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After having understood the light propagation
in planar waveguides in the previous section;
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now in this section we will analyse light
propagation in an optical fiber- the modes
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of light propagation in an optical fiber.
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So, if I look back to planar waveguide where
the refractive index variation is only in
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one direction x and the propagation direction
is z, then I had seen that the modes are given
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by these E and H. E is equal to E of x e to
the power i omega t minus beta z. And H is
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equal to H of x e to the power i omega t minus
beta z.
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You remember that these E and H are not constants,
if they are constants then they are plane
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waves. But here E is a function of x and H
is a function of x, so they are not plane
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waves propagating in z direction. But we had
seen that these are the superposition of two
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plane waves: one going in plus x-z direction
and another going in minus x-z direction.
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And that gives us a standing wave pattern
in x direction which flows in z direction.
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So, they are the modes.
If I look at rectangular channel waveguide,
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so now if I have confinement in x as well
as in y, so refractive index variation in
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x and y both. So, this is the kind of profile.
Then the modes would be given by E is equal
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to E of x,y e to the power i omega t minus
beta z and H is equal to H of x, y e to the
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power i omega t minus beta z. So, these E
and H would now be the functions of x and
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y both. So, the confinement is in both the
directions x and y, and those fields will
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propagate in z direction with certain propagation
constant beta.
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If now I convert this into a cylindrical geometry
then it becomes an optical fiber waveguide.
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I can represent the refractive index profile
in x and y; ns square x and y, but because
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of cylindrical geometry it is much more easier
to analyse the structure if I represent this
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refractive index profile in cylindrical coordinates:
r phi. So, r phi are the transfers coordinates
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and z is the longitudinal along which the
light is propagating.
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So, now what would be the modes of this kind
of a structure? Well, they can be given by
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Er phi e to the power i omega t minus beta
z and H r phi e to the power i omega t minus
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beta z; so this E which is a function of r
and phi, so this particular function propagates
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in z direction with certain propagation constant
beta. So these are the modes.
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In all the three cases what I observed is
the direction of propagation z is common.
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So, I would like to make use of this to analyse
optical fiber. Let me go back to the modes
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of the planner waveguide.
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So, I had TE-modes with non-vanishing components
Ey, Hx and Hz and TM-modes with non-vanishing
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components Hy, Ex and Ez. And these three
equations relate these three non-vanishing
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components here, and these three equations
relate the non-vanishing components of TM-modes.
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So what I have done actually, I found out
Ey and then I can get Hx and Hz from these
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equations. So, once I know Ey I had formed
the differential equation in Ey solved the
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equation for Ey got the modal fields. So,
once I get Ey then from there I could get
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Hx and Hz.
Similarly, here I get Hy first and then Ex
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and Ez and that is how I have the complete
solution. But, as I have seen in the previous
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slide that the common thing in all the three
structure is the longitudinal direction z.
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Can I utilize this here? Well, if instead
of solving for Ey and then getting Hx and
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Hz from here I can also do another thing that
I first form the equation in Hz solve it for
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Hz and then get Ey and Hx.
Here what I notice is that the only longitudinal
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component here is Hz, and Ez is 0. And in
this case I can do the same thing, and here
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I notice that the only longitudinal component
is Ez and Hz is 0. So, I can define my TE-modes
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in such a way that the fields for which Ez
is equal to 0 our TE-modes and if I have Hz
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is equal to 0 then they are TM-modes. If Ez
is equal to 0 then they are TE-modes and if
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Hz is equal to 0 then they are TM-modes.
Now let us look at an optical fiber.
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So, in optical fiber I have these E and H,
if I substitute these solutions; if I substitute
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these into Maxwell’s equations del cross
E is equal to minus mu naught del H over del
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t and del cross H is equal to epsilon del
E over del t then of course I will get 6 equations.
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The only thing is that I should now use cylindrical
polar coordinate system r phi and z and see
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what these 6 equations are. So, these 6 equations
will give me the transverse components Er
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and E phi in terms of longitudinal components
Ez and Hz.
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So, these are the 6 equations which I get
from those to Maxwell’s equations.
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What I see here is that these mathematical
equations from 1 to 6 these equations if I
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do mathematical manipulation of these 6 equations
then they can lead to these two equations:
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one in Ez and one in Hz. So, I can form a
differential equation in Ez or Hz. Solve these
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equations and then get Er, E phi, Hr, H phi
from these Ez and Hz using those 6 equations.
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However, what I see that if you look back
to those 6 equations; let me look back to
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those 6 equations. I cannot find an instance
where I can have only Ez or only Hz; that
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is I am not able to find out where Ez is not
equal to 0 and Hz is not equal to 0 at one
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time. I have that simultaneously both are
nonzero; which means that now the polarizations
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are coupled, I cannot separate two orthogonal
polarizations. So, two polarizations are coupled
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and in general what I will get hybrid modes.
I cannot get TE polarization and TM polarization
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in general.
That I could be able to do in case of planar
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waveguide, that this is TE polarization and
this is TM polarization, but here in general
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it is difficult. So, Ez naught is equal to
0 and Hz naught is equal to 0 simultaneously
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are there, so I cannot decouple to polarizations.
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However, there is one particular case where
I can still have this decoupling. And this
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case is when there is no phi dependence in
the solutions; when there is no phi dependence
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in the solution then let us see what happens.
So, if there is no phi dependence then I put
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these del phi terms 0 everywhere. So, these
del phi terms are not there then I can see
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that in these three equations I have only
Ez:, and there is no Hz. So, in these three
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equations if I considered these three equations
then Hz is equal to 0. And in these three
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equations there is no Ez, so Ez is equal to
0.
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So, only for this case I can separate out
the two polarizations; I can de couple two
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polarizations TE and TM.
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So now, for phi independent solutions we can
have two cases. One is Ez is equal to 0 and
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Hz is not equal to 0, which is nothing but
the case of TE-modes. And another is when
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Ez is not equal to 0 and Hz is equal to 0,
which is nothing but the case of TM-modes.
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So, in this way I can separate out two polarizations.
But in general what I will have? I will have
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Ez not equal to 0 and Hz not equal to 0 simultaneously.
So, Ez not equal to 0 Hz not equal to 0 simultaneously
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it means in these kinds of modes two polarizations
will be there and they are called hybrid modes.
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In these hybrid modes there can be two instances:
one is when Hz makes larger contribution than
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Ez, then we label them as HE-modes, and when
Ez component makes larger contribution than
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Hz then we label them as EH-modes. So, in
fiber I can have TE-modes, TM-modes HE-modes,
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EH-modes, ok.
And these modes are also known as vector modes
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of the fiber.
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Here I have plotted some typical vector modes
of the fiber and their electric fields. So,
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this is a typical TE01 mode, this is TM0 mode,
this is HE11 mode, this is HE21 mode.
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But the fiber that we practically use has
a small index contrast. The index contrast
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between the core and cladding.
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The relative index difference between the
core and the cladding is typically 0.3 percent
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which is very small. And if you look back
to the analysis of your symmetric planar waveguide
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or even asymmetric planar waveguide: we see
that if the index contrast between the high
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and low index regions is small then the propagation
constants of TE and TM polarizations they
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merge. It is very difficult to distinguish
between TE and TM polarization in terms of
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their propagation constant if index contrast
is a small.
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So, here we use that fact that if index contrast
is a small then the fiber will not distinguish
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between this polarization and this polarization.
So, in such kind of fibers we can still have
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the modes which are linearly polarized. That
is if one is polarized like this then another
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has to be like this or if there is any arbitrary
polarization then that arbitrary polarization
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can always be represented as the superposition
of this and this two orthogonal polarizations:
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horizontal and vertical.
So, in case of weekly guiding fiber what we
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have are linearly polarized modes and we can
have two orthogonal polarizations which have
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nearly the same propagation constants. And
now if I say that psi is the transverse component
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of electric field so you have a fiber and
psi is the transverse component of the electric
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field then it satisfies the equation: del
square psi is equal to epsilon naught n square
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mu naught del 2 psi over del TE square, this
psi is capital psi so it is a function of
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all the spatial coordinates as well as time.
In general, this ns square is a function of
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r and phi. And if this is a function of r
and phi then I can in general write this capital
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psi which is a function r phi z and t as psi
small psi of r phi e to the power i omega
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t minus beta z, because n square is not a
function of z. So, z part can be separated
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out. And this is how the solution of z part
and t part will come out. We have seen this
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several times.
So now, we want to find out this psi r phi.
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This function, and see how this propagates.
This function psi r phi represents the mode
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which travels with propagation constant beta.
So, the problem reduces to find out this now.
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For that what I do? I put this capital psi
back into this wave equation and when I do
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this then this equation becomes this: del
2 psi over del r s square plus 1 over r del
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psi over del r plus 1 over r s square del
2 psi over del phi square plus k naught square
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n square of r or r phi minus beta square psi
is equal to 0.
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In this course we will confine, we will limit
our discussion only to refractive index profiles
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which do not depend upon phi. So, we will
consider only n square of r and not n square
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of r phi.
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So, now this is the equation. And if n square
is a function of r only then this solution
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psi r phi can be separated out in r and phi.
I can separate out of r part, and I can separate
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out phi part; and this psi r phi as I have
said earlier that they are now the scalar
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modes or linearly polarized modes.
So, since n square is a function of r only
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naught of phi, so psi r phi can be written
as R of r and capital phi of phi. And if I
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now put this back into this equation then
I can separate out r apart from phi part.
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So, I get r square over capital R. d2R over
dr square plus small r over capital R. d capital
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R over d small r plus r square k naught square
n square r minus beta square is equal to minus
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1 over phi d 2 phi over d phi square. So,
I have separated them out, the usual procedure
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now is to equate it to some constant. And
since this is second order equation, so I
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will take the constant in the form of the
square.
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So, I take it as l square. So, first let me
solve this phi part and the phi solution comes
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out to be cosine l phi and sin l phi. I make
the use of the fact that in phi direction
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the solutions are periodic. So, you start
at some phi and then you make a round of 2
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phi you come back to the same point. So, capital
phi at phi plus 2 pi is same as phi at phi.
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So, that there is only single solution, then
this restricts the values of l to only integer
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values. So, l can only be 0, 1, 2 and so on.
So, this is the phi solution. Now for each
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value; and we see that in phi direction now
the solutions are discrete, the solutions
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are discrete because of this condition. So,
not all the values of l are allowed, only
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certain discrete values of l are allowed which
are integer values. Now for each of these
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integer values of l I can find out the solutions
r. So, let me now look at r part.
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So, the r equation is r squared d2R over dr
square plus rdR over dr plus k naught square
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n square of r minus beta square times R square
minus l square times R is equal to 0. So,
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this is the r equation.
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And I can solve this r equation for a given
n square of r. What is the n square of r?
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I am going to take in this course, I will
again limit myself to step index fiber and
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that to only two layer fiber; so which has
only two regions. So, nr is equal to n1 when
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r is less than a, and it is n2 when r is greater
than a. So, this is the core, this is the
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cladding: core has refractive index n 1 and
radius a. And this is infinitely extended
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cladding, ok. So, this is the r equation and
this is n of r.
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The procedure to solve this equation is exactly
the same as we have been doing for planar
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waveguides. What we will have to do? I will
have to write this equation in this region,
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in the core, and in the cladding. So, in the
core which is defined by the region r less
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than a, this equation becomes this. And for
r greater than a which is the cladding region
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the equation becomes this.
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We should pay attention again that for guided
modes our beta over k naught should lie between
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n2 and n1. So, beta lies between k naught
n2 and k naught n1.
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So, that is how I have arranged the terms
here in order to have this quantity positive.
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So, I have plus sign here and k naught square
n1 square minus beta square, and a negative
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sign here and beta square minus k naught square
n2 square here. So now, I have these two equations
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which I have to solve to find out the values
of beta for any given value of l: l can be
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0, 1, 2, 3 and so on.
So, let me define this k naught square n minus
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square minus beta square as some U square
over a square and this as some W square over
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a square.
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So, these equations now become r square d2R
over dr square plus rdR over dr plus U square
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r square over a square minus l square times
R is equal to 0 in the core. And in the cladding
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r square d2R over dr square plus rdR over
dr minus W square r square over a square p
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l square times R is equal to 0. So, now I
have these two equations, I have to solve
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them.
If I have a little background of mathematical
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physics then I can immediately see that these
are Bessel’s equations and solutions of
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these equations are Bessel functions. So,
where U square is this and W square this.
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So, I am going to get the solutions of these,
but before that I observe the fact that if
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I do U square plus W square then it becomes
k naught square a square times n1 square minus
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n2 square. And by this time I am able to recognise
this kind of term very well, this is nothing
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but the normalised frequency V. So, this is
V square.
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So, in case of optical fiber, I have V is
equal to 2 pi over lambda naught times a is
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the radius which is half the diameter times
n1 square minus n2 squares square root. This
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is normalised frequency. And again it contains
all the fiber parameters and the wavelength.
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Let us come back to the solutions of this.
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The solution of this equation is given by
Jl Ur over a; in terms of Jl Ur over a and
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Yl Ur over a which are Bessel functions. And
of this equation Kl Wr over a and Il Wr over
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a which are modified Bessel functions, ok.
So, if you go back to your planar waveguides
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I had in the region mod x less than d by 2
where nx was n1, I had the solutions cosine
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00:26:37,010 --> 00:26:51,450
kappa x and sin kappa x. And in the region
mod x greater than d by 2 where nx was n2.
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And of course, n1 was greater than n2. I had
solutions in the form e to the power minus
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00:26:59,240 --> 00:27:08,880
gamma x and e to the power plus gamma x.
So, I had oscillatory solutions in mod x less
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than d by 2 and exponential amplifying decaying
solutions in this. Here instead of sin cosine
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I have some Jl Yl, instead of e to the power
minus gamma x and e to the power plus gamma
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x I have some Kl Il. So, what these solutions
are, what these functions are, and how do
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I understand them and then how do I proceed
further I will see in the next lecture.
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00:27:43,570 --> 00:27:44,510
Thank you.