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In the previous lecture we were discussing
the directional coupler.
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We discussed about the physics and working
principle of directional coupler.
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In this lecture, we will continue the discussion
we would also look into few more components
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that can be made using directional coupler.
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And some other fiber optic components like,
polarization controller and fiber gratings.
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So, what we were discussing that if we have
a directional coupler with 2 non identical
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cores which have modes with propagation constants
beta 1 and beta 2.
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Then the fractional power that is in core
1 is given by 1 minus kappa square over gamma
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square sin square gamma z.
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And the fractional power in the core 2 is
given by kappa square over gamma square sin
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square gamma z, where gamma square is equal
to kappa square plus delta beta square by
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4 and delta beta is nothing but the phase
mismatch which is the difference between the
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propagation constants, beta 1 and beta 2.
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But we had seen that if delta beta over kappa
is small, then we can have maximum coupling
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of power from one core to another core this
zeta max can be close to 1 and if delta beta
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is equal to 0 then we can have 100 percent
coupling of power from core 1 to core 2.
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But if this phase mismatch delta beta is substantial
then we cannot have 100 percent transfer of
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power from core 1 to core 2 as shown here.
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So, for delta beta is over 2 kappa is equal
to 2 the maximum power that can be coupled
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from core 1 to core 2 is 20 percent.
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Let us work out an example, yet if we take
a typical value of kappa as 0.2 millimeter
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inwards.
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Then we find that for delta beta greater than
4 millimeter inwards zeta max would be less
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than one percent, this calculation is very
simple can be done very easily.
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How much is this delta beta will the, let
us work out some numbers and have a feel of
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delta beta how much this delta beta is.
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We know that delta beta is equal to k naught
delta n effective were k naught is 2 pi over
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lambda naught.
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So, if I fix a wavelength let us say 1300
nanometer, then delta beta is 2 pi over lambda
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naught times delta n effective and for delta
beta is equal to 4 millimeter inverse if I
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calculate delta n effective from here, then
it will come out to be 8 into 10 to the power
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minus 4 which means that if delta n effective
is larger than 8 into 10 to the power minus
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4, than the coupling of power from core 1
to core 2 would be less than one percent.
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How large this? delta n effective is?, You
can compare it with core cladding index difference
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in single mod fiber which is 3 into 10 to
the power minus 3.
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So, if you have a single mode fiber, if I
draw the refractive index profile r and this
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is nr this is n1 this is n2.
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So, this difference is typically 3 into 10
to the power minus 3.
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And here delta n effective which is the difference
between the propagation constants of 2 modes
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of 2 different wave guides or 2 different
fibers delta n effective is typically 8 into
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10 to the power minus 4.
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If it is greater than 8 into 10 to the power
minus 4 then there would be coupling of power
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which is less than one percent.
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So, if you compare it let us say there is
another fiber like this r nr.
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So, these are 2 non identical fibers.
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So, this is delta n effective 2 sorry, this
is n effective 2 and this is n effective 1.
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So, if you compare these 2 the difference
should be much smaller.
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And if you calculate the value for maximum
coupling of power for example, say for more
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than 99 percent coupling of power then this
delta n effective should be less than 8 into
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10 to the power minus 5.
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Now, we can use this directional coupler for
multiplexers and demultiplexers also.
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Here we utilize the fact that the coupling
coefficients kappa are wavelength dependent.
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So, if we send 2 wavelengths lambda 1 and
lambda 2, through this device then what will
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happen.
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Let us say at lambda 1 the coupling coefficient
is kappa 1, and at wavelength lambda 2 the
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coupling coefficient is kappa 2.
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We know that for phase matched case let us
consider phase matched case for convenience
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P1 is equal to Pi cosine square kappa L this
is the power in core 1, and P2 is Pi sin square
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kappa L power in core 2.
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Now, now if I choose this length L such that
kappa 1 times L is equal to m pi, and kappa
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2 times L is equal to m plus half pi, remember
that kappa 1 is the coupling coefficient at
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lambda 1 kappa 2 is the coupling coefficient
at lambda 2.
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In such a case at lambda 1 at wavelength lambda
1 P1(L) would be equal to Pi because kappa
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L is equal to m pi.
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And P2(L) would be 0 because, kappa 1 L is
equal to m pi, So P2 would be 0.
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So, this would be P2 and this would be Pi
and this would be 0 at lambda 1.
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At lambda 2 at lambda 2 since kappa 2 L is
equal to m plus half pi.
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So, this will give you P2 is equal to Pi and
this one will give you P1 is equal to 0 at
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length z is equal to L. Which means that,
which means that lambda 1 will come out from
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this core and lambda 2 will come out of this
core.
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If I plot normalized power that is: P1 over
Pi or P2 over Pi as a function of z then how
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it varies?
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P1 over Pi would vary like this and P2 over
Pi would vary like this.
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So, what I find that at this value of z, I
have power in core 1 at lambda 1 is equal
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to Pi and power in core 2 at lambda 2 is equal
to Pi.
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So, lambda 1 will come out from this port
and lambda 2 will come out form this port.
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So, this is how I can choose the coupling
length, multiplex de multiplexing length for
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such a coupler.
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So, what I am able to do with the help of
this?
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I am able to separate out 2 wavelengths into
2 different fibers.
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So, this works as wavelength de multiplexer.
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In the same device if I now launch lambda
1 from here and lambda 2 from here, then for
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a given length L which we can find from the
previous slide itself for that length L itself
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both the wavelengths will come out from a
single fiber.
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So, I can multiplex 2 wavelengths together.
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So, these kinds of couplers are known as WDM
coupler, because they are used in WDM systems.
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Let us know look at some practical parameters
of a directional coupler.
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A directional coupler is a 4 port device and
we can name the ports like this.
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This is the input port this is throughput
port, this is cross port or coupled port and
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this is reflection port.
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And important parameter is coupling ratio,
coupling ratio in percent can be defined as
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power in the cross port divided by the power
in cross port plus throughput port.
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So, power in cross port divided by total output
power, times 100 this is in percent.
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So, this is coupling ratio in percent in dB
I can define it as 10 log P c plus P t over
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P c.
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Then I can also define the excess loss, as
10 log Pi over Pc plus Pt, which basically
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tells you that how much loss this device will
introduce.
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What is the inherent loss in this?
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Because not all the power is coming out whatever
power you are inputting; so not all the power
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is coming out.
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Then we have insertion loss, insertion loss
is for a coupler or switch.
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So, it is defined as 10 log Pi over Pc, which
is nothing but summation of coupling ratio
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plus excess loss.
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Then another important parameter of a directional
coupler is the directivity which is defined
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as 10 log Pr over Pi.
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It is how it is the measure of how much power
comes back we want that all the power should
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go in one direction in the forward direction.
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So, this defines how much power comes back.
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So, larger the power comes back poor is the
directivity of the coupler.
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So, for a good directional coupler I should
have low insertion loss and high directivity.
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For a typical 3 dB commercial coupler excess
loss is less than or equal to 0.1 dB.
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Insertion loss says about 3.4 dB or less and
directivity is better than minus 55 dB.
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So now I can go into some other fiber optic
component.
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So, apart from this directional coupler, another
important component is polarization controller.
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We sometimes need to control the polarization
state in the fiber.
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And for that we require this polarization
controller.
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This polarization controller is based on bending
effect in the fiber.
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So, we will discuss that how bending of a
fiber can control the polarization state of
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the light which goes through the fiber.
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If I have a circular straight single mode
fiber, circular straight single mode fiber,
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since it is single mode fiber.
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So, and I am always working in weekly guiding
approximation.
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So, the modes are LP modes.
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So, in a single mode fiber I have LP01 mode.
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And this mode has 2 orthogonal polarizations.
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And these 2 orthogonally polarized modes have
same effective index.
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Because it is a weekly guiding fiber, but
if I bend this fiber.
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Then bending introduces a stresses in the
fiber, and this makes the fiber linearly birefringent,
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which means that which means that now this
polarization will travel with different velocity
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and this polarization will travel with different
velocity.
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That is this polarization will see different
refractive index of the fiber as compared
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to this polarization.
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And what happens is that the fast axis in
the plane of the loop and slow access in the
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plane perpendicular to the loop.
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So, so if I have a fiber which goes like this
if I make a loop like this in the fiber.
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And this is z direction and this is x direction
then this axis is the fast axis, which is
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in the plane of the loop and this axis is
slow axis which is perpendicular to the loop
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ok.
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So, here it is.
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So, if I have this kind of situation.
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So, the loop is in an exact plane.
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So, your x axis is the fast axis and y axis
is the slow axis.
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So, there is birefringence introduced in the
fiber.
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And the difference between the effective indices
of y polarized and x polarized mode would
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know be given by delta n effective is equal
to ney minus nex.
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And this is dependent on this depends on this
birefringence depends upon, what is the overall
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diameter of the fiber what is the cladding
diameter what is the thickness of the fiber
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and what is the bending radius?
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So, it is given by C times b square over r
square where b is the outer radius of the
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fiber and r is the radius of the loop.
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And it is it is intuitive because this birefringence
is produced by stresses induced in the fiber
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by bending.
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So, if the fiber is thicker.
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So, if you bend it if you bend a thicker fiber
then the stress is introduced would be more,
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a thin fiber would be very flexible.
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So, the stress is introduced would not be
very large.
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So, the birefringence would be smaller in
thinner fiber as compared to in thicker fiber
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then radius of the loop if the radius of the
loop is large.
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So, bending is very soft it is not very strong
bending then birefringence would be smaller.
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And if you give it a very tight bend that
is radius of curvature is very small then
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birefringence introduced was would be large.
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Typically for silica glass fiber this C; which
is the constant which depends upon the material
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of the fiber and electro-optic properties
of the fiber material.
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So, for silica glass fiber this is given by
C is equal to 0.133 typical value is 0.133
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for silica glass fiber, at wavelength 633
nanometer.
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Now, since there is a birefringence introduced
in the fiber.
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So, this polarization is travelling with different
velocity as compared to this polarization.
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Now if at the input end of the fiber you excite
both the modes simultaneously, both the polarization
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simultaneously.
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Then as they will propagate in the fiber of
phase shift would be accumulated between them.
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As they propagate in the fiber a phase shift
would be accumulated between them, and the
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state of polarization at any value of z inside
the fiber can now be found out by super posing
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these 2 polarizations with the accumulated
phase shift.
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So, so you can find out the state of polarization
at any distance z.
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And so, since there is a phase shift accumulated.
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So, this polarization state will slowly rotate
ok.
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So, let us say after one loop if I have one
loop of radius r here capital R and number
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of loops, let us say n is equal to 1.
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n is the number of loops, then how much length
does it take?
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Does it travel when it goes through this loop?
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So, the length would be 2 pi R times N and
if N is equal to 1 then it is 2 pi R. So,
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for so, this is 2 pi R for N is equal to 1
and for N number of loops this distance would
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be 2 pi RN.
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So, what would be the phase shift accumulated
if there are n number of loops?
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So, phase shift would be 2 pi over lambda
not times delta n effective times the length.
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Total length in the loops which is 2 pi RN,
if I now put the expression for delta n effective
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from here then it would be 4 pi square over
lambda not times C times b square over r times
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n.
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Now, since delta 5 is equal to this.
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So, I can now have a particular radius or
the number of loops, to introduce a particular
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value of phase shift.
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Let us say I take N is equal to 1, and then
I calculate what would be the radius of the
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loop to introduce a phase shift of pi.
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Or what is the radius of the loop to introduce
a phase shift of pi by 2.
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And I know I know that a phase shift of pi
by 2 corresponds a quarter wave plate, if
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I talk about bulk optics.
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In bulk optics if I have birefringent crystals
then out of these crystals I can make quarter
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wave plate half wave plate.
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Quarter wave plate means apart difference
of lambda by 4 or phase shift of pi by 2.
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So, I can immediately find out the radius
corresponding to quarter wave plate, if I
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put delta phi is equal to pi here.
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And find out correspondingly the radius.
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So, this radius comes out to be 8 pi over
lambda naught, times C times b square times
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N.
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For n number of loops, for half wave plate
I am sorry.
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There is a typographical error here.
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For half wave plate this quarter should not
be there delta phi is equal to pi and the
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corresponding radius is given by RHWP is equal
to 4 pi over lambda naught times C times b
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square times N. What are typical values of
these radii, if I consider a silica glass
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fiber with b is equal to 62.5 micrometer which
is the standard cladding radius.
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And I give it a bend give it a bend of about
3 centimeter radius.
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And I consider let us say C approximately
equal to 0.133 at lambda not is equal to 633
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nanometer, then for N is equal to 1 if I consider
N is equal to 1 then the radius of quarter
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wave plate would be about 2 centimeter.
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Radius corresponding to radius of the loop
corresponding to quarter wave plate is about
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2 centimeter.
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And the radius of the loop corresponding half
wave plate is about 1 centimeter ok.
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So, this much a loop of this band radius will
introduce a phase shift of pi by 2 or part
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difference of lambda by 4.
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And a loop of this band radius will introduce
a phase shift of pi or part difference of
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pi, part difference of lambda by 2.
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So, with the help of this with the combination
of these loops: I can virtually obtain any
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polarization state, how?
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Well, I take a fiber and make loops like this.
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Let us say the radius of the loops are such
that or radii of the loops are such that this
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loop introduces a part difference of lambda
by 4.
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So, this is the quarter wave plate this is
equivalent to half wave half wave plate and
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this is again quarter wave plate.
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Now, let me fix these points A, B, C and D.
So, these points are fixed.
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And now these loops these loops can be rotated.
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So, so if you have this fiber and there is
a loop here then this loop can be rotated
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like this.
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This loop can be rotated, this loop can also
be rotated, this loop can also be rotated.
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What do I achieve by rotating these loops?
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You see that if this is the loop, then I have
certain principle axis, I have this is the
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fast axis this is the slow axis.
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When I change the rotation if I rotate it
if I change the orientation of this loop the
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fast axis is this and slow axis is this.
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So, basically what I am doing?
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I am changing the principle axis of this of
the loop.
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00:25:12,679 --> 00:25:18,950
It means that it is equivalent to rotation
of your half wave plates or quarter wave plate
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in bulk optics.
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00:25:19,960 --> 00:25:28,269
So, by rotating these loops I am virtually
creating different orientations of half wave
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plates and quarter wave plate.
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00:25:32,080 --> 00:25:39,299
And so, the phase shift introduced now would
be different, I can control the phase shift
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00:25:39,299 --> 00:25:40,799
that can be introduced.
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00:25:40,799 --> 00:25:51,750
So, with the combination of these and different
rotations I can I can convert any input polarization
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00:25:51,750 --> 00:25:55,320
state into any other polarization state.
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00:25:55,320 --> 00:26:04,350
So, this comes out to be very, very efficient
device in terms of polarization controller.
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00:26:04,350 --> 00:26:11,799
The only thing is that since we are giving
tight bends here and bending always introduces
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00:26:11,799 --> 00:26:13,139
loss.
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00:26:13,139 --> 00:26:22,090
So, this kind of device would introduce certain
losses in the system.
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So, this is how I can rotate for example,
this and obtains any polarization state.
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00:26:34,809 --> 00:26:44,059
So, in the in this lecture we had seen how
directional couplers can be directional couplers
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00:26:44,059 --> 00:26:47,980
can be used as multiplexers and de multiplexers.
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00:26:47,980 --> 00:26:56,480
We had seen in this section how we can make
a power splitters how we can make switches
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using directional coupler.
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00:26:58,350 --> 00:27:09,299
We had also seen that with the help of introducing
certain loops in the fiber we can make polarization
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00:27:09,299 --> 00:27:10,470
controllers.
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00:27:10,470 --> 00:27:15,269
These polarization controllers are commercially
available also.
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00:27:15,269 --> 00:27:26,100
In the next few lectures we will study some
other components based on optical fiber.
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And these components involve gratings in the
core of the fiber, what you can do?
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You can in the core of the fiber you can introduce
periodic refractive index modulation, and
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00:27:41,929 --> 00:27:49,010
this periodic refractive index modulation
can alter the spectrum of the light, which
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00:27:49,010 --> 00:27:51,429
you pass through this fiber.
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00:27:51,429 --> 00:27:59,220
So, with the help of this we can make several
components we can have applications of these
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00:27:59,220 --> 00:28:09,159
components in different devices; in telecom
devices, as well as in sensing devices.
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00:28:09,159 --> 00:28:14,250
So, we will study these fiber gratings in
the next lecture.
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Thank you.