In the last lecture we had carried out the
analysis of asymmetric planar waveguide for
TE-modes. In this lecture we will look into
the TM-modes of the waveguide.
So this is the waveguide again, and we will
confine our analysis only to step index waveguides.
So, this is the step index asymmetric planar
waveguide.
So, for TM-modes the non-vanishing components
of electric and magnetic fields for this refractive
index profile and z propagation are Hy, Ex
and Ez.
Again for guided modes the beta over k naught
should lie between ns and nf, and if the beta
over k naught is below ns then field radiates
out. So, they are radiation mode. So, I again
write down the wave equation in three different
regions: the cover, the film, and the substrate.
And now the equations are in the form d2Hy
over dx square minus beta square minus k naught
square nc square Hy is equal to 0 for the
cover. And similarly in the film this is the
equation, and in the substrate this is the
equation.
What I see that these equations are exactly
the same. These equations are exactly the
same in TE case these equations were in Ey
and now these equations are in Hy. Again the
definitions of gamma c kappa f and gamma s
are the same.
And if I write the solutions the solutions
in the cover region is Hy of x is equal to
A e to the power minus gamma c x. In the film
it is B e to the power i kappa f x plus C
e to the power minus i kappa f x. In the substrate
it is D e to the power gamma s x, where gamma
c kappa f and gamma s are defined by these
expressions.
Again I will have to apply the boundary conditions
at the interfaces x is equal to 0 and x is
equal to minus d; to obtain the relationships
between A B C and D, and to obtain the eigenvalue
equation.
So, if I do this I again write down the solutions
in different regions. And now apply the boundary
conditions, remember again the boundary conditions
are the tangential components of E and H are
continuous. So, these boundary conditions
now become Hy and 1 over n square dHy over
dx are continuous at the interfaces x is equal
to 0 and at x is equal to minus d
So, when I now apply these boundary conditions
to these fields and do mathematical manipulations
similar to what we had done in the case of
TE-modes. I get the eigenvalue equation as
this. The only difference is now these factors
of nf square over ns square associated with
the substrate term, nf square over nc square
associated with the cover term. So, these
extra factors are there otherwise the equation
is similar.
So, this is the eigenvalue equation, after
solving this I can find out the modes, the
propagation constants of the modes and later
on the fields.
In normalized parameters if I define this
equation then it becomes tan 2 V square of
1 minus b nf square over ns square square
root of b divided by 1 minus b plus nf square
over nc square square root of b plus a over
1 minus b divided by 1 minus nf square over
ns square times square root of b over 1 minus
b and nf square over nc square times square
root of b plus a over 1 minus b.
What are the cut offs? The cut offs are again
defined by b is equal to 0. So, if I put b
is equal to 0 here the equation define the
cut offs for TM-modes are now tan 2Vc is equal
to nf square over nc square times square root
of a. So, if I compare it with the cut offs
of TE-modes then I have this extra factor
of nf square over nc square here. So, my cut
offs for m-th TM-modes are now given by m
pi by 2 plus half tan inverse of nf square
over nc square square root of a.
I can see from here that since nf is larger
than nf then these cut offs of TM-modes are
larger than the cut offs of corresponding
TE-modes.
So, let me now plot b-V curves here by solving
the transcendental equation for different
values of V and for the given value of a.
So here, I have plotted for two cases: one
is for symmetric waveguide and another is
for asymmetric waveguide. For comparison I
have also plotted the b-V curves for TE-modes.
So, the b-V curves for TE-modes are given
by solid line, for TM-modes they are given
by dashed line. If I look at these curves
now, for a is equal to 0 this is TE0 mode,
this is TM0 mode. Solid line is TE0 mode dashed
line is TM0 mode, and I see that both the
modes have the same cut off.
But if I take a naught equal to 0 that is
I consider the case of asymmetric waveguide
then the TE0 mode has cut off somewhere here,
and TM0 mode has cut off somewhere here. And
I can see that in this range, so this cut
off point is defined by half tan inverse of
square root a and this cut off point is defined
by half tan inverse nf square over nc square
square root of a.
So, in this region if the value of V lies
in this region then only TE0 mode is guided
and all the other modes included in TM0 are
cut off. So, in this range I guide strictly
one mode and that mode is TE0 mode. Or rigorously
speaking in this range I guide only one mode
and one polarization only TE polarization.
So, this range is also called SPSM range-
Single Polarization Single Mode range.
Let us look at the modal fields.
Modal fields are similar but there is one
difference we should mark, and that difference
is discontinuity at on the slope at the interfaces.
Because Hy is continuous so there is no discontinuity
in the field, but dHy over dx is not continuous
at the interface. So, there is discontinuity
in the slope at the interfaces.
The number of modes: how many modes are guided?
Well, now all the modes are shifted by this
much amount half tan inverse of nf square
over nc square square root of a. So, the number
of modes would now be an integer closest to
but greater than this number.
Let us look at how the number of TM-modes
vary when we change wavelength and compare
it with the TE case also.
So, here I have plotted the number of modes
as a function of wavelength, ok. If I find
out the cut off wavelength of TE0 mode then
for these waveguide parameters, then it comes
out to be 1.7187 micrometre. If I now find
out the cut off wavelength for the same waveguide
for TM0 mode then it comes out to be 1.5916.
So now, if I again start from a wavelength
two micron and start decreasing the wavelength,
then as I cross 1.7187 as I cross 1.7187 TE0
mode starts appearing this blue one; TE0 mode
starts appearing but TM0 is still not there.
And as soon as I go below 1.5916 then TM0
also starts appearing.
So, in this range only TE0 mode is there and
this is SPSM range, while if I go below this
value then I have both TE0 and TM0. Similarly
if I go below this then TM1 will also start
appearing. So, this is the range of wavelength
1.5916 to 1.7187, in this wavelength range
I have single polarization single mode operation
of the waveguide.
Let us workout few examples here.
Let me consider a dielectric asymmetric planar
waveguide with nf is equal to1.5 and ns is
equal to 1.48 and nc is equal to 1.
Now, I want to calculate the range of normalized
frequency for SPSM operation. So, I know that
SPSM operation involves asymmetry parameter
a, so first I calculate a for this waveguide
and a comes out to be about 20. Then SPSM
range is given by half tan inverse square
root a smaller than V smaller than half tan
inverse nf square over nc square square root
of a. So now, if I calculate these then this
range comes out to be 0.6753 less than V less
than 0.7358. So, this is the range of normalized
frequency V for SPSM operation.
And remember that the definition of V which
I have used is 2 pi over lambda naught times
d by 2 times square root of nf square minus
ns square. And I would like to bring out that
in the text books in several textbooks the
definition of V is 2 pi over lambda naught
times d times nf square minus ns square, while
I use d by 2. So, there would be a factor
of 2.
Second is for d equal to 1 micrometre what
is the wavelength range for SPSM operation?
So, I know that from the previous problem
I know that V should lie between 0.6753 to
0.7358 for SPSM operation. Now for this range
of V I can find out the corresponding range
of lambda if d is given. So, lambda in terms
of V and d and nf ns is given by 2 pi over
V d by 2 square root of nf square minus nc
square. So, I simply find out the value of
lambda naught corresponding to these values
of V.
So, it comes out to be lambda lies between
1.042 micrometre and 1.136 micrometer.
Third is if I choose a light source of wavelength
lambda naught is equal to 1 micrometre then
in what range of film thickness there would
be SPSM operation. So, again I have the SPSM
range of V from 0.6753 to 0.7358 and I simply
now find out the corresponding values of d
if lambda naught is given to me.
So, this comes out to be between 0.88 micrometre
and 0.96 micrometre. So, in this range in
this very small range of d I will have SPSM
operation.
How much is the power associated with the
modes. So, to find out the power associated
with the mode I will have to calculate the
pointing vector and then integrate it over
the entire cross section. Since it is a planar
waveguide, so I cannot integrate it over y,
because in y direction it is infinitely extended.
So, what I can have is power per unit length
in y direction.
So, for TE-mode I have already seen in case
of symmetric waveguides that P is given by
beta over 2 omega mu naught times integral
minus infinity to plus infinity Ey square
of xdx. So, if I now find out the integration
Ey square of xdx for the modes; for the modal
fields of asymmetric planar waveguide for
TE-mode then I can find out the power associated
with TE-modes and it is given by this.
Similarly for TM-modes the power per unit
length is given by beta over 2 omega epsilon
naught integral minus infinity to plus infinity
1 over n square Hy square xdx. And if I do
the same then the power corresponding to TM-modes
comes out to be like this.
The last thing that I would like to do in
this lecture is- how do I excite a particular
mode. Can I excite a particular mode? Selectively
excite; I know if I have waveguide and if
I launched the light from end then all the
modes would be excited in different proportion
depending upon what is the intensity profile
or what is the amplitude profile of the incident
light.
But, can I selectively excite one particular
mode? So for what is used is prism coupling
technique. So, what you do? You have a waveguide
this is an asymmetric planar waveguide whose
cover is air. So, this can be glass and this
can be polymer film for example, and this
is air. Now what I do I put a prism on top
of this and press it, clamp it, then what
I have here even though it is pressed hard
there is some air gap between the prism and
the film. And what happens is now I launched
light from here into the prism, this light
beam gets reflected into the prism.
And then depending upon this angle psi which
can be translated, which can be related to
this angle theta p with this beam makes with
the normal to the prism base then depending
upon this angle theta p this would get totally
internally reflected. So, I can have total
internal reflection at the prism base. I know
that a total internal reflection is always
associated with an evanescent tail. So, here
what basically I have a standing wave here
whose tail evanescent tail extends into the
film.
So, I have a standing wave here whose evanescent
tail extends into the field and it is this
evanescent field which excites the mode of
the waveguide. So, this is the standing wave.
And this is one typical guided mode. This
is you can see that it is TE0 mode if launched
polarization is TE. This guided mode is nothing
but a superposition of two plane waves and
these plane waves make angle plus minus theta
f from the waveguide axis. So, if this angle
is theta f then the propagation constant of
this plane wave is k naught nf, then the propagation
constant of the mode is k naught nf cos theta
f.
So, propagation constant of the mode is k
naught nf cos theta f, which is the horizontal
component of this. While if I look at this
standing wave pattern then, what is the horizontal
component of this plane wave. The horizontal
component of this plane wave is k naught np
sin theta p. So, if the horizontal component
of this standing wave is the same as the horizontal
component of this guided mode then well horizontal
component of the plane wave corresponding
to guided board then there would be phase
matching, because these two horizontal components
are now able to catch up each other. And that
is how this would be able to resonantly couple
energy into this mode.
So, if this condition is satisfied I have
excitation of that particular guided mode
which corresponds to a particular angle theta
f. And I know that this k naught nf cos theta
f is nothing but k naught n effective. So,
from here I can even find out the effective
index of the mode if I know angle psi. How
I have got this? Well, this angle psi can
be related to this theta p. If I look at this
triangle then A plus pi by 2 plus r where
r is the angle of reflection plus pi by 2
minus theta p should be equal to pi. Or I
get r in terms of psi from Snell’s law sin
psi is equal to np sin r and if I put it here
then I get theta p is equal to a plus sin
inverse sin psi over np.
And this gives me n effective in terms of
the incident angle, the prism refractive index,
prism angle A, and again prism reflective
index. So, if I know this then by just measuring
these psi I can find out the effective index
of the mode. And you can see that this resonant
excitation can take place when this condition
is satisfied and theta f is discrete; theta
f is discrete for different modes. For TE0
mode it is different for TE1 mode it is different
for TE2 mode it is different.
So, for different values of psi I will have
different values of theta p. And if they match
to these discrete theta f then I will excite
those particular modes. So, in this way I
can selectively excite, I can selectively
excite the modes of a planar waveguide.
This is the experimental setup. So, you have
a laser beam and this is a lens through which
you focus this onto the prism coupling arrangement.
This is a typical prism coupling arrangement,
this is the waveguide, and this is the prism.
And you can see that when I tune because this
assembly can be rotated with respect to the
beam, so I can change the angle psi. And I
can see that for a particular value of psi
the other input angle I see a mode is excited
and I see a streak going down the length of
the waveguide.
So, this represents the propagation of mode.
And I can see that as it goes then this becomes
feeble and feeble it becomes week because
of the losses. And in fact, you can see this
because there are scattering losses. There
are scattering losses that is how you can
see. So, more strong the streak is bad is
the quality of the waveguide. I can also do,
I can put another prism here and I can decouple
the modes in the same way as I have coupled
them. So, I put another prism here.
So, you can see that from this side I am coupling
light this is the streak and then I decouple,
then I decouple and when I put it here on
the screen then I see these lines which correspond
to the modes different modes. For example,
in this I can see 1, 2, 3, 4, 5, 6 modes,
and by changing the angle I can put light
into one particular mode. For example, here
it is coupled to third mode from the left.
These are known as m lines. So, by measuring
angles psi I can find out the effective index
of the mode n effective. And once I have the
values of n effective then using a technique
called inverse WKB method I can do the refractive
index profiling of the waveguide.
So, by measuring the refractive indices of
the modes I can find out the refractive index
profile of the waveguide. The only thing is
that I should have sufficient values of n
effective, I should have sufficient values
of n effective. For a single mode waveguide
it would not work, if you have only one value
of n effective it will not work. So, if it
is a single mode waveguide at one particular
wavelength then it will not work, you will
have to use different wavelengths to have
the values of n effective at different wavelengths
and then you can do it; otherwise, at a single
wavelength if it is highly multimode waveguide
then it is more accurate.
So, this is all in planar waveguides. And
in the next lecture we will go into cylindrical
geometry and find out how the modes are formed
in an optical fiber, and how do we analyze
the modes of an optical fiber, how do we find
out the modes of an optical fiber. I have
spent a lot of time in the analysis of planar
waveguide. And in the optical fiber I would
adopt all these results to cylindrical geometry,
because the physics is now clear.
And it was easier to understand all these
in planar geometry, because it was one dimensional
problem. And also the functions involved were
very simple: sin cosine functions and exponentially
amplifying and decaying functions. But in
case of cylindrical geometry these functions
would be different. However, the physics which
we have understood from the analysis of planar
waveguide would be applicable there also.
Thank you.
