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In the last lecture we had done the modal
analysis of optical fiber and we had obtained
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the transcendental equation satisfied by the
propagation constants of the modes, we had
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also seen the cut offs of various modes.
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Now in this lecture we will extend the analysis
and see how the modal fields look like.
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So, this is the step index fiber which we
are analyzing, which has a high index core
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of refractive index n1 of radius a and the
cladding of refractive index n2.
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And we have done this that the radial part
of the modal fields is given by A Jl Ur over
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a in the region r less than a which is the
core, and B Kl Wr over a in the region r greater
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than a which is the cladding.
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We had also defined the normalized frequency
V normalized propagation constant b and U
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and W are defined by these relations and you
can also express them in terms of normalized
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parameters V and b.
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So, we had seen that the propagation constants
of the linearly polarized modes of this fiber
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satisfy these transcendental equations or
the Eigen value equations.
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So, after solving these equations for a given
value of V, I can find out the propagation
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constants b of various modes of the fiber,
and if I plot them as a function of V, so
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they look like this.
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So, this we had done.
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Now, let us look at how the modal fields look
like.
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So, the total modal field would be given by
the radial part and the angular part or azimuthal
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part.
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In phi direction I can have solutions cosine
l phi and sin l phi and I label these 2 solutions
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by 2 different names I call cosine alpha solution
as even mode and sin alpha solution as odd
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mode.
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Now, let us look at the modal fields of LP01
and LP11 modes of this fiber.
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So, since we are looking at LP01 and LP11,
which means here l is equal to 0 here l is
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equal to 1.
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So, in the core I will have solution for LP01
mode as J0 Ur over a, and for LP11 mode J1
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Ur over a.
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So, let me plot J0 x and J1 x, they look like
this.
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In the cladding I will have the solutions
K0 Wr over a and K1 Wr over a.
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So, let me plot how K0 x and K1 x look like.
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So, when I combine this k.
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So, in the core I will have this solution
and in the cladding I will have this solution,
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and since these are the first modes of l is
equal to 0 and l is equal to 1 respectively.
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So, I will not have any 0 in this and I will
not have any 0 in this except a 0 at r is
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equal to 0.
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So, for LP01 mode in the core the field will
go like this, and in the cladding this function
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will take over.
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For LP11 mode the field in the core would
be given by J1.
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So, it would go like this without any 0 except
a0 at r is equal to 0 and then this K1 function
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will take over.
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So, this is how the radial part of the modal
fields would look like, what about the angular
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part?
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Angular part is simple for l is equal to 0
you have cosine alpha is equal to 1 and there
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is no contribution from sin l phi because
it would be 0 everywhere.
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So, I will have only this kind of solution.
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So, you take this in radial part and then
you go in angular part rotate it in phi direction
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and you will get this kind of variation.
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However, in for LP11 mode, I can have cosine
phi solution and sin phi solution.
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So, if I take cosine phi solution that is
at phi is equal to 0.
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I have a maximum, and then phi is equal to
pi by 2.
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I will have 0 and so on.
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So, if I rotate it like this in phi direction,
then I will get this kind of variation this
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kind of density plot or intensity pattern.
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If I take sin phi solution, then sin phi would
be 0 at phi is equal to 0 and then it would
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be maximum at phi is equal to pi by 2 and
then if I now rotate it.
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So, it would be 0 at phi is equal to 0 it
would be maximum at phi is equal to pi by
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2.
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So, it will give me this kind of solution.
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So, l is equal to 0 mode will only be of this
kind that is that is I cannot have 2 fold
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degeneracy for l is equal to 0 mode as I can
have for l is equal to 1 mode here or l is
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equal to nonzero mode l nonzero mode.
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I can look at the 3 D plot of these.
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So, if I look at the various modes.
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So, this is how the LP01 mode would look like
and if you view it from the top it would look
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like this.
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LP02 mode LP02 mode will admit 10 because
it is the second mode in l is equal to 0 series.
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So, it will admit 10 in the core.
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So, it will go down this is the intensity
plot this is the intensity plot.
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So, if you look at look at it from the top
it would look like this and there would be
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a 0 at r is equal to something here in the
core itself.
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If you look at LP11 mode it would look like
this, it would not have any 0 in r direction
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except a 0 at r is equal to 0 and these are
the this is the plot of even mode even LP11
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mode.
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So, if I again look at these plots, now I
have LP mode say or linearly polarized modes
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it means that I can have 2 independent or
orthogonal polarization states, I can label
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them as x polarized and y polarized because
that is the direction of propagation.
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So, I can label them as x polarized and y
polarized.
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So, if I look at LP01 mode, then it can be
x polarized or y polarized.
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So, this LP01 mode is twofold degenerate and
it does not have any 0 in r direction.
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If you look at LP11 mode then I can have cosine
phi solution, and in cosine phi solution itself
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I can have 2 polarizations y polarized and
x polarized and I can have sin phi solution.
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00:08:27,649 --> 00:08:31,419
So, there also I can have x polarized and
y polarized.
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So, it would be 2 it would be fourfold degenerate,
and again there would be no 0 in r direction
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except a0 at r is equal to 0.
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If I look at LP02 mode, then LP02 mode would
again be 2 fold degenerate you will have x
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polarized and y polarized.
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If you look at the second mode in l is equal
to one series that is LP12 mode, then it will
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00:09:02,720 --> 00:09:08,180
have one 0 in r direction, except a 0 at r
is equal to 0.
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So, the modal fields would look like this
and again this would again be fourfold degenerate.
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This is LP21 mode; now in LP21 mode because
l is equal to 2.
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So, l is equal to 2 though.
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So, the 5 solutions are cosine 2 phi and sin
2 phi and they will have 4 zeros in phi directions
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that you can see here
and in r direction there would not be any
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0 at except at r is equal to 0.
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So, so this is how the modal fields would
look like for even and odd modes.
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So, in general what I get for LPlm mode, the
number of zeros in phi direction would be
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2 l and the number of zeros in r direction
would be m minus 1 except any 0 at r is equal
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to 0.
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So, if I have various mode I can immediately
find out the number of zeros in r and phi
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direction for example, in LP21 mode there
would be 4 zeros in phi direction and no 0
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in r direction.
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I always exclude a0 at r is equal to 0, for
LP31 mode, I will have 6 zeros in phi direction
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and no 0 in r direction.
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For LP54 mode I will have 10 zeros in phi
direction, and 3 zeros 4 minus 1 3 zeros in
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r direction.
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So, if I know any mode if I am given any mode
then I can immediately find out the number
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of zeros in phi and r direction, and I can
also plot the intensity patterns of those
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modes.
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For example, these are the intensity plots
of various higher order modes, this is LP32
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mode LP32 mode will have 6 zeros in phi direction.
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So, I have 6 zeros 1, 2, 3, 4, 5, 6 zeroes
in phi direction, and 2 minus 1 that is one
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0 in r direction except a 0 at r is equal
to 0.
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LP43 mode will have 8 zeros in phi direction
and 2 zeros in r direction.
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LP54 mode 10 zeros in phi direction and 3
zeros in r direction 1 2 3.
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LP63 mode 12 zeroes in phi direction 2 zeroes
3 minus 1 2 zeroes in r direction and so on.
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So, if I am given any mode then I can immediately
draw the intensity pattern corresponding to
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that mode.
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Now, the question is how do I estimate the
number of modes here, in planar wave guide
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it was easy I just calculate the value of
V divide that value of V by pi by 2.
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And find out the closest, but greater integer
and that will give me the number of modes;
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here the cut offs are not evenly spaced and
there are several series corresponding to
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l.
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So, where do they fit I do not know immediately.
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So, what is the procedure of estimating the
number of modes here?
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So, the first step is the same you first calculate
the value of V.
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So, if you are given a fiber and the wave
length, first you calculate the value of V,
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then you find out the cut offs of various
modes using the cut off conditions and the
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zeros of Bessel functions.
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So, if you have the zeros of Bessel functions
and cut off conditions for various modes you
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know for l is equal to 0 mode, the cut off
condition for l not equal to 0 cut off conditions
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which are in terms of the zeros of Bessel
functions.
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So, you find the cut offs and then arrange
the modes in the increasing order of their
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cut offs.
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So, for example, the fundamental mode LP01
mode has 0 cutoff, LP11 mode has 2.4048 3.8317
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is for LP02 and LP21, 5.1356 is LP31, 5.5201
is LP12 and so on.
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So, you arrange them arrange the modes in
increasing order of their cut offs, and then
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locate the value of V here and find out how
many modes are above that value of V and those
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would be the number of modes supported by
the fiber.
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Let me work out an example if for a given
fiber and given wavelength I calculate the
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value of V and it comes out to be 5.3 then
I locate this 5.3 in this table which is here
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and I count the modes above this, 1, 2, 3,
4, 5.
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So, these modes LP01, 11, 02, 21 and 31 these
modes are guided.
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I can also find out the total number of modes
by including their degeneracies, I find that
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there are 2 modes which have l is equal to
0.
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So, there will be only 2 fold degeneracy.
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So, they comprise of 4 modes, and then I have
1 2 3 modes which are l is equal to non L
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as L is equal to not 0.
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So, so they will have 4-fold degeneracy and
they will comprise of 12 modes.
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So, I will have 12 plus 4.
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16 total modes including the degeneracies.
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So, this is how I can calculate the number
of modes.
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Of course, if V is less than 2.4048, then
there would only be one mode supported of
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course, there would be it is twofold degenerate.
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00:16:25,800 --> 00:16:33,059
So, rigorously it supports 2 modes corresponding
2 polarizations, but as a convention I call
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it single mode fiber because for a given polarization
it has only mode.
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So, for V less than 2.4048 I have a single
mode fiber and this is how the cut off condition
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or for single mode operation is coming; because
2.4048 is the cutoff of first higher order
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mode which is the LP11 mode.
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If V is much much larger than one typically
more than 10 or around 10, then the number
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of modes can be estimated using the approximate
formula which are given as for a step index
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fiber, the approximate number of modes are
V square by 2 and for a graded index fiber
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which power law profile which is characterized
by profile parameter q, the number of modes
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can be estimated by half of q over q plus
2 times V square.
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So, if it is parabolic index fiber then q
is equal to 2 then the number of modes would
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be V square by 4.
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Let me work out some examples, let me consider
a step index optical fiber with co refractive
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index n1 is equal to 1.45, cladding refractive
index n2 is equal to 1.44 and core radius
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8 micron, and I want to find out the total
number of modes including degeneracies supported
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by the fiber at lambda naught is equal to
1.5 micrometer.
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So, the procedure is very simple you have
to first find out the value of V, if you find
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out the value of V for these parameters then
it comes out to be 5.697.
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If you locate this value of V in the table
which you have created by arranging the modes
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in increasing order of their cut offs, then
you find that these modes are supported LP01,
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00:19:03,510 --> 00:19:09,180
11, 02, 21, 31 and 12.
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Now if you include the degeneracies of all
these modes and calculate the total number
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of modes then they come out to be 20.
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So, with this value of V the number of modes
would be around 20, you can also see that
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if you calculate using V square by 2, V square
by 2 would roughly be because it is 6 it is
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close to 6.
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So, 6 squares 36, 36 divided by 2 is approximately
18.
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So, little more than 18 and here you are getting20.
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So, they are close.
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Second is the wave length at which the fiber
is single moded, if I want to find out the
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wave length at which the fiber is single moded
or the wavelength range in which the fiber
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is single moded then I know the single mode
condition for a fiber is V should be less
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than 2.4048 and I have the expression for
V.
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Which goes as V is equal to 2 pi over lambda
naught times a times square root of n1 square
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minus n2 square.
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So, if I find lambda naught from here, then
I will see that for lambda naught greater
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than 3.55 micrometer, V would be less than
this and therefore, in this range of wave
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length the fiber would be single moded.
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Third part is approximate number of modes
at lambda naught is equal to 0.5 micrometer.
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So, I again calculate the value of V at 0.5
micrometer, and it comes out to be about 17
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which is much larger than one, then I can
find out the approximate number of modes by
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V square by 2 and it comes out to be about
146.
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So, this fiber will support 146 modes.
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Now, let me work out some examples on identification
of modes from they are given intensity patterns,
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how do I identify various modes from their
given intensity patterns.
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So, let us look at this, this one is easy.
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If I go in phi direction, I take any value
of r and go in phi direction I do not encounter
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any 0.
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So, the number of zeros in phi direction is
0 which means 12 is 0, which means l is equal
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to 0.
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Second thing I do is I count the number of
zeros in r direction.
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So, I take any value of phi and for that value
of phi I move in r direction and an end I
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00:22:35,760 --> 00:22:41,540
encounter 1 and 2 zeros, the number of zeros
in r direction is 2.
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So, m is equal to 3 because m minus 1 is equal
to 2.
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00:22:46,940 --> 00:22:58,450
So, this mode is LP03 mode, similarly for
this mode the number of zeros in phi direction
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00:22:58,450 --> 00:23:01,210
the number of zeros in phi direction is 4.
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00:23:01,210 --> 00:23:09,110
So, 2 l is equal to 4 l is equal to 2 and
number of zeros in r direction you exclude
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00:23:09,110 --> 00:23:10,110
0 here.
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00:23:10,110 --> 00:23:11,110
So, it is only one.
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00:23:11,110 --> 00:23:15,560
So, m minus 1 is equal to 1 which means m
is equal to 2.
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00:23:15,560 --> 00:23:19,650
So, this is LP22 mode.
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00:23:19,650 --> 00:23:27,440
Similarly here it is LP51 mode and this is
LP32 mode.
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00:23:27,440 --> 00:23:32,940
So, in this way I can identify any mode.
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00:23:32,940 --> 00:23:40,390
Next example is schematically draw the contour
intensity plots or density plots of these
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00:23:40,390 --> 00:23:46,140
modes, and I just want to draw the patterns
of even mode.
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00:23:46,140 --> 00:23:49,740
So, cosine alpha solution.
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00:23:49,740 --> 00:23:52,970
So, let me do it for LP21 mode.
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00:23:52,970 --> 00:23:59,960
So, for LP21 mode what I have l is equal to
two.
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00:23:59,960 --> 00:24:03,370
So, number of zeros in phi direction are 2
l.
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00:24:03,370 --> 00:24:05,180
So, they are 4.
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00:24:05,180 --> 00:24:10,290
So, now, where these 4 zeros are located?
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00:24:10,290 --> 00:24:30,280
I know that if you go in phi direction then
phi ranges from 0 to 2 pi, now you have solution
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00:24:30,280 --> 00:24:35,210
which has cosine 2 phi.
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00:24:35,210 --> 00:24:50,090
So, you now find out the zeros of this cosine
2 phi they will occur at pi by 2 then 3 pi
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00:24:50,090 --> 00:24:58,730
by 2 and where do you stop you stop because
phi goals phi goes up to 2 pi then 2 phi will
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00:24:58,730 --> 00:25:01,720
go up to 4 phi.
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00:25:01,720 --> 00:25:06,070
So, until you cross or reach 4 pi you do not
stop.
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00:25:06,070 --> 00:25:13,641
So, 3 pi by 2, phi pi by 2 and then 7 pi by
2, after that it would be nine pi by 2 which
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00:25:13,641 --> 00:25:15,330
is more than 2 pi.
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00:25:15,330 --> 00:25:16,750
So, you stop here.
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00:25:16,750 --> 00:25:26,060
So, here these 4 zeros would be located which
means phi is equal to pi by 4, 3 pi by 4,
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00:25:26,060 --> 00:25:31,850
5 pi by 4 and 7 pi by 4.
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00:25:31,850 --> 00:25:41,910
So, here you have 0 this is pi by 4, this
is 2 pi by 4, this is 3 pi by 4, 4 pi by 4,
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00:25:41,910 --> 00:25:45,340
5 pi by 4, 6 and this is seven pi by 4.
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00:25:45,340 --> 00:25:55,200
So, here you have 0 and in between you will
have the intensity and m is equal to 1.
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00:25:55,200 --> 00:26:00,210
So, there would not be any 0 in r direction
except a 0 at r is equal to 0.
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00:26:00,210 --> 00:26:06,020
So, you will have intensity something like
this.
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00:26:06,020 --> 00:26:11,520
So, this is how you will plot the intensity.
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00:26:11,520 --> 00:26:20,440
So, this is how it would look like similarly
for LP32 mode you will have 6 zeros in phi
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00:26:20,440 --> 00:26:24,580
direction and you can in the same way you
can calculate the values of phi where the
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00:26:24,580 --> 00:26:27,540
zeros will occur and since m is equal to 2.
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00:26:27,540 --> 00:26:33,780
So, there would be 1 0 in r direction, in
the same way you can plot for LP41 mode and
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00:26:33,780 --> 00:26:34,940
LP13 mode.
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00:26:34,940 --> 00:26:47,480
So, today we have learned about the mode modal
fields, in the next class we would look in
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00:26:47,480 --> 00:27:00,560
to what fraction of power of these modes is
confined in the core and of and about various
251
00:27:00,560 --> 00:27:03,630
parameters of a single mode fiber.
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00:27:03,630 --> 00:27:04,610
Thank you.