In the last lecture we have talked about propagation
of electromagnetic waves in free
space. We will continue with that today, but
we will be talking about what happens to a
harmonic electromagnetic wave, when it comes
to interface between two different
dielectrics, which is something which we of
course know from our schools that is we
know that the electromagnetic waves or what
we are more familiar with from in our
schools for instance the light wave, undergoes
a reflection and a refraction at the
boundary or the interface between two medium.
We will try to establish the well-known
laws of electromagnetic waves, well known
laws of reflection and refraction from the
electromagnetic theory that we have established
so far.
.
So, let us look at what happens in this picture,
what you find is a, this is an interface.
So,
below this interface is a medium second medium
and above is the first medium. So, I
will assume that the first medium is air or
vacuum and the second medium to be specific,
let us say glass. So, I have a an electromagnetic
wave incident on this interface, which I
.am taking as the x y plane namely the z equal
to 0 plane, at an angle of incidence of theta
i. The angle of incidence is defined the same
way as we have been doing always, namely
the angle that the incident ray makes with
the outward normal to this surface. And of
course, we know that there would be a reflected
wave and a transmitted wave.
So, I have represented incident as i, the
reflected as r and the transmitted with the
letter t.
So, this is the things, theta i, theta r,
the angle of reflection and theta t, the angle
of
refraction or the transmission. The the principle
that is used in obtaining the standard
laws of reflection and refraction are the
following, that the tangential component of
the
electric field at the interface between two
media must be continues. So, if you look at
this picture again, so you notice that the
the media or the medium for the incident ray
and
the reflected ray is the same and the medium
for the transmitted ray is different.
.
Therefore, what we do is to write down that
the tangential component, which I will
abbreviate by putting in, so the amplitude,
let us say E 0 I parallel because it is a
tangential component and I will use the complex
notation, but of course one could write
down the the harmonic the sine or the cosine
wave.
But it is much more convenient to do the algebra
in terms of exponential function and
later on take the real or the imaginary part
as we like. So, this is E to the power i k
dot r,
so I will write the propagation constant vector
propagation vector as k I dot r minus i
omega t. This is, I am taking forward moving
wave this must be, this plus because in the
.same medium I have got the reflected ray,
so I will write its E 0 R parallel again
exponential of E to the power i k R, k r is
the propagation vector for the reflected ray
dot
r minus i omega t.
This must equal the tangential component of
that transmitted ray, so which is E 0 T
parallel transmitted wave which is E to the
power i k T dot r minus i omega t. I have
intentionally use the capital letters, so
that reflection R, does not confuse with this
position vector r and similarly, the transmission
coefficient T does not confuse with the
time variation t. So, this is the equation
which tells me that the tangential component
of
the these vectors are the same. Now, notice
one thing that this type of an equation must
be valid at all time and at all points on
the interface.
So, interface vector is the vector r and time
is arbitrary, so as a result this type of
an
equation in order that they are valid at all
positions are on the interface and at all
time t,
the exponential factors must be the same.
Then we will look at what the other relations
should between E 0 I parallel, E 0 R parallel
and E 0 T parallel, but unless the
exponential factors are the same, I cannot
have such an equation satisfied at all time.
So,
let us write down out, this implies that I
must have K I dot r plus K R dot r or rather
I
must have the all the things are the same.
So, let me say in general it is K R dot r
plus let us say phi 1, which is equal to K
T dot r
plus phi 2. Now of course, I could have an
arbitrary phase, constant phase between the
three terms because E to the power I phi 1
or E to the power I phi 2 would simply
multiply with these complex amplitudes in
general. So, so let us look at, what do, what
does it imply? So, let us look at the first
equation.
..
The first equation is telling me that K I,
so I will write this down. K I minus K R dotted
with r, r as you remember is the position
vector on the plane that is equal to phi 1.
Now,
we are all familiar with vector equation of
a surface. We know that if the normal to a
plane is given by a vector n, then the n dot
r equal to constant defines an equation to
a
surface. So, in order that this represents
the surface the interface between the two
medium, so this tells me such a surface is
perpendicular to this K I minus K r, just
as n
dot r equal to constant is the equation pair
a vector equation to a surface. Therefore,
this
tells me this surface is normal to K I minus
K R.
Now, so suppose I say, that the K I minus
K R and the normal n to the surface.
Supposing they lie in x z plane, remember
that I have taken x y plane as the interface.
Therefore, I have a choice of the x y plane,
so let me choose this to be in the x z plane,
which is my incident plane. The definition
K I minus K R and the normal to the plane
of
incidence defines my incidence plane. Now,
since n and K I minus K R are in the same
plane, this tells me that K I minus K R vector
cross product with n, must be equal to 0.
Alternatively if the angle between the normal
as we had shown in the picture, let me
repeat that picture again.
So, this is my normal direction this is the
direction of K I, this is the direction of
K
reflected, this is the angle theta i and this
is theta R. So, this tells me that K I minus
K R
cross n is equal to 0, it implies that K I
sine theta because sine theta i. So, magnitude
of
.K I times sine theta i, must be equal to
magnitude of K R times sine theta R. Now,
since
the incident ray and the reflected ray are
in the same medium, the magnitude of the
propagation vector, their directions are different,
but the magnitude of the propagation
vector must be the same. So, magnitude of
the propagation vector K I is equal K R is
equal to omega divided by the velocity in
that medium.
Since, I have taken it to be the vacuum, so
let the velocity be the velocity of light
C.
Now, since these two magnitudes are the same,
it tells the sine theta i is equal to sine
theta R. Alternatively the angle of incidence
is equal to angle of reflection, which is
of
course, very well known lot to us. Now, we
will understand it on the basis of the
electromagnetic theory, let us try to prove
Snell’s law. This is almost similarly done
because we have said that K I minus K T dot
r just as we had proved that K I minus K R
dot r.
.
So, we have said K I minus K T dotted with
r, that is equal to phi 2, which is another
constant and once again this is an equation
to A. Surface which is the perpendicular to
that surface being K I minus K T vector. So,
as a result K I minus K T, just the way we
have done it earlier cross n is equal to 0.
This tells me that just as we had written
it
earlier magnitude of K I times cosine of sorry
sine of theta i must be equal to magnitude
of K T sine times sine theta T. Now, look
unlike the case of reflection, the two
.propagation vectors do not have the same
magnitude anymore, and that is because this
is
in medium one and that is in medium two.
So, I know that the magnitude of the propagation
vector is nothing but omega divided by
the velocity. So, let us write it K I let
us say, is omega divided by let us say, velocity
of
light it is in medium one and K T magnitude
will be omega. Remember that, when a
wave is incident at an interface, the frequency
does not change the, and so this is equal
this divided by the velocity in that medium
velocity in that medium T. Therefore, sine
theta i by sine theta T, sine of the angle
of incidence divided by angle of transmissions
is
magnitude of K T divided by magnitude of K
I, which is equal to, this sine theta i by
sine
theta T, is K T by K I. K T is omega by v
T, therefore this will be C by v T.
If you remember this is nothing but the definition
of the refractive index of the medium
two, that is the transmitted medium, with
respect to the refractive index of the medium,
incident medium or this is simply what we
call as the refractive index. In common
language when it is understood that the medium
with respect to which, we are talking
about this refractive index happens to be
air or the vacuum. Therefore, this is my Snell’s
law, sine i by sine r. r being the angle of
reflection refraction. So, this is Snell’s
law.
So, using electromagnetic theory, and the
fact that the tangential component of the
electromagnetic field on the interface between
the two media, must be the same. We
have been able to prove the two simple laws
which we have learnt right from our school
days. Now, having done that we still have
not talked about, what happen to the
amplitudes? The E 0 I, what is the relationship
between E 0 I, E 0 R and the E 0 T? The
exponentials we have taken into account, so
our next job will be to point out what these
relations are and before we do that, we will
be in a, we will require the, what we have
learnt earlier.
And the setup equations which will relate
the amplitude of the reflected wave with
respect to the incident wave or the amplitude
of the refracted wave with respect to the
incident wave. These set of equations are
known as Fresnel’s equations.
..
And let us look at how does one do it? In
order to do that, we need the other boundary
conditions that we have. The firstly, you
remember that from my Gaussian law of
magnetism, del dot of B is equal to 0, del
dot B equal to 0 we have seen by taking a
pill
box, a Gaussian pill box on the surface I
should be able to prove that the normal
component of b field is the same. Now, now
what we will doing actually, we will assume
that both the media they have the same magnetic
permeability. We are not discussing the
they are different dielectric, but we will
assume that the magnetic, they are not magnetic
material. So, the magnetic material permeability
of both the media will be taken to be
equal to mu 0. So, as a result if B 1 n is
equal to B 2 n, this is also implies that
H 1 n is
equal to H 2 n.
..
So, normal component of the magnetic field
H 1 n must be equal to H 2 n. Now, if you
refer back to this again, I have del cross
of E equal to minus D B by D t and this is
the
cross product. So, you remember that this
standard mechanism has been whenever I had
a del dot divergence, I have taken a Gaussian
pill box half in one medium, half in the
other medium. Now, whenever we had a cross
product, what we did is to take an
Amperian loop and then go by a surface integral.
A line integral because del cross of E
dotted with D s is related to a line integral
and this of course, B dot D s will give you
the
flux and this as we have seen several times
has given us E 1 t equal to E 2 t.
Now, del dot of D is equal to rho free, but
notice that unlike del dot of B which is equal
to 0, my del dot of D is not equal to 0, but
is equal to the surface charge density that
is
there. As a result there is a discontinuity
in the normal component of the D field. So,
D 1
n minus D 2 n is equal to sigma. So, let me
write it down D 1 n minus D 2 n is equal to
sigma. In a very similar way is if there is
a discontinuity, if the on the surface there
are
some surface currents then if I look at H
1 t minus H 2 t. Now, this gives me J
perpendicular, where J perpendicular is the
component of the surface density, which is
perpendicular to the direction of H, which
is being matched. Remember when I said
tangential direction all that I can, I know
is it is on a surface.
So, if your matching a particular direction,
J perpendicular is perpendicular to those.
Now, however what we are going to do, we are
going to be talking about perfect
.dielectrics. So, I do not have charge densities
on the surface, I do not have a current
density on the surface, as a result both the
tangential component and the normal
component of the both E field and H field
will be taken to be continues. So, these will
be
the issues that we will be using in our next
set of derivations. So, let us look at what
it
implies?
.
So, if you look at this picture here, so what
I have done is this. Now, we will be
discussing two different cases. The different
cases that we will be doing will be called
p
polarization, p standing for parallel. It
means the direction of the electric field
will be
taken to be parallel to the plane of incidence.
So, this is the picture of the plane of
incidence because so the interface actually
is perpendicular to the plane of the screen,
and so this ray, this ray and that ray, this
incident ray, the reflected ray and the
transmitted ray, this defines me and of course
the normal there are all in one plane. So,
this is the picture of that plane
So, what I have done is to draw a direction
which is perpendicular to the incident
direction and obtusely the magnetic field
direction. Then will be perpendicular to both
the electric field direction and the propagation
direction and this if you look at the fact
that electric field magnetic field, and the
direction of propagation turns out to be a
right
handed triad system. Then corresponding to
this, the direction of the magnetic field
is
coming out of the plane of the paper and so
what we have done is, actually to define take
.the direction of the H field coming out of
the plane of the paper. This is a convention
that
we have assumed and then decided what the
direction of the corresponding electric fields
will be.
This is fairly straight forwards to work out,
once you know that this is the direction of
the propagation and outward to the plane of
the paper is the direction of the magnetic
field, and then you can work it out. You can
check that if you take this theta to go to
0
then of course, what you find is that the
electric field directions will become oppose
to
each other at normal incidence. And so this
is what we have done, so let us look at this
p
polarization, the other polarization is s
polarization, which comes from a German word
centrate. The s polarization is squared the
electric field is taken perpendicular to the
plane of incidence. But let us first talk
about the p polarization. p standing for parallel,
so
let us look at what do I have, let me draw
this picture again here.
.
This is my normal, this is the direction of
the incident ray, this is the direction of
reflected ray and this is the direction of
the transmitted ray, and what you have done
is to
take the magnetic field this way out of the
plane, so that the electric field is like
this.
Now, in this picture I am going to now work
out that, what is the equation corresponding
to the tangential component of the electric
field being the same? So, you notice that
this
angle is theta, the tangential component is
like this. So, if this angle is theta, this
angle
happens to be the theta and like this that
is the same way here.
.So, these are in opposite direction, so as
a result what I get is E I cos theta i minus
E R, R
for reflection cos theta R is equal to this
is the direction, in which I have subtracted,
so is
equal to E T cos theta T. Now and similarly,
I have got a H I because H is being taken
perpendicular to the plane of the incidence.
So, I will have H I plus H R, remember the
normal components are also continues is equal
to H T. Now, I know that I am dealing
with propagation in dielectric, so as a result
my E and B or E and H have a relationship
and that relationship is given by, my H is
equal to square root of epsilon by mu 0,
actually mu if you like, but I will be taking
all mu to be equal to mu 0. So, mu let me
write it temporarily with the electric field,
now this is this is true in all medium. So,
you
have to take the corresponding epsilon, corresponding
to the medium in which you are in
and the correspondingly.
Similarly, mu of course, I have said that
I will take the mu is to be the same. So,
this is
this is fairly straight forward because I
know the magnetic field d is related to the
electric
field by magnetic field d being the electric
field divided by the velocity of
electromagnetic waves. And the velocity of
the electromagnetic waves is 1 over square
root of mu times epsilon and B and H are related
by B is equal to mu times H. So, if you
do that then you find h is equal to square
root of E by mu times epsilon. So, if you
plug it
into this equation, what you find is, so let
me write the pair of equations again.
.
.I have bought E I cos theta i minus E R cos
theta R is equal to E T cos theta T and I
have
got in terms therefore, let me write it down.
Incident medium, so I have got epsilon I by
mu I and reflected medium is the same, so
I will write it as E I minus E R and that
is
equal to square root of epsilon T by mu T
times E T. So, this is what we have got and
of
course, this tells me that I should be able
to solve these equations without any problem.
So, this is you just divide one by the other
this equation should have been a plus because
I had a H I plus this. So, just divide on
of the equations by other and you find that
this,
these are whole sort of algebra on the screen.
.
So, I have got E I plus E R divided by E I
minus E R, just divide one by the other and
sort of work out the algebra you get a rather
clumsy set of equations, but that is what
it
is. E R by E I is given by this and E T by
E I is given by this, just nothing, no grade
information from here, but just straight forward
solutions of these two equations.
..
So, let us look at these equations and look
at some special cases. This is simply rewriting
E Rr by E I, so and I have, I am also going
to be assume that the mu’s are the same
now.
Once we take the mu’s are the same, you
get E R B by E I in terms of the refractive
index of the second medium and refractive
index of the first medium. And if you
remember my Snell’s law, that is sine theta
i by sine theta R is n T by n I. So, use all
that
you can prove that E R by E I, can be written
as very straight forward algebra. I am not
going to do the trigonometry, it happens to
be tan theta T minus theta i by tan theta
plus
theta i.
And that is the amplitude reflection coefficient,
I I sort of alert to you, the old reflection
coefficient is sometimes used to indicate
the square of this quantity, but at this moment
I
am talking about the amplitude ratio E R by
E I. And similarly, E T by E I is given by
this expression and that is denoted by t p.
So, these are a pair of equations which are
going to be useful to us. So, let us just
look at what it tells me. First thing you
notice is
this, that incidentally this is very special
to the p polarization, that is parallel thing.
So,
you notice that E R by E I is given by tan
theta T minus tan of theta T minus theta i
by
tan of theta T plus theta i, so if it happens
that theta T plus theta i becomes equal to
pi by
2.
..
Now, you can look at the picture if this angle
plus this angle happens to be pi by 2,
elementary geometry will tell you because
the angle of reflection is the same as the
angle
of incidence. The angle between the reflected
ray and the transmitted ray is 90 degrees.
Now, if that happens, going back to this tan
of theta T plus theta i is infinite, so as
a
result the amplitude of the reflected ray
is 0. Now, what I have done in this picture
is the
following that I told you that I am talking
only about p polarization, but this is the
picture
which is of mix polarization.
That is supposing you take un-polarized light,
I can write down un-polarized light has an
equal mixture of light, which is polarized
in the parallel direction that is parallel
to the
incident plane and light which is polarized
perpendicular to the incident plane. And the
way we have shown it is these dots indicate
light polarized perpendicular to the incident
plane, and these lines arrows in the incident
plane. So, if you start with an un-polarized
light you have a mixture of both these black
dots and the orange arrows, but however at
this particular angle, we have seen that there
is no component of p polarized wave in the
reflected ray, the transmitted ray still as
both.
Now, this then tells you that if you take
un-polarized light, which is incident at this
angle, the angle for which theta 1 that is
theta i plus theta transmitted is 90 degrees,
this
is known as the Brewster’s angle. And if
you know the refractive index of the medium
then of course, you can easily calculate what
the Brewster’s angle. Therefore, that angle
.the reflected light does not have any p polarized
component. So, this I started with the
un-polarized light, the reflected light is
plane polarized. Now, this picture here gives
you
a the reflection coefficient the way the I
have defined it namely the ratio of the
amplitudes and this is the p polarization.
This is the picture of r s polarization i
have not yet taken up, but this will turn
a and you
can see that this red line crosses the x axis
which is the angle at an angle which is the
Brewster’s angle. So, at that angle the
reflected ray is plane polarized, this is
of course,
something which you have known from our earlier
days.
.
Now, the situation with respect to s polarization
is almost identical. So, what I have, I
have here is this, that in this case since,
the electric field is perpendicular to the
plane of
incidence, I have taken the electric field
to be coming out of the plane of the paper
and I
have using the fact that E, Hh and the direction
of propagation, for my triode I have
given the corresponding direction of the H.
So, I have got E I plus E R is equal to E
T
and the same way as I did before H I minus
H R cos theta i this is given by this and
if
you do that you get an expression for r s
you get an expression for t s.
..
And the, once again use, if you use mu I equal
to mu T, you find E R by E I is given by
minus sine theta i minus theta t by sine theta
i plus theta t, which is my r s and this type
of an expression is called my t s.
.
Now, if you take normal incidence, you can
take normal incidence means, the angle of
incidence is 0, so that if you can work it
out that r p is given by n minus 1 by n plus
1 and
r s is given by 1 minus n by n plus 1. Now,
incidentally you notice that there seems to
be
a a contradiction here because I have taken
normal incidence, both these results must
be
.the same, but here I got n minus 1, here
I have got 1 minus n. The reason is connected
with the fact that we had adopted two different
conventions of of taking the direction of
electric field and the magnetic field to be
in the same direction, and so this sort of
is not
very important, but one can sort of take different
convention for p polarization and s
polarization work it out, but otherwise they
they are the same.
So, let us return back to the p polarization
again.
.
Now, I had shown that the p polarization is
written by such an expression. Let me let
me
write it down.
..
r p is equal to n. n is the refractive index
of medium two with respect to medium 1. n
cos
theta I minus n cos theta T divided by n cos
theta I plus n cos theta T. Now, I am going
to 
look for a case for which, the second medium
has a lower refractive index then the
first medium. So, n T by n I is less than
1. Now, when this happens, I can rewrite this
expression, so this is, I will write this
as n cos theta I minus I use Snell’s law
because
Snell’s law gives me sine I by sine theta
T is equal to n. So, that sine theta T is
sine theta
I divided by n, so I can rewrite this as square
root of 1 minus sine square theta T, the
denominator is identical expression with a
plus n cos theta I plus root of 1 minus sine
square theta T, now what I am doing is this.
That I am, so this is this is nothing just
I have rewritten the cos theta T as root of
1 minus
sine square theta T, there should have been
an n there, let us put them back. Now, if
I use
this expression sine theta I by sine theta
T equal to n, then I will be able to write
this
expression in this fashion here. So, this
is if you plug it the same and we are going
to talk
about situation where remember n is less than
1, so I can think of an angle theta T for
which sine theta T becomes greater than n,
now when that happens this type of an
expression gives me an imaginary number.
So, I rewrite this as equal to, if you do
a little bit of an algebra it becomes n square
cos
theta I minus I times root of sine square
theta I sine square theta minus n square and
divided by the same thing with the plus. And
similarly, for r s now what is that if sine
.theta I exceeds n? If sine theta I exceeds
n then notice that, this is a quantity of
magnitude 1 because this is a minus I B by
A plus I B, then you can show that r s and
r p
is magnitude will be equal to 1. So, this
is true of course, if sine square theta is
greater
than n square, so that this is a real quantity
and so that the numerator and the
denominator here are complex conjugate of
each other.
.
So, r p is magnitude and r s is magnitude
works out to be 1, but let us look at what
is
happening to that wave. So, let us look at
what is happening to the transmitted wave.
Transmitted wave is E T 0 exponential i K
T dot r minus i omega t. Now, as I have taken
the incident plane in the x z plane, as I
have said earlier my K T dot r is K T times
x
times sine theta T plus K T times z times
cos theta T. Now, I can rewrite that, this
is K T
now, I have got a x, this sine theta T i simply
write as sine theta i by n Snell’s law,
and
exactly the same way cos theta T which will
be written this way.
So, this is my K T is equal to x sine theta
i by n and I have written this i times z because
i
am assuming that sine square theta i is greater
than n square. So, you notice that K T dot
r has two components, one is a real part which
is beta times x and what is beta? A beta is
simply K T sine theta i by n. And i times
z times alpha and this is quantity is my alpha,
alpha is k t times root of sine square theta
i by n square minus 1 and I write K T as
omega n T by C. Remember that omega by the
velocity in that medium times the same
.quantity and I have simply taken this n out
so that I write this as omega by C n I square
minus this.
So, I have got a part of this, beta is real.
So, when I plug this in, you find that I get
a
propagation vector which is beta which is
given by K T sine I by n which you can work
out is nothing but K T sine T, which is equal
to K I sine i.
.
And there is a term which when plugged into
the exponential will give you E to the
power minus alpha z. So, this is alpha and
the transmitted wave becomes E T 0 E to the
power minus alpha z times a propagating, let
me let me write it clearly, so that one can
sort of see what is happening there.
..
So, we have a propagation vector beta, which
is given by K T sine theta T and we have a
attenuation factor if you like or an absorption
factor alpha and which is given by omega
over C root of n I square sine square theta
I minus n T square. So, that the E T vector
becomes equal to E T 0 times E to the power
minus alpha Z times exponential i beta x
minus i omega t. Notice what is the picture?
The Z direction, Z positive direction is in
the second medium, remember that the, my picture
now is my second medium has a
refractive index, lower than the first medium.
So, Z greater than 0 is the transmitted medium,
so as you go into the transmitted medium
the amplitude of the wave decreases exponentially
with distance Z. There is a
propagating term, but if you notice this is
propagating along the plane, this is
propagating along the plane and we have seen
that the amplitudes of the reflection
coefficient and the transmission coefficient,
they have, they they r p and r s, the
reflection coefficient they have become one.
So, this is what is known as the total
internal reflection, the wave is not 0 in
the second medium, but it is a exponentially
decaying wave, this is known as an evanescent
wave. The propagation takes place on the
surface on the Interface.
..
We can do a little better, notice this propagation
vector has a very interesting
interpretation. The propagation vector we
have said is K T sine I by n which is K T
sine
theta T. Now, here I am showing a for example,
this is an incident direction, this is the
way which is incident at an interface. I remember
that we have talking about a
propagation vector, which is 2 pi by the propagation
vector is the wavelength. So,
suppose I draw perpendicular to the propagation
vector direction and I find out the
distance between two crests or two troughs
that would be my wavelength, and so as a
result 2 pi by K I, this is the wavelength
here.
But the this is provided the you measure the
wavelength in a direction perpendicular to
the direction of propagation, incident wave
vector, but notice that as it comes to the
interface. That is when I go to supposing
these were water waves, and this is the surface
then as it comes to the shore, the one would
be inclined to measure the distance like this.
And this is my 2 pi by beta, you can just
do an elementary there. Since, these are surface
waves propagating along this, the corresponding
wavelength is given by 2 pi by beta and
not 2 pi by this distance.
So, and and this is fast attenuating wave.
Naturally the question arises, is there a
total
internal reflection 1 for which is there a
transfer of energy to the second medium? Now,
what will prove is, on an average there is
no transfer of medium to the second medium
and let us look at it this way.
..
So, let us come back to our expression for
the electric field, we have seen that for
this
case E T is given by E T 0 the amplitude.
.
There is a amplitude d k factor E to the power
minus alpha z, then a propagation along
the surface which is exponential i beta x
minus i omega t, this is my E T. Now, what
I am
going to do is to derive the corresponding
magnetic field vectors or H factors from this.
And I do not have to, I could have actually
gone back to the solution of H and repeated
this, but I can do, use Maxwell’s equation.
So, del cross E that is is my minus d B by
d t
.which is if you recall this is minus mu 0
d H by d t. And remember that when you have
an expression like this d B by d t is nothing
but multiplying the things by minus I omega.
So, as a result come here, so let me take
what is d B x by d t. I am, I am working out
the
x component of this. So, let me write it down,
so minus d B x by d t, I will just work out
one, you can work out the remaining. This
is equal to del cross E T’s x component,
so
del cross E T x component, which is d by d
y of E T z, but I am taking the electric field.
Let me let me take this as a number and let
me take the electric field because it is p
polarized. So, let me take the electric field
along the y direction, now if I do, I am doing
del cross E T is x component.
So, this is minus d by d z of E y, this is
the only thing that survives. So, you can
take the
differentiation of this quantity with z and
from their you can find out what is B x this
is a
fairly straight forward because d by d t means
minus i omega and d by d z is simply
minus the alpha. So, as a result i get B x
and I get B z, both the components I can work
out, B x and Z z.
.
I already had E T, so E T is given by this
expression E T 0 E to the power minus alpha
z
along the y vector and B t is given by this
B t has A z component and it has an x
component. So, if I calculate now the pointing
vector which for complex E and H is half
of real part of E T cross H, you can calculate
the cross product of these 2 y cross z will
give you an x component, y cross x will give
you z component, but I need the real part
.therefore, this is not there, I am left with
only this part. But notice my interface is
perpendicular to the interface is the z direction.
And this is a vector the s average is vector
along the x direction, so as a result when
I
take the dot product of thus s vector with
my z vector, the normal direction there is
no
average energy flow into the second medium.
So, there is a wave which is the evanescent
wave, which amplitude of, which exponentially
decays the propagates along the
interface, but it does not carry any energy
to the second medium.
.
