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Till now we have studied the propagation characteristics
and salient features of optical fiber and
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optical wave guides. So, I think now it is
time to stop for a while, and have a recap
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of whatever we have done till now.
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So, in order to understand propagation characteristics
of optical fibers, we have first understood
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how an electromagnetic wave propagates in
an infinite extended dielectric medium. So,
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infinitely extended dielectric medium n square
is independent of x, y and z. So, it is homogeneous
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medium then an EM wave propagating in z direction,
has an electric field and magnetic field given
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by E(z, t) is equal to E0 e to the power i
omega t minus k z and H is equal to H0 e to
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the power i omega t minus kz. So, this has
x, y and z all the components, and we had
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shown that for such a case E0(z) and H0(z)
are 0 so, this is a transverse wave.
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For a linearly polarized wave which is polarized
in x direction and propagating in z direction,
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I can write the other components as E is equal
to x cap E0 e to the power i omega t minus
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kz and H is equal to y cap H0 e to the power
i omega t minus kz, where H0 and E0 are related
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by this. So, I see two things one you should
notice that this E0 and H0 they are constants
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they do not depend upon a spatial coordinates,
and also that this is purely real. So, H and
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E are in phase. I can also write it down in
terms of B, since B is equal to mu times H.
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So, B would be y cap B0 e to the power i omega
t minus kz. So, the magnitude amplitude of
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B is 1 over v times E. And so the amplitude
of B is really much smaller than the amplitude
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of E, then we had considered propagation in
wave guides.
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So, now if n square depends upon x and y,
and the wave is propagating in z direction,
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then the electric and magnetic fields associated
with the wave, light wave would be given by
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these and we notice here now these E0 and
H0 are not constants, but they depend upon
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x and y. So, there is some profile, there
is some electric field profile and magnetic
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field profile which propagates in z direction
with certain propagation constant beta. So,
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these are the modes of the system; if instead
of having n square x, y I have variation only
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in x direction.
So, n is a function of x only then it is a
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planar wave guide, and in such a case I can
write down the electric and magnetic fields
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as E0 of x e to the power i omega t minus
gamma y minus beta z, and H like this. So,
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here what I see that this E and H they depend
only on x. So, they are the functions of x
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and they form the mode of the planar wave
guide ok.
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And I can choose either beta here or gamma
depending upon in what direction the mode
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is propagating.
So, if I choose the direction of propagation
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is z then I can put gamma is equal to 0. Then
we considered planar optical wave guide with
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n square of x variation and propagation in
z direction then we had seen that we can categorize
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the modes of such a wave guide into two categories
two polarizations, one is TE modes, where
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the non-vanishing components of electric and
magnetic fields are Ey, Hx and Hz, and these
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three components are related to each other
by these three equations.
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I can have another category that is TM mode
TM polarization, where the non vanishing components
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are Hy, Ex and Ez and these components are
related to each other by these three equations.
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So, for TE modes for example, I can form a
wave equation I can eliminate Hx and Hz and
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get an equation in Ey, which tells me how
Ey changes with x and this is what I want
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to find out because I want to find out Ey
of x which is the mode.
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So, that satisfies this equation. So, for
any given n square of x can solve this equation
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find out the values of beta and corresponding
Ey. So, these are the modes.
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I work out first an example of planar mirror
wave guide where I have very thin layer of
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dielectric medium of refractive index n and
width d, which is coated with metal on top
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and bottom. So, the refractive index profile
looks like this and when we worked out the
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modes of this then the electric field associated
with this is like this, and the propagation
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constants are like this.
So, I have modes discreet modes of such a
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wave guide corresponding to different values
of m. So, they look like this they look like
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the modes of a string modes of a (Refer Time:
06:32) string; then we consider symmetric
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step index planar wave guide, dielectric wave
guide where the refract index profile is like
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this.
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So, I have a high index layer sandwiched between
two lower refractive index layers. The width
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of the high index layer is d which extends
from x is equal to 0 to x is equal to d by
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2 and 2minus d by 2.
If the direction of propagation again I considered
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z, then the electric and magnetic fields associated
with the light would be given by this and
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now my task is to find out this E0 of x and
H0 of x. Refractive index profile of this
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wave guide is given like this where n(x) is
equal to n1 for mod x less than d by 2, it
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is n2 for mod x greater than d by 2, and n1
is greater than n 2. We define some kappa
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and gamma as kappa square is equal to k naught
square n1 square minus beta square, and gamma
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square is equal to beta square minus k naught
square n2 square. So, when I find out the
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guided modes of this TE modes and TM modes.
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So, for TE modes again I have non vanishing
components Ey, Hx and Hz, and if I solve the
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wave equation that the differential equation
for Ey, then I find that I can further categorize
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these modes into symmetric and anti-symmetric
because their structure is symmetric about
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x is equal to 0. For symmetric modes the field
profiles are given by Ey of x is equal to
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a cosine kappa x in the in the guiding film
in the high index region, and it is exponentially
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decaying in the lower index surrounding. The
values of beta they satisfy this equation
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this is the transcendental equations.
So, we solve this transcendental equation
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get the values of beta which are the solutions
of this equation, and for those values of
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beta I find out Ey. So, in this way I get
the complete solution the mode field profiles
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as well as the propagation constants. Similarly
for anti-symmetric modes this is the field
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solution and this is the transcendental equation.
The same thing I can do for TM modes for which
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the non-vanishing components are Hy, Ex and
Ez, I have symmetric modes and anti-symmetric
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modes and these are the transcendental equations.
Now only difference that you would see in
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between this and this is a factor of n1 square
over n2 square, this comes from the boundary
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conditions.
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So, in a wave guide a very important diagram
is b-V diagram, where b is the normalized
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propagation constant defined like this, and
V is the normalized frequency defined like
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this which contains all the information which
you require about the wave guide and the wave
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length.
So, I can now translate the transcendental
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equations corresponding to TE and TM symmetric
and anti-symmetric modes into these normalized
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parameters like this, and then I can solve
these equations to generate these curves.
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The idea of and advantage of generating these
curves is that these curves are universal.
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So, if I talk about TE modes, then this is
the equation for symmetric and this is the
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equation for anti-symmetric. There should
not be this is the equation for TE is symmetric
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mode and this is the equation for TE anti
symmetric mode.
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So, let me make this correction here, TE symmetric
is V square root of 1 minus b tan V square
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root of 1 minus b is equal to V square root
of b, and TE anti symmetric is minus b square
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root of 1 minus b cot V square root of 1 minus
b, is equal to V root of b TM symmetric is
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V square root of 1 minus b tan V square root
of 1 minus b is equal to n1 square over n2
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square TM square root of b and TM anti symmetric
is equal to minus V square root off 1 minus
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b cot V square root of 1 minus b is equal
to n1 square over n2 square, V square root
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of b. So, I can have the transcendental equations
in normalized parameters, and by solving these
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equations for different values of V, I can
generate b-V curves. So, those b-V would look
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like this is V and this is b; for guided modes
b would vary from 0 to1. So, this would be
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TE0 mode TE0 this would be TM0 mode.
This would be TE1 mode; this would be TM1
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mode and so on. So, I can generate these universal
curves b-V diagrams, and if I now know the
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wave guide and the wavelength then I can just
calculate the value of V and go to these curves
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find out the value of V and find out the value
of propagation constants. The cut offs of
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these modes are here. So, this is 0 this is
pi by 2 and the cut offs are given by cut
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offs Vc m for TE and TM both is equal to m
pi by 2, where m is equal to 0, 1, 2 and so
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on. So, how the modal fields look like.
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So, if I take high index contrast wave guide,
I have taken high index contrast wave guide
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here in order to show the difference between
TE and TM modes for n1 is equal to this n2
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is equal to this, d half a micron lambda naught
2 micro meter. So, this is how the TE0 mode
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would look like and this is how the TM0 mode
would look like. You can see the discontinuity
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in at in the slope at the boundaries. This
is how TE1 mode look like this is how TM1
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mode would look like I also define penetration
depth, the extent up to which the field would
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penetrate into n2 region. So, it is given
by 1 over gamma. We have seen that these modes
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are nothing, but the superposition of two
plane waves one going in exact plane like
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this and one going in exact plane like this.
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So, upward and downward, making angles plus
minus theta from z axis. So, what we see that
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there is one plane wave going like this, another
plane wave going like this. So, if I take
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the components in x direction and in z direction,
then in x direction I get two counter propagating
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waves which give you standing wave pattern.
And these modes these modes are nothing, but
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the standing wave patterns. So, the field
stands in x direction does not propagate in
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x direction, but this is standing wave pattern
propagates in z direction with propagation
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constant beta, which can be given by k naught
n1 cos theta.
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And you had also seen that the condition for
guided mode can be translated to the condition
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for total internal reflection. So, this is
also we had seen then what is the power associated
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with the mode? To calculate the power associated
with the mode we found out the pointing vector
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which gives us the intensity, and since the
electric and magnetic field associated with
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light or fluctuating rapidly, k at a frequency
of about 10 to the power 14 or 10 to the power
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15 hertz. So, we need to take the average
time average of these.
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So, when a we do this, then we find that only
z component of S the pointing vector survives,
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and it is equal to beta over 2 omega u naught
Ey square of x, this is for TE modes. So,
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from here I can find out the power per unit
length in y direction, because y direction
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is infinitely extended for planar wave guide.
So, the power per unit length in y direction
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would be given by beta over 2 omega mu naught
integral minus infinity to plus infinity Ey
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square of xdx. So, when I calculate this for
symmetric planar wave guide, then for TE modes
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the power is given by this and for TM modes
the power would be given by this.
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These powers are in watts per meter then we
had extended the analysis to asymmetric planar
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wave guide, where I have asymmetry here this
is the sub straight this is the cover. So,
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the sub straight and cover refractive index
are different. So, ns here is greater than
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nc, and what we did we wrote down the fields
in all the three regions. So, these are the
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fields electric fields corresponding to modes
in all the three regions, and the beta the
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propagation constant will satisfy this transcendental
equation. For TM modes the fields would be
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given like this and the transcendental equation
becomes this.
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We had also defined normalized parameters
for asymmetric planar wave guide. So, V is
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defined with respect to ns. So, here you see
nf square minus ns square not with nc, because
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it is ns that will govern the guiding condition.
B is also defined with respect to ns. So,
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beta square over k naught square minus ns
square divided by nf square minus ns square,
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here we have another parameter which is asymmetry
parameter which tells you how different ns
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is with nc. In normalized parameters I can
have transcendental equation for TE modes
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and for TM modes like this, if I draw the
b-V curves here now then I see that the b-V
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curves for asymmetric wave guides they go
like this ok.
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What I notice that they do not have 0 cut
off. TE0 mode also does not have 0 cut off.
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If you look at the cut offs if you look at
the cut offs then you can see that for symmetric
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planar wave guide the cut offs were m pi by
2, but now these cut offs would be shifted
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by this amount half ten inverse of square
root of a because of this asymmetry parameter.
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If a is equal to 0 then it becomes the symmetric
planar wave guide. So, these I also notice
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that the cut offs of TE and TM modes are now
different. In case of symmetric planar wave
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guide the cut offs of TE and TM both the modes
were the same m pi by 2.
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But now the TE and TM modes have different
cut offs. So, this is TE0 mode this is TM0
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mode TE0 mode have cut off here, TM0 mode
has cut off here. Now if I have v which is
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somewhere here then I will guide only TE0
mode and TM0 mode is cut off. So, in the range
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of V from which goes from half tan inverse
square root a to half tan inverse nf square
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over nc square is square root of a, I have
single polarizations single mode wave guide,
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and this range is known as SPSM range. This
is how the modal fields would look like, this
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is for TE0 mode this is TM0 mode this is TE1,
TM1, TE2, TM2.
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We can notice that in the sub straight the
field extends more because index contrast
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with the film is less while in the cover,
the field very quickly decreases goes down
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to 0, because the index contrast is high.
Then we worked out the modes of a step index
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optical fiber. So, in a step index optical
fiber I have a core and cladding I have assumed
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to be extended to infinity.
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So, here I have the modes which are function
of r and phi. So, this is the r solution A
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Jl Ur over a, and phi solution can is cosine
l phi and sin l phi this is in the core, in
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the cladding the r solution is Kl Wr over
a, and phi solution is again cosine l phi
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and sin l phi. Then we again we had defined
TE normalized frequency and normalized propagation
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constant, and worked out the transcendental
equations for different modes corresponding
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to l is equal to 0 and l greater than or equal
to 1, because now I have phi directions. So,
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there would be discreet modes in phi direction
also. So, which are governed by the values
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of l and the l assumes only integral values
0, 1, 2, 3. So, for those I can then find
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out what are the r solutions, and the modes
propagation constant would then be given by
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these two transcendental equations.
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So, if I do that. So, for different values
of b different values of V I can find out
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the propagation constants and I can plot them.
So, this is LP01 mode, this is LP11 mode this
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is LP21, 02, 31, 12 and so on. The cut offs
of the modes are not straight forward they
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are the solutions of these equations they
are given by this. So, which a where j are
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the Bessel functions. So, by solving these
by finding out the zeros of these Bessel functions
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I can find out the cut offs.
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So, these are the cut offs of various modes
first few modes. This is how the modal fields
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would look like this is LP01 mode LP02 mode
LP11 mode. So, these are 3-D plots and these
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are the plots.
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When you see them from the top these are the
contouring intensity plots. What I see that
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l is equal to 0 modes are 2-fold degenerate,
because phi solution is cosine l phi and sin
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l phi. So, l is equal to 0 then there is no
phi dependence there the degeneracy comes
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only from polarization, while L naught equal
to 0 mode are 4-fold generate 2-fold degeneracy
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comes from phi and two fold degeneracy comes
from the polarization. Then we had seen the
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single mode fiber and a single mode fiber
we had defined three very important parameters,
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one is cut off wavelength which is given by
2 pi over 2.4048 times a square root of n1
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square minus n2 square.
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And what is the cut off wavelength of the
fiber? It is the wavelength corresponding
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to the cutoff of LP11 modes.
So, if lambda is greater than lambda c then
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the fiber is single moded, and if lambda is
smaller than it is multi moded. The mode propagation
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constants of the fiber can be fitted with
an empirical relation, if V lies between 1.5
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and 2.5, and this empirical relation is given
by this with these values of a and b. The
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spot size the modal field profile of a single
mode fiber if you look back to this, it resembles
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a Gaussian and it can be very well fitted
with Gaussian. So, we can fit a Gaussian to
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this profile then we define this spot size
as Gaussian spot size where the field intensity
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drops down to 1 over e square of it is value
at the center.
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So, w is Gaussian spot size and 2 w is mode
field diameter, we can also fit an empirical
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relationship between excuse me w over a and
V which is given by this. So, if I have a
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fiber I just calculate the value of V for
a given wavelength, and I can get the value
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of w over a from this relationship. Then another
important parameter for a single mode fiber
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is what is it is bend loss.
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00:26:44,890 --> 00:26:52,799
So, bend loss is given by this, then their
instances there are instances where we need
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to join two fibers together, and when we join
two fibers together then there would be losses.
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If two fibers are not identical, then even
if there are no misalignment while joining
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the fibers there would be a loss which is
known as mode mismatch loss, which is purely
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due to different spot sizes of the modes of
two fibers. If you join two identical fibers
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and then there can be three kind of misalignments
transverse, angular and longitudinal, and
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these are the misalignment losses or splice
losses due to different kind of misalignments.
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So, this we had seen. Then in the end we had
seen the wave guide dispersion.
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So, ultimately we need to use this fiber in
telecom system, and when we send pulses through
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this optical fiber then dispersion happens
we had seen that in a fiber there is inter
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modal dispersion and material dispersion,
when you use single mode fiber inter modal
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dispersion is diminished.
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So, we get rid off inter modal dispersion,
but then there is another dispersion which
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is wave guide dispersion which comes out because
of the dependence of mode propagation constant
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of fiber mode on wavelength ok.
So, the wave guide dispersion coefficient
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is given by this, and we can use empirical
formula between b and V to calculate this
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vector, or we can use a more accurate empirical
formula given by marquees for V times d2(bV)
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over dV square. So, in this way we had understood
the salient features of optical fiber, how
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light is guided into optical fiber, what are
the important parameters of a single moded
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fiber.
So, with all this background about optical
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fibers and the propagation characteristics
of optical fiber, now we are ready to use
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this fiber in a system or to use this fiber
to make components and devices, which we will
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do in the subsequent lectures.
Thank you.