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In the last lecture we had seen that if for
a planar waveguide V is less than pi by 2,
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then the number of modes is one. It supports
only one mode and when V lies between pi by
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2 and pi there are 2 modes and if it is between
pi and 3 pi by 2, then there are 3 modes and
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so on. If I keep on increasing the value of
V then the number of modes would increase,
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but I also see that the propagation constants
can be approximated by certain expression,
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and we can find out the propagation constants
even without solving the transcendental equation
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for first few modes. So, let us see how.
So, if it is a highly multimode waveguide
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where V is much larger than 1 typically V
is more than 10 or so then What I see if I
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now again plot these transcendental equations
this is this is eta is equal to psi tan psi
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and the blue one is eta is equal to minus
psi cot psi.
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Now if I plot the sections of circles for
large values of V let us say V is equal to
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10 or V is equal to 18 or 20, what do I see
there for the first few modes the points of
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cross sections are somewhere here.
So, for first mode that is m is equal to 0
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mode the point of intersection is close to
pi by 2. And for m is equal to 1 the point
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of intersection is close to pi and so on.
So, and these are only for first few modes,
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if you go to higher order modes than these
points of intersections are much further from
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these values. So, what I can say that for
first few modes, the points of intersections
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can be given psi m is equal to m plus 1 pi
by 2, where m is equal to 0 1 or 2. So, for
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m is equal to 0 it is pi by 2 for m is equal
to 1 it is pi and so on.
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So, in this case in this case I can find out
the beta in this way, I know psi m is equal
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to kappa m d by 2. So, it is equal to m plus
1 pi by 2 and kappa m is k naught square n1
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square minus beta m square is square root.
So, so this gives me beta m square is equal
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to k naught square n1 square minus m plus
1 square pi square by d square. Now I can
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increase the value of V by 2 ways I can have
large value of V by 2 ways - one is that if
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I have very large value of d. If I have large
value of d then you can see there that is
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beta m will approach to k naught n1 beta m
will approach to k naught n1. And it is obvious
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it if I start if a start from a very thin
waveguide then there are discrete modes and
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they have propagation constants which lie
between k naught n2 and k naught n1.
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But when I increase this width of the waveguide
and the width is very large as compared to
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lambda then it is as good as infinitely extended
medium. So, so the light will propagate with
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propagation constant k naught n1 where n1
is the refractive index of the medium. So,
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I will approach towards infinitely extended
medium if d is very large. Now if I obtain
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the large value of V by having very large
index contrast, that is very large difference
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between n1 and n2, then what happens is there
is a huge index contrast and this very large
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index contrast will give very high reflection
a very high reflection ok.
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So, it is it would be difficult for the wave
to penetrate into n2 region. So, it would
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be something like it would be very close to
what is planar mirror waveguide and in that
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case you can see the expression for these
beta closely resemble to the expression of
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beta for planar mirror waveguide. Let us work
out some more examples here.
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I consider a step index planar symmetric waveguide
with n1 is equal to 1.5, n2 is equal to 1.4
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and d is equal to 4 micro meter. And let us
calculate the number of modes at 0.5 micrometer
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and effective refractive indices or effective
index of modes of TE0 and TE1 modes.
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So, again to find out the number of modes
it is the same I find out the value of V.
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So, when I find out the value of V. It comes
out to be 13.53 at lambda naught is equal
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to 0.5 micrometer if I divide this value of
V by pi by 2. This number comes out to be
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8.61.
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So, the number of modes supported would be
9. I can see this value of V is very large.
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So, it would be now easier for me to find
out the propagation constants of TE0 and TE1
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mode, even without solving the transcendental
equation.
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Since it is much larger than 1. So, for the
first few modes the values of beta would be
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given by this, and for TE0 mode m is equal
to 0 and for TE1 mode m is equal to 1 and
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in this way if I find out the values of beta,
then they come out to be 18.83 and 18.78 correspondingly
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the effective index of TE0 mode comes out
to be 1.498698, and for TE1 mode 1.494783.
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You can see the very close to the refractive
index of the guiding film which is 1.5 ok.
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Now, let us do extend all this analysis by
using normalized parameters. So, we had symmetric
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modes antisymmetric modes.
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And the transcendental equations were in terms
of psi and V. Now I already have one normalized
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parameter which is normalized frequency V
which contains all the waveguide parameters
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and wavelength. Let me also define normalized
propagation constant, that is let me normalized
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beta and how do I normalized beta yet I take
beta over k naught square minus n2 square
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divided by n1 square minus n2 square. If I
normalized beta in this way then for guided
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modes the value of b lies between 0 and 1.
So, for guided modes the value of b lies between
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0 and 1. So, this is now normalized propagation
constant. Now my task is to replace that psi
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with b and V. So, how do I do this? So, this
is the transcendental equation for symmetric
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mode this is the transcendental equation for
antisymmetric mode. Let me find out how can
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I how can I get psi in terms of b and V. So,
psi square is equal to kappa square d square
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by 4. So, this is nothing but d square by
4 and kappa square is k naught square n1 square
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minus beta square. I do a little mathematical
manipulation I add and subtract k naught square
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in 2 square here, and then I regroup these
2 terms and these 2 terms.
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So, if I group these 2 terms and these and
also I take this term outside. Why I am doing?
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So, because I can immediately identify that
if I club this term with this I get d square.
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So, I do that. So, this becomes psi square
is equal to k naught square d square by 4
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n1 square minus n2 square times one and from
here I take minus outside and I write it down
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as beta over k naught square minus n2 square
divided by n1 square minus n2 squares. So,
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this is nothing but b. So, this is b this
is V square.
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So, what I have got now from here psi square
is equal to V square and this is 1 minus b.
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So, this gives me psi is equal to V times
square root 1 minus b. So, as soon as I get
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the psi in terms of V and b I put it in these
equations and I get the transcendental equation
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in terms of b and V only.
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So, for symmetric mode this transcendental
equation will transform to V times square
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root 1 minus b tan V is square root of 1 minus
b is equal to V times square root of b ok.
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Because from here V square minus psi square
V square minus psi square would be V times
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square root of b. And for antisymmetric mode
this equation will become this. From here
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I can find out the cut off condition of a
mode. A mode is said to be cut off when it
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is propagation constant approaches k naught
n2. I know that if an effective for guided
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modes I know for guided modes an effective
or beta over k naught beta over k naught beta
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over k naught or an effective lies between
n2 and n1.
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So, if this beta over k naught approaches
n2 then a particular mode is cut off. So,
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mode is guided when the when the effective
index of the mode lies between these 2, but
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as soon as the effective index goes below
this then it is no more guided. So, this value
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of an effective will give me the cutoff. This
beta over k naught is equal to n2 gives me
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b is equal to 0. So, this is the cut off condition
in terms of b. So, from here I can immediately
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find out for what values of V for what values
of normalized frequency a mode would be cut
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off. So, I simply put b is equal to 0 here
then the solutions of the equation that I
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get will give me the cutoffs of the modes.
So, this will give me if I put V is equal
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to 0 sorry b is equal to 0 and V is equal
to Vc which symbolizes the cutoff frequency
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cutoff normalized frequency, then it gives
me Vc tan Vc is equal to 0 for symmetric modes.
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And for antisymmetric modes it will be Vc
cot Vc is equal to 0. So, these 2 together
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will give me the cutoff of mth mode is m pi
by 2 where m is equal to 0 1 2 and so on.
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So, TE0 mode will have no cut off m is equal
to 0, TE1 mode will have a cutoff of pi by
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2 TE2 mode will have cut off pi and so on.
So, this is how I get the cutoffs of the modes.
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Now very beautiful thing about these normalized
parameters is that they are independent of
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waveguide parameters. I get these 2 equations
and these 2 are purely mathematical equations
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2, purely mathematical transcendental equations.
What I can do? I can solve these equations
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for I can solve these equations for different
values of V. And I find out the roots of these
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and plot these roots as a function of V.
So, I start with I start with a very small
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value of V. When I do this as long as the
value of V is less than pi by 2, then what
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I see that there is no root coming out from
this equation and there is only one root coming
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out of this equation. I find out that root
for different values of V and plot it, it
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go something like this. As soon as the value
of V crosses pi by 2 1 root starts one root
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starts appearing from this equation. And the
value of V up to pi I will get one root from
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here, one root from here I keep plotting them.
And then I further increase more and more
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roots start appearing and I plot them ok.
The very first root which comes from here
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which is closer to which is closer to 1 the
first root which is closer to 1 is TE0 mode.
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The first root from here is TE1 mode the second
root from here is TE2 mode and so on. So,
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I plot them, once I have plotted them then
these curves are these curves are universal
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curves. So, once I have done this exercise
once I have done this exercise then it is
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forever these are universal curves, now what
I can do using these curves?
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If there is a given waveguide and wavelength
then I simply calculate the value of V this
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should be V naught and there should not be
any 0 here. So, so I just calculate the value
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of V, and then I go to these curves and read
the value of b corresponding to that value
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of b for example, if I get this value of V
let us say V is equal to 1.4 then I draw a
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vertical line corresponding to V is equal
to 1.4 and read the value of b corresponding
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to this from b-V curves. Once I get the value
of b then I can find out the effective index
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of the mode from here, this is coming out
from the very definition of the normalized
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propagation constant.
If I am somewhere here, if I am somewhere
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here if the value of V is let us say 3.5.
So, it will go like this I will have 3 points
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of intersection and corresponding to that
I can read the value of b. So, I will get
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the propagation constants of all the 3 modes.
So, in this way I can find out the propagation
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constants for any symmetric planar waveguide.
Let us do some examples. So, if I have a waveguide
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with n1 is equal to 1.5 n2 is equal to 1.48.
d is equal to 2 micrometer and I operate it
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lambda naught is equal to 1.5 micrometer.
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If I calculate the value of V from here it
comes out to be 1.0226. I go back to those
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b-V curves and read the value of b from here
corresponding to this value of V it comes
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out to be 0.4636. And then I find out the
value of an effective, which comes out to
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be 1.4893053 ok.
Now, I change my waveguide parameters a little.
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So, I now have n1 is equal to 1.5. n2 I have
change to 0.418. D is equal to 1 micrometer
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and lambda naught is equal to 1.5. I have
artificially change these parameters for achieving
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the same value of V actually. What I want
to demonstrate is that, if my waveguide is
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different d one if my waveguide is different,
but it gives the same value of V then the
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value of b is always the same. For same value
of V even if the waveguide is different the
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value of b normalized propagation constant
is the same, but the effective index would
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be different now, because n1 and n2 are different
n1 and n2 are different. So, in this case
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the effective indexes this while in this case
the effective index would be this.
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So, different waveguides can have same value
of V and hence the same value of b which means
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that these b-V curves are universal and they
do not depend upon waveguide parameters. Let
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us now look at modal fields, how do the modal
fields look like. Let us plot the field for
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TE0 mode TE0 is symmetric mode.
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So, the modal field will be given by Ey of
x is equal to A cosine kappa x in the region
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mode x less than d by 2, and Ce to the power
minus gamma mod x for mod x greater than d
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by 2. I know for TE0 mode psi would lie between
0 and pi by 2, because V would lie between
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0 and pi by 2. So, the point of intersection
the value of psi for point of intersection
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will always lie between 0 and pi by 2.
It will not exceed pi by 2. What is psi? Psi
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is kappa d by 2. So, kappa d by 2 will lie
between 0 and pi by 2 now let us look at the
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waveguide. And the modal filed in the region
mode x less than d by 2 the solution is cosine
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function. So, I will have cosine kappa x what
would be the value at the boundary? What would
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be the value at the boundary? At the boundary
I have kappa x is equal to kappa d by 2 and
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minus kappa d by 2, but kappa d by 2 is always
less than pi by 2, if kappa d by 2 is always
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less than pi by 2 then cosine kappa d by 2
will never cross any 0. The first 0 of cosine
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function will occur at kappa x is equal to
pi by 2, but I have even at the boundary kappa
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x is always less than pi by 2. So, I would
not have any 0 crossing.
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So, Ey in the guiding film will be like this.
And then in the lower refractive index region
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the exponential decaying part exponentially
decaying solution will take over and it will
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go like this. So, I will have the modal field
which would look like this all right. What
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I see that there is oscillatory solution here
and decaying solution here. The field extends
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field extends to lower indexed region also.
How much does the how much does this extend?
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Into this region I can quantify this, I find
that at x is equal to d by 2 plus 1 over gamma
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because this field is going as e to the power
minus gamma times x. So, at x is equal to
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plus x is equal to 1 over gamma this field
will decay to 1 over e of it is value at the
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boundary.
So, if I go if I start from here and mover
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distance 1 over gamma from the boundary then
the field will decay to 1 over e of it is
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value at the boundary. Then this is how I
can quantify this distance up to which this
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field extends into the lower index region.
So, it is defined by 1 over gamma, and it
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is known as penetration depth into lower index
region. So, 1 over gamma is penetration depth
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into lower indexed region let us plot the
field now for TE1 mode. TE1 mode is antisymmetric
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mode and the solutions would be given by sin
function in the in the guiding film and exponential
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decaying function again in the lower index
medium.
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Now for TE1 mode, psi would lie between pi
by 2 and pi psi would lie between pi by 2
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and pi, because this the green curve will
cut the blue curve only in this region. Now
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if I which means that kappa d by 2 would lie
between pi by 2 and pi, and now in the same
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way if I plot the field in the guiding film
first. So, there would be a sin function ok.
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But at the boundary it has to be less than
pi. So, kappa d by 2 has to be less than pi.
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So, it would not cross another 0 there would
only be 1 0 which is at x is equal to 0. So,
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in the guiding film the solution would be
like this and in the lower index region the
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00:25:04,000 --> 00:25:11,620
exponential decaying term will take over and
this is how the modal field would look like.
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So, I can see that if I go back then here
for TE0 mode there was no 0 in the modal field
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there was no 0 crossing for TE1 mode there
is only one 0 crossing.
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And if I go on then TE2 mode has two 0 crossings
TE3 mode has three 0 crossings which means
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that if it is TEm mode then it will have m
number of 0s ok. It will have m 0 crossings,
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this is one thing another thing that I see
is when I solve this, that with the number
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of modes the propagation constant decreases
right, if you remember the b-V curves.
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So, you have this is V which is normalized
frequency this is b which is normalized propagation
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constant, and b always lies between 0 and
one for guided modes. I had seen that TE0
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mode goes something like this. This is TE0
mode TE1 mode it goes something like this,
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this is TE1; TE2 mode like this. So, if I
have a value of V somewhere here, then the
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propagation constant of TE0 mode is here,
propagation constant of TE1 mode is here,
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propagation constant of TE2 mode is here.
So, the propagation constant decreases as
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the ode number increases.
I also see that the lower order modes are
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more confined into guiding film. So, you can
see that TE0 mode it has more power here.
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And as I and the and the tail into n2 region
is a smaller, but as I go to higher order
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modes this tail extends more and more. So,
if the number of mode increases if the number
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of mode increases then the corresponding confinement
in the mode decreases.
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I can see the effect of this penetration depth
I can see this penetration depth. So, for
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a given waveguide if beta decreases because
penetration depth is defined by 1 over gamma
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and gamma is equal to beta square minus k
naught square n2 square it is a square root.
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So, if beta decreases then, then if beta decreases
then gamma decreases which means penetration
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depth increase and beta decreases with mode
number. So, that is why the penetration depth
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increases. So, if I have this waveguide and
this wavelength and I see the mode profiles
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of different modes. So, this is TE0 mode it
has penetration depth which is about 0.47
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micrometer for TE1 mode this is 0.52 micrometer
and for TE2 mode it is 0.65 micrometer.
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What is the effect of index contrast on penetration
depth? Well again, the penetration depth is
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given by this, so for a given n1d and lambda
naught. Now if I change the value of n2 and
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hence the index contrast, for example, if
I decrease n2 if I decrease n2 then the value
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of V increases. The value of V increases because
n1 square minus n2 square increases. If V
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00:29:30,610 --> 00:29:38,380
increases the propagation constant if V increases
the propagation constant increases, beta increases
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if beta increases gamma increases and penetration
depth decreases.
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00:29:44,380 --> 00:29:54,370
So, this is how it looks like. So, if I have
n2 is equal to 1.48. Then penetration depth
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is 0.47 micrometer. For TE0 mode for the same
mode if I change the refractive index n2 from
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one point 48 to 1.3 then you can see the penetration
depth decreases the field is pushed more towards
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the guiding film. And if I change to one then
it is even more confined into the high index
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region and penetration depth decreases to
0.1.
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What is the effect of wavelength if for a
given waveguide if I decrease the wavelength?
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If I decrease the wavelength then I am increasing
the value of V which means I am increasing
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the value of propagation constant ok.
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If I am increasing the value of b that is
beta square minus k naught square n2 square
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increases gamma increases. So, penetration
depth decreases. So, this is the field of
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a given waveguide when operated at 1.55 micrometer
you can see how much the field extends outside
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and when I decrease the wavelength to 1.3
the field is pushed more towards n1 region
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and when it is 0.7 micrometer than the penetration
depth is much smaller.
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So, this is the; this is how this is how the
modal field would vary with different parameters
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of the waveguide. And wavelength in the next
lecture we will have more inside into the
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modes and we will understand what these modes
basically are. We will have physical understanding
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of modes.
Thank you.