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In this lecture we will study the periodic
refractive index modulation in the fiber core,
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these periodic structures are called grating.
So, we are going to study fiber gratings.
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So, what we have is in a fiber if we create
periodic refractive index modulation, then
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these kind of structures can be used as components
or devices in telecommunication and sensing
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systems. Depending upon the periodicity of
this modulation we can have short period gratings
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or long period gratings. Short period gratings
are also referred to as fiber Bragg gratings,
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because the phenomenon is equivalent to Bragg
reflection or Bragg diffraction in crystals.
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Typical refractive index modulation amplitude
is 2 into 10 to the power minus 4 and length
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of the grating is few millimeters it can be
a couple of millimeters to 10 or 15 millimeter.
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So, let us understand how does it work first
we will consider fiber Bragg gratings. So,
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we have this refractive index modulation,
we have low refractive index, high refractive
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index then low refractive index periodically,
now when a light beam is incident here because
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of this index contrast a part of the beam
gets reflected from this interface, part of
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the beam goes out and then it gets reflected
from the rear interface. Similarly from all
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the other layers there would be reflections.
What happens is that when all these reflections
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are add up are added up in phase then we can
have a very strong reflection and this will
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happen at a particular wavelength.
So, in order to understand this let us zoom
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this, and consider different waves here. So,
when this wave is incident here this is the
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incident wave, and we have the periodicity
lambda. So, for convince let us divide it
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into lambda by 2 for high refractive index
region and lambda by 2 low refractive index
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region. Let us say in high refractive index
region the refractive index is n0 plus delta
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n0 and in this region it is n0 minus delta
n0.
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Now when this wave is incident here a wave
gets reflected from this interface, let us
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say it is a part of the wave gets reflected
from this interface let us say it is b then
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whatever portion of the wave whatever fraction
of the intensity goes out here that gets reflected
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from other interfaces. If all these three
are added up in phase if they have a phase
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shift of 2 pi or integral multiple of 2 pi,
then we will have strong reflection.
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Let us find out the condition for this strong
reflection and wavelength at which this will
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happen. So, if I consider these 2 waves a
and b, then the phase difference between a
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and b would be pi minus 2 pi over lambda not
times n0 plus delta n0 times lambda, because
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this is the path difference from here to here;
so lambda by 2 and lambda by 2 in the refractive
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index region of n0 plus delta n0. So, this
is the part difference and this pi phase shift
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occurs because there is a reflection from
rarer to denser interphase media.
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So, I have delta phi a and b the phase difference
between these 2 waves, similarly I can have
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delta phi bc the phase difference between
these 2 waves, and if all the three waves
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have to have constructive interference and
let us say this happens at wavelength lambda
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b, then if I take the total phase shift the
phase shift between these and these there
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it should be integral multiple of 2 pi and
for m is equal to 1 if I take the value of
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that integer as 1then it should be equal to
2 pi, and if I put the expressions from here
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the lambda the value of lambda b will come
out to be 2 n0 times capital lambda.
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Where I have assumed that delta n0 is much
smaller than n0, and which is the case because
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n0 is typically 1.5 and delta n0 is about
2 into 10 to the power minus 4. So, I can
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neglect n0 with respect to sorry I can neglect
delta n0 with respect to n0.
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I can also understand this interaction as
coupling between 2 modes. If I have a fiber
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let us say I have a single mode fiber. So,
in a single mode fiber if I launch light from
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this side the mode is excited and mode propagates
in this direction. If this is ideal fiber
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and there are no scattering centers, and if
I have index matching liquid at this output
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end then there is no way that the light will
travel back because there is no back reflection
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from here, there are no scattering centers.
So, there is no mechanism by which the wave
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can travel back.
So, in general even if I do not put index
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matching liquid here and there are some scattering
centers, then the light which comes back is
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very small. So, most of the light goes in
forward propagating mode, this grating here
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acts as periodic refractive index perturbation
and this perturbation can couple light from
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forward propagating mode to backward propagating
mode. So, if I have forward propagating mode
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whose propagation constant is beta plus, and
backward propagating mode whose propagation
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constant beta minus and of course, since it
is single mode fiber. So, these modes are
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the same they their propagation constants
magnitude are the same only their directions
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are different. So, this magnitude beta plus
would be equal to magnitude beta minus let
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us say it is equal to beta, and if n effective
is the effective index then I can write it
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down as 2 pi over lambda not times n effective.
If the wave vector corresponding to this periodic
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refractive index modulation or the spatial
frequency is K is equal to capital K is equal
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to 2 pi over capital lambda, then if this
K is such that this length is equal to this
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then if a mode goes like this, and then capital
K takes it here. So, the resultant would be
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beta minus. So, in this way I can couple power
from beta plus to beta minus, and the condition
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for this mathematically can be given by twice
of this magnitude that is 2 times 2 pi over
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lambda b times n effective, it should be equal
to the magnitude of this K it is 2 pi over
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capital lambda.
So, this gives me lambda b is equal to 2 times
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n effective times capital lambda. So, this
particular wavelength will satisfy this condition,
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and all the power in this spectral component
can be reflected back. Typically if I take
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a silica glass fiber n effective is of the
order of 1.5 typical value 1.5, and if I consider
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the Bragg wavelength as 1.5 micrometer. So,
this periodicity required is about 0.5 micrometer.
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So, these are sub-micron gratings. So, this
periodicity is really very small that is why
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these Bragg gratings are also known as short
period gratings.
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We can study the power evolution in the forward
and backward propagating modes by using coupled
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mode theory and in coupled mode theory, the
amplitude of the forward propagating mode
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varies with z according to dA over dz is equal
kappa B e to the power i gamma z and that
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of backward propagating mode varies as dB
over dz is equal kappa A e to the power minus
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i capital gamma z, where capital gamma is
2 beta minus K which is nothing, but phase
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mismatch if gamma is equal to 0 then there
is phase matching between the forward propagating
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mode and backward propagating mode.
Intuitively, I can say that it is clear that
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for efficient coupling between the forward
propagating mode and backward propagating
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modes, you should have phase matching conditions
satisfied. So, gamma should be equal to 0.
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If I look at these 2 equations the equation
of A contains B and equation of B contains
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A. So, these are coupled equations and this
theory is known as coupled mode theory. Now
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this coupling between the 2 modes occurs due
to periodic refractive index modulation, and
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in the grating region if I write down the
refractive index then I can write it down
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as delta n g square g refers to grating xyz
is equal to n square of xy plus delta n0 square
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sign kz, where I have considered this periodic
refractive index modulation is sinusoidal.
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So, this is sinusoidal refractive index modulation
with special frequency K or wave vector K.
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So, if I use this then I can show that this
kappa which is known as coupling coefficient
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is given by omega epsilon naught by 8 integral
over x and integral over y psi star del n
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square x y psi dxdy where psi is normalized
modal field it is power normalized modal field.
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Now if I consider a single mode fiber and
consider Gaussian approximation that is I
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approximate my mode, mode of the single mode
fiber by a Gaussian of width w, then this
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integral can be written as I is equal to 1
minus exponential minus 2 a square over w
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square and then this kappa can be represented
in terms of this integral as pi delta n0 times
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I over lambda b, where a is the core radius
and w is the Gaussian spot size of the mode.
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So, in order to find out kappa I need to find
out overlap integral which is this, and you
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can see that it depends only on fiber parameters.
What is the modes spot size Gaussian spot
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size and what is the core radius. It involves
minus a square over w square, and I know that
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for a given fiber w over a varies with V like
this, this is empirical formula given by marques.
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So, I can now find out how this overlap integral
with vary with V it is very simple now to
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do this. So, I can see that as V increases,
this overlap integral increases typically
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around between 2 and 2.4, the value of this
integral is around 70 percent to 80 percent
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ok.
So, there is almost 78 percent to 80 percent
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overlap between the modes via this short period
grating or fiber Bragg grating. So, this with
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the help of this I can find out what is the
strength of interaction. So, the strength
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of interaction which is defined by coupling
coefficient and given by pi delta n0 I over
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lambda b, now I already know how I varies
with V then I can find out how kappa varies
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with V. If I take typical value of delta n0
as 2 into 10 to the power minus 4 and Bragg
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wavelength as 1550 nanometers then kappa varies
with V like this.
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Again I can see that that typical value of
kappa for the values of V from 2 to 2.4 is
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about 0.3 millimeter inverse. So, it is around
0.3 millimeter inverse if you go to shorter
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values of V then it can be around 0.2.
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Now, with the help of this since I know how,
a which is the amplitude of forward propagating
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mode, and b which is the amplitude of backward
propagating mode. So, the equations describing
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the variations of a and b with z, I already
know. So, if I solve these equations for given
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kappa for given fiber for giving grating,
then I can find out how much would be the
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reflectivity. Because from these amplitudes
I can find out the powers and if I find out
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the ratio between the reflected power at the
input end of the fiber, and the incident power
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at the input end of the fiber, then I can
know what is the reflectivity.
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If I work that out then reflectivity comes
out to be kappa square, sin hyperbolic square
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gamma z, divided by gamma square cosine hyperbolic
square gamma L plus capital gamma square by
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4 sin hyperbolic square gamma L, where gamma
square small gamma square is kappa square
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minus capital gamma square by 4. And this
expression has been obtained using the conditions
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that at the input end of the fiber that is
at z is equal to 0 A is equal to 1 that is
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all the power is in forward propagating mode,
and B at z s equal to l is 0. So, when light
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reaches to the end of the grating then at
that value of z the power in reflected mode
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or the backward mode is 0, which means that
all the power gets reflected at that wavelength
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lambda B phase matching condition is satisfied.
So, 2 beta is equal to kappa which means capital
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gamma is equal to 0, if capital gamma is equal
to 0 then small gamma is equal to kappa.
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If I use this in this expression, then the
reflectivity of the grating comes out to be
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ten hyperbolic square kappa L. If I plot this
reflectivity as a function of kappa L, then
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what I see that as I increase kappa L the
reflectivity of the grating increases, and
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eventually for large values of kappa L it
will eventually reach the value unity, there
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would be hundred percent reflection, but that
will occur at the value of kappa L is equal
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to infinite. But what I can have is that at
kappa L is equal to three if I work this out
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here, read this out then at kappa L is equal
to 3 reflectivity is almost 99 percent.
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So, this value of kappa L is sufficient for
me, this happens when I am at Bragg wavelength.
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So, if I choose let us say kappa l is equal
to 3 then at Bragg wavelength the reflectivity
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is 99 percent, now if I slightly move away
from Bragg wavelength on either side then
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phase matching condition is not satisfied
and the reflectivity will drop down. So, what
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I anticipate that at lambda is equal to lambda
b there should be a peak in the reflection
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spectrum, and as I move away from this lambda
b the reflectivity would fall down.
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Here, I show how the power in the transmitted
and reflected modes vary as a function of
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z, what is the evolution of power in the forward
and backward propagating modes for different
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values of kappa. For a given grating length
L is equal to let us say 5 millimeters I can
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see that when kappa is the small value say
0.1 milliliter inverse, then even at l is
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equal to 5. I have the transmitted power,
transmitted power drops down only by 20 percent
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and it will almost saturate, and the reflected
power is maximum it is around 20 percent,
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it will take very long length for this power
to go beyond this.
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However, if I increase kappa then I can increase
this power, the transmitted power drops down
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and the power in reflection mode increases
and when I have the value of kappa around
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0.4, then I can see that the power in reflected
mode is or backward propagating mode is more
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than 90 percent. And if I have the combination
of kappa and L is equal to three then of course,
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the reflected power would be 99 percent here
kappa times l is only 2.4 times 5.
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Let us find out what is the optimum grating
length, if I want to have reflectivity let
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us say 99 percent. I can always find out the
grating length required for a desired value
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of reflectivity, because I know that the reflectivity
varies at tan hyperbolic square kappa L. So,
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L would be tan hyperbolic inverse square root
of R over kappa. This plot is for delta n
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is 0 is equal to 2 into 10 to the power minus
4 lambda B is equal 1550 nanometers and for
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the value of R 0.99. So, in order to have
99 percent reflection.
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So, if I vary my fiber parameter that is I
vary the value of V then this is how the length
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of the grating has to be tuned to achieve
99 percent reflectivity. I can see that if
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the value of V is large, then the grating
length required is the smaller, and for smaller
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values of V. I require much larger grating
length. It is obvious because if I increase
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the value of V my mode is more and more confined
in the core and the overlap integral increases.
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While if I go to shorter wavelength then the
field spreads out from the core and the overlap
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integral mind it that this overlap integral
since grating is only in the core region.
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So, it is the power in the core region that
matters, so, if power in the core region is
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small then the overlap integral would be smaller,
and hence you will require much longer grating
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to reflect back 99 percent power.
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Let us work out few examples, if I have a
typical fiber Bragg grading with delta n0
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is equal to 2 into 10 to the power minus 4,
and I assume 75 percent overlap between the
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forward and backward propagating modes via
fiber Bragg grating, then let us calculate
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coupling coefficient at 1550 nanometer wavelength
and the required grating length for 99 percent
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peak reflectivity at 1550 nanometer wavelength.
I know the coupling coefficient is given by
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pi delta n0 times I over lambda B and delta
n0 is 2 into 10 to the power minus 4, I is
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0.75 and lambda B is 1550 nanometer. So, this
gives me the value of kappa about 0.3 millimeter
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inwards.
What is the required grating length for 99
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percent reflect peak reflectivity. So, I know
the reflectivity is given by tan hyperbolic
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square kappa L. So, l would be tan hyperbolic
inverse square root of R over kappa is 0.99
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to have peak reflectivity 0.99 percent, and
the value of kappa I have obtained at 1550
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nanometer wavelength as 0.0 0.3040. So, if
I put these values here, I will get the grating
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length about ten millimeters or 1 centimeter.
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Let us now look at reflection and transmission
spectrum of the grating. if I incident this
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spectrum which is broadband spectrum, then
out of all these wavelengths only one wavelength
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which satisfies the condition 2 n effective
capital lambda is equal to lambda B, that
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wavelength will be reflected and all the other
wavelengths will pass through. So, in the
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transmitted spectrum this wavelength lambda
B would be missing and in the reflection spectrum
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this wavelength will appear.
Then what is the spectral width of this? This
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spectral width of this reflection spectrum
is given by delta lambda is equal to lambda
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B square divided by n effective times L, times
one plus kappa square l square over pi square
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to the power half. So, if I know all these
values n effective kappa L is given for given
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reflectivity, and then it will depend upon
what is the value of n effective what is the
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value of L.
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This is typical reflection spectrum for delta
n0 is equal to 2 into 10 to the power minus
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4,grating period about 0.54 micro meter and
L is equal to 11 millimeter for a typical
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single mode fiber.
So, this grating period with for a given for
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the given single mode fiber gives me lambda
B is equal to 1550 nanometers. So, I get the
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peak reflectivity at 1550 nanometers and look
at the value of delta lambda. Delta lambda
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is about 0.2 nanometers, delta lambda is the
width at which the power falls down to 50
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percent of its peak value. So, it is about
this width. So, delta lambda here in reflection
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spectrum of fiber Bragg grating is really
very small, it is 0.2 nanometer or even less
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depending upon the length of the grating depending
upon other grating parameters and fiber parameters.
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If I look at the spectral width for a typical
fiber given by n1 is equal to 1.447 and n2
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is equal to 1.444 then if I change the value
of V, then delta lambda changes according
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to this. For reflectivity 99 percent I know
for reflectivity 99 percent kappa L is equal
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to 3. So, all the values here in this bracket
is fixed now it will depend only on the grating
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length, and n effective n effective means
the fiber and the Bragg wavelength which also
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depends upon fiber parameters.
So, what I see that that as I vary V that
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is I change my fiber, then delta lambda varies
between let us say 0.15 to 0.25 nanometers,
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it is in nanometers. Why do I obtain this
kind of variation how can I understand this
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kind of variation? If I increase V then L
decreases this I have already seen that if
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I increase V the length required for 99 percent
reflection decreases, if I increase V. n effective
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also decreases. So, if L decreases an n effective
decreases then delta lambda would increase.
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So, this is how, I can understand this curve.
In the next lecture I will work out some more
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examples, and look into some applications
of fiber Bragg grating.
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Thank you.