Polarization

helpful linksPolarization

Polarition as Electromagnectic Waves

Recall the general solution of electromagnectic waves progating in the z-direction. The oscillating electric field components are generally given by

\[\begin{align*} \vec{E}: \, E_x(z,t)=&E_{0x}\cos(kz-\omega t+\phi_x)\\ E_y(z,t)=&E_{0y}\cos(kz-\omega t+\phi_y)\\ E_z=&0 \end{align*}\]

In general, we can condiser two such harmonic lightwaves of the same frequency, moving through the same region of space, in the same direction \(\hat{z}\)

\[\begin{align*} \vec{E}_x(z,t)=&\hat{i}E_{0x}\cos(kz-\omega t)\\ \vec{E}_y(z,t)=&\hat{j}E_{0y}\cos(kz-\omega t+ \epsilon) \end{align*}\]

Linear Polarization

If \(\epsilon\) is zero or an integral multiple of \(\pm 2\pi\), \(\epsilon = 2m\pi, m \in \mathbb{Z}\) the resultant wave is

\[\begin{align*} \vec{E}=&\vec{E}_x+\vec{E}_y\\ =&\left(\hat{i}E_{0x}+\hat{j}E_{0y}\right)\cos(kz-\omega t) \end{align*}\]

Circular Polarization

right-circular polarized

When both constituent waves have equal amplitudes \((E_{0x}=E_{0y}=E_{0})\) and their relative phase difference \(\epsilon=-\frac{\pi}{2}+2m\pi\), where \(m\) is an integer, the resultant wave is

\[\begin{align*} \vec{E}=&\hat{i}E_0\cos(kz-\omega t)+\hat{j}E_0\cos(kz-\omega t-\frac{\pi}{2})\\ =&E_0\left[ \hat{i}\cos(kz-\omega t)+\hat{j}\sin(kz-\omega t) \right] \end{align*}\]

The resultant \(\vec{E}\) is rotating clockwise at an angular frequency of \(\omega\), as seen by an observer toward whom the light is moving. This is referred to as right-circular polarized .

The \(\vec{E}\)-vector makes one complete rotation as the wave advances through one wavelength.

left-circular polarized

When their relative phase difference \(\epsilon=\frac{\pi}{2}+2m\pi\), where \(m\) is an integer, the resultant wave is

\[\begin{align*} \vec{E}=&\hat{i}E_0\cos(kz-\omega t)+\hat{j}E_0\cos(kz-\omega t+\frac{\pi}{2})\\ =&E_0\left[ \hat{i}\cos(kz-\omega t)-\hat{j}\sin(kz-\omega t) \right] \end{align*}\]

The amplitudes is unaffected, but \(\vec{E}\) now rotates counterclockwise , and the wave is left-circular polarized.

A Math Description of Polarization

Monochromatic Light and Natural Light

An idealized monochromatic plane wave must be depicted as an infinite wavetrain. If this disturbance is resolved into two orthogonal components perpendicular to the direction of propagation, they, in turn, must have the same frequency, be infinite in extentm, and therefore be mutually coherent (i.e. \(\epsilon=constant\))

\[\begin{align*} \vec{E}_x(z,t)=&\hat{i}E_{0x}\cos(kz-\omega t)\\ \vec{E}_y(z,t)=&\hat{j}E_{0y}\cos(kz-\omega t+\epsilon) \end{align*}\]

A perfectly monochromatic plane wave is always polarized.

Natural light is also known as unpolarized or randomly polarized light. We can mathematical represent natural light in terms of two arbitrary, incoherent, orthogonal, linearly polarized waves of equal amplitude (i.e. waves for which the relative phase difference varies rapidly and randomly).

More often, the electric-field vector varies in a way that is neither totally regular nor totally irregular, and such an optical disturbance is partially polarized. One useful way of describing this behavior is to envision it as the result of the superposition of specific amounts of natural and polarized light.

Polarizing Sheets

Unpolarized Light

Electric field oscillations of unpolarized light can resolve into two components with equal intensity. Therefore, the intensity \(I\) of the polarized light emerging from a polarizing sheet is then half the intensity \(I_0\) of the original light, i.e.

\[\begin{align*} I=\frac{I_0}{2} \end{align*}\]

Polarized Light

For polarized light, only the component parallel to the polarizing direction of the sheet \((E_y=E\cos\theta)\) can be transmitted. Therefore, the intensity of the emerging wave is

\[\begin{align*} I=I_0\cos^2\theta \end{align*}\]

Polarization by Reflection

Consider a ray of unpolarized light incident on a glass surface. The field \(\vec{E}\) of the incident light can be decomposed into two components of equal magnitude, one perpendicular and another parallel to the plane of incidence. In general, the reflected light is partially polarized.

Brewster angle: reflected light is fully polarized

When the light is incident at a particular incident angle, caleld the Brewster angle \(\theta_B\), the reflected light is fully polarized. One finds experimentally that at the incident angle \(\theta_B\) the reflected and refracted rays are perpendicular to each other:

\[\begin{align*} \theta_B+\theta_r=90^{\circ} \end{align*}\]

According to Snell's law

\[\begin{align*} n_i\sin\theta_B=&n_r\sin\theta_r\\ =&n_r\sin(90^{\circ}-\theta_B)\\ =&n_r\cos\theta_B \end{align*}\]

or

\[\begin{align*} \frac{n_r}{n_i}=\tan\theta_B \end{align*}\]

If the incident and reflected rays travel in air, we can approximate \(n_i\) as unity, so

\[\begin{align*} n_r=\tan\theta_B \end{align*}\]