Problems on uniform plane wave in a meduim

# Problems on uniform plane wave in a meduim

In this session let us solve some problems
related to uniform plane waves traveling in arbitrary direction in a medium and also the
problems related to the polarization of an electromagnetic wave. We will also solve a
problem related to the dielectric boundaries and so on. So let us consider a problem here
that the complex vector electric and magnetic fields in a dielectric medium are given by
this. So e is equal to minus j root 3 unit vector
x plus 2 unit vector y plus j unit vector z with a phase function which is e to the
power minus j 0.02pi x plus root 3z and the magnetic field is given as 1 upon 30pi root
3 x plus j 2 y minus z cap with a phase function which is same as this which is e to the power
minus j 0.02pi x plus root 3z. And what is asked is show that these fields represent
fields of a uniform plane wave. Also you have to find the frequency and velocity of the
wave. Also find the phase velocities along the x and y direction and z direction. Also
find the state of polarization of this wave. Then first we have to show that these fields
represent the uniform plane wave. So we know from the property of uniform plane wave that
the direction of wave propagation the electric field and the magnetic field they are perpendicular
to each other and also at every instant of time the ratio of the magnitude of electric
field and the magnitude of the magnetic field is equal to the intrinsic impedance of the
medium. So essentially these are the properties we are going to use to establish the uniform
plane wave nature for these fields. So essentially we have to show that E is perpendicular to
H and is also perpendicular to the direction of the wave which we denote by the unit vector
n. Now we know that if we write down the fields
in the standard form that any electric field is having some amplitude E 0 with the phase
function which is e to the power minus j beta into n dot r and then we can compare this
phase function with the phase function which is given for these fields, so this phase function.
So essentially we can get from here that this quantity n dot r is n dot r is equal to as
we know is cosine of phi x into x plus cosine of phi y into y plus cosine of phi z into
z and in this case it is given this quantity here which is x plus root 3z. So first of
all we can get this quantity for this and compare with this so that essentially it gives
you this quantity as x plus root 3z and as we know this quantity should have… the n
should have the magnitude which is unity. So, if we can write down this quantity appropriately
we can take this as 2 times x 1 upon 2 into x plus root 3 upon 2 into z so that is the
quantity essentially which represents the n dot r so because this quantity here corresponds
to cos of phi x so what that means is from here if I compare these two we get phi x is
equal to 60 degrees phi x equal to 1/2 so that is equal to pi by 3 we get phi y, there
is no terms for y here that means this cos of phi y should be zero so phi y is equal
to pi by 2 and then phi z which is the angle which the wave vector makes with the z axis
that is equal to cos of phi z is root 3 by 2 so that is equal to 30 degrees which is
equal to pi by 6. So one one thing we can note from here that
since the phase function has only x and z terms in this essentially the wave is propagating
in the xz plane and that is what essentially we see here that the wave normal makes an
angle 60 degrees with the x axis, 30 degrees with the z axis and 90 degrees with the y
axis that means the vector lies in the xz plane. So essentially the wave is traveling… we
write down the coordinate axis: this is x, this is y, this is z, the wave essentially
is traveling in this xz plane so it is traveling since in that, this is your unit vector n
where this angle is 60 degrees, this angle is 30 degrees and this angle with y that is
90 degrees. So from here essentially we get the unit vector n as we got that is 1 upon
2 x cap plus root 3 upon 2 z cap. Now you have to show that the electric and
magnetic fields are perpendicular to each other and they are also perpendicular to the
direction of the wave vector which is n cap. So for this we can take the dot product of
E H and e and the unit vector so we get E dot H that is equal to… if I take from here
we will get the dot product of these so this will become minus j into 3 so that that gives
me minus j 3 plus 4 into j plus 4 into j minus j minus j so that is equal to 0. So that means
the electric field of this is perpendicular to the magnetic field. Same thing we can do for the unit vector and
the electric field. we take electric field and the dot product of that to the unit vector
that is equal to minus j root 3 x plus 2 y plus j z dot unit vector n which is 1 upon
2 x cap plus root 3 upon 2 z cap so that is equal to minus j root 3 upon 2 minus j root
3 upon 2 plus j root 3 upon 2 that is equal to 0. So now we have got E is perpendicular
to H, E is also perpendicular to n that means E H and n are perpendicular to each other.
That is one of the properties of uniform plane waves. So it satisfies this property and therefore
we say that the fields represent the uniform plane wave. Now, to find out the frequency at the velocity
of the wave essentially we have to find out the propagation constant beta and for that
again we we take this term which is the phase term, this quantity this quantity here and
we do some rearrangement so that we can get this quantity propagation constant beta. So, the phase function which is given there
which is e to the power minus j beta into n dot r if I take this quantity now from here
we will get beta into n cap that will be equal to 0.02pi this phase function here into x
cap plus root 3 into z cap. Writing this as unit unit vector this will be 0.02pi into
2 and this will be 1 upon 2 x cap plus root 3 upon 2 z cap so you multiply by 2 and divide
it by 2 so this becomes the unit vector. So this quantity then essentially represent the
the phase constant beta. So from here we get this phase constant beta which is equal to
0.04pi radians per meter. Now, to find out the dielectric constant…
for finding out the velocity essentially we have to find out the medium properties, still
we do not know because the frequency is not given; in fact we have to find the frequency
of this wave so first we have to find out what is the medium property or what is the
velocity of the wave and then we can find out essentially from the property of uniform
plane wave that is the ratio of mod E to mod H that should be equal to the intrinsic impedance
of the medium and since we are talking about a dielectric medium the permeability of a
medium is same as the free space permeability, only the permittivity of the medium is different. So we know that mod E upon mod H that should
be equal to the intrinsic impedance of the medium which is square root of mu upon epsilon
and since the medium is only dielectric it is equal to mu 0 upon epsilon 0 into epsilon
r. Now, to find out the ratio of this magnitude,
since the electric and magnetic fields are having this component which are j so essentially
it means that the fields are complex. So what we can do is we can find out the magnitude
of electric and magnetic fields at some instant of time and since we know that the ratio of
electric and magnetic field should be equal to the intrinsic impedance at every instant
of time the ratio can be taken at some instant of time and the easier way to do that is take
the time t equal to 0. So what these components are showing you is
that these two components are 90 degrees out of phase in time with respect to the y component
so if I take t equal to 0 essentially these two components will be 0 at that instant of
time because they are 90 degrees out of phase and at that same instant on time if I take
the magnetic field this component will be 0 this will be 90 degrees out of phase, so
at t equal to 0 if I take the electric and magnetic fields that will be corresponding
to this term for electric field and these two terms for the magnetic field. So if I take at t equal to 0 sometime then
the magnitude the electric field for this is given as 2 into y cap with of course a
multiplier by the phase function and the magnetic field at that instant of time will be 1 upon
30pi into root 3 x minus z cap. So that instant of time now we can find out the magnitude
of electric and magnetic fields, so this gives me that the magnitude of electric field mod
of E that is equal to 2 it is oriented in y direction and mod of H will be mod of this
so that is that will be equal to 1 upon 30pi into square root of 3 plus 1 so that will
be mod of H from here mod H will be equal to 2 upon 30pi. Then we can take the ratio of these two and
from there we can find out the dielectric constant of the medium. So from here essentially
we get mod E upon mod H that is equal to 2 upon 2 upon 30pi that is equal to 30pi that
is equal to square root of mu 0 upon epsilon 0 into 1 upon square root epsilon r. we just
take here the square root of epsilon r how? So we have this quantity which is square root
of mu 0 upon epsilon 0 which is nothing but the intrinsic impedance of the free space
and that number we know is equal to 120pi. So this quantity square root of mu 0 upon
of 1 epsilon 0 is 120pi from here we can get square root of epsilon r that is equal to
1 upon 30pi into this is which is 120pi which is 120pi that is equal to 4. So the square root of the dielectric constant
which we also call as a refractive index of the medium that is equal to 4 and therefore
the dielectric constant epsilon r for this medium will be equal to 16. Once we get the dielectric constant then we
can find out the velocity of this wave in this medium and that is the velocity of the
wave v will be equal to 1 upon square root mu 0 epsilon 0 epsilon r is equal to 1 upon
square root mu 0 epsilon 0 square root epsilon r and 1 upon mu 0 square root mu 0 epsilon
0 is nothing but the velocity of light in the free space which is 3 into 10 to the power
8 meter per second so we can substitute here 3 into 10 to the power 8 meters per seconds
for this and square root of epsilon r is 4 so the velocity will be 3 into 10 to the power
8 divided by 4 is equal to 0.75 into 10 to the power 8 meters per second. So, for this wave the velocity in whatever
dielectric is given which is having a dielectric constant 16 the velocity will be 0.75 into
10 to the power 8 meters per second. Once we know the velocity then now we can find
out the frequency of the wave because we know that the velocity for the wave is omega upon
beta or from here since you want to find out the frequency which is equal to 2pi into frequency
f divided by beta so the frequency from here will be equal to velocity into beta v into
beta divided by 2pi. So we have found out this velocity which is
0.75 into 10 to the power 8 meters per second, we also have found out the value of beta which
is 0.04pi radians per meter, we can substitute this, this is equal to 0.75 into 10 to the
power 8 multiplied by beta which is 0.04pi divided by 2pi that if we simplify that will
turn out to be 0… 1.5 megahertz. So the frequency of this wave is 1.5 megahertz. The next thing we have to find out is the
velocity of the wave along different directions so we are supposed to find out the phase velocities
along x, y and z directions. So as we know that the velocity v x the phase velocity along
x direction is equal to v divided by cos of phi x, the velocity along y direction is v
divided by cos of phi y and the velocity along z direction is v divided by cos of phi z and
we have calculated this phi x, phi y and phi z so phi x is 60 degrees, phi y is phi by
2 so 90 degrees and phi z is 30 degrees so from here we get the velocity along the x
direction that is 0.75 into 10 to the power 8 divided by cos of phi x which is 1 upon
2 you get 1 upon 2 so that will be equal to 1.5 into 10 to the power 8 meters per second. The phi y is 90 degrees so cos of phi y is
0, the phase velocity along the y direction is infinity and phase velocity along z direction
will be 0.75 into 10 to the power 8 divided by cos of phi z which is cos of 30 degrees
which is root 3 by 2 so this will be root 3 by 2
so that is equal to 0.866 into 10 to the power 8 meters per second. So, since the wave is
traveling in the xz plane the phase velocity in the y direction is infinity and these are
phase velocities which are in the two directions x and z. Now, to find the state of polarization of
this wave one way is to find out the equation of the ellipse which the electric field vector
draws but in this case what we can do is we can look at the electric field and draw…
find out the magnitude of electric field at different times. So let us say if I write down the electric
fields… in this the electric field has three components: the x component is given by this,
the y component is given by this and the z component is given by that. So here the magnitude
is root 3 and there is a minus j that means this x component can be written as: E of x
can be written as root 3 cos of omega t minus pi by 2 that corresponds to minus j, the y
component has an amplitude 2 and it has a zero phase so we can write down the y component
E y that is equal to 2 cos of omega t at the z component E z that is equal to this j so
it has magnitude 1 and phase plus pi by 2 so that is cos of omega t plus pi by 2 so
this gives us root 3 sin of omega t and this will give us minus sin of omega t. Now we can calculate the the field magnitude
at different instant of time. So let us consider the time which is t equal to 0, t equal to
quarter period t by 4 t equal to t by 2, t equal to 3 t by 4 and so on and find out what
is the magnitude of the electric field. So at t equal to 0 if I take then this this field
will be 0, this field will be 2 and this will be again 0. So you will have E x equal to
0, E y is equal to 2 and E z is equal to 0 giving mod of E equal to 2. If I go after a quarter period which is t
by 4 then this quantity will be plus 1, this will be minus 1 and this quantity will be
0 so in this case we will get E x which is equal to root 3, E y which is equal to 0 and
E z is equal to minus 1 that again gives the mod of E which is square root of E x square
plus E z square that will be equal to 2. Similarly, if I go one more time which is
which is t by 2 so again this will be equal to pi, you will get the magnitude which will
be the same so let us take a time which is not multiples of pi by 2 but let us say take
a time which is t by 8. So if I substitute for this now this angle will be equal to 45
degrees so you get root 3 upon root 2 so for this time you get E x is equal to root 3 upon
2, E y will be equal to 2 upon root 2 2 upon root 2 and E z will be equal to minus 1 upon
root 2. Again if we calculate the magnitude of the
electric field that will be equal to 2. So what that indicates is as the time varies
from 0 to t by 8 to t by 4 the magnitude of the electric field remains constant which
is equal to 2 that means this wave represents a circularly polarized wave. Now, to find the sense of rotation essentially we
can look at this diagram and the wave is traveling essentially in this direction. So if I consider
at t equal to 0 the electric field is in the in the y direction so the electric field is
along y direction which is like that and that is what is given by this. If I go at a t by
time little later the E x is root 3, E y is 0 and E z is is minus 1. So initially the
field the electric field was it was like that and after the time t by 4 the E x is become
root 3 and the z component is negative. So essentially the electric field vector has
moved like that. So since this since this quantity z is negative in this so the point
which has moved here the electric field vector it will be root 3 in this direction and minus
1 in the z direction so the point will come somewhere here and that is where the electric
field would be so essentially the electric field has gone from here to here something
like that in this direction. Now since the wave is moving this way if I
now look in the direction of the wave the electric field vector is rotated like this
from here to here that means this wave essentially represents a left handed rotation. So we have
the sense of rotation for this wave which is left handed. So this problem clearly demonstrates how knowing
the electric and magnetic fields one can find out the various parameters regarding the medium,
the velocity, the frequency, the state of polarization and so on. So if we systematically
proceed with the concept which we have developed for uniform plane wave, then by knowing the
fields we can essentially find out the various parameters of the medium, as well as the properties
of the electromagnetic wave. Let us us take a second problem. Here we have a uniform plane wave having 10
watts per meter power density is normally incident on a 5 centimeter thick dielectric
sheet with epsilon r equal to 9. If the frequency of the wave is 1 gigahertz, find the power
density of the wave transmitted through the sheet.
So essentially what we have here? We are having a medium which is having a finite thickness
and then we have to find out how much field is transmitted on the other side of the sheet
which is having a dielectric constant nine and a thickness of 5 centimeter. So this problem
essentially can be can be solved as follows: We have a sheet which is like that whose dielectric
constant epsilon r is equal to 9, on this side you have dielectric constant epsilon
r equal to 1 and on this side also you have epsilon r equal to 1. Now the wave is incident
from this side is normally incident on the dielectric slab which is having a power density
of 10 watts per meter. So the power density p is equal to 10 watts per meter square for
this wave and it is having a thickness of 5 centimeter. This problem essentially can be solved in
many ways. one is which… since this is a normal incidence we can treat it like a one
dimensional problem. So we can treat it as if you are having a medium here which is having
certain impedance since it is like a like a long transmission line, then you are having
another section here whose medium parameter is this so it has a different characteristic
impedance and then in this side again you are having a infinite medium so you have another
line which is having a impedance which corresponds to epsilon r equal to 1. So one of the ways to solve this problem is
by analogy with the transmission line which says this is equivalent to the transmission
line then there is another transmission line here and then the same transmission line again
which continues up to infinity. Now the characteristic impedances of these
lines would be the intrinsic impedances of this media. So here we have the intrinsic
impedance which will be eta which is same as eta 0 which is equal to 120pi. Here again
you will have a impedance eta which will be eta 0 which will be 120pi and in this case
you have an impedance eta which will be eta 0 divided by square root of epsilon r. So
we will have here eta which is equal to 120pi divided by square root of 9 so that is equal
to 40pi. So now the wave is incident from this side,
you have a section here which is having an impedance characteristic impedance of 40pi
and you have the length of this transmission line which is 5 centimeter so this is since
this lies infinite now it will see an impedance which is equal to a characteristic impedance
of this so I can I can find out… so this is equivalent to, this section is equivalent
to… we have a transmission line of 120pi then we have a transmission line which is
40pi characteristic impedance and this is now this line is terminated in an impedance
which is equal to 120pi. So one of the ways to solve this problem now
is we have to find out what is the power transmitted through this sheet on the other side? So essentially
we want to find out what is the power density which is gone to this resistance which is
120pi. So if you go by transmission line approach
essentially what we can do is we can find out what is the impedance which would be seen
at this location, then how much power has been transmitted transferred to that impedance
that is the power which is finally will be delivered to this impedance so we can get
the power and from there we can find out the power which is transmitted to this medium.
For this we have to find out now this length which is 5 centimeter which we have to find
in terms of the wavelength and now the wavelength depends upon the dielectric constant of the
medium so here the velocity we can calculate and from there we can calculate the wavelength
of the wave in that medium. So since the epsilon r is 9 the velocity of
the wave in that medium on the sheet
that is equal to the velocity in the air divided by the square root of epsilon r which is root
of 9 so that is equal to 10 to the power 8 meters per second and the frequency is given
for this problem is 1 gigahertz so the f given this way is equal to 1 gigahertz that is equal
to 10 to the power 9 hertz. So from here as we have calculate the value of lambda which
is the velocity divided by frequency that is equal to 10 to the power 8 divided by 10
to the power 9 so that is equal to 0.1 meters or that is equal to 10 centimeter. So in this medium the wavelength lambda is
10 centimeters. That means this length here 5 centimeter essentially is the distance of
lambda by 2. So in this case the problem becomes very simple that now this distance is equal
to lambda by 2 and the impedance which you see the normalized impedance is transformed
here at a same value at a distance of lambda by 2. So essentially we can we can calculate
this this impedance and transform this impedance on the other side so you can get essentially
the impedance which is 120pi which is on the other side. Once you get that impedance which is transformed
here then you can find out what is the power which is delivered with this load and from
there then you can find out the power density which is gone to the second medium; that is
one approach. The second approach is which is more like
a wave approach and that is… consider now this medium and the wave incident on this.
So let us say this boundary is is boundary… this medium is 1, this medium is 2 and this
medium is 3. So we have now the two media interfaces: one is from 1 to 2 then the other
one is from 2 to 3 so the wave is incident on this, on this boundary it sees the reflection
and part of the energy is transmitted here, this energy travels all the way up to this
boundary then part of the energy is reflected from here, part gets transmitted, this energy
then travels all the way up to this point since this is again medium discontinuity part
of the energy is transmitted here and reflected here. This energy again reaches at this point
it gets reflected or part gets transmitted, this when reaches here, again part gets reflected
part gets transmitted so what we can find essentially that the net reflected wave will
be sum of all these reflections; this first reflection plus this second reflection plus
the third reflection, if we sum up all of them together that gives me the total reflected
wave in this direction. Similarly, the transmitted wave would be this
transmitted wave plus this plus this. If I sum up all these waves together that essentially
give us the total transmitted wave on the other side. So now what we can do is we can essentially
find out the the reflection which are coming from here so we can write down the reflection
coefficient for each of these boundaries and the transmission coefficient on each of the
boundaries and then find out the total summation and from there essentially we can get the
total reflection and the total transmission. Now if I say that the reflection at i j boundary
so that means 1 2 boundary if I take that would be gamma 1 2, so from here let us say
reflection I write down at an interface which is given by i j that is nothing but impedance
of the medium j minus impedance of the medium i divided by impedance of the medium j plus
impedance of the medium i and from here essentially we get gamma ij is equal to minus of gamma
ji. So, if the wave is going from here to here
the reflection coefficient is gamma 1 2. However, when the wave comes here the reflection coefficient
this would be gamma 2 1. So this one is gamma 1 2, this will be gamma
2 1, this will be gamma 2 3 and so on. Similarly the transmission coefficient you
can get tau ij that is equal to two times eta j divided by eta j plus eta i. So this
transmission coefficient here would correspond to eta 2 divided by eta 1 by eta 2, this transmission
coefficient would correspond to eta 3 2 times eta 3 divided by eta 3 plus eta 2 and so on. So now if I if I write down and you are having
this distance d and the phase constant in this medium let us say is given by beta so
you are having the phase change when the wave travels from here to here which is equal to
corresponding to beta d. So this wave travels a distance which is beta d phase bet beta
d, when it comes here it has traveled a distance 2 times beta d so now what we have here is
the first reflection plus the second quantity here which is one reflection here one transmission
one reflection and one transmission which is this wave. So, if I write down now the the the total
reflected field which is going to come from here that would correspond to your E reflected
that is equal to gamma 12 E incident plus tau 12 then gamma 23 and tau 21. So we can
write here tau 12 E i gamma 23 tau 21 with a phase term of which is e to the power minus
j 2 beta d. The next term would be whatever the whatever has come here this will be tau
12 plus gamma 23 gamma 21 gamma 23 and tau 21. So we can we can write that so that is
equal to tau 12 E i gamma 23 gamma 21 gamma 23 tau 21 e to the power minus j 4 beta d
plus and so on. So essentially you see if I take this quantity
common you will have now a geometric series for this. so we can write down now the reflected
wave E r that is gamma 12 E i plus tau 12 gamma 23 tau 21 e to the power minus j beta
d into 2 and summation of the geometric series which we get for this term will be multiplied
by E i upon 1 minus gamma 12 gamma 23 e to the power minus j 2 beta d.
So if I know now the field which is incident on this on this wave then i can find out what
would be the the net reflected wave which will be essentially given by that. So precisely
same thing we can do for the transmitted wave also which is going to going to this. So this
transmitted wave will be… this transmitter multiplied by this transmitter that will be
two reflections and then this transmitted. If I sum up again all these fields essentially
we will get the transmitted wave also on the line similar to this that will be equal to
tau 12 tau 23 e to the power minus j beta d into E i divided by 1 minus gamma 21 gamma
23 e to the power minus j 2 beta d. So I can substitute now this value for beta
d for 2 beta d and I can substitute the value for gamma 21 and gamma 23 and from here then
I can find out what is the transmitted wave. So the power density which will be transmitted
will be proportional to mod t square so I can find out this quantity and can get actually
the the power transfer. So in this case since now the quantity which are given are… the
situation is this that this is equal to epsilon r is 1, epsilon r is 1, and this is epsilon
r is 9 essentially we have the eta 1 which is equal to eta 0, eta 2 is eta 0 is square
root of epsilon r is 3 eta 3 is equal to eta 0 again. So from here we can get essentially these
quantities which is gamma 12 which is eta 0 by 3 minus eta 0 upon eta 0 by 3 plus eta
0 so that is equal to 1 upon 3 minus 1 upon 1 upon 3 plus 1 so that will be equal to minus
2 upon 4 is equal to minus 1 upon 2 and gamma 23 will be eta 0 minus eta 0 upon 3 minus
eta 0 plus eta 0 upon 3 that will be equal to 2 upon 4 equal to 1 upon 2.

## 2 Replies to “Problems on uniform plane wave in a meduim”

1. prasant rocking says:

Sir I really like your approach of teaching…
but I have a doubt at 36:37 if we try to solve problem using transmission line analogy…
medium 1 and medium 3 are same..
medium 2 has thickness  λ/2 and so input impedance at the interface between medium 1 and 2 will be equal to intrinsic impedance of medium 3.
so there should be 100% transmission of power but..using the shown method in this lecture..transmission is 9/25 of incident..