
structure and as soon as i cross this go below
go below this wavelength then t zero mode
starts appearing ok and t zero 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 zero point four nine micrometer
t e one mode starts appearing and below these
this wavelength i will have both t e zero
and t e one and so on so this is how the waveguide
would guide different modes if i change the
wavelength how the thickness affects the the
guided modes so from here i can find out the
cut off thickness of mth t e mode from here
itself ok 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 zero thickness it means no waveguide
no mode ok if i start increasing the thickness
then up to zero point five eight micron there
is no mode guided because t zero mode has
finite cut off and as soon as i cross this
then t e zero mode starts appearing and when
i cross two point zero three micron then t
e one mode starts appearing and so on so so
as i increase the waveguide thickness as i
increase the waveguide thickness the the number
of modes start increasing i should have the
label here which is d in micron 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 n f is equal to one point five and n s
is equal to one point four eight and n c is
equal to one and d is equal to four micrometer
now the first thing is to calculate the number
of modes at lambda naught is equal to zero
point five micron so i first calculate the
value of asymmetric parameter a which comes
out to be about twenty then i find out what
is what is the value of v at lambda naught
is equal to zero point five micrometer and
this comes out to be about six point one three
then i find out this number v minus half tan
inverse of square root a divided by pi by
two so this comes out to be three point four
so the number of modes are four the second
is wavelength range in which the waveguide
does not support anymore ok so i know there
would not be any more support it if v is less
than half tan inverse of square root a which
if i put the value v here the expression for
v here then this gives me the condition in
terms of lambda naught as lambda naught should
be greater than this and if i plug in all
these numbers i find out for lambda naught
greater than four point five four two seven
micrometer the waveguide would not support
anymore
third is what is the cut off wavelength of
t e two mode so i find out what is the v value
for cut off of t e two mode so which is given
by this with m is equal to two so if i put
m is equal to two then then v c is equal to
this and the cut off v c for cut off v for
t e two mode is this if i translate it to
the wavelength it comes out to be zero point
eight zero three eight micrometer what is
the range of d so that only t e zero and t
e one modes are guided at lambda naught is
equal to one micrometer ok
so i know that that for mth mode the cut off
thickness is this and i want t e zero and
t e one both the modes guided so i find out
this d c for both the modes and i also know
this v c m for mth mode is given by this so
so i find out v c zero for t zero mode v c
is this for t e one mode v c is this correspondingly
if i now find out d c for for these two modes
then for t e zero mode cut off thickness is
zero point eight eight micrometer and for
this is two point nine three micrometer so
so if t e zero and t e one modes are to be
guided then thickness should be between this
and this
so this is all in this ah lecture in the next
lecture we will extend the analysis to t m
modes
thank you
