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In the last lecture we had carried out the
analysis of asymmetric planar waveguide for
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TE-modes. In this lecture we will look into
the TM-modes of the waveguide.
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So this is the waveguide again, and we will
confine our analysis only to step index waveguides.
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So, this is the step index asymmetric planar
waveguide.
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So, for TM-modes the non-vanishing components
of electric and magnetic fields for this refractive
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index profile and z propagation are Hy, Ex
and Ez.
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Again for guided modes the beta over k naught
should lie between ns and nf, and if the beta
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over k naught is below ns then field radiates
out. So, they are radiation mode. So, I again
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write down the wave equation in three different
regions: the cover, the film, and the substrate.
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And now the equations are in the form d2Hy
over dx square minus beta square minus k naught
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square nc square Hy is equal to 0 for the
cover. And similarly in the film this is the
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equation, and in the substrate this is the
equation.
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What I see that these equations are exactly
the same. These equations are exactly the
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same in TE case these equations were in Ey
and now these equations are in Hy. Again the
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definitions of gamma c kappa f and gamma s
are the same.
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And if I write the solutions the solutions
in the cover region is Hy of x is equal to
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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
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e to the power minus i kappa f x. In the substrate
it is D e to the power gamma s x, where gamma
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c kappa f and gamma s are defined by these
expressions.
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Again I will have to apply the boundary conditions
at the interfaces x is equal to 0 and x is
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equal to minus d; to obtain the relationships
between A B C and D, and to obtain the eigenvalue
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equation.
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So, if I do this I again write down the solutions
in different regions. And now apply the boundary
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conditions, remember again the boundary conditions
are the tangential components of E and H are
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continuous. So, these boundary conditions
now become Hy and 1 over n square dHy over
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dx are continuous at the interfaces x is equal
to 0 and at x is equal to minus d
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So, when I now apply these boundary conditions
to these fields and do mathematical manipulations
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similar to what we had done in the case of
TE-modes. I get the eigenvalue equation as
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this. The only difference is now these factors
of nf square over ns square associated with
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the substrate term, nf square over nc square
associated with the cover term. So, these
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extra factors are there otherwise the equation
is similar.
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So, this is the eigenvalue equation, after
solving this I can find out the modes, the
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propagation constants of the modes and later
on the fields.
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In normalized parameters if I define this
equation then it becomes tan 2 V square of
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1 minus b nf square over ns square square
root of b divided by 1 minus b plus nf square
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over nc square square root of b plus a over
1 minus b divided by 1 minus nf square over
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ns square times square root of b over 1 minus
b and nf square over nc square times square
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root of b plus a over 1 minus b.
What are the cut offs? The cut offs are again
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defined by b is equal to 0. So, if I put b
is equal to 0 here the equation define the
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cut offs for TM-modes are now tan 2Vc is equal
to nf square over nc square times square root
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of a. So, if I compare it with the cut offs
of TE-modes then I have this extra factor
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of nf square over nc square here. So, my cut
offs for m-th TM-modes are now given by m
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pi by 2 plus half tan inverse of nf square
over nc square square root of a.
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I can see from here that since nf is larger
than nf then these cut offs of TM-modes are
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larger than the cut offs of corresponding
TE-modes.
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So, let me now plot b-V curves here by solving
the transcendental equation for different
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values of V and for the given value of a.
So here, I have plotted for two cases: one
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is for symmetric waveguide and another is
for asymmetric waveguide. For comparison I
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have also plotted the b-V curves for TE-modes.
So, the b-V curves for TE-modes are given
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by solid line, for TM-modes they are given
by dashed line. If I look at these curves
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now, for a is equal to 0 this is TE0 mode,
this is TM0 mode. Solid line is TE0 mode dashed
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line is TM0 mode, and I see that both the
modes have the same cut off.
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But if I take a naught equal to 0 that is
I consider the case of asymmetric waveguide
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then the TE0 mode has cut off somewhere here,
and TM0 mode has cut off somewhere here. And
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I can see that in this range, so this cut
off point is defined by half tan inverse of
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square root a and this cut off point is defined
by half tan inverse nf square over nc square
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square root of a.
So, in this region if the value of V lies
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in this region then only TE0 mode is guided
and all the other modes included in TM0 are
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cut off. So, in this range I guide strictly
one mode and that mode is TE0 mode. Or rigorously
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speaking in this range I guide only one mode
and one polarization only TE polarization.
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So, this range is also called SPSM range-
Single Polarization Single Mode range.
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Let us look at the modal fields.
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Modal fields are similar but there is one
difference we should mark, and that difference
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is discontinuity at on the slope at the interfaces.
Because Hy is continuous so there is no discontinuity
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in the field, but dHy over dx is not continuous
at the interface. So, there is discontinuity
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in the slope at the interfaces.
The number of modes: how many modes are guided?
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Well, now all the modes are shifted by this
much amount half tan inverse of nf square
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over nc square square root of a. So, the number
of modes would now be an integer closest to
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but greater than this number.
Let us look at how the number of TM-modes
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vary when we change wavelength and compare
it with the TE case also.
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So, here I have plotted the number of modes
as a function of wavelength, ok. If I find
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out the cut off wavelength of TE0 mode then
for these waveguide parameters, then it comes
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out to be 1.7187 micrometre. If I now find
out the cut off wavelength for the same waveguide
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for TM0 mode then it comes out to be 1.5916.
So now, if I again start from a wavelength
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two micron and start decreasing the wavelength,
then as I cross 1.7187 as I cross 1.7187 TE0
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mode starts appearing this blue one; TE0 mode
starts appearing but TM0 is still not there.
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And as soon as I go below 1.5916 then TM0
also starts appearing.
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So, in this range only TE0 mode is there and
this is SPSM range, while if I go below this
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value then I have both TE0 and TM0. Similarly
if I go below this then TM1 will also start
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appearing. So, this is the range of wavelength
1.5916 to 1.7187, in this wavelength range
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I have single polarization single mode operation
of the waveguide.
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Let us workout few examples here.
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Let me consider a dielectric asymmetric planar
waveguide with nf is equal to1.5 and ns is
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equal to 1.48 and nc is equal to 1.
Now, I want to calculate the range of normalized
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frequency for SPSM operation. So, I know that
SPSM operation involves asymmetry parameter
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a, so first I calculate a for this waveguide
and a comes out to be about 20. Then SPSM
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range is given by half tan inverse square
root a smaller than V smaller than half tan
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inverse nf square over nc square square root
of a. So now, if I calculate these then this
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range comes out to be 0.6753 less than V less
than 0.7358. So, this is the range of normalized
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frequency V for SPSM operation.
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And remember that the definition of V which
I have used is 2 pi over lambda naught times
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d by 2 times square root of nf square minus
ns square. And I would like to bring out that
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in the text books in several textbooks the
definition of V is 2 pi over lambda naught
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times d times nf square minus ns square, while
I use d by 2. So, there would be a factor
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of 2.
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Second is for d equal to 1 micrometre what
is the wavelength range for SPSM operation?
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So, I know that from the previous problem
I know that V should lie between 0.6753 to
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0.7358 for SPSM operation. Now for this range
of V I can find out the corresponding range
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of lambda if d is given. So, lambda in terms
of V and d and nf ns is given by 2 pi over
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V d by 2 square root of nf square minus nc
square. So, I simply find out the value of
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lambda naught corresponding to these values
of V.
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So, it comes out to be lambda lies between
1.042 micrometre and 1.136 micrometer.
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Third is if I choose a light source of wavelength
lambda naught is equal to 1 micrometre then
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in what range of film thickness there would
be SPSM operation. So, again I have the SPSM
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range of V from 0.6753 to 0.7358 and I simply
now find out the corresponding values of d
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if lambda naught is given to me.
So, this comes out to be between 0.88 micrometre
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and 0.96 micrometre. So, in this range in
this very small range of d I will have SPSM
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operation.
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How much is the power associated with the
modes. So, to find out the power associated
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with the mode I will have to calculate the
pointing vector and then integrate it over
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the entire cross section. Since it is a planar
waveguide, so I cannot integrate it over y,
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because in y direction it is infinitely extended.
So, what I can have is power per unit length
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in y direction.
So, for TE-mode I have already seen in case
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of symmetric waveguides that P is given by
beta over 2 omega mu naught times integral
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minus infinity to plus infinity Ey square
of xdx. So, if I now find out the integration
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Ey square of xdx for the modes; for the modal
fields of asymmetric planar waveguide for
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TE-mode then I can find out the power associated
with TE-modes and it is given by this.
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Similarly for TM-modes the power per unit
length is given by beta over 2 omega epsilon
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naught integral minus infinity to plus infinity
1 over n square Hy square xdx. And if I do
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the same then the power corresponding to TM-modes
comes out to be like this.
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The last thing that I would like to do in
this lecture is- how do I excite a particular
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mode. Can I excite a particular mode? Selectively
excite; I know if I have waveguide and if
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I launched the light from end then all the
modes would be excited in different proportion
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depending upon what is the intensity profile
or what is the amplitude profile of the incident
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light.
But, can I selectively excite one particular
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mode? So for what is used is prism coupling
technique. So, what you do? You have a waveguide
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this is an asymmetric planar waveguide whose
cover is air. So, this can be glass and this
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can be polymer film for example, and this
is air. Now what I do I put a prism on top
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of this and press it, clamp it, then what
I have here even though it is pressed hard
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there is some air gap between the prism and
the film. And what happens is now I launched
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light from here into the prism, this light
beam gets reflected into the prism.
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And then depending upon this angle psi which
can be translated, which can be related to
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this angle theta p with this beam makes with
the normal to the prism base then depending
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upon this angle theta p this would get totally
internally reflected. So, I can have total
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internal reflection at the prism base. I know
that a total internal reflection is always
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associated with an evanescent tail. So, here
what basically I have a standing wave here
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whose tail evanescent tail extends into the
film.
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So, I have a standing wave here whose evanescent
tail extends into the field and it is this
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evanescent field which excites the mode of
the waveguide. So, this is the standing wave.
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And this is one typical guided mode. This
is you can see that it is TE0 mode if launched
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polarization is TE. This guided mode is nothing
but a superposition of two plane waves and
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these plane waves make angle plus minus theta
f from the waveguide axis. So, if this angle
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is theta f then the propagation constant of
this plane wave is k naught nf, then the propagation
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constant of the mode is k naught nf cos theta
f.
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So, propagation constant of the mode is k
naught nf cos theta f, which is the horizontal
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component of this. While if I look at this
standing wave pattern then, what is the horizontal
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component of this plane wave. The horizontal
component of this plane wave is k naught np
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sin theta p. So, if the horizontal component
of this standing wave is the same as the horizontal
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component of this guided mode then well horizontal
component of the plane wave corresponding
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to guided board then there would be phase
matching, because these two horizontal components
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are now able to catch up each other. And that
is how this would be able to resonantly couple
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energy into this mode.
So, if this condition is satisfied I have
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excitation of that particular guided mode
which corresponds to a particular angle theta
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f. And I know that this k naught nf cos theta
f is nothing but k naught n effective. So,
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from here I can even find out the effective
index of the mode if I know angle psi. How
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I have got this? Well, this angle psi can
be related to this theta p. If I look at this
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triangle then A plus pi by 2 plus r where
r is the angle of reflection plus pi by 2
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minus theta p should be equal to pi. Or I
get r in terms of psi from Snellâ€™s law sin
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psi is equal to np sin r and if I put it here
then I get theta p is equal to a plus sin
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inverse sin psi over np.
And this gives me n effective in terms of
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the incident angle, the prism refractive index,
prism angle A, and again prism reflective
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index. So, if I know this then by just measuring
these psi I can find out the effective index
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of the mode. And you can see that this resonant
excitation can take place when this condition
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is satisfied and theta f is discrete; theta
f is discrete for different modes. For TE0
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mode it is different for TE1 mode it is different
for TE2 mode it is different.
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So, for different values of psi I will have
different values of theta p. And if they match
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to these discrete theta f then I will excite
those particular modes. So, in this way I
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can selectively excite, I can selectively
excite the modes of a planar waveguide.
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This is the experimental setup. So, you have
a laser beam and this is a lens through which
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you focus this onto the prism coupling arrangement.
This is a typical prism coupling arrangement,
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this is the waveguide, and this is the prism.
And you can see that when I tune because this
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assembly can be rotated with respect to the
beam, so I can change the angle psi. And I
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can see that for a particular value of psi
the other input angle I see a mode is excited
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and I see a streak going down the length of
the waveguide.
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So, this represents the propagation of mode.
And I can see that as it goes then this becomes
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feeble and feeble it becomes week because
of the losses. And in fact, you can see this
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because there are scattering losses. There
are scattering losses that is how you can
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see. So, more strong the streak is bad is
the quality of the waveguide. I can also do,
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I can put another prism here and I can decouple
the modes in the same way as I have coupled
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them. So, I put another prism here.
So, you can see that from this side I am coupling
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light this is the streak and then I decouple,
then I decouple and when I put it here on
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the screen then I see these lines which correspond
to the modes different modes. For example,
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in this I can see 1, 2, 3, 4, 5, 6 modes,
and by changing the angle I can put light
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into one particular mode. For example, here
it is coupled to third mode from the left.
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These are known as m lines. So, by measuring
angles psi I can find out the effective index
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of the mode n effective. And once I have the
values of n effective then using a technique
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called inverse WKB method I can do the refractive
index profiling of the waveguide.
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So, by measuring the refractive indices of
the modes I can find out the refractive index
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profile of the waveguide. The only thing is
that I should have sufficient values of n
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effective, I should have sufficient values
of n effective. For a single mode waveguide
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it would not work, if you have only one value
of n effective it will not work. So, if it
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is a single mode waveguide at one particular
wavelength then it will not work, you will
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have to use different wavelengths to have
the values of n effective at different wavelengths
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00:25:56,860 --> 00:26:04,100
and then you can do it; otherwise, at a single
wavelength if it is highly multimode waveguide
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then it is more accurate.
So, this is all in planar waveguides. And
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in the next lecture we will go into cylindrical
geometry and find out how the modes are formed
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in an optical fiber, and how do we analyze
the modes of an optical fiber, how do we find
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out the modes of an optical fiber. I have
spent a lot of time in the analysis of planar
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waveguide. And in the optical fiber I would
adopt all these results to cylindrical geometry,
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because the physics is now clear.
And it was easier to understand all these
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in planar geometry, because it was one dimensional
problem. And also the functions involved were
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very simple: sin cosine functions and exponentially
amplifying and decaying functions. But in
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case of cylindrical geometry these functions
would be different. However, the physics which
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we have understood from the analysis of planar
waveguide would be applicable there also.
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Thank you.