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Now, in this lecture we will start the analysis
of optical waveguides using electromagnetic
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theory. Let us first look at types of waveguides,
this is the bulk medium.
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So, this is not a waveguide, when the dimensions
are very large as compared to wavelength.
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We can have a waveguide like this where we
have a channel or a material in this geometry
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and the dimensions are comparable to the wavelength
and it is surrounded by a lower index medium
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like this. Then in this way I create a channel
for light to flow along this. This is a rectangular
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channel waveguide. I can have this channel
in the form of a cylinder high index refractive
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index cylinder which is surrounded by lower
refractive index medium. Again all are comparable
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to wavelength the dimensions are comparable
to wavelength and this is optical fiber waveguide.
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And then I can also have a configuration like
this we are this width of the channel can
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be infinitely extended.
So, instead of having this kind of channel,
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I can have a slab thin slab of high refractive
index. It is sandwiched between 2 other thick
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slaps of lower refractive indices. Then this
is known as planner waveguide. In these 2,
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I have confinement in 2 dimension and propagation
in this longitudinal direction. Here I have
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confinement only in one direction and propagation
in this longitudinal direction. So, we are
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going to now do the electromagnetic wave analysis
of such kind of media. We had seen that in
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infinitely extended medium where there were
no such refractive index discontinuities,
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the solutions of the wave equations were plane
waves.
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If you remember that the electric field associated
with that wave in an infinitely extended medium
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if it is infinitely extended medium.
I had seen that E was given by E0 e to the
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power i omega t minus kz and H was H0 e to
the power i omega t minus kz and I had these
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vectors.
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So, they have components Ex, Ey, Ez, Hx, Hy,
Hz, this is Hz. And the point to be notice
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here was that, these E0 and H0 they were constants.
S0, E0, H0 purely constants; what we could
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see that in case of infinitely extended medium
our direction of propagation was z. So, I
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had z part coming out like this and then time
part coming out like this. Here also what
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I will do? I will take this direction which
is propagation direction is z and z part would
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be like this itself. So, in order to now find
out the propagation of light waves in such
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kind of medium where we have refractive index
discontinuity refractive index is not uniform,
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when I consider Maxwell’s equations in homogeneous,
linear isotropic charge free current free
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dielectric medium.
In the previous case it was homogeneous medium
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now it is inhomogeneous medium. And again
I write down all the Maxwell’s equations
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Maxwell’s equations go like this. But now
if I look at constitutive relations, when
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D is equal to epsilon E this epsilon is now
in general a function of x, y and z, which
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is nothing but epsilon naught times n square
of x, y and z. So, this is not a constant
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scalar now, which was the case for infinitely
extended medium. And again I am talking about
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dielectric. So, B is equal to mu H which is
closely equal to mu naught times H because
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it is non-magnetic medium. So, what should
I do? Know again I should form the wave equation.
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So, that I can find out what are the electric
and magnetic field solutions. So, the usual
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procedure to find out the wave equation is
I take the curl of this equation and use other
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Maxwell’s equations into it.
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So, I get del of del dot E minus del square
E is equal to minus del t of mu times del
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D over del t, thus in the same way as I had
done in the previous case. But now here I
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should check that del dot E is not 0 I should
be careful. In the previous case del dot E
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was 0, but now I have del dot D and D is equal
to epsilon E epsilon depends upon x, y and
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z.
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So, I cannot take that epsilon out of this
del operator. So, del dot E would not be 0
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if del dot E is not 0 then how this wave equation
is going to change? So, let me know find out
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what is del dot E for that I take del dot
D is equal to 0 and put D is equal to epsilon
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not n square En square is a function of x,
y and z. So, I write it down as epsilon naught
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del of n square dot E plus n square del dot
E and that should be equal to 0. This gives
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me del dot E is equal to minus gradient of
n square over n square dot E. I put this back
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into this equation and rearrange the terms
to get a wave equation in this particular
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form.
What it is del square E plus del of 1 over
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n square del n square dot E minus mu naught
epsilon naught n square del 2 E over del t
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square. So now, you can notice an extra term
here, this is an extra term that we have got
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it as compared to the case of infinitely extended
medium.
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So, this is a wave equation in an inhomogeneous
medium. What is the implication of this term
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now? What complicates it can introduce in
our analysis? Well if I expand this then I
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write down this x, y and z components and
if expand this what I see that now it is not
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possible for me to separate out x, y and z.
t can be separated out, that is not a problem.
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But I cannot separate out x, y and z. x, y
and z solutions cannot be separated or which
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I could do in case of infinitely extended
medium. So, what do I do now? Well similarly
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if I do it for the magnetic field I will obtain
an equation in H something like this, again
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there would be a middle terms which makes
it impossible to separate out x, y and z solutions.
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So now let me consider a case where refractive
index where is only transfers direction which
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is the case of optical waveguides and optical
fibers. So, so I take in general a case where
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n square is a function of x, y which can be
a channel waveguide or optical fiber. So,
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n square does not vary with z, in this case
I can separate out z and t parts. And if I
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can separate out z and t parts then the solution
z and t solution can be written in the same
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way as this. So, I write down z and t solution
like this. And x, y solution is still remains.
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So, I put it with E0. So, the associated electric
field I can write as E(x, y, t) is equal to
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E0 of x, y times e to the power i omega t
minus beta z Similarly H.
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Now what I have got? I have got that the solutions
here have this form. So, this is a function
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of x and y and this is the propagation in
z direction. So, as if some function of x
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and y is propagating in z direction with some
propagation constant beta, similarly for H.
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So, these are the modes of the system. And
as I will find out that there can be only
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certain such functions possible which sustain
their shape and propagate with certain propagation
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constant beta these are the modes of the system.
So now, my problem reduces to find out these
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functions, E0 of x, y and H0 of x, y and their
corresponding propagation constants beta.
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Let me start with doing a very simple problem
where I remove any index discontinuity even
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in y direction. So, I take the simplest case
here the variation of refractive index is
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only in x direction.
So, I have n square as a function of x only
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and this is the case of say planner waveguide
something like this, where you have this is
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x this is y and this is z. So, where y is
infinitely extended, z is infinitely extended
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and you have index cn discontinuity only in
x direction. So, here you have different refractive
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index here different and here different. So,
n square is the function of x only. Now if
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it is a function of x only then I can separate
out y part also. And the solution I can write
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as e to the power i omega t minus some gamma
y minus beta z. And x part will be now associated
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with E0. So, E0 is not a constant is it is
a function of x some function of x. Similarly,
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H is equal to H0x e to the power i omega t
minus gamma y minus beta z.
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So, these are the form of solutions now. What
I can do? I can always choose my direction
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of propagation if the light is propagating
in this direction I can label it as z or I
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can label it as y it is up to me to choose
my access. So, what I do? I choose z axis
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as the direction of propagation then without
loss of any generality I put gamma is equal
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to 0. So, you can see that if this is infinitely
extended, this is infinitely extended index
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discontinuity is only in x. So, you can launch
light into this in this direction and let
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the light propagate along y or you can let
the light propagate along z you can launch
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it from here. So, I choose to take the direction
of propagation as z. So, I put gamma is equal
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to 0.
Then I can write the solutions as E0 of x
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e to the power i omega t minus beta z and
x 0 of x e to the power i omega t minus beta
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z. So, I have got similar form of solution,
but it is not the same. Here E0 and H0 are
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constants; here E0 and H0 are the functions
of x and these functions now I want to find
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out.
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I need to find out how E varies with x and
how H varies with x, that is what I need to
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know and they will give me the modes. So,
I have the solutions here and do not forget
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that I have vector sins here which means this
E is nothing but Ex x cap plus Ey y cap plus
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Ez z cap and similarly H. So, basically I
have 6 such equations 3 in E and 3 in H. So,
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if I write down the components of this then
I can write them as Ej is equal to Ej x e
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to the power i omega t minus beta z and similarly
Hz and similarly Hj where j can be x, y or
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z.
Now, let me put these solutions into curl
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equations. Why I am doing this? Ultimately
I want to find out how E varies with x and
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how H varies with x. So, I need to form a
differential equation in E with respect to
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x. In order to do that and I know from here
I will get del E over del x del H over del
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x terms. So, that is why I put these into
these equations now. When I do this then this
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will give me 3 equations, one corresponding
to Hx then Hy and Hz. And this will also give
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me 3 equations, Ex, Ey an Ez x, y and z components
here, let me do it. The x component from here
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we will come out to be if you expand this,
because you know that del cross E del cross
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E you can write as x cap y cap z cap del del
x del del y del del z. And then you have Ex,
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Ey Ez is equal to So, this is del cross E
minus mu naught del del t and then you have
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3 components here Hx, Hy and Hz.
So, from here you can find out 6 equations,
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the first one would be i beta Ey is equal
to minus i omega mu naught Hx. The x equation
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from here would be i beta Hy is equal to i
omega epsilon naught n square of x times Ex.
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The second one from here would be minus i
beta Ex minus de Ez over del x is equal to
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minus i omega mu naught Hy. And here it would
be minus i beta Hx minus del Hz over del x
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minus i omega epsilon naught n square of x
Ey.
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Third one would be del E by over del x minus
i omega mu naught Hz and here it would be
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del Hy over del x, i omega epsilon naught
n square of x Ez. So, I have got 6 equations
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which relate the electric and magnetic field
components Ex, Ey and Ez and Hx, Hy and Hz.
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What do I do with these equations? Well, what
I notice one thing that I can simplify the
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situation to certain extent. And how can I
simplify the situation? Well if I have if
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I have a waveguide and I launch light into
this when launching light into this I have
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some control on light and that is I can launch
this polarization or this polarization.
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This is x axis this is y axis. So, if I decide
to launch this polarization that is Ey is
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non 0 and Ex is 0 then let me see which equations
do I invoke. Do invoke all the 6 equations
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or I invoke only a few of them?
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And what I find that the equations which have
E y non 0 and E x 0 are this one has E y non
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0 E x is equal to 0. This one has E y non
0 and this one has E y non 0. So, so if I
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if I launch y polarized wave then I invoke
these 3 equations and when I launch exploitation
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I invoke these 3 equation. So, 3 equations
can be involved at a time. So, this gives
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me this gives me a room to simplify the problem,
because I need to now consider only 3 equations
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at a time. These 3 equations the blue ones
are these and what I see there is they have
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only 3 non vanishing components of E and H
and they are Ey, Hx and Hz.
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In these 3 I get that there is only one component
of E and that is transverse. Then these modes
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are also known as TE modes or transverse electric
modes or transverse electric polarization.
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While the other 3 have non vanishing components
of E and H as Hy Ex and Ez and I see that
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there is only one component of magnetic field
and that is transverse component then they
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are known as transverse magnetic modes or
TM polarization and this will correspond to
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these 3 equations. So now, let me do the analysis
of TE modes first. So, what I want to do?
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Again do not forget I want to find out how
E and H vary with x and I need to find out
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a differential equation in E or H with respect
to x. So, for TE modes I write down these
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3 equations.
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And what I can do now since these 3 equations
relate Ey, Hx and Hz then if I know one of
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them then I can find out the others. So, for
example, if I know Ey I can find out Hx from
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here and Hz from here ok.
So, what I do let me find out Ey. So, I substitute
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for Hx and Hz from these 2 equations into
the third equation. And when I do this I form
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a differential equation in Ey. And this comes
out to be d2 Ey over dx square plus k naught
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square n square of x minus beta square times
Ey is equal to 0. Where I have use the effect
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that k naught is omega be c, which is also
2 pi over lambda naught where lambda naught
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is free space wavelength. So now, I have got
now I have got a differential equation in
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Ey for a given n square of x. So, if I know
my planner waveguide that is I know n square
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of x then I can solve this equation for the
given n square of x and obtain Ey. And that
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will give me the modes, that will tell me
how electromagnetic wave propagates in that
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medium of n square x refractive index variation.
So, let me apply it to a very simple waveguide,
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which I call planner mirror waveguide. What
is a planner mirror waveguide?
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You take a very thin slab of refractive index
and let us say glass. It has got a width d
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and refractive index n. And I polish and I
sorry not polish and I deposit metal here
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and here. When I deposit metal on top and
bottom and if I launch any light then that
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light will be reflected back and forth from
this mirror and from this mirror and should
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be guided.
So, this is the simplest waveguide I can think
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of let me do that. So, I deposit metal on
top and bottom if I look at the refractive
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index profile. Then I find that in this region
between 0 and d. I have refractive index n
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and here at the boundaries I have metal. When
it boundaries I have metal then the electric
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field at the metal boundary should be 0 that
is what I know. So, what I do know I write
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down the wave equation the equation which
I obtained in the previous slide. And I put
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n square of x as n square and I write it down
in the region between 0 and d.
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So, this would be the equation. So, I from
here I can find out how Ey varies in this
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layer. And I know that the fields has to be
0 here. Now let me defined this k naught square
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n square minus beta square as some kappa square
since I know that beta which is the propagation
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constant of the wave in this region has to
be less than k naught n it cannot be greater
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than k naught n, because propagation constant
cannot be greater than the propagation constant
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of the medium itself k of infinitely extended
medium.
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So, beta is less than k naught n. So, kappa
square is always greater than 0 which means
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that the solution of this equation would be
Ey is equal to A sin kappa x plus B cosine
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kappa x. Now my feel has to be 0 here and
here. So, I apply these boundary conditions
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Ey is equal to 0 at x is equal to 0 and at
x is equal to d and this gives me B is equal
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to 0 and kappa d is equal to m pi. So, since
B is equal to 0. So, this term goes off and
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kappa is equal to m pi over d.
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00:30:43,120 --> 00:30:49,820
So, I put kappa is equal to m pi over d. So,
my solution now becomes E y is equal to a
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00:30:49,820 --> 00:31:01,640
sig m pi x over d where m can take integer
values now what are. So, I have got E y of
191
00:31:01,640 --> 00:31:09,890
x what is left corresponding beta from where
beta are coming from here because I know kappa
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00:31:09,890 --> 00:31:18,020
d is equal to m pi and kappa square is equal
to k naught square n square minus beta square.
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00:31:18,020 --> 00:31:27,150
So, this gives me that there would be only
certain discrete values of beta defined by
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00:31:27,150 --> 00:31:34,250
beta m and given by beta m square is equal
to k naught square n square minus m pi over
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00:31:34,250 --> 00:31:43,930
d whole square.
So, I have got for a planer mirror waveguide
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00:31:43,930 --> 00:31:53,520
only certain functions only certain functions
which has certain propagation constants they
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00:31:53,520 --> 00:32:03,410
will be sustained. If I plot them then for
m is equal to one it will look like this for
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00:32:03,410 --> 00:32:11,570
m is equal to 2 it would look like this m
is equal to 3 like this. They are nothing
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00:32:11,570 --> 00:32:21,390
but if you if you look carefully they look
like as the modes of vibrations of a stretched
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00:32:21,390 --> 00:32:28,340
string modes of vibrations of a stretched
string. So, it is similar to that. So, in
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00:32:28,340 --> 00:32:39,620
the next lecture I am going to understand
what do they exactly represent. I know that
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00:32:39,620 --> 00:32:48,860
in a waveguide in a waveguide if I launch
if I launch ray like this then it will we
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00:32:48,860 --> 00:32:54,210
reflected back and forth or if I launcher
wave then wave would be reflected back and
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00:32:54,210 --> 00:33:07,240
forth, but how do these represent the guidance
in an a waveguide? So, let us understand it
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00:33:07,240 --> 00:33:09,370
in the next lecture.
Thank you.