PHYS212 : Lectures

5.3 Polarization

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

  $$\hat{\wp} = \frac{\vec{E}}{E}$$ (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.

  
• 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),
  

  $$\vec{E}_1 = E_o e^{i(kx - \omega t)} \hat{\jmath}$$ (5.14)

  
  
  and
  

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

  $$\vec{E} = E_o \cos(kx - \omega t) \hat{\jmath} + E_o \sin(kx - \omega t) \hat{k}$$ (5.16)

  
  
  with magnitude $E_o$ and polarization
  

  $$\hat{\wp} = \cos(kx - \omega t) \hat{\jmath} + \sin(kx - \omega t) \hat{k}$$ (5.17)

  
  

• At a given position $x$, the polarization vector rotates about $\vec{k}$ with constant angular speed $\omega$. The magnitude of the electric field
  

  $$\hat{\wp} \cdot \vec{E}_o = E_o$$ (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.

Copyright © 2003-2009, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006