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Let us continue our analysis on planar dielectric
optical waveguides. If you remember that we
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were doing the analysis for confinement in
x direction and propagation in z direction.
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And we had seen that if the confinement is
in x direction and the propagation is in z
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direction. Then the electric and magnetic
fields associated with the light wave can
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be given as E (x, y, z, t). In fact, here
you need not to take why because I am taking
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the planar waveguide. So, it is simply E (x,
z, t) vector is equal to now it is E e 0 or
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E of x e to the power i omega t minus beta
z.
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Similarly, H is x z t is equal to H of x this
is H of x e to the power i omega t minus beta
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z. So, they represent the modes they represent
the modes E of x and H of x and we know that
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since this is this is propagating in z direction.
Then what we have is basically Ex, Ey, Ez
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all these components will be there and Hx,
Hy and Hz all these components of E and H
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would be there. This is a mode propagating
inside direction and you should remember that
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these E and H are not constant. So, these
are not plane waves, these are not plane waves
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that is why I will have all the components
here ok.
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What I want to find out now is what are the
values of beta and what are the corresponding
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E and H, they define the modes. And we were
doing the analysis of this; we were doing
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the analysis for symmetric step index planar
waveguide where this n of x is defined like
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this. And this waveguide is symmetric about
x is equal to 0, it has high refractive index
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region n1 and low refractive index regions
n2 and the width of the high index region
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is t.
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So, what we did we categorize the modes of
this waveguide using the symmetry of the structure
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into symmetric and antisymmetric modes and
for symmetric modes.
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The field the electric field we were doing
the analysis of TE modes. So, for TE modes
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the field is given by this, A cosine kappa
x for mod x less than d by 2 that is in the
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guiding film. And Ce to the power minus gamma
mod x for mod x greater than d by 2, which
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is in these 2 regions since it is TE mode.
So, the non-vanishing components of E and
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H are Ey, Hx and Hz and Ey Hx and Hz are related
by these 3 equations. So, we apply the boundary
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conditions at the interfaces x is equal to
plus minus d by 2, and the boundary conditions
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are the tangential components of E and H should
be continuous at x is equal to plus minus
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d by 2, and the tangential components at x
is equal to constant are Ey and Hz.
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So, and since Hz is related to Ey if some
constant dEy over dx that is why Ey and dEy
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over dx should be continuous at x is equal
to plus minus d by 2. So, when we applied
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this then we got the transcendental equation
kappa d by 2 tan kappa d by 2 is equal to
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gamma d by 2. Now our task is now our task
is to solve this equation, let us see how
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we can solve this equation and get physical
insight into the modes of a waveguide.
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So, to do that what we do when we define we
what we do we take this transcendental equation
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kappa d by 2 tan kappa d by 2 is equal to
gamma d by 2. And here we see there are 2
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terms appearing which contain kappa d by 2
and gamma d by 2 where kappa square is given
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by this and gamma square is given by this.
So, if I add these 2 up kappa d by 2 square
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plus gamma d by 2 square then I get k naught
square n1 square minus n2 square times d by
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2 square. What does it have? It has all the
waveguide parameters and the wavelength because
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k naught is 2 pi over lambda naught. So, so
by adding these 2 up I get something which
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contains only the waveguide parameter waveguide
parameters, and the wavelength which are the
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inputs basically.
So, let me define this since everywhere there
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is a square. So, let me define this as V square
and where V is k naught d by 2 square root
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of n1 square minus n2 square or 2 pi over
lambda naught times d by 2 square root of
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n1 square minus n2 square. And this I call
normalized frequency, because it contains
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lambda in the denominator. So, I call this
normalized frequency or generally I also call
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it V parameter. So now, in order to in order
to solve the transcendental equation graphically
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what I do I define this kappa d by 2 this
kappa d by 2 as some parameters psi, some
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variable psi.
Because this is the transcendental equation
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in beta kappa contains beta the only unknown
here is beta. So, let me define kappa d by
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2 is equal to psi and therefore, if I put
this kappa d by 2 is equal to psi here then
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gamma d by 2 automatically becomes a square
root of V square minus psi square. So, if
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I put these 2 here in this equation the then
the transcendental equation becomes psi tan
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psi is equal to V square minus psi squares
square root.
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So, in this equation I have all the input
parameters in the form of V. So, if I have
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all the waveguide parameters with me that
is n1, n2 and d. And I have the wavelength
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at which I am operating my waveguide then
I can calculate V from here and for that value
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of V. I can solve this transcendental equation
and get the values of psi and hence the values
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of beta. Similarly, if I do the analysis for
antisymmetric modes in the same way then it
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leads to the transcendental equation which
is given by minus psi cot psi is equal to
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square root of V square minus psi square.
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So, the guided modes of a planar waveguide
where a symmetric or antisymmetric there it
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fall in the range of an effective line from
n2 to n1, and they are given by Ey of x is
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equal to A cosine kappa x for mod x less than
d by 2 and Ce to the power minus gamma x for
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mod x greater than d by 2 for symmetric modes
this is the symmetric solution. And the values
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of beta which are allowed are given by psi
tan psi is equal to square root of V square
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minus psi square.
So, these are symmetric modes, and similarly
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the antisymmetric modes satisfy the equation
minus psi cot psi is equal to square root
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of V square minus psi square and the model
field is given by in terms of sin function
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sin kappa x in the guiding film and again
exponentially decaying function in the in
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the n2 region or lower index surrounding.
So, by solving these equation first I need
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to get the values of beta, only certain values
of beta or certain values of psi are allowed,
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I first need to find out those values of beta
and from those values of beta I can find out
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kappa and gamma those values I put here in
the field and then I get the model field that
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is how I will get the complete solution for
the modes of planar dielectric waveguide.
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So, let us solve it. So, I have again symmetric
modes antisymmetric modes in order to solve
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these 2 equations what I do I equate both
the LHS and RHS of these equations to some
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variable eta. So, in this way I will have
sets of 2 simultaneous equations for symmetric
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modes as well as for antisymmetric modes,
and there are eta is equal to psi tan psi
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from here and if I equate this to eta then
it gives me psi square plus eta square is
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equal to V square. These 2 equations are for
symmetric modes and similarly for antisymmetric
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modes I get the equations eta is equal to
minus psi cot psi and psi square plus eta
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square is equal to V square.
Now, I can obtain the solutions of these equations
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by graphical method. So, what I need to do
I just plot these 2 equations on psi eta plane
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and look for the points of intersection. Those
point of intersection will give me the allowed
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values of psi and hence the allowed values
of beta ok.
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So, let us plot these. So, here the red line
shows the equation eta is equal to psi tan
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psi, which corresponds to symmetric modes.
And these blue lines they show the plot for
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eta is equal to minus psi cot psi they correspond
to antisymmetric modes. And these circles
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for different values of V are basically psi
square plus eta square is equal to V square.
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So, for a given V for a given value of V,
I take the circles. And see the points of
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intersections of these circles with these
blue and red curves and they will give me
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the allowed values of psi. So, let us examine
these. So, V is this. So, for a given waveguide
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that is n1 n2 and d and given wavelength that
is lambda or lambda naught, V is known. If
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this V lies between 0 and pi by 2 if this
V lies between 0 and pi by 2 we can see that
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there would only be one point of intersection.
And that would be with the red curve that
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is corresponding to symmetric modes.
So, this point of intersection will give me
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the value of psi for this symmetric mode.
So, in this range if V lies in the range 0
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to pi by 2, then there is only one mode, which
is possible and this mode is symmetric mode.
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Because it is obtained from the intersection
of red curve with the circle and red curve
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corresponds to symmetric modes. Now if V lies
between pi by 2 and pi for example, if V is
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equal to 2.5. I see 2 points of intersection.
One is with red and one is with blue. So,
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one is with symmetric and another is with
antisymmetric. So, I get 2 modes one symmetric
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and one antisymmetric. Then if I look in the
range from pi to 3 pi by 2 if V lies from
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pi to 3 pi by 2 for example, V is equal to
4 then I get 2 symmetric modes and one antisymmetric
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mode ok.
So, in this wave for any given value of V.
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I can find out the number of modes the number
of symmetric modes the number of antisymmetric
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modes and their propagation constants.
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So, in general the number of guided modes
I can find by looking at the value of V if
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V lies between m pi and m plus half pi where
m is an integer 0, 1, 2 and so on, then I
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will get m plus 1 symmetric modes and m antisymmetric
modes. On the other hand if V lies between
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m plus half pi and m plus 1 pi then I will
get m plus 1 symmetric modes and m plus 1
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antisymmetric modes.
In general, I can get the modes for any value
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of V and I labeled these modes as TEm mode
where m is equal to 0 1 2 and so on. So, the
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very first mode is T E 0 mode than there is
TE1, TE2 TE3 and so on. So, that is how the
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modes are labelled. Now the number of modes
I can also find out by looking them at V line.
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So, if I have V line and I divided into the
sections of pi by 2, then what I see that
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from 0 to pi by 2 there would be TE0 mode
if V lies somewhere here than TE0 and TE1
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if V lies somewhere here than TE0, TE1, TE2
and so on.
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So, every pi by 2 I am adding one mode. So,
in general if I know the value of V then the
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total number of modes can be found out by
dividing this V by pi by 2. So, I divide this
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V by pi by 2 and get a number then the number
of modes would be an integer closest to, but
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greater than that number closest to, but greater
than that number. So, if this V over pi by
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2 comes out to be 2.1, then the number of
modes would be 3. Now what is the single mode
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waveguide?
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If V lies between 0 and pi by 2 that is V
is less than pi by 2, then there is only one
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mode supported that is TE0 mode and this TE0
mode is the fundamental mode of the waveguide,
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I give you an example, if n1 is equal to 1.5,
n2 is equal to 1.48 and d is equal to 3, then
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this waveguide can be single moded or multi-moded
it depends upon the wavelength I am using
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because V contains wavelength also apart from
these 3 parameters.
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So, so by just looking at waveguide parameters
I cannot say whether the waveguide a single
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moded or multi-moded I will have to say it
is single moded at this wavelength or it is
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single moded in this range of wavelengths.
So, I can find out that range. I know that
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V is equal to 2 pi over lambda naught times
t by 2 times square root of n1 square minus
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n2 square and if V is less than pi by 2 then
the waveguide a single moded than this gives
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me that for these parameters of waveguide
if lambda naught is greater than 1.46 micrometer
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then the waveguide is single moded. That is
for all the wavelengths longer than 1.46 micrometer
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the waveguide is single moded and for all
the wavelengths which are smaller than 1.46
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micrometer the waveguide is multi-moded.
For all the wavelength shorter than 1.46 micrometer
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the waveguide is multi-moded. Let us do few
more examples. So, let us consider a dielectric
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step index symmetric planar waveguide, with
n1 is equal to 1.5 n2 is equal to 1.46 and
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d is equal to 4 micrometer. For this waveguide
let us calculate the number of symmetric and
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antisymmetric modes at wavelength lambda naught
is equal to 1 micrometer.
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First part and second part the waveguide thickness
for waveguide to be single moded at lambda
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naught is equal to 1 micrometer.
Well, so let us first solve the first part,
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I know for the number of modes what I should
do? I should find out the value of V and divided
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by pi by 2. So, I should calculate V over
pi by 2. V over pi by 2 comes out to be 2
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d over lambda naught times square root of
n1 square minus n2 square, because we know
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V is equal to V is equal to 2 pi over lambda
naught timesd by 2 times square root of n1
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square minus n2 square. So now, I put the
value of d lambda naught n1 n2 here and I
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find out that this V over pi by 2 comes out
to be 2.7527.
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Which means that the number of modes should
be an integer closest to, but greater than
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this and there it is why the number of modes
would be 3. If there are 3 modes what modes
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are supported? Well we start from 0. So, TE0
then TE1 and TE2; TE0, TE1 and TE2 and we
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see that all the even number modes are symmetric
modes and odd number modes are antisymmetric
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modes that is how we got the solutions also.
So, so here we have 2 even number modes and
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one odd number modes. So, so we have 2 symmetric
modes and one antisymmetric mode.
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Let us do the second part, waveguide thickness
for waveguide to be single moded. We had seen
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that that if the waveguide thickness is 4
micrometer, then at lambda is equal to 1 micrometer
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the waveguide supported 3 modes. Now if I
still want to operate at lambda naught is
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equal to 1 micrometer and want my waveguide
to be single moded, then what should be the
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value of d? That is what I need to find out.
So, I know again for single mode waveguide
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V should be less than pi by 2, which means
2 pi over lambda naught times d by 2 times
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square root of n1 square minus n2 square should
be less than pi by 2 or d should be less than
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lambda naught divided by 2 square root of
n1 square minus n2 square.
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Now, I just put the values of lambda naught
and n1 n2 it gives me for all the values of
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d is smaller than 1.5431 micrometer this waveguide
would be single moded at lambda naught is
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equal to 1 micrometer. I can also find out
the number of modes with wavelength. So, if
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I have a given waveguide let us say defined
by n1 is equal to 1.5 n2 is equal to 1.46
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and d is equal to 4 micrometer.
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Then, if I change the wavelength the number
of modes will change. If I start if I start
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from a wavelength which is say 2 micrometer
then there would be only one mode supported,
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let us say and that would be TE0 mode. If
I decrease the wavelength now gradually then
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up to about little less than 1.4 micrometer
there would be only one mode supported, that
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is TE0 mode.
As soon as I cross this then TE1 mode will
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also come into the picture. So, here I will
have TE0 and TE1 up to this point let us this
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is about 0.9 micrometer as I further decrease
the wavelength below 0.9 micrometer, then
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TE2 mode will also come into picture now here
you will have TE0, TE1 and TE2 mode up to
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about let us say it is little less than 0.7
micrometer. And then we I will have TE3 mode
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and so on. So, so as I decrease the wavelength
as I decrease the wavelength of operation
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in a given waveguide the number of modes will
change.
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So, I go to shorter wavelengths I increase
the number of modes. So, this is how I can
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find out how many modes would be supported
and how the number of modes will change with
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wavelength, how the number of modes will change
with waveguide parameters. What is still remains
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to find out is the values of beta, the values
of propagation constant, the values of the
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model fields the expression for model field
the field plots. So, for that what I need
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to do now I have already got the value of
beta, but approximate value of beta. This
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approximate value of beta I can refine by
using numerical techniques. I can solve the
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transcendental equation by various numerical
techniques and I can find out the much more
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accurate value of beta and from those values
of beta I can find out the model field ok.
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So, in the next lecture we would look into
the model fields and before that. In fact,
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I would also like to define these transcendental
equations in terms of normalized parameters.
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We have already defined one normalized parameter
which is normalized frequency V. I would like
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to also define a normalized parameter with
respect to propagation constant. So, normalized
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propagation constant and the idea of defining
these normalized parameters is to have certain
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universal curves certain universal solutions
which do not depend upon waveguide parameters
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or the wavelength. So, in the next lecture
we will do extend all this analysis by using
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the normalized parameters and then we will
look into the model fields also.
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Thank you.