In the preceding discussions we have ah extensively
used this plane waves particularly ah in the
discussion of electromagnetic waves, in isotropic
medium, an isotropic medium and also ah mediums
like absorbing medium, conducting mediums.
And in almost every part of optics and photonics,
it is ah the plane waves which is mostly used
and the formaly formulation is also ah very
ah useful in making analytical ah formulation
for the various propagation characteristics
of electromagnetic waves.
So, the content of these discussion is the
electric field of plane waves. We will try
to learned ah how simply and lucidly we can
express electromagnetic plane electromagnetic
waves. ah And then, we will take the superposition
of waves which will give the resultant ah
field in terms of the electric field, ah we
will take a concept of this how the superposition
gives rise to interference. Then making use
of the plane wave formulation of ah using
this electric field ah we will ah bring out
the concept of polarization. Then we will
talk about the linear circular and elliptical
polarization.
So, let us first consider a plane wave which
is propagating along positive x direction.
ah Then the wave vector is k equal to i k.
The phase fronts are parallel to the yz plane.
The electric field is polarized along y direction.
Let us suppose that the electric field is
polarized along y direction in that case,
then we can write the electric field as E
equal to unit vector j then E naught, which
is the amplitude of the electric field and
e to the power of i omega t minus kx.
Because, the propagation direction is along
the x positive x, you can write this expression
E as j E naught e to the power of i kx minus
omega t as well. You can see that kx minus
omega t or omega t minus kx both represent
electromagnetic waves propagating in the positive
x direction.
Let us see that ah let us assume that a plane
ah phase of the ah wave plane wave which is
propagating along positive x direction, ah
that can be represented by omega t minus kx
.We can also represent this as kx minus omega
t, as I said before and we will see that they
will actually a point to the same electromagnetic
wave with an initial phase; ah which is ah
which is not very important and relevant.
Take a wave function E of x t E 0 ah sin kx
minus omega t. Then add a phase pi, then we
will get the phase factor kx minus omega t
plus pi. You can write this respected as E
0 sin pi minus omega t minus kx. Therefore,
this will give you E 0 sin omega t minus kx.
You can see that omega t minus kx which started
with kx minus kx minus omega t, but you get
omega t minus kx. So that means, they at the
same, but there is an initial phase which
we have added constant phase pi; expect the
initial phase the two waves are same and propagating
in the positive x direction.
So, this is very useful to understand that
whether we write kx minus omega t or omega
t minus kx, in general when it is a three-dimensional
ah wave three-dimensionally, three-dimensional
propagation then we write k dot r minus omega
t or omega t minus k dot r as well. So, this
is one very useful aspect whenever we rewrite
the plane waves. So, the plane wave E of x
t E 0 sin kx minus omega t, we are started
from here at x equal to 0. The amplitude is
the the the wave is 0, field is 0 and then
it is increasing at x equal to at t equal
to 0, but we have plotted at against x.
So, this is this is the one which represent
that there is a phase of pi initial phase,
but both of them are representing the same
wave. But, you can write this wave in terms
of cosine because, you can translate sin omega
t minus kx as E 0 cosine omega t minus kx
plus pi by 2; in that case it will also represent
the same wave. So, these are the various ways
of writing a plane wave and this plane wave
formulation is very useful almost everywhere
in the in the branch of optics and photonics.
The conclusions are the initial phase is just
a constant contribution. Maybe this initial
phase may be attached to the phase at the
source or the generator where ah there is
already because, you start counting the phase
when it is already pi. And this initial phase
is independent of the propagation in space
and time.
Now, electric field of a plane electromagnetic
waves ah electromagnetic wave which is propagating
along positive x direction, we can write that
is equal to E 0 sin kx minus omega t or you
can write E of x t equal to E 0 sin omega
t minus kx. These are the various ways of
writing the electric field of a plane wave.
You can write as E 0 cosine kx minus omega
t or you can write E 0 cosine omega t minus
kx. These are the various ways of writing
this and E 0 E to the power of this if you
write as the complex ways ah the plane wave
along negative x direction, which is propagating
along the negative x direction you can write
that.
In place of minus it will be plus kx plus
omega t and all other things will remain the
same. So, this is very useful and very interesting
to know if you have kx plus omega t or omega
t plus kx, this is a positive negative x propagating
wave. Whereas, if it is kx minus omega t or
omega t minus kx then it is a positive x propagating
electromagnetic wave.
Now, let us consider a plane wave which is
propagating in the xy plane and the direction
of propagation makes 30 degree, 60 degree
with x and y axis. Then we can represent the
wave vector k equal to i k cosine 30 degree,
j k cosine 60 degree because, these are the
angles which are made with the x and y axis.
So, you can write this k vector in this form.
The electric field ah is let us suppose that
it is polarized along the y direction x direction,
then you have to use this i cap E 0 e to the
power of i k and this phase factor.
For the above wave that is for this wave,
if the electric field vector lies in the x
y plane then the field expressions because,
it can have two polarizations E x and E y.
So, we will have to modify this equation in
this from. So, we are trying to write the
electric field in different configurations
different situations .
Now, will take up one example that, let us
consider two coherent plane waves with the
wave vectors k 1 equal to this i cos 30 degree
and j sin 30 degree whereas, this k 2 is k
i sin 30 degree j cos 30 degree. So, these
are the the waves then given that k equal
to 1.2 10 power minus 6, this is the propagation
vector. The waves superpose on a screen which
is perpendicular to the axis. So, under this
condition under this condition we see that
this is i cos 30 degree, this is i sin 30
degree, j sin 30 degree, j cos 30 degree.
So, they are just opposite. So, we will have
to look at the configuration how the electric
fields, the waves are incident on the screen.
Write down the electric field of the plane
electromagnetic waves. Hence, take the superposition
of the electric fields at the plane of the
screen. Then let us, calculate the width of
the straight line fringes which will be observed
at the screen.
So, to do this let us look at the configuration
of the electromagnetic waves. So, k 1 will
constitute this because, this is i cosine
30 degree and j sin 30 degree and k 2 so,
they will actually look like this. These are
the two waves which are meeting on the on
the screen at this point and let us consider
this is the origin of the coordinate system.
Then the electric fields are E 1 will be E
0 e to the power of i, this phase factor and
for E 2 this will be the phase factor. We
have just replaced this i cosine 30 degree
and j sin 30 degree in terms of the numerical
values. Here E 2 also we have replaced in
terms of the numerical values.
Therefore, electric field on superposition
we just have to add this because, these are
the figure addition, the vector addition.
So, E 0 e to the power of i k this phase and
so, this is E 1 and E 2. If we add this because,
e to the power of i k under root 3 by 3 x
by 2 this quantity, we just break them into
their individual index components. Then considered
that the screen is perpendicular to the x-axis.
As I have mentioned this this the screen is
perpendicular to the x-axis and therefore,
at this point because this is the x y plane
so, the value of x at this point the value
of x on the screen is constant. If it is if
it is the origin then it is 0, but otherwise
if it is x equal to a x equal to b some constant,
with this constant will be throughout the
screen. So, we take the value of x as constant
and which will add to a phase in the case
if it is the origin x equal to 0, this delta
1 and delta 2 will also become 0. These are
actually come from this part of this, from
this part of this.
So, the I take out this constant part and
then add them together which will which will
give you that the intensity will be E and
mode of E square which is E star E. So, I
get E 0 square this quantity into this quantity.
So, on multiplication I get 1 plus 1 into
this quantity and this quantity. Therefore,
if I therefore, if I ah summarize this E 0
square 2 plus 2 cosine delta plus k by 2 y
1 minus under root 3, which will give you
that ah which will give you if you take 2
out outside because, this will be 1 plus cosine
2 theta .
So, that gives you 4 E 0 square cosine square
phi; where phi I have used to this value of
phi equal to half delta plus k by 2 this,
which is actually the consequence of this,
this quantity divide by 2 will be called phi.
So, that is what half of this of this quantity
is equal to phi. Now, for the position y m
of the mth fringe, let us go back to this
screen. So, this is the y direction any position
any position, let us call this position is
y mth position and for y mth position the
the free the for the y mth position of the
fringe you have this phase will be equal to
delta by 2 plus k by 4 y m 1 minus under root
3 equal to m into pi.
This phase will be equal to m pi because,
this will be the integral multiples of pi
and for the position y m plus 1 of the m plus
1 th fringe, you will have this phase is equal
to m plus 1 pi. Now, if I subtract these 2
quantities then I should get the width of
each of the fringes because, this is for mth
fringe and this is m plus 1 th fringe which
is very known and well understood.
So, subtracting 1 from 2 1 from 2 so, we will
end up with only 1 pi on the right hand side.
So, this y m plus 1 minus y m which will be
equal to the fringe with delta y and that
is equal to 4 pi by this. So, if I substitute
the value of k into this equation then I will
get delta y equal to 14.3 micrometer. So,
and the fringes will look like this. So, this
is the very beautiful example that we started
with a plane wave which is incident here.
We have another plane wave which is incident
here and these plane waves are meeting on
the screen, they are overlapping on the screen
and therefore, at different point the phases
will be different. And therefore, there will
be interference fringes on the screen the
fringes will be along the y axis.
So, let us try to do this with some alternative
way of writing the electric field; let us
write the complex electric field as E 1 complex
amplitude A 1 e to the power of i omega t
minus k 1 dot r. This is for the wave k 1,
with propagation vector k 1 and this is for
the wave which is ah having a propagation
ah vector k 2. And then, the real part of
each of the complex phase complex a electric
field can be written as A cosine omega t with
a with an initial phase.
And here also, the real part of the wave can
be written with an initial phase phi 2. The
screen lies in the yz plane that we have seen.
So, the width of a fringes corresponds to
a phase-shift of 2 pi. If the fields E 1 and
E 2 superpose with the same phase at some
point on the screen, then at another point
on the fringe ah on the ah ah on the screen
one fringe-width away the phase difference
will be twice pi.
So, that is what we will use to see how the
fringe width comes out. So, the delta delta
delta the phase difference between the points
where the phase difference is 0, then we can
write this equation equal to this. So, it
is the same phase at some point r on the screen
and then will consider at some other point
on screen which is delta r apart, but again
it will have the same phase, but this is the
minimum distance. So, this distance will correspond
to one fringe width that is the concept.
So, when the when at point r when the phase
difference is 0 we write in this form, which
is equal to 0. So, that gives you the initial
phase difference is k 2 minus k 1 dot r and
the other point which is r plus delta r away
from this point, we will have this phase difference
which will be equal to this minus this quantity.
I just have replaced r by r plus delta r r
by r plus delta r and then we write down this
equation. But, phi 1 minus phi 2 we have already
seen that this is equal to k 2 minus k 1 dot
r.
Therefore, at this point the phase difference
is also this , but this phase difference must
correspond to twice pi. And this, gives you
the condition of the fringe width.
So, the phase difference at point r 1 will
be this, phase difference at point r 2 will
be this and the phase difference corresponding
to the points r 1 and r 2 will come out to
be the same. Because, we have done it in in
like three different ways that we considered
the phase equal to 0 at some point, at some
other point the phase will be twice pi. Then
we look at the look at the condition between
this distance and the difference of k 2 minus
difference of k 1 and k 2.
The under this consideration the phase difference
at point r 1, which could be anything and
then the phase difference at point r 2 which
could be anything. These differences if we
make the condition equal to twice pi, this
is twice n pi and this is twice n pi plus
2 pi. So, effectively the phase difference
is only 2 pi, if I plug in that condition
then again we will get k 1 minus k 2 dot delta
r is equal to twice pi.
Now, we will use this condition to to calculate
the fringe width. So, we plug in the values
of this cosine 30 degree, sin 30 degree and
so on and delta r equal to delta y j plus
delta z k because, it is in the yz plane.
So, k sin 30 degree and putting all these
things we will get and again if you substitute
the value of k equal to this will get this
value. So, effectively there are few ways
of taking the superposition of this effectively,
it gives you k 2 minus k 1 dot delta r which
will be twice pi will give you the condition
for the same phase points on the screen . The
fringes are alound ah ah are aligned along
the z direction with a width of y along the
y direction.
So, delta y equal to this and then we will
consider another superposition of two optical
disturbance. That is, given by E x equal to
this quantity we have seen whether you write
kz minus omega t or omega t minus kz; obviously,
this wave is z propagating and the ah polarization
is along x. This wave is also z propagating,
but the polarization is along y. So, we have
considered two waves which are which are ah
orthogonal to each other, one is x polarized
another is y polarized.
And we take the superposition of these, analyze
the state of the resultant wave for this waves
as it propagates along the z direction. So,
the first case is when you take delta equal
to 0 or delta equal to twice pi, what happened
and when you take delta equal to plus minus
pi then what happens .
So, if delta equal to 0 or delta equal to
twice pi the two disturbances are in phase.
The resultant disturbance is the vector sum
of these two waves. Therefore, therefore,
therefore, you have this you have this E equal
to E 0 cosine of kz minus; I just get the
superposition of these two waves. Now, if
you put delta equal to 0, then i of E 0 x
plus j of E 0 y that quantity is the common
quantity and because, delta equal to 0 so,
we can write cosine kz minus omega t. Now,
this i E 0 x and plus j E 0 y is nothing,
but r E x y.
So, because this gives your fixed amplitude
E x y and the variation is the same as the
individual waves, the phase variation the
phase is the same as the individual ah so,
waves component waves. Therefore, the resultant
wave has a fixed amplitude this and if I if
I look at the the amplitude is given by E
of x y and the direction is tan theta which
is equal to E y; you can look see this E 0
y equal to this, E 0 x is equal to this.
So, the angle the resultant wave makes an
angle ah angle with this which is given by
tan theta with the x axis with the x axis,
this angle tan theta will be equal to E 0
y by E x. So, this wave is plane polarized.
Let us take the other possibility that delta
equal to plus pi or minus pi, the two waves
are out of phase by pi. In that case what
happens, we will write this E y E y becomes
because you have taken the phase you have
taken the the phase delta equal to now plus
pi or minus pi. In the in this case you have
taken delta equal to plus pi so, you can write
that sin will come out to be this, ah cosine
there is a minus sin and the resultant wave
will now be represented by this equation.
Now, because cosine kz minus omega t cosine
kz minus omega t this factor appearing in
in both the terms so, you take them bracketed
out and you can write the amplitude part like
this i k 0. So, again you get a fixed amplitude
which is again this r of ah r unit vector
E x y. So, the resultant wave has a fixed
amplitude r of E x y. The direction of this
wave you can see this E x y E 0 x is along
this direction, E 0 y is along this direction.
So, the resultant wave E x y is along this
direction. So, this is again a plane polarized.
Now, we have another possibility that if the
waves are out of phase by pi by 2 plus or
minus, then what happens. In that case because,
these are again the plane waves they are ah
when they are superpose ah at some point on
the space and ah both of them are travelling
along the z axis. Then, we can write we can
write the equation in this form with an initial
phase difference of delta. If delta equal
to minus pi by 2 then, what happens this putting
this delta equal to minus pi by 2 will make
this this quantity sin this plus pi by 2.
So, sin in the second quadrant will be positive.
So, I write this equation as positive j E
0 sin kz minus omega t , but the first quantity
it remains the same as it is. So, cosine kz
minus omega t and this becomes sin kz minus
omega t so, that is what I have written here.
Therefore, the resultant wave if I add these
two waves i E 0 j E 0; one is sin another
is cosine then it will have a fixed amplitude.
I just have to square the amplitudes and add
and it will be a time varying because, you
cannot directly add them you have to look
at the the phase at various time and along
various z.
So, let us considered that at the point z
equal to 0, at the point z equal to 0 what
happens to this. So, this quantity become
0, this quantity also become 0 as a result
it becomes i E 0 cosine of minus omega t which
is equal to cosine omega t and this will become
E 0 sin minus omega t. So, this will become
minus E 0 sin omega t that is what I have
written here, i equal to E 0 cosine omega
t minus E 0 sin omega t.
Now, we look at the time evolution at z equal
to 0 at z equal to 0 z equal to 0 and at all
at time t equal to 0 this equation will give
you E 0 minus j of 0 because, at time t equal
to 0 this will give you 1, this will give
you zero. So, that is what is written here.
So, it gives you the entire field is now along
the x axis along the x axis and the resultant
is also x axis. But if you if you considered
the time t equal to pi upon 4 omega, then
E equal to i E 0 by under root 2 for this
cosine omega t cosine omega t we will now
corresponding to 45 degree and j this will
also give you this 45 degree. As a result,
both the amplitudes along x and y are now
existing ah with a factor of 1 by root 2 this
and this, but this is j is in the negative
direction and i is along the positive direction.
So, the resultant vector will be along this
that is what is here. So, you can see that
initially the field was along this direction
now, the resultant field has rotated through
an angle of 45 degree and it is now along
this direction. And, in this way if you considered
twice pi by 4 omega t this time then this
quantity will become 0 and this quantity will
become equal to E 0 because sin of pi by 2
will become 1. So, the entire amplitude is
now along the minus y direction. You can see
this the component is here minus y direction
and this is the resultant. So, we have the
resultant field along this direction.
So, effectively we started at time t is equal
to 0, when the resultant field was here then
it has come down to this position and then
it is moving in this direction. So that means,
it represents in a wave whose electric field,
the tip of the electric field vector is now
rotating along this ah in the in the clockwise
direction and as it is at the wave is propagating
forward along the z axis.
So, the resultant vector rotating in the clockwise
sense with the frequency of omega which is
the frequency of the component waves individual
waves where those were under superposition;
against the and hm when viewing towards the
source this is in the clockwise sense. The
tip of the resultant vector describes a circle
of radius E 0. And this situation, we call
the right-handed circularly polarized electric
field.
Next, we consider this situation that is delta
equal to plus by 2 plus pi by 2. The resultant
disturbance is the vector sum of these two
waves and in that case we can write this equation
because, you had a you had a delta here. So,
in place of delta I write a pi by 2 and it
comes down it becomes this equation, this
equation. Therefore, the resultant wave has
a fixed amplitude again and the direction
will be time varying. So, we have to again
check the time evolution of the wave. So,
the time evolution is at time t equal to 0,
you can see cosine omega t will become 1 so,
you have entire amplitude which is along the
x direction. So, the resultant is alone ah
also along the x direction.
For time t equal to pi by 4 omega, then you
have ah this resultant components are along
this and this, but the amplitude is reduced
by ah 1 by root 2 because of the 45 degree
cosine 45 and sin 45 degree. So, the resultant
amplitude is now along this direction and
at time t equal to twice pi by 4 omega that
is the next. So, this amplitude because this
because of the pi by 2 phase factor this become
0, but this becomes equal to 1. So, the entire
you know electric field amplitude is along
the positive y direction. So, you have the
resultant field is also along the positive
y direction.
This means that we started at time t equal
to 0, when the electric field was entirely
along the x direction. Then it has rotated
through an angle of 45 degree, find at this
time it has rotated to an angle of 90 degree.
So that means, if you continue with time then
the tip of the electric field vector is rotating
anticlockwise, as it is advancing along the
z direction. So, this is how we see that the
superposition of simple plane waves can give
rise to various phenomena like interference
and the the polarization evolution along the
z direction. So, this we have already discuss
that the field is rotating in the anticlockwise
sense. The tip of the resultant vector describes
a circle of radius E 0 and this situation
is called a left-handed circularly polarized
right.
Now, we consider these optical disturbances
that delta is arbitrary then you can write
E 0 of x is not equal to so; that means, this
initial amplitudes the the component wave
amplitude are not same and delta is also arbitrary.
Then you can write E y by E 0 y and E x by
E 0 x, if you square an add you will get this
equation and now you square either sides,
then you can write down this equation like
this. So, you have the generally equation
of an ellipse in that case. So, which the
in the earlier case it was circularly polarized
light .
Now, this will correspond to elliptically
polarized light, the generally equation of
ellipse making an angle of alpha with E x
and E y. So, and the angle will be equal to
tan of 2 alpha which corresponds to this quantity;
which is again a function of delta delta can
assume plus minus pi by 2 plus minus 3 pi
by 2 which is then you get the familiar form
of the ellipse familiar form of the ellipse.
Else there could be any inclination depending
on the delta values.
So, this is the delta initial phase that is
going to decide the inclination of the orientation
of the ellipse in the in the x y plane. The
state of polarization can be left handed and
right handed and this rules are the same as
we have seen in the case of the circularly
polarized light. So, through this discussion
we try to ah understand how we can simply
write the the electric field for hm an electromagnetic
waves, the plane waves. And the plane wave
is very useful because, throughout in optics
and photonics ah we use this plane wave formulation.
Then to understand how the superposition of
this plane waves can be mathematically tackled,
we look at the resultant waves of the coherent
source will give rise to interference. So,
we try to understand that inclination factor
ah that gives you the ah fringe width ah which
is ah very interesting. Then we will look
at the superposition of two orthogonal plane
waves, the polarizations are orthogonal in
that case how you ah you encounter this linear,
circular and elliptical polarization.
Thank you .
