PHY294H - Lecture 39

Polarisation

Linear polarisation - Electric field oscillates in one direction.

Unpolarised - Random mixture of polarisation in all directions. If unpolarised light is passed through a polariser then 1/2 of the intensity passes through and the outgoing light is linearly polarised.

Circularly polarised - Electric field vector rotates as the wave propagates. If we add two waves which are in phase, but which are linearly polarised in the x and y-directions respectively we get a wave which is linearly polarised at 45^o. If we add the same two waves but now with a phase difference $\phi$ between them, we can have elliptically polarised light. The special case of $\phi = 90^o$is circularly polarised light and righthanded. The case $\phi = -90^o$ is lefthanded. The case $\phi = 180^o$ is again linearly polarised.

Polaroid - Aligned long molecules of polyvinyl alcohol(PVA) impregnated with Iodine. The Iodine acts as an electron donor and gives the PVA molecules electronic conductivity.

Use of polarisation in wireless communication

Linearly polarised light has an electric field which oscillates in one direction. Long distance communications are sent with horizontal polarisation. This is better as vertically polarised light is more susceptible to attenuation by treetops. In contrast local broadcasts are sent with vertical polarisation. This is due to the fact that vertical receivers are more convenient on cars and rooftops. The fact that local and long distance communications have opposite polarisation directions also helps reduce interference.

Polarisation by scattering

To understand scattering it is convenient to think in terms of charges which are accelerated by the electric field in the EM wave. Accelerated charges emit radiation. They emit the radiation predominantly perpendicular to their direction of acceleration.

We consider scattering of an unpolarised beam of light. Scattering can be visualised as the instigation of oscillation of the charges from which the light scatters. This oscillation leads to a dipole radiation field which we observe as the scattered light. If we observe this scattering along the direction of the incident light, then the oscillations of the charges in the scatterer are always perpendicular to our line of sight. In contrast if we observe the scattering at 90^o to the direction of propagation, we will observe light that is vertically polarised. The horizontal component is eliminated. the scattered radiaion is preferentially polarised perpendicular to the plane of the scattering.

Reflection off a surface is an important example. Unpolarised light which is scattered off a surface is preferently polarised perpendicular to the plane of the scattering. The most extreme case is where the transmitted and reflected rays are at 90^o. This is the Brewter angle $\theta_B$The light obeys Snell's law which states that $n_1 sin(\theta_1) = n_2 sin(\theta_2)$. In addition for maximum polarisation, we need the scattered and transmitted rays to be at 90^o. This implies that $\theta_1 + \theta_2 = 90^o$. The special angle at which this occurs is called the Brewster angle $\theta_B$ where we have,

$$n_1 sin(\theta_B) = n_2 sin(90-\theta_B)$$ (1)

which implies that,

$$\theta_B = tan^{-1}({n_2 \over n_1})$$ (2)

Dispersion - wavelength dependence of refractive index

Dispersion: n depends on frequency. At high frequency (blue) n is larger.

To discuss dispersion effects, we first have to expand our discussion of light. Our discussion so far has been focused on light of one wavelength. This light is called monochromatic light. White light has a mixture of wavelengths (colors) and is called polychromatic or polydisperse. The key new feature we shall use is that the refractive index of many optical materials depends on the wavelength of the incident light. This is called dispersion and is the reason that light produces colors when it is passed through a prism. For example the refractive index of light varies from n=1.528 in the blue to n=1.51 in the red. This small variation leads to easily observed splitting of the colors of the spectrum. Examples are refraction through a prism and the physical origin of rainbows.

Rayleigh scattering - the wavelength dependence of scattering

Rayleigh Scattering: Scattering intensity is larger at higher freqencies (shorter wavelength). $I(\lambda) = 1/\lambda^4$.

If there was no atmosphere, the sky would be black except if we looked straight at the sun. The fact that the sky is not black is due to scattering of light from air molecules and particles in the atmosphere. The color that we see is due to the fact that light scattering is stronger at short wavelengths, because it is easier for the electrons in small particles to be excited by short wavelength light. This means that we see a predominantly blue color in the sky.

Atomic origin of dispersion

The simplest model is to consider a mass spring system as a model for the electron/nucleus system. This system has a resonant frequency $\omega_0 = k/m$. When an electric field of frequency $\omega$ is incident on this system, resonance occurs when $\omega = \omega_0$. Light in this regime is at much lower frequency than the resonant frequency, so we have, $\omega << \omega_0$. The diplacement of the electron is approximated by,

$${dz\over dt} = {1\over \omega_0^2 - \omega^2} cos(\omega t)$$ (3)

The average acceleration is then,

$${d^2 \over dt^2} = {\omega^2 \over \omega_0^2 - \omega^2} \sim {\omega^2\over \omega_0^2}$$ (4)

The radiated energy is proportional to the square of the acceleration, which leads to the $1/\lambda^2$ which was first found by Lord Rayleigh. The response of the system also leads to a dielectric constant which depends on frequency. Since the dielectric constant and the refractive index are related, we also expect the refractive index to depend on frequency. In fact the refractive index, within the model above, depends on the frequency as

$${1\over n^2} = 1 - {C\over \omega_0^2 - \omega^2}$$ (5)

This indicates that the refractive index increases as the frequency of light increases. In most atoms and molecules the above model is not aquequate as there are terms where the frequency of light is similar to the vibration frequency of the electrons. In that case resonant terms occur.

Properties of light

As we have seen, the classical theory of optics allows us to vary a variety of properties of light

- Wavelength

- Polarisation

- Relative phase

- Amplitude

These properties all have quantum interpretations. The light over which we have most control is laser light, where we control the phase relation of photons as well as the other properties above. However the photon has spin 1 and we would like to know how that is related to the properties above. The correspondence is that circularly polarised light has a net angular momentum. The right circularly polarised case corresponds to spin -1, while the left polarized light corresponds to spin 1. Spin zero corresponds to in phase x and y components of the electric field.