Download Module 11: The vector nature of electromagnetic radiation

Module 11: The vector nature of electromagnetic radiation
Lecture 11: The vector nature of electromagnetic radiation
Consider a situation where the same electrical signal is fed to two mutually
perpendicular dipoles, one along the y axis and another along the z axis as
shown in Figure 11.1.
E
y
x
z
Figure 11.1: The resultant electric field at x
We are interested in the electric field at a distant point along the x axis.
The electric field is a superposition of two components
~
E(x,
t) = Ey (x, t)bj + Ez (x, t)kb
(11.1)
one along the y axis produced by the dipole which is aligned along the y axis,
and another along the z axis produced by the dipole oriented along the z axis.
11.1
Linear polarization
In this situation where both dipoles receive the same signal, the two components are equal Ey = Ez and
b cos(ωt − kx)
~
E(x,
t) = E(bj + k)
(11.2)
If we plot the time evolution of the electric field at a fixed position (Figure
11.2) we see that it oscillates up and down along a direction which is at 45◦
to the y and z axis.
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74CHAPTER 11. THE VECTOR NATURE OF ELECTROMAGNETIC RADIATION
z
ο
45
θ
y
Figure 11.2: Linear Polarization
The point to note is that it is possible to change the relative amplitudes of
Ey and Ez by changing the currents in the oscillators. The resultant electric
field is
~
E(x,
t) = (Ey kb + Ez bj) cos(ωt − kx)
(11.3)
The resultant electric field vector has magnitude E =
q
Ey2 + Ez2 and it
y
oscillates along a direction at an angle θ = tan−1 E
with respect to the z
Ez
axis (Figure 11.2).
Under no circumstance does the electric field have a component along the
direction of the wave i.e along the x axis. The electric field can be oriented
along any direction in the y − z plane. In the cases which we have considered
until now, the electric field oscillates up and down a fixed direction in the y − z
plane . Such an electromagnetic wave is said to be linearly polarized.
11.2
Circular polarization
The polarization of the wave refers to the time evolution of the electric field
vector. An interesting situation occurs if the same signal is fed to the two
dipoles, but the signal to the z axis is given an extra π/2 phase . The electric
field now is
h
~
E(x,
t) = E cos(ωt − kx)bj + cos(ωt − kx + π/2)kb
h
= E cos(ωt − kx)bj − sin(ωt − kx)kb
i
i
(11.4)
(11.5)
~
If we now follow the evolution of E(t)
at a fixed point, we see that the tip
~
of the vector E(t) moves on a circle of radius E clockwise (when the observer
looks towards the source) as shown in Figure 11.3
We call such a wave right circularly polarized. The electric field would
have rotated in the opposite direction had we applied a phase lag of π/2. We
would then have obtained a left circularly polarized wave.
75
11.3. ELLIPTICAL POLARIZATION
z
z
E
E
y
y
Right Circular
Left Circular
Figure 11.3: Circular polarization
11.3
Elliptical polarization
Oscillations of different amplitude combined with a phase difference of π/2
produces elliptically polarized wave where the ellipse is aligned with the y − z
axis as shown in Figure 11.4. The ellipse is not aligned with the y − z axis for
an arbitrary phase difference between the y and z components of the electric
field. This is the most general state of polarization shown in the last diagram
of the Figure 11.4. Linear and circularly polarized waves are specific cases of
elliptically polarized waves.
z
z
E
z
E
y
y
E
y
Right elliptical
Left elliptical
Figure 11.4: Elliptical plarization
Problems
1. Find the plane of polarization of a light which is moving in the positive x
direction
√ and having amplitudes of electric field in the y and z directions,
3 and 3 respectively in same units. The oscillating components of the
electric field along y and z have the same frequency and wavelength and
the z component is leading with a phase π.
2. Find the state of polarization of a light which is moving in the positive x
direction with electric field amplitudes same along the y and z directions.
76CHAPTER 11. THE VECTOR NATURE OF ELECTROMAGNETIC RADIATION
The oscillating components of the electric field along the y and z have
the same frequency and wavelength and the z component is lagging with
a phase π/3.
(Ans: Left elliptically polarized and the major axis is making an angle
π/4 with the y axis.)
3. Find the state of polarization of a light which is moving in the positive x
direction and having amplitudes of electric field in the y and z directions,
1 and 31/4 respectively in same units. The oscillating components of the
electric field along y and z have the same frequency and wavelength and
the y component is leading with a phase π/4.
(Ans: q
Left elliptically polarized and the major axis is making an angle
√
−1
tan
1 + (2/ 3) with the y axis.)
4. Find out the maximum and minimum values
point x for
√ of electric 2field at√
2
= ( 3−1)/2.)
the previous problem. (Ans: Emax
= (3+ 3)/2 and Emin