Polarisation Masterclass · Vidya720
Physics · Class 12 Wave Optics · NEET

Polarisation, from a sunglasses stall at the lake to a NEET question

A full lesson: a cartoon movie to start, then a wave on a string through a slot, polaroids and Malus' law, chains of three polaroids, why the blue sky is polarised, Brewster's angle for reflected glare, NEET shortcuts, problems to solve yourself, and 12 MCQs.

NCERT Class 12 Physics Part II · Ch 10 Wave Optics · §10.7 Polarisation · pp. 269–271 · plus polarisation by scattering and by reflection (Brewster's law, NEET extension) · about 60 min

Voice
About 6 minutes. The animations start by themselves as the teacher explains them. Each stage's Listen button uses the same language.

Glare from water vibrates mostly sideways. A polaroid lets through only one direction of vibration, so it removes the glare. Two crossed polaroids block all light. That only makes sense if light is a transverse wave.

Stage 1 · The hook (5 min)

The sunglasses stall

Level

Priya and Arjun are at the lake on a sunny Sunday. The water throws so much glare into Priya's eyes that she cannot see the fish. Raju bhaiya at the stall hands her a pair of "polaroid" sunglasses for ₹299. She puts them on and the glare is gone, while the trees and the sky look almost the same, just a little darker.

Then he does a trick: he holds two lenses one behind the other and slowly turns one. The overlap gets darker and darker and at 90° becomes completely black. Arjun turns his phone while wearing the glasses: the screen goes dark too. Ordinary tinted glass never does this. What is special about these lenses, and what does it tell us about light?

Try at home: look at your phone or laptop screen through polarised sunglasses (or 3D-cinema glasses) and slowly tilt your head. Also look at the glare on a wet road or a table top and tilt the glasses.

Teacher says

"A polaroid is like a fence with vertical gaps. It lets through only the vibration that fits the gaps. If light can be blocked like this, its vibration must be sideways: light is a transverse wave."

Pause and predict

"Could you make 'polaroid earplugs' that block sound the same way?"

Answer (Stage 2 shows it)

No. Sound in air is longitudinal: the air moves back and forth along the direction of travel. There is no sideways direction to select, so sound cannot be polarised.

Stage 2 · The concept (10 min)

A wave on a string through a slot

Level

Shake the end of a long string. The wave y(x, t) = a sin(kx − ωt) travels along x, while each bit of string moves along y: a transverse wave. Because the vibration stays in the x–y plane, it is linearly (plane) polarised. Now thread the string through a slot in a board and turn the slot.

Polarised

vibration in one fixed plane

y-polarised: vibrates in the x–y plane. z-polarised: in the x–z plane. Also called linearly or plane polarised.

Unpolarised

direction changes randomly

Light from the sun or a bulb: the vibration direction changes about 10⁸ times a second, with no favourite.

Only transverse

longitudinal ⇒ no polarisation

Sound in air passes any slot unchanged. This is how polarisation proves light is transverse.

Teacher says

"For a light wave, the thing that vibrates is the electric field E. When we say 'the direction of polarisation', we mean the direction in which E oscillates."

Stage 3 · The law (10 min)

Polaroids and Malus' law: I = I₀ cos²θ

Level

A polaroid is a plastic sheet with long-chain molecules all lined up in one direction. Electric fields along the chains drive electrons along them and are absorbed. The component perpendicular to the chains passes: that direction is the pass-axis. Drag the orange analyser P₂ (or use the slider) to turn it.

  1. Unpolarised light through P₁

    I₁ = I₀ / 2

    The direction of E is random, and the average of cos²θ over all directions is ½. Turning P₁ changes nothing.

  2. Only the component along the pass-axis passes

    E = E₀ cos θ

    θ is the angle between the polarisation (P₁'s axis) and P₂'s pass-axis.

  3. Intensity ∝ (amplitude)²

    I = I₁ cos²θ  (Malus' law)

    θ = 0°: all passes. θ = 90° (crossed): nothing passes. Two dark positions per full turn of P₂.

Teacher says

"Malus' law is for light that is ALREADY polarised. For unpolarised light, the first polaroid always gives one half. Apply the half only once, then cos² for every polaroid after it."

Pause and predict

"Through which angle must the analyser be turned from the bright position to cut the light to one quarter?"

Answer

cos²θ = 1/4, so cos θ = 1/2 and θ = 60°.

Stage 4 · Maths (10 min)

Three polaroids: light through a "closed" door

Level

P₁ and P₃ are crossed, so no light gets through. Now slide a third polaroid P₂ between them at angle α. Surprisingly, light comes back! Turn P₂ and watch the intensity bars.

  1. After P₁

    I₁ = I₀/2

    The one and only "half".

  2. After P₂ at α

    I₂ = (I₀/2) cos²α

    Malus with the angle between P₁ and P₂.

  3. After P₃ at 90°

    I₃ = (I₀/2) cos²α cos²(90° − α) = (I₀/2) cos²α sin²α

    The angle between P₂ and P₃ is 90° − α.

  4. Use sin 2α = 2 sin α cos α

    I₃ = (I₀/8) sin²2α

    Maximum I₀/8 at α = 45°. Zero at α = 0° or 90° (P₂ lines up with P₁ or P₃).

Teacher says

"Each polaroid only cares about the polaroid just before it. Write the angle between neighbours, square the cosine, and multiply along the chain."

Stage 5 · The big idea (10 min)

Why the blue sky is polarised

Level

Syllabus note: polarisation by scattering is enrichment; it is not in the rationalised NCERT text.

Sunlight shakes the electrons in air molecules, and these re-radiate (scatter) light in all directions. An oscillating charge sends no light along its own line of vibration. Drag Priya around the molecule to change the scattering angle ψ, and turn her polaroid.

Sunlight in

E in the page (↔) and out of it (•)

Unpolarised: both directions of vibration, equally.

Seen at 90°

only • survives → plane polarised

The in-page oscillation points straight at the observer, and a charge does not radiate along its own line of motion.

Check it

polaroid + sky at 90° to the sun

Turning the polaroid makes that part of the sky darker and brighter. Bees and some birds navigate using this.

Teacher says

"Scattering explains two facts at once: the sky is blue because short wavelengths scatter much more, and it is polarised because of the 90° rule."

Stage 6 · Real life (5 min)

Glare and Brewster's angle: tan i_B = n

Level

Syllabus note: Brewster’s law is in the NEET syllabus but no longer in the rationalised NCERT text. Learn it for NEET.

When light reflects from water or glass, the vibration in the plane of incidence (blue arrows ↔) reflects less than the vibration perpendicular to it (red dots •). At one special angle, the Brewster angle, the ↔ part is not reflected at all: the reflected light is completely plane polarised. Drag the lamp or use the slider.

Brewster's law

tan i_B = n

Glass 56.3°, water 53.1°, diamond 67.5°. (Not the critical angle: that is sin i_c = 1/n.)

The 90° rule

i_B + r = 90°

Reflected and refracted rays are perpendicular. From Snell: sin i_B = n cos i_B ⇒ tan i_B = n.

Uses of polaroids

sunglasses · LCDs · 3D films · camera filters

Polaroid glasses have a vertical pass-axis, which blocks the horizontally vibrating glare from roads and water.

Stage 7 · NEET level (10 min)

Shortcuts and tricks

Level

The one-time half

unpolarised → ×½

Only at the first polaroid. Every later polaroid: × cos² (angle from the previous one).

Malus

I = I₀ cos²θ

cos²30° = 3/4, cos²45° = 1/2, cos²60° = 1/4, cos²90° = 0.

Crossed pair + middle

I = (I₀/8) sin²2α

Maximum I₀/8 at 45°. For a middle polaroid rotating at ω: output varies as sin²2ωt.

Amplitude vs intensity

E ∝ cos θ, I ∝ cos²θ

A question asking for the amplitude ratio wants cos θ, not cos²θ.

Brewster

tan i_B = n · i_B + r = 90°

If the refracted angle is given, i_B = 90° − r first, then n = tan i_B.

Scattering

at 90°: fully polarised

Along the beam (ψ = 0°) the scattered light is unpolarised.

Teacher says

"In polaroid chains, NEET hides the angle: they give the angle of each polaroid from the FIRST one. Always convert to the angle between neighbours before using cos squared."

Stage 8 · Solve it yourself

Five worked problems

Level

Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute.

Problem 1 · Malus' law

Plane polarised light of intensity 32 W/m² falls on an analyser. Find the transmitted intensity when the angle between the polarisation and the pass-axis is 30° and when it is 60°.

Hint

The light is already polarised: no factor ½. I = I₀ cos²θ.

Full solution
30°: I = 32 × cos²30° = 32 × 3/4 = 24 W/m²
60°: I = 32 × cos²60° = 32 × 1/4 = 8 W/m²
24 W/m² and 8 W/m²
Problem 2 · Two polaroids

Unpolarised light of intensity 64 W/m² passes through two polaroids whose pass-axes are at 60°. Find the final intensity.

Hint

½ at the first polaroid, then cos²60° at the second.

Full solution
after P₁: 64/2 = 32 W/m²
after P₂: 32 × cos²60° = 32 × 1/4 = 8 W/m²
8 W/m²
Problem 3 · Polaroid between crossed polaroids (NCERT style)

Unpolarised light of intensity 80 W/m² falls on two crossed polaroids. A third polaroid is placed between them with its pass-axis at 30° to the first. Find the final intensity. At what angle would it be largest, and how large?

Hint

Angles between neighbours: 30° and 60°. Or use (I₀/8) sin²2α.

Full solution
I = 80/2 × cos²30° × cos²60° = 40 × 3/4 × 1/4 = 7.5 W/m²
check: (80/8) sin²60° = 10 × 3/4 = 7.5 W/m²
largest at α = 45°: 80/8 = 10 W/m²
7.5 W/m²; maximum 10 W/m² at 45°
Problem 4 · Brewster angle for glass

Find the polarising (Brewster) angle for glass of refractive index 1.5 and the angle of refraction at this angle of incidence.

Hint

tan i_B = n, and i_B + r = 90°.

Full solution
i_B = tan⁻¹ 1.5 = 56.3°
r = 90° − 56.3° = 33.7°
(check with Snell: sin 56.3° / sin 33.7° = 0.832 / 0.555 = 1.5)
i_B ≈ 56.3°, r ≈ 33.7°
Problem 5 · Glare from a lake

At what height of the sun above the horizon is the sunlight reflected from a calm lake (n = 4/3) completely polarised?

Hint

i_B is measured from the vertical normal; the sun's height is measured from the horizontal.

Full solution
i_B = tan⁻¹(4/3) = 53.1° from the vertical
height above horizon = 90° − 53.1° = 36.9°
When the sun is about 37° above the horizon
Stage 9 · Practice ladder

12 MCQs, from easy to NEET-level

Level

First decide: is the incoming light polarised or unpolarised? Then click.

Score: 0 / 12

Stage 10 · Exam corner

How NEET asks this

If the question says…Write this firstThe trap
"unpolarised light through a polaroid"I = I₀/2Using cos²θ for unpolarised light
"plane polarised light", "analyser at θ"I = I₀ cos²θAlso dividing by 2
"two crossed polaroids"I = 0Answering I₀/2 or I₀/4
"third polaroid between crossed ones"(I₀/8) sin²2α, max I₀/8Using the angle from P₁ for both steps
"polaroid rotated with angular speed ω"θ = ωt, I ∝ cos²ωt (or sin²2ωt between crossed)Forgetting the square
"polarising angle", "Brewster angle"tan i_B = nUsing sin i = 1/n (critical angle)
"reflected and refracted rays"perpendicular at i_B: i_B + r = 90°Answering 180° or 0°
"which waves cannot be polarised"longitudinal (sound in air)Picking radio or X-rays (they are EM, transverse)
"sky light at 90° to the sun"plane polarisedSaying unpolarised because sunlight is

Mistakes to avoid

Where this connects next

This completes Wave Optics: Huygens wavefronts, interference, diffraction and polarisation all treat light as a transverse electromagnetic wave. The next chapter, Dual Nature of Radiation and Matter, shows where the wave picture fails and light behaves like particles (photons).

Values used: n(water) = 4/3 ≈ 1.33, n(glass) = 1.5, n(diamond) = 2.42; cos²30° = 3/4, cos²45° = 1/2, cos²60° = 1/4.

0:00 / 0:00

Well done. What next?

Follow the four steps: understand it, remember it, try it, then prove it with past papers.

Try it: the Chapter Challenge ›✍️ Practise this lesson: Wave Optics Challenge Q21–23: polarisationRemember it: 5-Minute Revise‹ Previous: Lesson 5: Single-slit diffraction‹ Back to chapter🏠 Physics Part 2 home