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After having evaluated the modal fields and
propagation constants of the modes of asymmetric
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planar dielectric waveguide, now in this lecture
let us find out how much power is associated
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with a mode, how much energy these modal fields
carry as they propagate along the waveguide.
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So, we will do the analysis for TE-modes we
are talking about. And for this waveguide,
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this refractive index profile and propagation
direction is z the non,-vanishing components
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of electric and magnetic field for TE-modes
are so Ey, Hx and Hz.
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We know that the intensity of an EM wave is
given by pointing vector. So, we now need
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to find out what is the pointing vector corresponding
to these fields; the modal fields. The pointing
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vector is given by S is equal to E cross H.
And since we are talking about electromagnetic
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waves in optical frequency range, so E is
fluctuating with a frequency of something
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like 10 to the power 15 hertz and so the magnetic
field that is. Therefore, S is also fluctuating
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at a very rapid rate. Any detector even though
it is very fast detector cannot record such
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rapid fluctuations and so our eye. So, what
we record is basically the average value-
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time averaged value.
So, we will find out what is the average intensity
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by taking the average of E cross H time average
of E cross H. And since intensity is a real
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quantity, so while calculating intensity we
must take the real parts of E and H. So, for
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TE-modes I know that Ey is defined as Ey of
x e to the power i omega t minus beta z. So,
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it is real part would be Ey of x cosine omega
t minus beta z.
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So, I have Ey what is left is Hx and Hz so
that I can find out the pointing vector and
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therefore, the intensity.
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So Ey is this, how do I find out corresponding
Hx and Hz? Well, I know how H and E are related
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through Maxwellâ€™s equations. It is del cross
E is equal to minus mu naught del H over del
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t. So, if I expand this in matrix form it
would look like this. And in E I know Ex is
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equal to 0 and Ez is equal to 0 only Ey is
non-vanishing for TE-modes. So, from here
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I can find out what is Hx, Hy and Hz. Hy is
not there of course, in case of TE-modes so
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Hx and Hz I can get from here. So, from here
I get if I take the x component del Ey over
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del z is equal to with a negative sign is
equal to minus mu naught del Hx over del t,
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and Ey is given by this. This gives me del
Hx over del t is equal to beta over mu naught
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Ey sin omega t minus beta z.
If I integrate this I can get Hx; integrate
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this with respect to time so I get Hx is equal
to minus beta over omega mu naught Ey cosine
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of omega t minus beta z. Similarly, if I take
z component from here then I get del Ey over
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del x is equal to minus mu naught del Hz over
del t. And if I differentiate this with respect
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to x. I get del Hz over del t is equal to
minus 1 over mu naught dEy over dx cosine
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omega t minus beta z integrating it over time
will give me Hz; which comes out to be minus
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1 over omega mu naught dEy over dx sin omega
t minus beta z.
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So, I have all the three components in place
corresponding to TE-modes. So now, we are
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ready to calculate the pointing vector.
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Pointing vector is given as E cross H, so
I fill in the values here now. So, Sx, Sy,
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Sz in matrix form I can write it like this;
these are the non-vanishing components of
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electric and magnetic fields for TE-modes
Ey, Hx and Hz. And from here I can find out
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what is Sx, Sy and Sz; that is x, y and z
components of the pointing vector. From here
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I get Sx is equal to Ey, Hz minus 0. So, average
value of Sx would be average value of Ey,
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Hz. And if I put Ey, Hz as I had calculated
in the previous slide so this comes out to
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be like this. And in this, what I see of vector
here cosine omega t minus beta z multiplied
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by sin omega t minus beta z, ok.
So, this is something like half of sin 2 omega
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t minus 2 beta z. So, this is fluctuating
and if I average it over a complete cycle
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for any value of z the average value would
be 0; the time average would be 0. So, this
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gives me 0. So, x component of pointing vector;
the average value of x component of pointing
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vector comes out to be 0.
Now let me evaluate Sy: Sy is clearly 0 from
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here itself and Sz if I calculate Sz, Sz will
give you minus Ey, Hx. So, now I substitute
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for Ey and Hx. So, I get beta over omega mu
naught E square of x cosine square omega t
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minus beta z. I know for any value of z the
time average of cosine square omega t would
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be equal to half. So, this gives me beta over
2 omega mu naught Ey square of x.
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So, what do I see? I see that for TE-modes
the average value of Sx and Sy is 0 and I
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get the average value of only z component.
And this is obvious also, this is understandable,
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because my mode is propagation propagating
in z direction so it should carry energy along
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z direction. So, this is the intensity. So,
if I know the modal field that is Ey of x
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then I can find out the intensity. If I integrated
over the entire area then I can get the power
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associated with the mode.
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So, intensity is this, ok. To find out power
I should integrate it over the area, over
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the transverse cross section. The mode is
propagating in z direction so the transverse
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plane is x-y plane, but why is also extended
to infinity. So I cannot integrate it over
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y, I can integrate it only over x. So, in
the case of planar waveguide I cannot have
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power in terms of watts, but I can have only
power per unit length in y direction, because
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I can integrate it only over x.
So, I get P is equal to beta over 2 omega
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mu naught integration Ey square xdx from minus
infinity to plus infinity. And this will give
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me power per unit length in y direction in
the units of watts per meter. So, now, if
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I know Ey of x in the entire region I can
find this out. So, let us consider the case
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of symmetric modes to evaluate this. The modal
field for symmetric modes is given by A cosine
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kappa x for mod x less than d by 2 and C e
to the power minus gamma mod x for mod x greater
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than d by 2.
So, let me substitute this into this expression.
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So I get, and I also make use of the fact
that this is symmetric mode. So, this integral
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from minus infinity to plus infinity can be
written as 2 times 0 to infinity. So, I make
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use of that and then substitute Ey of x, then
I get beta over 2 omega mu naught 2 times
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0 to d by 2 A square cosine square kappa xdx
plus d by 2 to infinity C square e to the
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power minus 2 gamma xdx.
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So, let me evaluate this integral while using
the boundary conditions also, because I need
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to relate C to A. So, here I simplify this
as- I take A square outside and cosine square
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kappa x can be written as 1 plus cosine 2
kappa x by 2 and the C square comes out to
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C square over A square integral d by 2 to
infinity e to the power minus 2 gamma xdx.
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And since, from boundary conditions I know
A cosine kappa d by 2 would be equal to C
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times e to the power minus gamma d by 2. So,
from here I will get C over A which I substitute
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here and evaluate this integral which is very
simple e to the power minus 2 gamma x divided
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by minus 2 gamma and then I put the limit.
So, when I simplify this what I get; I get
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d by 2 from here and sin kappa d over 2 kappa
from here and 1 over gamma cosine square kappa
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d by 2 from this term. This I can further
simplify. So, I take this vector to outside,
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so this becomes d and this becomes 2 over
gamma; cosine square kappa d by 2 can be written
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as 1 minus sin square kappa d by 2. So, this
2 over gamma which is associated with 1 comes
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out here and the rest of the terms I can write
as sin kappa d as 2 cosine kappa d by 2 sin
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kappa d by 2, and this I take common.
So, in the bracket inside I will be left with
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the term which goes as gamma minus kappa 10
kappa d by 2. It would be clear if you do
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this little mathematics. And why I have done
in this fashion because I can see this is
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nothing but the transcendental equation. So
this has to be 0, because gamma is equal to
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kappa tan kappa d by 2. So, if this is 0 this
whole thing goes out and I get a very neat
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expression for power associated with the symmetric
modes. And it comes out to be beta over 2
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omega mu naught half A square times d plus
2 over gamma.
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I can remember it in a very interesting way.
And it is interesting to see that this term
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comes out to be the area of triangle under
this curve. So, how you see that P is equal
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to beta over 2 omega mu naught times integral
Ey square of xdx.
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So, P is beta over 2 omega mu naught integral
minus infinity to plus infinity Ey square
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of xdx. So, if I find out the area under this
Ey square of curve then let us see what do
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I get. So, if this is Ey square as a function
of x then this is nothing but A square, this
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is waveguide width. And field extends into
n2 regions by distance 1 over gamma on either
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side. So, if I make a triangle which has height
A square and base as d plus 2 over gamma,
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then the area of this triangle is simply half
A square times d plus 2 over gamma. So, this
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is interesting that this comes out to be like
this.
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For TM-modes I can do the same analysis and
find out the power associated with TM-modes
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and it is given by this. Although, I had found
out the power by taking the example of symmetric
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modes this expression, this expression, and
this expression, these expressions are valid
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for antisymmetric modes also. This can be
proved and this can be evaluated. So, this
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is how I can get power associated with the
mode and these powers are in watts per meter;
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power per unit length in y direction.
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Let us work out some examples. This is the
example adopted from introduction to fiber
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optics by Ghatak and Thyagarajan. Where I
have a planar symmetric waveguide with n1
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is equal to 1.5, n1 is equal to 1.48, and
d is equal to 3.912 micrometer. At lambda
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naught is equal to 1 micrometer beta for TE0
mode is this and beta for TE1 mode is this;
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it supports two modes at 1 micrometer wavelength.
If at z is equal to 0 the electric field in
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the guiding film is given by this.
So, you see I have TE0 mode and TE1 mode,
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added z is equal to 0. I excite both the modes
with different amplitudes. This mode is excited
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with this amplitude, TE0 mode is excited with
this amplitude and TE1 mode is excited with
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this amplitude. So, the total field at z is
equal to 0 is this much volts per meter.
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Then what is the power carried by each mode?
You can take mu naught is equal to this. So,
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I know the power carried by a mode is given
by this. So, what I need to know; I need to
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know what is the beta for that mode, what
is the amplitude of that mode, what is gamma
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for that mode. And of course, I need to know
what is omega and omega is nothing but 2 pi
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C over lambda naught, since lambda naught
is given to you so you can immediately calculate
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the value of omega.
Then you find out gamma beta is already given
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to you, A is already given to you. Gamma you
can find out from square root of beta square
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minus k naught square n2 square. So, for TE0
mode this gamma comes out to be 1.4126 micrometer
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inverse, for TE1 mode gamma comes out to be
0.9979 micrometer inverse. And if you put
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these values into this expression you will
find that for TE0 mode the power associated
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comes out to be 1 watt per meter and for TE1
mode also you find out that the power comes
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out to be 1 watt per meter. In fact, these
amplitudes have been adjusted in such a way
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that both the modes carry unity power.
When the amplitudes are adjusted in such a
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way then they are power normalized. If you
remember that when we found out the modal
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fields we had retained A only and v related
C to A and said that A can be found out by
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normalization. This is one way of normalization
that you find out the value of A in such a
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way that that the modal field carries unity
power. So, these are power normalized modes.
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Let me take another interesting example of
again a symmetric planar waveguide which supports
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two modes: TE0 and TE1. There propagation
constants are beta 0 and beta 1 and electric
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field amplitudes are A0 and A1.
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If the difference in their propagation constants
is defined by delta beta and z is the direction
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of propagation, then what would be the total
intensity in the guiding film at different
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values of z. First one at z is equal to 0,
second z is equal to pi over delta beta, and
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third z is equal to 2 pi over delta beta.
So, what do I see here at z is equal to 0
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the total field would be A0 cosine kappa 0
x, where kappa 0 you can find from beta 0
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plus A1 sin kappa 1 x where kappa 1 can be
found out from beta 1. So, at z is equal to
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0 this would be the total field.
As these fields propagate it will go with
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propagation constant e to the power i beta
0 z, this will go as e to the power i beta
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1 z. So, as they propagate in z direction
they would be a phase shift accumulated between
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them; a phase difference accumulated between
them. And that phase difference would be delta
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beta z. So, you will have e to the power i
delta beta z. And I know that at z is equal
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to pi over delta beta then this e to the power
i delta beta z would simply be e to the power
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i pi; which means that these two modes would
be pi out of phase. And if we are doing in
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the way e to the power i omega t plus then
it should be minus, so we can have this form
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of expression. So, instead of plus i beta
z, because I am doing it in such fashion so
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I can retain the same convention.
So, the thing is that the two modal fields
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will be pi out of phase when they traverse
this distance. The intensity is nothing but
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the field square, so intensity would be some
alpha times Ey square. So, this would be at
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z is equal to 0, but at z is equal to pi over
delta beta the total field would be this minus
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this, this minus this because they are pi
out of phase. So, the intensity would be this
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much. While at z is equal to 2 pi over delta
beta it would be 2 pi phase shifted 2 pi phase
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shifted means there is no phase shift. So,
your intensity would again be this.
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And the analysis of guided modes of a planar
symmetric waveguide with food for thought
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I had said that.
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In a symmetric planar waveguide the guided
modes are the superposition of plane waves
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making angles plus minus theta m from the
waveguide axis. Where the angles are given
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by cos theta m is equal to beta m over k naught
n1. So, if I launched two plane waves at angles
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plus minus theta 0 then I excite the TE0 mode;
then TE0 mode is excited and the pattern corresponding
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to TE0 mode is found and it goes along the
waveguide and it sustains its shape.
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Similarly, if I excited plus minus theta 1
TE1 mode is there. The question is what happens
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if the waves are launched at angles which
do not correspond to these guided modes; the
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angles corresponding to these guided modes
please think about it.
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And in the end I complete this analysis of
planar symmetric waveguide by briefly mentioning
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the radiation modes.
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I have seen that if beta lies between k naught
n2 and k naught n1 or n effective lies between
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n2 and n1 then I have guided modes. If beta
is less than k naught n2 they should be less
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than n2. So, n effective is less than n2 or
beta is less than k naught n2 then what will
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happen. If I write down the equation wave
equation for TE-modes and then I write it
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down in both the regions, in this region and
in this region then for mode x less than d
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by 2. I have this equation beta is less than
k naught n2 and hence beta is also less than
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k naught n1. So, kappa square is positive.
Now, in the region for mod x greater than
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d by 2. I would have d2Ey over dx square plus
k naught square n2 square minus beta square
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is equal to 0. And since beta is smaller than
k naught n2 then if I define this as delta
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square and delta square would be positive.
And what I will have? I will have oscillatory
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solutions here as well as here. So, everywhere
I will have oscillatory solutions; which means
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that energy is carried up to infinity in x
direction that is energy radiates out in the
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n2 region. And these kinds of modes are known
as radiation modes because the energy radiates
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out corresponding to these modes, and they
form continuum they are not discrete modes.
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So, they form continuum of modes.
So, with this finish the analysis of planar
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waveguide, symmetric planar waveguide, whatever
we have learned about the modes, modal fields,
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and the procedure of finding out the modes
would be very useful when we will do a more
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complicated structure such as optical fiber.
Thank you.