5.3 Polarization

Electromagnetic waves are transverse waves as illustrated by Fig. 5.1. The wavenumber vector $\vec{k}$ specifies the plane in which

and

oscillate, and $\vec{E}$ and $\vec{B}$ must be mutually perpendicular, but any absolute orientation in the plane is possible.

We define the polarization of an electromagnetic wave is the orientation of $\vec{E}$ , described by the polarization vector

$\begin{displaymath} \hat{\wp} = \frac{\vec{E}}{E} \end{displaymath}$

(5.13)

**Figure 5.2:** (a) The electric fields of two plane waves at right angles to each other and $\frac {\pi }{2}$ out of phase and (b) their left-circularly polarized superposition.
$\scalebox{0.7}{ \includegraphics{circular.eps} }$

A superposition of two plane waves with perpendicular electric fields, $\frac {\pi }{2}$ out of phase with each other, is circularly polarized. Consider for example two plane waves propagating in $\hat{\imath}$ , shown in Fig. 5.2(a),

$\begin{displaymath} \vec{E}_1 = E_o e^{i(kx - \omega t)} \hat{\jmath} \end{displaymath}$

(5.14)

and

$\begin{displaymath} \vec{E}_2 = E_o e^{i(kx - \omega t - \frac{\pi}{2})} \hat{k} \end{displaymath}$

(5.15)

Noting that $e^{-i\frac{\pi}{2}} = -i$ and taking the real part explicitly, their superposition, shown in Fig. 5.2(b), is

$\begin{displaymath} \vec{E} = E_o \cos(kx - \omega t) \hat{\jmath} + E_o \sin(kx - \omega t) \hat{k} \end{displaymath}$

(5.16)

with magnitude

and polarization

$\begin{displaymath} \hat{\wp} = \cos(kx - \omega t) \hat{\jmath} + \sin(kx - \omega t) \hat{k} \end{displaymath}$

(5.17)

At a given position

, the polarization vector rotates about $\vec{k}$ with constant angular speed $\omega$ . The magnitude of the electric field

$\begin{displaymath} \hat{\wp} \cdot \vec{E}_o = E_o \end{displaymath}$

(5.18)

is constant for circularly polarized light. Hence, the trajectory of the tip of the polarization vector of circularly polarized light is a helix, as shown in Fig. 5.2(b).

With $\phi_2 - \phi_1 = -\frac{\pi}{2}$ , the electric field vector rotates counterclockwise for observers which the light approaches. This is known as left-circular polarization. With $\phi_2 - \phi_1 = \frac{\pi}{2}$ , we obtain clockwise rotation and right-circular polarization.

Changing the phase angle of $\vec{E}_2$ from $-\frac{\pi}{2}$ to $\frac {\pi }{2}$ changes the direction of the angular rotation of the polarization.

A nonzero phase difference other than $\pm \frac{\pi}{2}$ gives elliptical polarization.

Updated Wed Jan 18 09:51:28 2006