Question

(a) Explain polarisation by scattering and by reflection.

(b) State Brewster’s law connecting angle of polarization and refractive index. Prove that µ = tan ip.

Answer 1

(a) Polarisation by scattering

The incident sunlight is unpolarised. The dots stand for polarisation perpendicular to the plane of the figure. The double arrows show polarisation in the plane of the paper. When this light becomes incident on nitrogen molecules of the atmosphere, electrons of the molecules are set into vibration in these two perpendicular directions and are accelerated.

The observer is looking in a direction at 90° to the direction of sun. The electrons accelerating parallel to the double arrows do not radiate energy towards the observer. The electrons accelerating along dot radiate energy towards the observer. The radiation scattered towards the observer is therefore, represented by dots. The radiation is polarised perpendicular to the plane of the paper Fig. 

It is polarised perpendicular to the plane of the figure.

A plane at right angle to the plane of vibration of light wave is called plane of polarisation.

Polarised light by reflection. Fig shows unpolarised light, incident on a transparent (water) surface. The light is reflected as well as refracted. The angle of incidence is such that reflected light and refracted light proceed in perpendicular direction. It happens when i + r = 90°. In incident light, double arrows represent radiations polarised in plane of paper and dots represent radiations polarised perpendicular to the plane. In refracted wave, double arrows are parallel to the direction of reflected wave. These radiations are not transmitted alongwith the reflected light. It has only those radiations which are represented by dots. Hence reflected light becomes polarised perpendicular to the plane of the paper. The polarisation can be checked by an analyser.

(b) Brewster’s law: It state that the tangent of polarising angle is equal to the refractive index of transparent material.

i.e. tan i_p = µ

or µ = tan i_p

where i_p is the polarising angle.

Proof: As shown in Fig. a if ip is the angle of incidence (polarising angle) are r the corresponding angle of refraction, then according to

Brewster’s law.

∠POP' = 90°

∴ i_p + r = 90°

or i_p = 90° - r

or cos i = cos (90° - r) = sin r.

Now, if µ is the refractive index of the medium,

then

µ = \(\frac{sin\ i_p}{\sin r} = \frac{sin\ i_P}{\cos i_p}\)

Hence the tangent of the polarising angle is equal of the refractive index of the medium at which reflection takes place.