๐๏ธ Note: We will assume $\mu=\mu_0$ ie. the material doesn't affect the magnetic field of the light wave
Linear polarization properties:
For a wave travelling in the $z$ direction
$$ \hat {\mathrm{x}}E_{0x}\cos(kz-\omega t+\phi)+\hat {\mathrm y}E_{0y}\cos(kz-\omega t+\phi) $$
Electric field remains in the same plane as the waves propagates
The plane containing electric field vectors is at an angle $\arctan(E_{0y}/E_{0x})$ to the $x$-axis
The maximum $E$-vector amplitude is $\sqrt{E^2_{0x}+E^2_{0y}}$
There is no phase shift between $\hat{\mathrm x}$ and $\hat {\mathrm y}$ components of the wave
Circular polarization
The electric field vector remains the same size but rotates around the direction of propagation as the wave advances
There is a $\pm90\degree$ between $x$ and $y$
For a wave travelling in the $z$ direction
$$ \hat {\mathrm x}E_{0x} \cos(kz-\omega t+\phi)+\hat {\mathrm y}E_{0y}\cos(kz-\omega+[\phi\pm \pi/2]) $$
๐๏ธ Note: one can also use $\sin$ or $-\sin$ functions for $x$ and $y$
Right hand circular (RHC) when looking at the photon coming towards you
๐๏ธ Note: this convention is not used by everyone
If we add LHC and RHC waves of the same amplitude, we get linear polarization with twice the original amplitude
Elliptical Polarization
Travelling in the $z$ direction
$$ \hat {\mathrm x}E_{0x} \cos(kz-\omega t+\phi_1)+\hat {\mathrm y}E_{0y}\cos(kz-\omega+\phi_2) $$
Linear polarization if $\phi_1=\phi_2$
Circular polarization if $E_{0x}=E_{0y}$ and $\phi_1-\phi_1=\pm \pi/2$
$E$-vector describes an ellipse with time at any given point
Unpolarized light
- Wave train composed of very small fragments of polarized light where $E_{0y}$, $E_{0x}$ and $\phi_2-\phi_1$ vary rapidly and randomly
($\theta_i=\theta_b)$ is known as the Brewster angle โ
๐ผ Consider: If the angle between the reflected and refracted rays is $90\degree$ then there is no reflected ray in one of the polarizations because dipoles do not radiate along their length
Since $\theta_i+\theta_r=\pi/2$ then $\sin(\theta_r)=\cos(\theta_i)$ combining with Snell we get
$$ \tan(\theta_i)=n_r/n_i $$
๐๏ธ Notes: