Polarization of Light — Brewster's Law

Polarization of Light — Brewster’s Law

Polarization is one of those topics that confuses students until they realise it’s about direction, not intensity. Light is a transverse wave with the electric field oscillating perpendicular to the direction of travel — and “polarization” simply asks: in which perpendicular direction?

For NEET, polarization is a steady $4$ -mark area. For JEE Main, expect $1$ direct question. Brewster’s law is the most-asked single result of the entire wave optics chapter, so we’ll spend serious time on it.

Key Terms & Definitions

Unpolarized light: Light in which the electric field vector oscillates in all directions perpendicular to the direction of propagation, rapidly and randomly. Sunlight and bulb light are unpolarized.

Plane-polarized (linearly polarized) light: Light in which the electric field oscillates in a single fixed plane.

Polarizer: A device (e.g., Polaroid) that transmits light with $\vec{E}$ along its transmission axis and blocks the component perpendicular to it.

Brewster’s angle ( $\theta_B$ ): The angle of incidence at which reflected light is completely polarized perpendicular to the plane of incidence.

Polarization by reflection, scattering, refraction, double refraction: Four standard mechanisms — be ready to name examples for each.

Malus’s Law

When a polarized beam of intensity $I_0$ passes through a second polarizer (called the analyzer) whose transmission axis makes angle $\theta$ with the first, the transmitted intensity is:

I = I_0 \cos^2\theta

For unpolarized light passing through a single polarizer, the transmitted intensity is $I_0/2$ (averaged over all orientations).

Example: Unpolarized light of intensity $I_u$ passes through two polarizers with their axes at $30°$ . The first reduces intensity to $I_u/2$ . The second applies Malus’s law: $I = (I_u/2)\cos^2 30° = (I_u/2)(3/4) = 3I_u/8$ .

Polarization by Reflection — Brewster’s Law

When unpolarized light reflects off a transparent surface (water, glass), the reflected light is partially polarized in general. But at one special angle of incidence — Brewster’s angle — the reflected light is completely polarized, and the reflected and refracted rays are perpendicular to each other.

\tan\theta_B = n

where $n$ is the refractive index of the second medium relative to the first.

Why It Works

At Brewster’s angle, the reflected and refracted rays are exactly $90°$ apart. The component of $\vec{E}$ in the plane of incidence cannot radiate along its own direction of oscillation — so it doesn’t get reflected. Only the component perpendicular to the plane of incidence reflects.

Numerical example: For air-glass interface ( $n = 1.5$ ):

\tan\theta_B = 1.5 \implies \theta_B = \tan^{-1}(1.5) \approx 56.3°

For air-water ( $n = 1.33$ ): $\theta_B \approx 53°$ . This is exactly the angle at which photographers use polarizing filters to cut surface glare from water and glass.

Polarization by Scattering

Sunlight scattered by atmospheric molecules at $90°$ from the original direction is plane-polarized. This is why the blue sky is partially polarized — and why looking through Polaroid sunglasses while rotating them makes parts of the sky alternately bright and dark.

Polarization by Double Refraction

Some crystals (calcite, quartz) split an unpolarized ray into two: the ordinary ray (O-ray) that obeys Snell’s law and the extraordinary ray (E-ray) that doesn’t. The two rays are polarized perpendicular to each other.

This is the principle behind Nicol prisms and Polaroid sheets: physically separate the two rays, block one, transmit the other.

Worked Examples

Example 1 (CBSE) — Malus’s Law in Sequence

Unpolarized light of intensity $32 \text{ W/m}^2$ passes through three polarizers. The transmission axes of consecutive polarizers are at $30°$ to each other. Find the final transmitted intensity.

Unpolarized → polarized at half intensity:

I_1 = \tfrac{32}{2} = 16 \text{ W/m}^2

I_2 = 16 \cos^2 30° = 16 \times \tfrac{3}{4} = 12 \text{ W/m}^2

I_3 = 12 \cos^2 30° = 12 \times \tfrac{3}{4} = 9 \text{ W/m}^2

Final intensity: $\mathbf{9 \text{ W/m}^2}$ .

Example 2 (JEE Main) — Brewster + Snell

Light is incident on a glass slab at Brewster’s angle. The refractive index of glass is $\sqrt{3}$ . Find $\theta_B$ and the angle of refraction.

\tan\theta_B = \sqrt{3} \implies \theta_B = 60°

At Brewster’s angle, reflected and refracted rays are perpendicular: $\theta_B + r = 90°$ , so $r = 30°$ .

$\sin 60° / \sin 30° = (\sqrt{3}/2)/(1/2) = \sqrt{3} \checkmark$ .

Example 3 (JEE Advanced) — Crossed Polaroids with Insertion

Two crossed polaroids transmit zero intensity. A third polaroid is inserted between them with its axis at $45°$ to the first. Find the transmitted intensity in terms of the unpolarized input intensity $I_0$ .

$I_1 = I_0/2$ .

$I_3 = I_1 \cos^2 45° = (I_0/2)(1/2) = I_0/4$ .

$I_2 = I_3 \cos^2 45° = I_0/8$ .

So inserting a polariser between crossed polarisers actually increases transmission from zero to $I_0/8$ — a counter-intuitive result that JEE loves.

Exam-Specific Tips

NEET: $1$ MCQ from this chapter is almost guaranteed. Most common: applying Malus’s law twice, or finding $\theta_B$ given $n$ . Memorise $\tan\theta_B = n$ and the half-intensity rule for unpolarized light.

JEE Main 2023 Shift 1 had a question on three polaroids in sequence — exactly Example 1’s pattern. Practise the $\cos^2\theta$ chain so it’s automatic.

CBSE Class 12: Boards usually ask: state Brewster’s law, derive the relation $\theta_B + r = 90°$ , and solve a numerical. Worth $3$ – $5$ marks.

Common Mistakes

Mistake 1: Applying Malus’s law to unpolarized light directly. Unpolarized light through a polariser becomes polarised at half intensity — first halve, then apply $\cos^2\theta$ for subsequent polarisers.

Mistake 2: Confusing the polarization direction. The reflected ray at Brewster’s angle is polarised perpendicular to the plane of incidence, and the refracted ray is partially polarised in the plane. Reading the wrong direction loses one mark in NEET.

Mistake 3: Forgetting that Brewster’s law uses the relative refractive index. For light going from medium $1$ to medium $2$ , $\tan\theta_B = n_2/n_1$ , not the absolute index of either medium.

Mistake 4: Treating polarization as an intensity reduction, not a direction restriction. Unpolarized light of intensity $I$ contains all directions; a polariser keeps half the energy because only the component along its axis transmits — not because the polariser is “absorbing 50%”.

Mistake 5: Forgetting that the angle in Malus’s law is measured between transmission axes — not between $\vec{E}$ and the axis.

Practice Questions

At what angle should two polarizers be set so that an unpolarized beam passes with one-quarter of its original intensity?
The refractive index of a medium is $\tan 60°$ . Find Brewster’s angle for light going from air into this medium.
Why is sky light along the horizon less polarized than at $90°$ from the sun?
Two polaroids are set at $30°$ to each other. Polarized light of intensity $I_0$ falls on the first along its transmission axis. Find the output.
A polaroid passes $40\%$ of incident polarized light. Find the angle between the polarization direction and its axis.
State and prove that at Brewster’s angle, the reflected and refracted rays are mutually perpendicular.
Calculate the intensity reduction factor when unpolarized light passes through two polaroids whose axes are parallel.
The polarizing angle for a transparent material is $58°$ . Find its refractive index.

Answers:

$\cos^2\theta = 1/2 \implies \theta = 45°$ .
$\theta_B = \tan^{-1}(\tan 60°) = 60°$ .
Sun is roughly along the horizon at sunrise/sunset, so light from the horizon is in a forward-scattering geometry — minimum polarization.
$I_0\cos^2 30° = 3I_0/4$ .
$\cos^2\theta = 0.4 \implies \theta = \cos^{-1}\sqrt{0.4} \approx 50.8°$ .
From Brewster $\tan\theta_B = n = \sin\theta_B/\cos\theta_B$ . From Snell $n = \sin\theta_B/\sin r$ . Equating: $\sin r = \cos\theta_B = \sin(90° - \theta_B)$ , so $r = 90° - \theta_B$ , hence $\theta_B + r = 90°$ .
Halved overall — first polariser cuts unpolarized light to $I/2$ , second (parallel) transmits all of that.
$n = \tan 58° \approx 1.60$ .

Frequently Asked Questions

Q1: Can sound waves be polarized?

No. Polarization is a property of transverse waves only. Sound in air is a longitudinal wave (oscillation along the direction of propagation), so there’s no perpendicular direction to polarize.

Q2: Why do Polaroid sunglasses cut glare from water?

Light reflected off water at angles near Brewster’s angle is heavily polarised horizontally. Polaroid sunglasses are oriented to block horizontally polarised light, eliminating most of the surface glare while letting through vertically polarised light from objects.

Q3: What is the difference between linear, circular, and elliptical polarization?

Linear: $\vec{E}$ stays in one fixed plane. Circular: $\vec{E}$ rotates while keeping constant magnitude. Elliptical: $\vec{E}$ traces an ellipse. JEE Main mostly tests linear; NEET tests linear; JEE Advanced occasionally hints at circular through quarter-wave plate questions.

Q4: What is a half-wave plate?

A birefringent crystal cut to a thickness that introduces a $\pi$ phase difference between the O-ray and E-ray. Its effect on a linearly polarised input is to rotate the polarization plane by $2\theta$ where $\theta$ is the angle between input polarization and the plate’s optic axis.

Q5: Does total internal reflection produce polarised light?

Generally no — TIR keeps the polarization state intact (with a phase difference between components, leading to elliptical polarisation in some cases). Brewster’s effect needs partial reflection at a refracting surface.

Q6: Why is the sky blue and polarised?

Both effects come from Rayleigh scattering by air molecules. Blue light scatters more (Rayleigh’s $\lambda^{-4}$ dependence) and the scattered light at $90°$ from the sun is plane-polarised because only the components of $\vec{E}$ perpendicular to the scattering direction contribute.