Good morning, in the last class we had started
discussing a situation where, we have an electric
dipole which is essentially a rod or a wire.
And we have fed the situation we considered
in the last class we had a dipole like this.
And we had connected an oscillating voltage
source to this dipole. So, there were charge
particles oscillating up and down this metal
wire metal rod like this. The oscillation
of the charge we represented as yt where,
y is displacement of the charge along the
y axis is equal to y naught cos omega t.
So, the charges move up and down with an angular
frequency omega. And so, we consider this
situation and for this situation we calculated
the electric field at a large distance away
from the dipole. And we saw that at a large
distance the electric field along the x axis
at large distance the electric field also
oscillates parallel to the y axis. So, it
is in the same direction as the direction
in which the electron oscillates up and down
which is along the y axis.
So, at a large distance x the electron well
everywhere x along everywhere along the x
axis the electric field oscillates up and
down parallel to the dipole. And at a large
distance we can treat this oscillating electric
field as a sinusoidal plane wave; it behaves
like a sinusoidal plane wave. So, at a large
distance if you look at the electric field
at the fixed instant of the time it will have
this kind of a sinusoidal pattern. And with
time this whole pattern moves forward which
is precisely what we called a sinusoidal plane
wave.
Then we also have the magnetic field which
is perpendicular to the electric field and
it oscillates in exactly the same phase as
the electric field. In any arbitrary direction
over here we have exactly the same thing we
have a sinusoidal plane wave at a large distance
from the dipole. The electric field is now
along the so, the electric field over here
is to be calculated by taking the component
the projection of the dipole. Normal to the
line of sight which is what I have shown here
and the sinusoidal plane wave the wave propagates
along this direction; the radial direction.
And we also have a magnetic field which oscillates
in the same phase as the electric field, but
which is perpendicular to both direction of
the wave and the electric field vector.
Now, in the last class having discussed the
field the electric field pattern, in the last
class we calculated the energy density and
then we were calculating the energy flux.
I told you that the average energy flux is
the average energy density the time by average
we mean the time average. So, if you look
at the energy density average it over a time
period which is much larger than the time
period of the oscillations of the dipole.
The oscillation of electric field the energy
flux is the average of the energy density
into the speed of the light.
So, the whole wave propagates in a direction
at the speed of light. So, the energy density
also moves forward at that direction if I
put a surface normal to the direction in which
the wave is propagating. The energy density
the energy per unit volume into c is the amount
of the energy that will cross unit area of
this surface in a unit time. And we had calculated
this; this comes out to be half epsilon naught
c E square where, E is the electric field
at this point.
So, let me remind you from over here the energy
flux is a vector along the direction of the
wave which is in this direction. if I am over
here, the energy flux will be a vector in
the direction of the wave which is in this
direction. And if I put, a surface area surface
normal to this surface of unit area normal
to this.
Then the energy that will cross that surface
in a unit time is half c epsilon naught E
square.
So, putting in the expression for E which
we had calculated we get the flux the energy
flux vector. And this is given by this expression.
So, we have the put in the expression for
the electric field at a any point over here.
And we have also done the time average.
So, to just let me just remind you that the
electric field at any point was q into y naught
omega square where, y naught was the amplitude
of the oscillations of the charge particle.
This divided by four pi epsilon naught c square
into the distance into cos omega t minus r
by c into this whole thing into sin theta.
So, we have to square this and take the time
average and which gives us an extra factor
of half.
So, when I square this I get all these factor
over here q square y naught square omega 2
the power of 4 by 32 pi square epsilon naught
square into c square epsilon naught c square
into sin square theta by r square into the
unit vector r. So, this is the expression
for the energy flux. The point to note is
that the energy flux vector points in the
radial direction. So, if I am over here it
will be pointing radially outwards along.
So, the radial direction is defined by the
position of the dipole and my positions.
If I am looking here I have to draw the radial
line from the position of the dipole to this
point here. And the energy flux vector points
in this direction and this gives you the magnitude
the magnitude of the energy flux vector falls
as 1 by r square. So, the flux falls as 1
by r square this is the feature which we see
quite commonly.
So, if I have any source of radiation could
be a bulb or anything of that sort. The flux
or the radiation flux falls is known to fall
as 1 by r square which is what we see over
here. So, if I put a unit area at a distance
r and if I put a unit area at a distance to
r, then the flux over here is going to be
1 fourth r square. The energy crossing this
area per unit time falls as the inverse square
of the distance which is what we have just
which we can see here in the expression that
we have just derived.
And the energy flux also depends on the angle.
So, if the dipole is oscillating like this
the energy flux is proportional to sin square
theta. And it is maximum in this direction
and as I move up theta will decrease theta
is 90 here as I move up theta decreases. The
amount of the energy flux goes down and if
I am located just in the same direction along
the dipole I will get no energy coming over
here.
The other feature which you should note in
this expression is as follows: if this is
the energy flux vector. And if I put a surface
unit if I put a surface area like this, then
the amount of energy that crosses this is
given by the S over there. If I put the surface
certain angle then you have to put in an extra
factor of cos theta which is another point
which you have to keep in mind when, interpreting
when applying this expression.
Now let us move ahead. So, we have calculated
the energy flux the energy flux tells us that
if I am located at a distance r away from
the dipole. So, my dipole is over here and
I am located over here it tells with the amount
of energy that crosses across a surface of
unit area per unit time over here. Now, let
us now go on to a slightly different question
the question which we shall take up next is
I have this or same oscillating dipole.
What is the radiation pattern of the dipole?
So, when we talk about the radiation pattern
we change our point of reference, we shift
the point of reference to the origin of the
dipole. So, let us go to this picture. We
have this dipole oscillating up and down.
And we have changed our reference point to
the origin of the dipole and we ask the question.
How much energy does the dipole send out per
solid angle?
So, in a solid angle d omega what is the radiation
sent out by the dipole when the solid angle
is in a particular direction. So, in a given
direction in a fixed direction I put a solid
angle d omega and ask the question how much
energy does the dipole send out in this direction;
in this solid angle. So, we can calculate
this as follows corresponding to this solid
angle over here there will be an area.
So, the area corresponding to this solid angle
the solid angle is d omega the area corresponding
to this solid angle the area subtended by
this solid angle with reference to the dipole
is r square d omega where, the r square is
the distance to this point. So, this is the
area corresponding to this solid angle. And
if I write a vector normal to; so, this area
I can represent as a vector where, the direction
of the vector will be along the normal to
this surface.
So, the normal to this surface of the radial
vector r cap. So, the area corresponding to
this solid angle is the normal is the vector
r cap into r square d omega. And the power
that goes into this solid angle is the energy
flux into the area corresponding to this solid
angle which is what we are over here.
Now, putting in the expression for the energy
flux vector that we just calculated; putting
in this.
Putting in the expression for the area which,
we have here the area corresponding to the
solid angle
We can calculate the radiation pattern. So,
the radiation pattern is the amount of the
power that is that is emitted in the solid
angle d omega where, the solid angle is located
at an angle theta this.
So, notice that the solid angle the area corresponding
to the solid angle is proportional to r square.
The flux the flux vector S is also proportional
to 1 by r square. So, these factors of r square
is cancel out.
And we get the power per solid angle the power
per unit solid angle to be an expression which
is independent of r and this is what is called
the radiation pattern of a dipole. So, the
radiation the point the important point over
here is the radiation pattern of the dipole
the amount of energy the amount of power the
dipole sends into a solid angle d omega per
solid angle d omega. Depends only of the direction
at which we place the solid angle it does
not depend on how the far away the solid angle
is located. This is an important feature of
the radiation pattern.
So, the radiated power per solid angle depends
only on the direction and it with respect
to the dipole and it goes as sin square theta.
So, this picture over here shows us the radiation
pattern the length of the distance from the
origin. So, the dipole is located here and
the length of this line from the dipole to
this point on the curve. So, for a particular
theta I will have a different length for theta
equal to 90 degrees the length is going to
be maximum for theta equal to 0 or theta equal
to pi the length is going to be minimum.
This length tells me the magnitude of the
power that is emitted in this direction power
per solid angle that is emitted as a function
of theta. So, the maximum power per solid
angle is emitted towards theta is pi by 2
perpendicular to the direction in which the
dipole oscillates. And as you move in this
direction or in this direction the power per
solid angle that is emitted false as sin square
theta.
So, there is no power emitted in the direction
in which the dipole actually oscillates. The
maximum power is emitted in the direction
perpendicular to the direction in which the
dipole oscillates.
You should also remember that, the whole pattern
is symmetric around the dipole. So, if the
dipole oscillates up and down like this, the
whole pattern is symmetric around the dipole.
So, it is if this is the plane perpendicular
to the dipole the dipole is over here this
is the plane. So, the whole radiation pattern
is symmetric around this. So, it is maximum
over here in all directions in this plane
it is maximum and then if I consider a circle
over here, which makes an angle. The amount
of energy that is sent out per solid angle
in this direction will fall by a factor sin
square theta. We could now, move a little
further and calculate the total power which
is radiated by the dipole.
So, to calculate the total power we have to
integrate over all solid angles this expression
tells as the power that is emitted per unit
solid angle per solid angle d omega. So, to
calculate the total power that is radiated
you have to integrate this expression over
d omega. Now, if you look at this expression
for the power emitted per unit solid angle
it has sin square theta. This is the only
term that comes in when you integrate this
over solid angle.
In spherical polar coordinates this solid
angle d omega is sin theta d theta d phi.
So, let me explain this a little bit here.
So, we have we have the dipole oscillating
like this along the y axis and you can choose
a coordinate system which has got an angle
theta. And an angle phi, phi is over here
in this case and this is the z axis this is
the y axis this is the x axis. So, when you
integrate over solid angle you have to do
an integral over. And the solid angle can
be expressed in terms of these coordinates
theta and phi which is the expression that
I have over here.
So, the d omega is sin theta d theta d phi.
And I have to integrate the power per solid
angle.
Over all solid angles so, I have the integral
sin cube theta d theta d phi the range of
theta is zero to pi the range of d phi is
zero to 2 pi.
If I do this integral I will get a factor
of 8 pi by 3.
So, if I take the expression for this power
emitted per solid angle. And add up the contribution
overall solid angles.
Then we get the total power that total power
is q square y naught square omega to the power
4 divided by 12 pi c cube into epsilon naught.
So, this gives us the total power that is
emitted by the dipole. Now, the point so this
is this gives us a total power. An interesting
point which you should note is than the total
power does not depend on the total power emitted
does not change with the distance from the
dipole.
Let us just go back and ask the question why
is why do we have this kind of a behaviour?
How come the total power radiating by the
dipole does not depend or how far away we
are from the dipole?
So, this you can trace back the origin this
feature to the behaviour that the electric
field E. The radiation part of the electric
field is proportional to 1 by r. Remember,
when we had the full expression for the electric
field which is emitted by a charge. There
were terms which had 1 by r dependence there
were also terms which had 1 by r square dependence.
And you so, you could have combination of
such terms. And there are situations where
you have terms which are 1 by r cube which
have also have 1 by r cube dependence.
For example, if you have a static dipole remember
you can recollect that the electric field
falls as 1 by r cube. We have not discuss
it here, but I am sure all of you would have
learnt this in earlier courses that, if I
have a static dipole the electric field fall
as 1 by r cube. So, you could also have a
situation where you have electric field a
part of electric field going as 1 by r. And
you could also have components in the electric
field which fall as 1 by r square 1 by r cube
etcetera.
Now, we had all of these terms all of these
possibilities. But I focused only on the part
which falls as 1 by r and I told you that
this is the only thing which corresponds to
radiation, the rest of them do not correspond
to radiation. Let me elaborate a little on
this point again. So, when you calculate the
power the total power that is emitted you
have to essentially look at E square and then
multiply it. So, E square gives us the energy
flux the energy flux is proportional to E
square and you have to multiply it with r
square when you do the integral over all solid
angle.
You have to multiplied by r square and then
do integrate over all solid integrate over
d theta and d phi. So, the crucial point is
that you have to look at that r dependence
of this combination E square into r square.
Let me just remind you again what we are talking
about when I calculated it that the total
radiation pattern that is when I ask the question.
How much energy goes out per solid angle?
We looked at the energy flux S the energy
flux is proportional to E square right.
So, the energy flux as you can see here is
proportional to E square.
So, you have to look at E square.
And then when you ask the question how much
energy is emitted per solid angle You have
to multiply it by the area corresponding to
the solid angle which has a factor of r square.
Now, you have when you do the solid angle
integral that r does not come into the picture
anymore. So, all the r factors are here you
have to and the r factors have a dependence
which is E square into r square.
So, if you ask the question what is the power
that comes out you have to look at the r dependence
of E square into r square. Now, if you have
a 1 by r electric field then for the 1 by
r electric field this E square into r square
is a constant. So, it is for this particular
electric field it is a constant it does not
depend on r. whereas, for a 1 by square electric
field if I look at E square into r square
the whole combination.
So, when I square 1 by r square I will get
1 by r to the power four multiplied by r square.
So, the whole combination will fall off as
1 by r square. And if I had 1 by r cube then
the whole combination would fall off as 1
by r to the power this will fall as this and
this will fall as this.
So, what we see from this is that if I had
a source some kind of a charge particle doing
some kind of a motion over here. And ask the
question what is the power which comes from
this charge, which produce by this charge
which is oscillating or doing something that
crosses this surface at a distance r. Then
we find that if the electric field is 1 by
r the power that crosses this is also the
same as the power that crosses this and it
does not depend on how the large I make the
sphere.
So, if I have a 1 by r electric field and
I take the limit of r going to infinity I
will find that, there is some power being
transmitted there also. So, if I have a 1
by r electric field there is power being radiated
sent out all the way to infinity. And it is
this power that goes out to infinity which
we refer to as radiation. So, let me repeat
this if I have a situation where, charges
combinations of charges produce a 1 by r electric
field. The power which is being sent out does
not depend on the distance from these charges
it is a constant.
It does not depend on the distance and; however,
large I make this sphere the amount of power
that crosses it is a constant. So, if even
if I make it infinitely large there will be
the same amount of power crossing it. So,
this set of charges which are which are producing
a 1 by r electric field or putting or sending
out energy all the way to infinity. And it
is this that we referred to as radiation.
So, a radiating set of charges send out power.
The power which is send out the total power
which is send out does not decay away as I
move further and further away from this set
of oscillating; the set of charges which are
producing the electric fields. And it is this
phenomenon that we referred to as radiation.
On the contrary if I have a set of charges
which produce a 1 by r square electric field
the 1 by r square component of the electric
field the power which is being sent out from
this. The power which is being sent out or
from this depends on the size of the sphere
across which, I am measuring the power.
And it falls as 1 by r square. So, if I keep
on looking at larger and larger spheres after
sometime I will find that very little power
is coming out. So, such a set of charges do
not send out power actually the power is the
energy is confined to a certain region. The
power which comes out from this is confined
to a certain region and no power is lost basically
to infinity.
So, electric fields which fall faster than
1 by r do not correspond to radiation, we
do not referred to these as radiation they
do not carry away power to infinity .It is
only the 1 by r electric field which carries
away power to infinity which we referred to
as radiation. So, this tells you why we ignored
all the terms which fell off faster than 1
by r. Because, they do not give rise to any
power being transmitted to infinity and there
is no radiation; they do not corresponds to
radiation. So, having explained to you the
reason why we took this 1 by r term only.
Now, let us go back to the issue that we were
looking we looking at the total power that
comes out from this oscillating dipole, the
total power that comes out from oscillating
dipole. If you write it in terms of the magnitude
of the oscillation of the charge which moves
up and down.
So, remember that the charge we had assumed
at the charge moves up and down the dipole
the displacement of the charge was given by
y 0 cos omega t. And the total power that
is emitted depends on this amplitude y naught
square.
So, if I double the amplitude the total power
will go up 4 times it also depends on the
angular frequency to the power 4. So, if I
make the charge oscillate twice as faster
the total power that is emitted will go up
by 2 to the power 4 which is 2 4 8 16 so it
will go up 16 times. Now, we could also express
the total power that is emitted in terms of
the current.
So, if I have this charge particle going up
and down this dipole the charge particle moves
up and down this dipole. Then there will be
a current. So, if I express the amount of
total power which comes out, this is radiated
out; in terms of the magnitude of the currents.
So, if I express the current like this 
and if I write the total power which is emitted
from the dipole in terms of the magnitude
of the current and the angular frequency omega.
Then the expression for the power assumes
this form shown over here.
It depends on the magnitude of the current
square and it depends on the square of the
angular frequency. So, if I maintain the magnitude
of the current fixed and increase the angular
frequency twice. Then the power will go up
4 times whereas, if I maintain the displacement
of the charge fixed and increase the angular
frequency 2 times the power will go up 16
times.
So, this is the point to bear in mind that
the angular frequency or the frequency dependence
depends on the variable which in terms of
which I have expressed the power. There is
also another convenient expression for the
power. So, note that the power that is emitted
by this oscillating dipole is proportional
to the current squared 
is proportional to the square of the current.
So, if I have a current flowing in a resistance
R the power emitted the power lost in the
resistance is half I square R. In the resistance
the power is dissipated as heat and the amount
of power dissipated at as heat is half into
the current magnitude of the current square
into the resistance. Whereas, here we see
that if I have a dipole and I feed a current
into this.
Then again there is some power emitted and
this power is proportional to the currents
squared. So, I could define an equivalent
effective resistance corresponding to the
dipole or an impedance of the dipole equivalent
resistance of the dipole in this case. So,
corresponding to this dipole the dipole converts
some of the energy. So, dipole actually also
dissipate some power, but here the power is
dissipated as radiation not as heat. So, the
power that is lost as radiation I could represent
using an effective resistance.
So, that the power is half I square R and
if you put in the numerical values of all
of these constants. And if you express the
angular frequency omega using the wavelength
you get an a very convenient expression for
the resistance.
The resistance comes out to the l by lambda
square 790 ohms. So, let me repeat again what
we mean by this if I have a dipole of length
l and I send a current through this. Then
the dipole will dissipate away energy will
dissipate power in the form of radiation.
The power which comes out in the form of radiation
we have seen is proportional to the magnitude
of the current squared. So, I could write
it as an effective resistance into the current
magnitude of the current squared this effective
resistance can be calculated from the expression
for the power.
The expression for the power notice depends
on omega square.
The dispersion relation for radiation where
omega is the angular frequency of the radiation;
the dispersion relation for the radiation
is omega by k is equal to c. So, it basically
tells us that omega by 2 pi lambda is equal
to c. So, I can replace omega and write it
in terms of lambda. So, omega is inversely
proportional to lambda. Which is why you see
that, the resistance the value of the resistance
is comes out to be inversely proportional
to lambda squared.
Because the expression of the power has a
omega square over here. So, if you express
this omega in terms of the wavelength of the
radiation that comes out. You get you can
write this expression in this form.
The equivalent resistance turns out to be
l the length of the dipole divided by lambda
square into 790 ohms. So, we shall come back
to applications of this formula when we discuss
problems on the radiation that comes out from
a dipole. Today, let us go on to discussing
another aspect of the radiation that comes
out from the dipole. And the aspect that we
are going to discuss let me let a before going
on to this, let me again just recollect the
quantity that we are the situation that we
are discussing.
So, the situation we have been discussing
until now is that we have a dipole aligned
along the y axis and we are feeding in a sinusoidal
voltage. So that, the currents so that, the
charges move up and down the y axis as yt
is equal to y naught cos omega t. And this
produces a sinusoidal plane wave far away.
And along the x axis if you look at this sinusoidal
plane wave along the x axis you will see the
electric field the oscillating up and down
along the y axis.
So, the electric field here oscillates in
the same direction as the dipole. And the
magnetic field also oscillates perpendicular
in the same in the same phase as the electric
field. But, it is perpendicular to the direction
of propagation of the wave and the direction
of the electric field.
So, here I have the expression for the equivalent
resistance.
Now, let us come to the situation that we
now on to discuss the situation that we would
now like to discuss we have 2 crossed dipoles.
So, instead of having 1 dipole along the y
axis we now have 2 dipoles 1 along the y axis
and another along the z axis. So, we have
these 2 dipoles and we feed exactly the same
current to both of these dipoles. So, let
us first consider a situation a simple situation
where, we have 2 dipoles. They are perpendicular
to each other that is why we referred to them
as crossed dipoles.
So, we have 2 dipoles 1 along the y axis another
along the z axis. We will study the electric
field pattern at a point which is a large
distance away along the x axis direction.
So, it the point we are looking at is perpendicular
to both the y the dipole along the y axis
and the dipole along the z axis. So, the question
is what do we see at this point over here
which is far away along the x axis. Now, as
we have already discussed the dipole along
the y axis will produce an electric field
along the y axis. And the dipole along the
z axis is going to produce an electric field
along the z axis.
So, the quantity which you measure over here
a large distance away is going to be a superposition
both of both of these. So, you will have the
electric field along the y axis produced by
the dipole along the y axis. And you will
have the electric filed along the z axis produced
by the dipole along the z axis.
The situation which, we are considering their
both being fed exactly the same current the
same signal is being fed into both them. So,
the magnitude of these 2 electric fields is
the same and they both oscillate; the 2 electric
fields are going to be oscillating in the
same phase. So, we have cos omega t minus
kx. So, it will only change as I move along
the x axis and with time both of these are
going to be oscillating in the same phase.
So, I can take that factor E outside Ex is
equal to Ey I can take the factor E outside
and write it as the unit vector j plus the
unit vector k into a constant E into cos omega
t minus kx. So, this the total electric filed
at the point x Now, let us fix the value of
x and let us fix it so that, kx is a multiple
of 2 pi. So, if kx is the multiply 2 pi I
can ignore it for the discussion
So, we will we will consider a fix the behaviour
of the electric field at the fixed value of
x and we will choose x. So that, k into x
is a fixed number which is a multiple of 2
pi.
Then ask the question how does the electric
field vary over there? So, let us take the
time instant t equal to 0. At the time instant
t equal to 0 Ex and Ey both of them have the
same value and they will have a value equal
to E. So, at t equal to 0 the electric field
vector is going to be magnitude E into j plus
k. So, it is going to point at in a direction
at 45 degrees to the y axis this is the y
axis, this is the z axis.
So, the electric field vector at the point
x at a fix value of x is going to point at
45 degrees to the y axis when t is 0. When
t is 0 this whole think is 1 cos omega t is
1. And this has the maximum value which it
can assume. Now, as t increases this cos omega
t is going to come down. So, both the y component
of the electric field and the z component
of the electric field is going to come down
by exactly the same amount. And then when
omega t is pi 2 it is both of them are going
to be 0 and then when it crosses pi by 2 both
of them are going to be negative.
So, the electric field is going to become
like this and its going to oscillates back
and forth like this. So, this kind of behaviour
of light is called linear polarization. So,
the light is said to be linearly polarized
in this case the light is linearly polarized
at 45 degrees to the y z to the y z y axis
and z axis. And it will the light the electric
field vector will oscillate at 45 degrees
to these axis.
So, let me again just remind to you of the
situation that we are discussing we are discussing
a situation where exactly, the same signal
is fed to these 2 dipoles which are crossed.
And in this situation the electric field over
here is going to oscillate at 45 degrees to
the y z axis.
Now, let us next consider a situation where
the amplitude of the currents being fed to
the y and z axis differ but, the phase of
the current being fed here and here or exactly
the same.
So, this is the next situation that we are
going to differ we are going to discuss. So,
in this case the electric field at a fix point
x is Ez this should be Ez into k plus Ey into
j. Let me just write it correctly over here.
So, this is actually this should be is Ez
into k and Ey into j there was a small mistake
over there. And the magnitude of Ez and Ey
are different because that the currents being
fed into 2 dipole magnitude of the current
being fed into the 2 dipoles is different.
But, the phase is the same. So, I have the
same factor cos omega t plus kx multiplying
both the y component of the electric field
and the z component of the electric field.
So, the electric field at that point can be
written as like this. Now, if I fix the value
of x again so that, kx is a multiple of 2
pi. So, I can ignore this term. So, we have
this factor Ez into k plus Ey into j multiplying
cos omega t. So, when cos omega t is 0 I have
the electric field vector its y its y component
is Ey its z component is Ez.
So, the electric field vector at this instant
of time t equal to 0 is along the direction
theta where, theta is tan inverse of Ez by
Ey.
It has a magnitude theta has a magnitude which
is Ez square plus Ey square the square of
that. Now, as time increases cos omega t is
going to go down and both these vectors are
going to go up by this go down by the same
amount. When omega t is pi by 2 it is going
to be at 0 and then again it is going to increase.
So, you can easily see that the electric field
vector is going to oscillate back and forth.
Then it is going to oscillate along this line
and the oscillation is going to be at an angle
theta to the y axis where theta is tan inverse
of Ez by Ey. So, such a situation the situation
that we have just discussed when you feed
current of the same phase, but possibly different
amplitudes to the 2 crossed dipoles. The electric
field at that point x which is a distance
away oscillates along the line; at a fix position
the electric field oscillates along the line.
Such a situation is referred to as linearly
polarized electromagnetic wave the electromagnetic
wave is said to be linearly polarized because;
the electric field oscillates along the line.
And you have this situation when both the
dipoles are fed with the same current with
possibility different amplitudes, but the
same phase. Now, this is also referred to
as plane polarized light because if you look
at the electric field along the entire x axis.
Then the electric field at different x points
will appear to oscillate along the plane the
plane may be at an angle to the y z y z directions.
So, this is called plane polarized electromagnetic
radiation or plane polarized light.
Now, let us now move on to considering a slightly
a different situation. The situation which
we are going to consider next is as follows
we have now changed the phase of the current
that is being fed in. So, the current that
is being fed in to the z dipoles is given
an extra phase of pi by 2 the same current
the magnitude of the current being fed in
to that 2 dipoles. The 1 along the y axis
and the 1 along the z axis are exactly the
same. So, currents of the same magnitude are
fed to both the dipoles, but the dipole along
the z axis has an extra phase of pi by 2.
So, now let us again study the behaviour of
the electric field at this point x over here.
So, the electric field at this point x is
a superposition of the electric fields produced
by these 2 dipoles. The dipole along the y
axis produces an electric field E cos omega
t minus kx into j. The dipole along the z
axis produces the same electric field along
the z direction, but with an extra phase of
pi by 2. So, the electric field along the
z direction has an extra phase of pi by 2.
Now, this extra phase of pi by 2 in this cosine
term over here can be a used to write this
term in terms of sin. So, the same expression
is written over here and we have just replaced
cos omega t minus kx plus pi by 2 with minus
sin omega t minus kx. So, what we see is that,
if I put an extra phase of pi by 2 to the
current being fed into this dipole which is
along the z direction. Then, the electric
field along the y axis oscillates as cos omega
t minus kx whereas, the electric field along
the z axis oscillates as sin omega t minus
kx with the minus sign over here.
So, now let us look at the behaviour of the
electric field with time at a fix point x
and again we will choose x so that, kx is
a multiple of 2 pi. So, what you have is the
electric field is E cos omega t into j minus
sin omega t into k.
So, let us plot the behaviour of this. So,
we have written here again the expression
for the electric field at a fix point x. And
we have chosen kx to be a multiple of 2 pi
So, at t equal to 0 the electric field is
along the is has a is this cos omega t minus
kx has value 1. So, we have j here sin omega
t minus kx has a value 0 at t equal to 0.
So, the electric field is aligned totally
along the y direction. So, the electric field
at t equal to 0 is aligned over here.
So, let me draw that so at t equal to 0 the
electric field is aligned like this. Now,
what happens as time increases. So, if you
increase omega t cos omega t the values going
to go down is going fall from 1. If you if
you increase omega t sin omega t is going
increase from 0. But, you have a minus sign
here, so, you are going to have a negative
z component and the y component is going to
come down. So, the y component is going to
go down and going to get a negative z component.
So, the electric vector is going looks something
like this. So, this is t greater than 0 and
as time evolves the electric field is going
to the goes is going to go around like this.
And we have the electric field going around
in a circle the magnitude of the electric
field is fixed at a value E. And let us now
discuss briefly the direction of the circle.
So, in this situation the electric field goes
around like this. The arrow over here shows
you the direction in which the electric field
rotates.
Now, let us ask the question in which direction
is the wave propagating? Let us go back to
the picture of the wave, so the wave, just
remember that the dipoles are here and we
are looking at the electromagnetic field along
the x axis. The wave is also propagating along
the x axis.
So, in this picture this is y and this is
z. So, if you do y cross z you will get a
vector pointing out of the out pointing write
out towards you. Now, so the wave is actually
propagating outwards I could draw it over
here. So, this is propagating out of the screen.
So, the wave is propagating forward out of
the screen. And the electric field is going
around in the direction shown by this arrow.
Now, we will adapt a convention for the circularly
polarized light we will adapt a convention
where, if the wave propagates in the direction
of the thumb. And if the electric field goes
around like this. So, if the wave propagates
like this and if the electric field goes around
like this, so the electric field goes around
like this. We will call this right circularly
polarized light whereas, if the wave is going
like this.
If the electric field rotates in this direction
we will call this left circularly polarized
light. Now let us just go back to the situation
which we have over here.
Let me draw it for you over here it will help
us understand it. This is the y axis this
is the z axis and the electric field is going
around in a circle like this and the waves
is coming out of the screen. So, it is coming
out this way which is the x direction. So,
the question is, is this light is this radiation
left or right circularly polarized.
Now, if I put my right hand with the thumb
pointing out towards the direction in which
the wave is going you see that, this does
not correspond to right circularly polarized
light. Whereas, if I put my left hand here
with my thumb pointing outwards towards the
direction in which the wave is propagating
it matches with the direction in which the
electric field is rotating.
So, the situation that we have so the situation
over here situation which we have studied
corresponds to left circularly polarized light.
So, the situation where we have put an extra
phase of pi by 2 to the current to the oscillations
along the z direction gives us left circularly
polarized light or left polarized light electromagnetic
radiation.
You could also have right circularly polarized
electromagnetic radiation. For in right circularly
radiation electromagnetic waves the electric
magnetic field will go around in exactly the
opposite direction. Question is how will you
produce right circularly polarized radiation?
How will you produce right a circularly polarized
electromagnetic wave in the situation which
we have been discussing namely, 2 crossed
dipoles. I am sure you can work out the answer
to this question. So, let me stop here for
today and resume the discussion in the next
lecture.
