11: Wave Phenomena 11.5 Polarisation

Polarisation When a charged particle loses energy, a tiny disturbance or ripple in the surrounding electromagnetic field is created – an e-m wave. The wave will be in the same plane as the plane of movement of the particle. If all the disturbances in a ray of light were occurring in the same plane, the light is said to be plane polarised. In reality most sources of light produce unpolarised light because the electrons in the source move in all directions. Electromagnetic waves electron

Polaroids A Polaroid filter (a polariser) polarises light rays. If an unpolarised ray of light passes through the filter, about 50% of its energy will be absorbed and 50% pass through. Planes before: Plane after:

The original light waves exist in many different planes. The polarising filter only allows components of the wave in one plane to pass through. E.g. - One wave in its original plane: - This has vertical and horizontal components: - Wave after passing through filter with a vertical plane of polarisation (horizontal component absorbed):

Polarisation by Reflection When an unpolarised ray of light is reflected off a non metallic surface, the reflected ray becomes partly polarised. E.g. Reflections of light from cars, the sea etc. The degree of polarisation is dependant upon the angle of incidence of the light.

Brewster’s Angle When light reflects off water, polarisation will be 100% when the angle of incidence is equal to Brewster’s angle ( ϕ – phi). This occurs when the angle between the reflected and refracted rays is 90°. ϕ r ϕ

On the right hand side of the normal: ϕ + r = 90 but... so... but...sin (90 – ϕ ) = cos ϕ so Brewster’s angle is given by... n = sin i sin r n = sin ϕ sin (90 – ϕ ) n = sin ϕ = tan ϕ cos ϕ ϕ = tan -1 n

E.g. Calculate Brewster’s angle for diamond (n = 2.4). ϕ = 67°

Malus’ Law Polarised light of intensity I 0 can be split into two parallel components which in turn may be polarised by a second polarising filter (called an analyser). The resulting intensity (I) is given by Malus’ Law: I = I 0 cos 2 θ Intensity I 0 Intensity I θ = angle of rotation between the two polarisers. I = intensity, the power per unit area (Wm -2 )

E.g. A sheet of Polaroid is used to reduce the intensity of a beam of polarised light. What angle should the transmission axis of the Polaroid make with the plane of polarisation of the beam in order to reduce the intensity of the beam by 50%? so... I = ½ I 0 and I = I 0 cos 2 θ  ½ I 0 = I 0 cos 2 θ ½ = cos 2 θ cos 2 θ = √(0.5) θ = cos -1 √(0.5) θ = 45°

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