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Till now we have carried out the analysis
of symmetric planar wave guides where the
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high index region is sandwiched between two
lower index regions of same refractive index.
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Now in this lecture we will extend the analysis
to asymmetric planar waveguide asymmetric
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planar waveguides are more practical wave
guides. So, we will extend this analysis to
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asymmetric planar waveguide and see how the
waveguide hence is done in these kind of waveguides.
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So, what is an asymmetric planar waveguide
you for example, take a glass of substrate
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of refractive index ns, and you deposit a
film of refractive index nf of thickness d,
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this film can be for example, of a polymer
material, the refractive index nf is higher
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than the refractive index ns and then it is
surrounded by air. So, air works as cover;
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also you can put a cover of polymer material
itself of refractive index nc, where nc is
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smaller than ns and ns is smaller than nf.
So, this kind of waveguide we can see that
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is has asymmetric structure, because ns is
not equal to nc here. If I look at the refractive
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index profile of this waveguide, then this
is the substrate an example is glass, this
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is the guiding film the example is polymer,
and this is the cover region which can be
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air or another polymer of lower refractive
index. Here at x = 0, I have made the interface
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between the film and the cover, and at x = -d
represents the interface between the substrate
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and the film.
Since the refractive indices of the individual
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layers are uniform. So, this waveguide is
step index waveguide. I can also have a waveguide
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in which the refractive index in the guiding
film is not uniform, but it varies with x.
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So, how I can make this waveguide well I take
a substrate of material glass, and then I
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diffuse certain ions into this pay over a
certain distance. So, I will have a diffusion
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profile for example, I can diffuse silver
ions into glass and that will change the refractive
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index of the material near the surface.
So, now this nf would be a function of x now,
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and then I can have cover as air or cover
of a polymer layer. So, now, this nc is smaller
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than ns, and then the refractive index of
the film at the surface of the film nf at
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f = 0. So, ns is smaller than that. If I look
at the refractive index profile of this structure
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now, so, I have a substrate and this is the
guiding film. This guiding film now has refractive
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index with varies which varies with x and
ultimately it merges into the refractive index
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of glass.
So, this is the surface index nf(0), and this
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nf(0)is larger than ns and which is larger
than nc.
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So, this is a typical graded index ion exchange
waveguide. Let us do the modal analysis of
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this step index waveguide, the step index
planar asymmetric waveguide. So, this is the
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refractive index profile and I can write down
the refractive index in various regions as
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nx = nc, which is defined by the region x
> 0 which is the cover, then in the region
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minus –d < x <0 the refractive index is
nf which is the film region, and for x < -d
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which represents the substrate region, the
refractive index is ns.
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Now, the modes of this waveguide are here
guided modes, whose refractive index is lie
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between ns and nf. So, these are some representative
guided modes, and for, the field would radiate
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out in the substrate region and forthe field
would radiate out in the cover region as well.
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So, in this region you will have radiation
modes.
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We are interested in the guided modes of the
structure, which are defined in this range,
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and let us first do the TE mode analysis of
the waveguide, and as I can see that the refractive
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index variation is in x-direction, and I am
considering propagation in z-direction. So,
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the non- vanishing components corresponding
to TE modes are Ey, Hx and Hz.
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I know from my previous analysis that the
wave equation satisfied by TE modes of such
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a waveguide is given by . So, the procedure
is again the same I write down this equation
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in all the three regions, I write down the
solutions of these three equations and then
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apply boundary conditions to match the solutions.
So, this is the equation in cover region x
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= 0, which
is , because in this region nx = nc, and I
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have taken minus sign outside and represented
it in this form because , so that this quantity
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is positive for guided modes.
In the film region which is defined by minus
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-d < x < 0, the equation becomes, again for
guided modes this quantity in the square brackets
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is positive. In the substrate region defined
by x < -d, the equation is
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and again for guided modes this quantity is
positive.
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So, now I can represent this as , this as
and this as and then find out what are the
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solutions of these three equations.
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So, these are the three equations now, where
, , and .
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So, what are the solutions? The solution of
this differential equation is. And as you
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know that we are interested in guided modes.
So, we cannot have exponentially amplifying
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solution. So, I am only considering the exponentially
decaying solutions in cover regions and substrate
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region.
Now, in the film this equation will give you
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the oscillatory solutions, which I write as
,
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and in the substrate region it is because
x < -d. So, it is negative. So, this again
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gives you exponentially decaying solutions.
Now a b c and d are constants and these constants
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are determined by the boundary conditions.
So, let us apply boundary conditions and obtain
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relationships between A, B, C and D obtain
the Eigen value equation or characteristic
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equation which is satisfied by the propagation
constants of the guided modes.
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So, these are the solutions in different regions
in substrate in the film and in the cover
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and what are the boundary conditions? Boundary
conditions are again the tangential components
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of E and H are continuous, and the tangential
components here are Ey and Hz. So, they give
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me that Ey andbecause Hz is related to Ey
as. So, they are continuous at the interfaces
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x = 0 and at x = -t.
So, let us apply these boundary conditions.
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So, if I look at the continuity ofat. So,
I apply it here. So, if I approach from the
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film and if I approach from the cover towards
x = 0 is equal to 0. So, I will get B + C
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= A. Now continuity of at x = -d which is
the substrate film interface, this gives me
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.
Now, let
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us see the continuity of the derivatives at
x = 0 and at x = d. So, at x = 0 gives meand
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continuity ofat x = -d interface gives me
.
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So, I
have now four equations relating these four
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constants A, B, C and D and by mathematical
manipulation of these four equations. I can
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do two things one is obtain B, C and D in
terms of A, and then eliminating A, B, C,
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D all together and form a transcendental equation
inwhich is also known as Eigen value equation
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or characteristic equation. So, that equation
comes out to be:.
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So, this is the equation which is satisfied
bybecause the only unknown here is, appears
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in and . So, if I solve this equation and
find out the roots of this equation then I
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can know what are the propagation constants
of the modes, what are the modes, which are
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supported by this asymmetric step index planar
waveguide.
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As we have done earlier also that it is always
a good idea to obtain everything in normalized
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parameters. So, that we are more or less independent
of waveguide parameters, here we cannot have
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complete independence from waveguide parameters,
but to certain extent it is still we can draw
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some universal curve. So, how do we define
normalized parameters here? First is normalized
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frequency V.
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If you remember that in symmetric planar waveguide
we had only 2 refractive indicesand , . So,
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there we had defined it in terms of .
Now, now in place of, I have, but low refractive
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indices are now up toand which one we should
take in order to define V. So, it is very
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simple to choose becausedefines the cut-off
of guided modes. So, I should choosebecause
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as soon as the effective index of a mode falls
belowit is no more guided. So, I chooseto
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define the normalized frequency v here, and
similarly for defining the normalized propagation
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constant b I choose.
So, b is defined as . There is another parameter
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here in this structure, which tells you how
asymmetric this waveguide is. The asymmetry
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in the structure is introduced by the different
values ofand. So, what is the difference between
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and? So, let us define by asymmetry parameter
a, which is given as.
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So, now I have here three normalized parameters
in place, and now I can represent my transcendental
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equation or Eigen value equation in terms
of these normalized parameters. If you remember
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that in the transcendental equation the three
terms appear which are, and , which contain.
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So, now I will have to represent these , and
in terms of V, b and a. And we can see that
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.
and .
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So, keeping these in mind and looking at the
expressions here I would now try to obtain
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, 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
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I can obtain this in the numerator of this.
So, b is given by this. So, if I do 1- b it
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becomes, and .
So, if I multiple here by d /2 and then I
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can havehere or I havedirectly here and this
is nothing, but V2d/2. So, this becomes . So,
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this gives me.
If I look atgamma s square, . So, it is directly
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appearing herethis thing. So, this will immediately
give me . What about now. So, I can see from
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here that if I do b + a then I can only obtainin
the numerator. So, I do b + a. So, it becomesdivided
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by this. So, which isdivided byand this gives
me .
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So, in this way I have now obtained, and in
terms of V, b and a. So, I can now represent
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my eigen value equation in normalized parameters.
So, this equation simply becomes: .
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And of course, since and for cut offguided
mode,lies between ns and nf. So, b would lie
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as usual between 0 and 1. So, I solve this
equation for a given value of V, V is if you
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remember. So, this V contains all the waveguide
parameters and the wavelength. So, for a given
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waveguide and wavelength I have V and for
that value of V I can solve this equation
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to obtain the normalized propagation constant
b.
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What are the cut offs? Cut offs are defined
byor b = 0. So, if I put it there then the
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cut off equation becomes or cut off for TE
mode mth TE mode is now given by . You can
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see here that the cut off of even TE0 mode
is finite.
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So, in a particular range of V which is defined
by V even the TE0 mode is not guided.
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This is the difference as compared to the
symmetric planar waveguide. In symmetric planar
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waveguide TE0 mode has 0 cut off. So, TE0
mode was always guided, but here it is not
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the case. Now let us solve this equation the
characteristic equation or Eigen value equation
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for different values of V, and plot the roots
as obtained as a function of V. So, here I
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have plotted the roots for both the cases
a = 0 which represents the symmetric planar
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waveguide, and then for a very high value
of a which is 40 for asymmetric planar waveguide.
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These solid lines are corresponding to symmetric
planar waveguide a = 0, and these dash lines
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are corresponding to asymmetric planar waveguide
And what I can see that TE0 mode has cut off
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here, TE1 mode has cut off here, TE2 mode
has cut off here. In case of symmetric planar
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waveguide this was the cut off was the corresponding
cut off. Now all the cut offs have been shown
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shifted by this much amount which is . So,
this is one thing another thing that I see
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here is that if I take a particular value
of V then the propagation constant of the
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mode is now smaller than the propagation constant
of the corresponding symmetric waveguide mode.
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And it is it is understandable that because
asymmetric is introduced by introducing the
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cover region. So, if this is the symmetric
waveguide, this is the symmetric waveguide.
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So, this is ns this is nf this is again the
level ns, but now in asymmetric planar waveguide
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I have nc. So, I am reducing the refractive
index in the cover region. So, the effect
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is to pull down the effective indices of the
modes towards lower side so that we can see
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from here itself that the propagation constants
of asymmetric planar waveguides are smaller
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than those corresponding to symmetric planar
waveguide.
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Let us look at modal fields. So, these are
the modal fields of TE0, TE1 and TE2 modes
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of a typical asymmetric planar waveguide.
So, I can see that the modal fields are asymmetric
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and you can see that the penetration gap is
more in the substrate and less in the cover
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which is understandable, because here at this
interface the index contrast is smaller as
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compared to the index contrast at this interface
ok.
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So, the field extends more into the substrate
region as compared to in the cover region.
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How many modes are supported? If you go back
to symmetric planar waveguide the number of
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modes are the integer which is closest to,
but greater than , but now all the cut offs
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are shifted by this much amount. So, the number
of modes would now be. So, you obtain this
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number and find out the integer closest to,
but greater than this number.
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What are the cut off wavelengths of various
modes and if I change the wavelength how the
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number of modes would change. So, I can see
from here that the cut off wavelength for
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mth TE mode would be given bywhich comes directly
from the definition of normalized frequency
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V. So, I can write it down also which we will
quite often use.
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. So, from here I get these cut off wavelengths,
now if I find out the cut off wavelengths
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corresponding to various modes then I see
that for TE0 mode the cut off wavelength is
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1.7187 for these parameters of waveguide,
and for TE1 mode it is 0.4914 and for TE2
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mode it is 0.2867.
So, if I start from a longer wavelength let
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us say 2, then until I cross this TE0 mode
is not guided. So, from here to here there
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is no mode guided by the structure, and as
soon as I cross this go below this wavelength,
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then TE0 mode starts appearing and TE0 mode
would be guided now for all the wavelengths
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is smaller than this.
If I further reduce the wavelength and as
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soon as I go below 0.49TE1 mode starts appearing
and below these this wavelength I will have
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both TE0 and TE1 and so on. So, this is how
the waveguide would guide different modes
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if I change the wavelength. How the thickness
affects the guided modes. So, from here I
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can find out the cut off thickness of mth
TE mode from here itself. So, if I find out
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d in terms of V now and put the cut off of
various modes in terms of V, then I can find
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out the cut off of thickness ok.
So, if I have 0 thickness it means no waveguide
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no mode. If I start increasing the thickness
then up to 0.58 there is no mode guided because
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TE0 mode has finite cut off and as soon as
I cross this then TE0 mode starts appearing,
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and when I cross 2.03then TE1 mode starts
appearing and so on. So, as I increase the
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waveguide thickness the number of modes start
increasing I should have the label here which
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is d in. So, the label here is t in micron.
Let us work out few examples. So, I take a
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dielectric step index asymmetric planar waveguide.
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Defined by nf =1.5 and ns =1.48 and nc = 1,
and d = 4. Now the first thing is to calculate
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the number of modes at lambda naught is equal
to 0.5. So, I first calculate the value of
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asymmetric parameter a which comes out to
be about 20, then I find out what is the value
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of V at = 0.5, and this comes out to be about
6.13.
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Then I find out this number. So, this comes
out to be 3.4. So, the number of modes are
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4.
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The second is wavelength range in which the
waveguide does not support anymore. So, I
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know there would not be any more support it
if , which if I put the value V here the expression
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for V here. Then this gives me the condition
in terms ofasshould be greater than this,
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and if I plug in all these numbers I find
out for > 4.5427 the waveguide would not support
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00:26:41,380 --> 00:26:42,380
anymore.
193
00:26:42,380 --> 00:27:00,960
Third is what
194
00:27:00,960 --> 00:27:38,100
is the cut off wavelength of TE2 mode? So,
I find out what is the V value for cut off
195
00:27:38,100 --> 00:27:56,160
of TE2 mode. So, which is given by this with
m/2. So, if I put m /2 then Vc is equal to
196
00:27:56,160 --> 00:28:34,910
this, and the cut off Vc for cut off V for
TE2 mode is this if I translate it to the
197
00:28:34,910 --> 00:28:42,520
wavelength it comes out to be 0.8038.
198
00:28:42,520 --> 00:29:02,060
What is the range of
d so, that only TE0 and TE1 modes are guided
199
00:29:02,060 --> 00:29:13,860
at.
So, I know that for mth mode the cut off
200
00:29:13,860 --> 00:29:39,820
thickness is this, and I want TE0 and TE1
both the modes guided. So, I find out this
201
00:29:39,820 --> 00:30:06,890
dc for both the modes, and I also know thisfor
mth mode is given by this. So, I find outfor
202
00:30:06,890 --> 00:30:23,060
d0 mode, Vc is this for TE1 mode Vc is
203
00:30:23,060 --> 00:30:41,480
this correspondingly if I now find out dc
for these 2 modes, then for TE0 mode cut off
204
00:30:41,480 --> 00:31:08,700
thickness is 0.88, and for this is 2.93. So,
if TE0 and TE1 modes are
205
00:31:08,700 --> 00:31:23,610
to be
guided
206
00:31:23,610 --> 00:33:56,990
then thickness should be
207
00:33:56,990 --> 00:34:15,490
between
208
00:34:15,490 --> 00:34:28,350
this
209
00:34:28,350 --> 00:35:57,440
and this.
So, this is all in this lecture in the next
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00:35:57,440 --> 00:36:01,580
lecture, we will extend the analysis to TM
modes.
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00:36:01,580 --> 00:36:02,650
Thank you.