8 min readConcept sheetSection B very short answer (2 marks) numerical on the polarising angle; short answer (3 marks) deriving μ = tan iₚ

Brewster's law and everyday uses of polarisation

At the polarising angle iₚ, reflected light is fully plane polarised and at 90° to the refracted ray. Brewster's law: μ = tan iₚ.

Brewster's lawμ = tan iₚ (μ = refractive index of the surface, iₚ = polarising angle, also called the Brewster angle)

At the polarising angleThe reflected and refracted rays are at 90°, so iₚ + r = 90°

Typical valuesWater (μ ≈ 1.33): iₚ ≈ 53°. Glass (μ = 1.5): iₚ ≈ 56°

Polarisation by reflection

Light reflected from glass, water or a polished road is usually partly plane polarised. At one angle of incidence, the reflected light is completely plane polarised. This angle is the polarising angle iₚ, also called the Brewster angle. In the reflected light, the vibrations are parallel to the surface. At this angle, the reflected ray and the refracted ray are at 90° to each other. The polarising angle depends on the refractive index of the material, as Brewster's law shows. So glare from a lake or a wet road is largely polarised, and a Polaroid can absorb it.

Derivation: Brewster's law

1
Unpolarised light falls on a glass surface at the polarising angle iₚ. Let r be the angle of refraction.
2
Experiment shows that the reflected and refracted rays are then at 90° to each other.
3
The reflected ray makes iₚ with the normal above the surface, and the refracted ray makes r with the normal below it. With the 90° between them, they make a straight angle: iₚ + 90° + r = 180°, so r = 90° − iₚ.
4
Snell's law: μ = sin iₚ / sin r = sin iₚ / sin(90° − iₚ).
5
Since sin(90° − iₚ) = cos iₚ, μ = sin iₚ / cos iₚ.
6★
So, μ = tan iₚ: this is Brewster's law.

★ key step

How to draw polarisation by reflection

Draw a horizontal line for the glass surface and a dashed normal at the point O.
Draw the incident ray coming to O at the angle iₚ. Mark it with arrows and dots; label it unpolarised.
Draw the reflected ray leaving O at the same angle iₚ on the other side of the normal. Mark it with dots only; label it plane polarised (vibrations parallel to the surface).
Draw the refracted ray inside the glass at the angle r; label it partly polarised.
Mark 90° between the reflected ray and the refracted ray.

Worked example: the polarising angle

Problem

(a) Find the polarising angle for glass of refractive index 1.5, and the angle of refraction at that angle. (b) The polarising angle for a transparent material is 58°. Find its refractive index. (c) For water, μ = 1.33. How high above the horizon must the Sun be for its glare from a calm pond to be fully polarised?

Working

(a) tan iₚ = μ = 1.5, so iₚ = tan⁻¹ 1.5 ≈ 56.3°.
r = 90° − iₚ = 90° − 56.3° = 33.7°.
(b) μ = tan 58° ≈ 1.60.
(c) iₚ = tan⁻¹ 1.33 ≈ 53°. This angle is measured from the vertical normal.
So the Sun's height above the horizon = 90° − 53° = 37°.

Answer

(a) iₚ ≈ 56.3°, r ≈ 33.7°; (b) μ ≈ 1.60; (c) about 37° above the horizon

Everyday uses of polarisation

Polaroid sunglasses have vertical transmission axes. They absorb the horizontally polarised glare from water, wet roads and snow.
Polarising filters on cameras remove reflections from water and glass, so that details under the surface show up.
Polarimeters use polarised light to measure the amount of sugar in a solution, for quality control in sugar factories.
LCD displays in calculators, watches and some screens work with polarised light.
Measuring the polarising angle gives the refractive index of a material, using μ = tan iₚ.

In real life

Fishermen and boat drivers on the backwaters of Kerala wear Polaroid sunglasses to cut the glare from the water and see below the surface. The glare from a calm pond is fully polarised when the Sun is about 37° above the horizon.

Exam angle

2 marks: “The polarising angle for a transparent medium is 57°. Calculate its refractive index (tan 57° = 1.54).” By Brewster's law, μ = tan iₚ = tan 57° ≈ 1.54.

3 marks: “State and derive Brewster's law.” At the polarising angle iₚ the reflected light is completely plane polarised and μ = tan iₚ. At iₚ the reflected and refracted rays are at 90°, so r = 90° − iₚ; Snell's law gives μ = sin iₚ / sin(90° − iₚ) = sin iₚ / cos iₚ = tan iₚ. Diagram with the 90° marked.

2 marks: “Why do Polaroid sunglasses reduce the glare from a wet road?” Light reflected from the road is largely plane polarised, with vibrations parallel to the road. The Polaroids in the glasses have vertical transmission axes, so they absorb these horizontal vibrations.

Remember it

Brewster: μ = tan iₚ, and reflected ⟂ refracted.