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In the last lecture using Maxwell’s equations
we formed a wave equation and I had seen how
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these EM waves propagate in an infinite extended
dielectric medium. In this lecture we will
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further looked into the propagation of EM
waves in dielectrics.
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The outline of the lecture is we would again
look into electric and magnetic fields associated
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with light waves, what is polarization, what
is the energy associated with these waves
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and we will define a vector called pointing
vector for this, then if the EM wave encounters
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boundary if they go from one dielectric to
another dielectric, then what are the conditions
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they should meet at the boundary of two dielectric
media. And then we will look into some entry
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cases in total internal reflection which will
become very handy when we will do wave analysis
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or EM wave analysis of optical waveguides.
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So, what are the electric and magnetic fields
associated with the light beam? Well we have
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seen that for a linearly polarized wave which
is polarized along x and propagating along
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z, I can write the associated electric field
as x cap E0 e to the power i omega t minus
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kz, and H is equal to y cap H0 e to the power
i omega t minus kz, where the amplitude of
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H can be related to the amplitude of E by
this expression. If I have electric field
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then I can find out magnetic field from this
expression, and if I have magnetic field I
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can find out electric field by this expression.
So, if I know one other can be found out because
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they are interrelated.
When we work out some real quantities then
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instead of using phaser notation, it is useful
to use the real part of these and then I can
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write E as x cap E0 cosine omega t minus kz,
and H is equal to y cap H naught cosine omega
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t minus kz. When we plot them they look like
this since E is along x, then E goes like
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this the red curve and the direction of vibration
of electric field vector is given by these
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green arrows the electric field is vibrating
along x, and correspondingly the magnetic
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field is vibrating along y. k and it is represented
by this blue curve. What I can notice here
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is that this E naught and H naught they are
real and since they are real, because it is
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dielectric medium lossless medium then since
they are real then the electric and magnetic
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fields are perfectly in phase. So, when E
has a maximum H also has a maximum, and E
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goes down to 0 at the same instant H also
goes down to 0. So, they are perfectly in
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phase.
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What is the velocity of this wave? In the
previous lecture we had seen that omega over
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k is nothing, but 1 over mu epsilon naught
and this is the velocity with which the surfaces
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of constant phase are moving. So, these are
there. So, this v is nothing, but the velocity
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of the wave and let me find this out for free
space, for free space mu is equal to mu naught
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and epsilon is equal to epsilon naught and
I can approximate mu naught by 4 pi times
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10 to the power minus 7 and epsilon naught
by 10 to the power minus 9 divided by 36 pi,
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and if I work this out it comes out to be
3 into 10 to the power 8 meters per second
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which is nothing, but the velocity of light
and that is how it has been established that
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light is an electromagnetic wave.
What is the refractive index of dielectric?
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When these waves move in if you go if these
wave move in free space and then the wave
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go in dielectric medium, we find that they
have different velocities. So, this difference
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in velocities can be associated with what
is known as a certain property of the medium
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which is known as refractive index of the
dielectric.
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And it can be given by c by v which is a square
root of mu epsilon over mu naught epsilon
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naught, and since for all the dielectrics
they are non-magnetic. So, I can approximate
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mu by mu naught. So, n comes out to be square
root of epsilon over epsilon naught or I can
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write epsilon of a dielectric medium as epsilon
naught times n square where epsilon naught
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is the permittivity of free space. We see
that electric and magnetic fields they are
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interdependent, you have time varying electric
field with generates time varying magnetic
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field, and this time varying magnetic field
intern generates time varying electric field.
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So, there is mutual generation of electric
and magnetic fields, and it is due to this
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mutual generation of electric and magnetic
fields, EM wave propagates even in free space
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or in vacuum they do not require any medium
to propagate.
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Now, let us look at what is the meaning of
polarization, if the electric field vector
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associated with an EM wave vibrates in x-direction
the oscillations are in x-direction, and there
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is a propagation in z-direction, it is something
like this if I take a string I attach one
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end of by string on the front wall, and then
one and I shake in x-direction. So, I am vibrating
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the listing in x-direction and generating
the wave in z-direction. So, this x which
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is the direction of vibration of my hand or
this end of the string or here it is the electric
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field, it is the direction of polarization.
And the electric field in general now can
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be given by of x polarized wave as x cap E0
cosine omega t minus kz plus phi, where phi
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is an arbitrary phase it decides where you
start your clock. So, I can also write this
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down as in terms of x y z components of electric
field, as Ex is equal to E0 cosine omega t
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minus kz plus phi, Ey is equal to 0 and Ez
is equal to 0. Ez is of course, 0 because
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they are transverse waves.
Correspondingly I can get the components of
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magnetic fields, now Hy would be H0 cosine
omega t minus kz plus phi, Hx would be 0 and
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Hz would be 0. So, this is x polarized wave
if I can also shake by string in y direction,
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instead of x direction I can shake that string
in y direction and then the wave goes something
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like this. It again travels in z-direction,
but now the displacement is in y-direction.
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So, here E is now y cap E0 cosine omega t
minus kz plus phi or you can again write them
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down in terms of the components of electric
and magnetic field and these are the components
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you will get corresponding to this y-polarized
wave.
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Now, if I take a transverse plane, here it
is x-y plane and notice the vibration of electric
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field vector then in x polarized.
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I see that the vibration is like this for
y polarized the vibration is like this, but
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I can also shake the string in this direction
this is x this is y, but I can shake it in
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this direction also. Then I will create a
wave which goes like this, the direction of
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vibration is now slanted, it make certain
angle say theta from x axis then this is this
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I called a linearly polarized wave which has
both the components x and y. In this case
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you have only x component in this case you
have only y component, but you can have x
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and y both components. And the amplitude of
these components we will depend upon what
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is the direction of vibration of electric
field vector.
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So, these are some examples of linearly polarized
light, I have another possibility I can shake
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the string like this. Again the movement of
my hand is confined to x-y plane and I am
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shaking the string like this and I generate
a wave like this, then it is known as circularly
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polarized light.
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So, if I trace the variation of the tip of
electric field vector on a transverse plane,
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it will go like this. So, how can I represent
it are since it is a circle. So, institutively
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I can say that if Ex is equal to E0 cosine
omega t, then Ey should be E0 sin omega t.
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So, that Ex square plus Ey square gives me
some E naught square, which is the equation
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of a circle. If I now find out how the tip
of electric field vector goes in x-y plane,
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I plot this t is equal to 0 a t is equal to
0 I will have Ex is equal to E 0, Ey is equal
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to 0. So, I will be somewhere here. At t is
equal to pi over 2 omega I will come here
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t is equal to pi over omega here, t is equal
to t pi by 2 omega here.
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So, as time passes it moves in this direction
this is clockwise. So, the direction if the
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direction of propagation I have fixed as plus
z and rotation is clockwise, then I can call
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this as circular right circularly polarized
light or RCP it is a convention whether you
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call it right circularly or left circularly.
So, if you call this as right circularly then
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that one which is the anticlockwise and direction
of propagation is plus z always then it would
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be left circularly. So, I call it RCP.
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Then we often here term un polarized light,
what is un polarized light; well if it one
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instant I look at the direction of electric
field vector if it is x and the next incident
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it becomes like this, and another instant
it becomes like this, and then like this then
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I do not have any correlation between the
direction of electric fields at different
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instances, they change their direction randomly
then such kind of light is known as randomly
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polarized light or un polarized light people
call it un polarized light, but I prefer to
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use randomly polarized light.
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Then what is the energy associated with an
EM wave. For that let me considered taking
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divergence of this E cross H, you may think
why suddenly I am taking the divergence of
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E cross H; well I want to find out energy
associated with an EM wave, and right now
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I am considering medium which is isotropic
medium and in an isotropic medium what I have
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the energy flows in the direction of wave.
Wave is propagating like this and energy also
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flows like this, and I know that E cross H
represents the direction of flow direction
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of propagation of wave; it is in the same
direction as the wave vector.
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And I also notice that E is I take E cross
H, E is volts per meter and H is amperes per
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meter. So, v times a volts times ampere will
give you power. So, it is power per meter
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square. So, it is kind of intensity. So, it
has got the units of power per area power
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per unit area.
Let us say S, and if I want to find out power
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then I should multiply it by S. So, what I
should do if I take the entire cross sectional
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area. So, I do this and integrated over the
whole area. This E cross H dot ds from divergence
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theorem, I know that it is equal to divergence
over integral over whole volume v, which is
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enclosed by this surface times divergence
of E cross H dv, v is the volume.
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So, this basically will give me the power
flowing out, that is why I consider taking
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divergence of E cross H and see by using now
Maxwell’s equations, whether it really represents
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the power or not. So, I take divergence of
E cross H which I can expand know as H dot
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del cross E minus E dot del cross H, and from
Maxwell’s equations I know del cross E is
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minus del B over del t, and del cross H is
equal to J plus del D over del t. So, this
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del dot E cross H can now be written as because
it is minus del B over del t. So, minus mu
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times del H over del t mu comes here. So,
it becomes H dot del H over del t and here
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you have J here you have e. So, E dot J or
minus J dot e here minus epsilon times E dot
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del E over del t. So, this I can also write
as H times del H over del t can be written
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as half of del H square over del t.
So, this del dot E cross H can now be written
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something like this.
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Now let me take this del, del t outside, then
it is half mu H square plus half epsilon E
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square minus J dot E. What is this we know
that half mu H square is nothing, but the
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density of energy stored in magnetic field
and this is nothing, but the density of energy
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stored in electric field, what is J dot E?
J is current per unit area this is current
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density and E is volts per meter potential
divided by the distance. So, this is nothing,
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but power v times I is power and a times d
is volume. So, this is nothing, but joule
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loss per unit volume. So, if you have a resistor.
So, P would be the power dissipated in register
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per unit volume. So, this is ohmic loss or
joule loss per unit volume.
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Now, let me represent this E cross H as S
some vector S, and this energy density as
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u which is the energy density of a stored
energy in electromagnetic field, then I have
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this and let me integrated over the whole
volume then I get this is equal to minus del
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del t u dv minus J dot E dv all right.
So, this is what it is again here this is
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nothing, but the rate of decrease of energy
stored in EM field, u times dv will give you
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energy and this is nothing, but the rate of
Joule loss.
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So, this SD what is S now? I can now interpret
S. So, this S is nothing, but here if I look
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in this form S dot dS it is nothing, but the
flux of this vector S through closed surface
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s which encloses this volume V; if I look
at this. So, if this is the volume and this
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is the surface area S and I have some power
P in some power, P out then this is the net
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output flux through this volume and what is
happening here? I have a stored electric energy
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is stored magnetic energy and joule loss.
So, this S can now be interpreted as the energy
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per unit time per unit area which flows out.
So, this is this vector has a magnitude which
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shows you energy per unit time per unit surface
area or power per unit area or intensity as
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I have seen here and its direction is perpendicular
to E and H, which is in the direction of k
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and this vector is known as pointing vector,
and represents intensity associated with an
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electromagnetic wave. So, this would be useful
when we will calculate the energy associated
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with modes. Another useful thing would be
which we will use in our further analysis
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of optical waveguides and fiber or boundary
conditions, what are the boundary conditions.
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If I have two media medium 1 and medium 2,
these are dielectric media medium 1 has dielectric
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permittivity epsilon 1 medium 2 has dielectric
permittivity epsilon 2, and if there is an
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electromagnetic field here then if I know
the field here I can find out the field here
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or I can relate the field in region 1 to field
in region 2 using some condition ok.
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With what condition I can relate the fields
here with the fields here. So, if I consider
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that the electric field here is E1 let us
in this direction, and electric field is E2
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let us say in this direction. Then these fields
I can always break into two components, one
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is tangential to this surface interface and
one is normal to this. So, for E1 as well
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as E2 and then I can try to find out the relationships
between these tangential and normal components
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for that what I do? I consider a closed loop
something like this abcda and I know that
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E dot dl over this closed loop should be equal
to 0. If I apply this I go from a to b, I
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need to take tangential component to the surface.
So, it is E1 t times delta w, delta w is the
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path this the very small loop. So, that it
can represent a point when delta w and delta
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h tend to 0 and a surface and interface when
delta h tends to 0, then I have from here
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to here let us say it is delta h by 2 here
delta h by 2 here.
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So, E1 normal components time delta h by 2
then minus E2n delta h by 2, E2 t delta w
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E2n delta h by 2, E1 and delta h by 2. So,
if I do this I find that E1t should be equal
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to E2t. So, the tangential component of the
field here and the field here they should
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be continuous, the tangential component of
electric field should be continuous what about
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normal component? For normal component I consider
a pill box like this very small pillbox let
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me take initially the height delta h and it
has got some surface area delta s. And I consider
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D1 and D2 vectors here, they are dielectric
displacement vectors I again resolve them
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into the components which are parallel to
the surface and perpendicular to the surface.
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Now, in this a small pill box I apply Gauss’s
law, which gives me D dot ds should be equal
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to 0. When I do this then I find is they it
would be D1n times delta s, this is D2n times
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delta s with the negative sign and what is
the flux through this curved surface? This
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flux through this curved surface would be
0 as delta h tends to 0 to obtain this boundary
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to close down to this interface. So, this
gives me D1n is equal to D2n, which means
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that normal component of D should be continuous.
So, I have got tangential component of E should
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be continuous and normal component of D should
be continuous at the interface of two dielectric
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media provided that there are no free charges
at the interface and this is true in case
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of dielectrics there are no free charges no
free currents.
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In a similar way I can show that for magnetic
fields the tangential component of H and normal
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component of B should be continuous.
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So, these are the boundary conditions that
I will use whenever I will encounter an interface
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between two dielectric media. Now lastly let
me revisit total internal reflection and understand
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some intricacies. You remember that when I
did the ray theory, ray theory simply tells
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me that a ray goes like this and it comes
back there is no energy which flows out into
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this medium.
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If this is the medium of refractive index
n1 this is of n2, where n2 is a smaller than
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n1 then ray theory tells me if I launch array
which makes an angle larger than the critical
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angle, then it this ray will come back into
the same medium and there would be no transmitted
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ray in this. But if I talk about waves and
with waves, these are EM waves associated
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electric and magnetic fields.
Let me assume that there is some translated
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wave also, there is some transmitted wave
and then find out what would be the nature
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of that. So, first of all I would ask would
there be any transmitted wave at if yes what
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would be the nature of that. So, what I do?
I like that there be a transmitted wave which
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is coming out at an angle I 2.
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So, if this is n1 this is n2, this is the
interface and this is the normal, this is
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the incident one at i1 and let me say here
I have at i2 this is the transmitted one.
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So, the field associated with this would be
given by E2 let us say E2 is the field here,
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E2 is the field associated with this wave
these are waves not raise would be now E 20,
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E to the power i know since it is in x-z plane,
this is z this is x. So, this is an x-z plane.
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So, I will have a component along x of wave
vector, which is k2x, k2 is like this. So,
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here you will have k2x and here you will have
k2z. So, I will have E to the power i k2x
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times x plus, k2z times z minus omega t. What
is k2x? k2x is nothing, but k2 cosign i2,
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this is i2. So, that is what I have written
k2 cosign i2 times x and similar this is k2
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sign i2 time z minus omega t. If I apply a
smell snow here then I know that sin i2 would
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be n1 over n2 sin i1. So, I will get sin i2
and correspondingly cosine i2 would be 1 minus
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n1 square over n2 square sin square i1 and
from here if I take this n1 over n2 out. So,
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this would be n2 is square over n1 square
minus sin square i1, and since this i1 is
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greater than phi c for total internal reflection.
So, sin i1 should be greater than n2 over
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n1, if sin i1 is greater than n2 over n1 then
this quantity would be imaginary. So, I will
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have this cosine i2 as some i gamma times
n1 over n2, let me represent this as i gamma
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which is imaginary now. So, now, if I put
these values of I sin i2 and cosine i2 back
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into this, what I get that all right corresponding
to z it is find k2z, n1 over n2 sin i1 minus
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omega t, but corresponding to this I have
i gamma vector and this i gamma when it is
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multiplied by this i, it becomes exponentially
decaying.
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So, I will get some E20 e to the power minus
alpha x times this where alpha is this. So,
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what I have got? I have a transmitted wave
whose amplitude decreases with x exponentially.
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So, I have a transmitted wave whose amplitude
decreases exponentially in x direction, this
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is known as evanescent wave. So, with total
internal reflection there is an associated
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transmitted wave in lower index medium, whose
amplitude decreases exponentially and this
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wave is known as evanescent wave. This is
going to be useful when we will do the analysis
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of optical waveguides and optical fibers.
Thank you.