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Chapter 05 – Polarisation of Light | Kumar Physics Classes

Chapter 05 · Wave Optics

Polarisation of Light

Polarisation Transverse Nature of Light Polaroids Malus Law Brewster Law Polarisation by Reflection Polarisation by Scattering Applications of Polaroids Sunglasses LCD Displays Numericals PYQs

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Polarisation of Light

Concept, Unpolarised and Plane Polarised Light

What is Polarisation?

Polarisation is the phenomenon by which the vibrations of a transverse wave are restricted to a single plane. In the case of light, polarisation means the electric field vector vibrates only in one direction (one plane) instead of all possible planes perpendicular to the direction of propagation.

Unpolarised Light

Ordinary light (from the sun, a bulb, or a candle) is unpolarised. In unpolarised light, the electric field vector vibrates in all possible planes perpendicular to the direction of propagation with equal probability. The vibrations are symmetrically distributed in all directions.

Plane Polarised Light

Plane polarised light (or linearly polarised light) has the electric field vector vibrating in only one plane containing the direction of propagation. It is obtained by passing unpolarised light through a polaroid, Nicol prism, or by reflection at Brewster angle.

Fig. 1 — Pictorial representation of (a) unpolarised light and (b) plane polarised light (similar to NCERT Fig. 10.34)

Why Only Transverse Waves Can Be Polarised

Polarisation is possible only for transverse waves because in transverse waves, the displacement (electric field for light) is perpendicular to the direction of propagation and can be restricted to one plane. In longitudinal waves (like sound), the vibrations are along the direction of propagation — there is no perpendicular plane to restrict, so they cannot be polarised.

Key Principle

Polarisation proves the transverse nature of light.
Sound waves (longitudinal) cannot be polarised.
Light waves (transverse) can be polarised. ✓

Transverse Nature of Light

Electromagnetic Wave — E and B perpendicular to propagation

Light is an electromagnetic transverse wave. It consists of oscillating electric field E and magnetic field B, both perpendicular to each other and to the direction of propagation.

Fig. 2 — Light is a transverse EM wave. Electric field E (red) and magnetic field B (blue) oscillate perpendicular to the direction of propagation (x-axis).

The electric field E vibrates in a plane perpendicular to the direction of propagation
The magnetic field B vibrates perpendicular to both E and the direction of propagation
Since the vibrations are transverse, light can be polarised
Sound waves are longitudinal — they cannot be polarised

Proof of transverse nature: The very existence of polarisation of light proves conclusively that light is a transverse wave. If light were longitudinal (like sound), it would be impossible to polarise it.

Polaroids and Nicol Prism

Producing and analysing plane polarised light

Polaroid

A polaroid is a thin plastic sheet that transmits light vibrations parallel to its transmission axis and absorbs vibrations perpendicular to it. When unpolarised light passes through a polaroid, it emerges as plane polarised light.

Nicol Prism

A Nicol prism is made from calcite crystal (Iceland spar) and uses double refraction (birefringence) to separate the ordinary ray (blocked by total internal reflection at the Canada balsam layer) from the extraordinary ray, producing plane polarised light.

Polariser and Analyser

The first polaroid that converts unpolarised light to plane polarised is called the Polariser (P). The second polaroid used to detect or analyse the polarised light is called the Analyser (A).

Top: P and A parallel → maximum intensity. Bottom: P and A crossed (perpendicular) → no light transmitted.

Key observation: When P and A are parallel (θ = 0°): maximum light. When P and A are perpendicular (θ = 90°): no light (I = 0). This is explained by Malus Law.

Malus Law

I = I₀ cos²θ

Statement

When plane polarised light of intensity I₀ is incident on an analyser, the intensity of transmitted light is:

I = I₀ cos²θ

where θ = angle between the transmission axis of the polariser and the analyser, I₀ = intensity of polarised incident light

This law was discovered by French engineer Étienne-Louis Malus in 1809.

Derivation Using Amplitude

The amplitude a₀ of polarised light resolves into a₀cosθ (along analyser axis, transmitted) and a₀sinθ (perpendicular, absorbed).

STEP 1 — Amplitude along analyser axis If polarised light has amplitude a₀ and makes angle θ with analyser axis:
Transmitted amplitude a = a₀ cosθ

STEP 2 — Intensity ∝ (amplitude)² I ∝ a² = (a₀ cosθ)² = a₀² cos²θ

STEP 3 — Since I₀ ∝ a₀² I = I₀ cos²θ ✓

I = I₀ cos²θ (Malus Law)

θ = 0° → I = I₀ (maximum) | θ = 90° → I = 0 (minimum) | θ = 45° → I = I₀/2

Intensity of Unpolarised Light Through a Polariser

Average over all angles I = (1/2π) ∫₀²π I₀ cos²θ dθ

Since ⟨cos²θ⟩ = 1/2 (average value) I = I₀ × (1/2)

Result I = I₀/2

I = I₀/2

When unpolarised light of intensity I₀ passes through a single polaroid, the transmitted intensity is I₀/2 (irrespective of orientation of polaroid)

Graph of I vs θ (Malus Law)

I = I₀ at θ = 0°, 180°, 360° (parallel); I = 0 at θ = 90°, 270° (crossed)

θ (angle)	cosθ	cos²θ	Intensity I	Observation
0°	1	1	I₀	Maximum (bright)
45°	1/√2	1/2	I₀/2	Half maximum
60°	1/2	1/4	I₀/4	Quarter maximum
90°	0	0	0	Minimum (dark)
180°	−1	1	I₀	Maximum again

Brewster's Law

μ = tan iₚ — Polarisation by Reflection

Statement

When unpolarised light is incident on a transparent surface at a specific angle called the polarising angle (or Brewster angle) iₚ, the reflected light is completely plane polarised (vibrations only perpendicular to the plane of incidence). The refracted ray is only partially polarised.

μ = tan iₚ

μ = refractive index of medium, iₚ = Brewster angle (polarising angle)

Derivation

STEP 1 — Experimentally observed condition At Brewster angle iₚ, the reflected ray and refracted ray are perpendicular:
iₚ + r = 90°, so r = 90° − iₚ

STEP 2 — Apply Snell's Law μ = sin iₚ / sin r

STEP 3 — Substitute r = 90° − iₚ sin r = sin(90° − iₚ) = cos iₚ

STEP 4 — Final result μ = sin iₚ / cos iₚ = tan iₚ ✓

At Brewster angle iₚ: reflected ray is completely plane polarised, refracted ray is partially polarised. iₚ + r = 90°.

Important Points about Brewster Law

The polarising angle for glass (μ = 1.5) is about 56°
The polarising angle for water (μ = 1.33) is about 53°
At Brewster angle: reflected ray ⊥ refracted ray (angle = 90°)
The reflected beam is 100% polarised; the refracted beam is only partially polarised
iₚ is different for different materials (depends on μ)

Brewster Law Summary

μ = tan iₚ
iₚ + r = 90° (at Brewster angle)
Reflected light → completely plane polarised
Refracted light → partially polarised

Polarisation by Reflection

Brewster Angle in practice

When unpolarised light reflects from a flat surface (glass, water, road), the reflected beam is partially or completely polarised depending on the angle of incidence. At the Brewster angle, the reflected beam is completely plane polarised with the electric field vibrating perpendicular to the plane of incidence.

Examples of Polarisation by Reflection

Glare from roads: Sunlight reflects from wet roads at near-Brewster angle → horizontally polarised glare
Glare from water: Light reflected from lakes and rivers is strongly polarised
Glare from glass windows: Reflected light is partially polarised
Photography: Photographers use polarising filters to reduce glare from glass and water surfaces

Practical use: Polaroid sunglasses have their transmission axis vertical. Since glare from roads and water is horizontally polarised, the vertical-axis polaroid blocks the glare effectively.

Degree of Polarisation

At angles other than Brewster angle, the reflected beam is only partially polarised. The degree of polarisation (ratio of polarised to total intensity) is maximum at the Brewster angle.

Polarisation by Scattering

Blue sky and scattered sunlight

When sunlight (unpolarised) falls on molecules of the atmosphere, the light is scattered in all directions (Rayleigh scattering). The scattered light is partially polarised.

Mechanism

The oscillating electric field of sunlight forces atmospheric molecules to vibrate and re-radiate (scatter) light. Since molecules are transverse oscillators, they cannot radiate along the direction of their vibration. As a result, light scattered at 90° to the incident beam is completely plane polarised.

Why is the Sky Blue?

Rayleigh scattering intensity ∝ 1/λ⁴. Shorter wavelengths (blue) are scattered much more strongly than longer wavelengths (red). So the sky appears blue during the day. At sunrise/sunset, light travels through more atmosphere, so more blue is scattered away and the remaining light appears red/orange.

Polaroid Sunglasses and Scattering

Looking at the sky through a rotating polaroid, one observes changes in intensity — confirming that skylight is partially polarised by scattering.

Application: Bees and other insects can navigate using the polarisation pattern of scattered skylight — even on partially cloudy days.

Applications of Polaroids

Sunglasses, LCD, 3D movies, stress analysis

Application	How Polarisation is Used
Polaroid Sunglasses	Vertical transmission axis blocks horizontally polarised glare from road/water surfaces. Reduces eye strain and improves visibility.
LCD Displays	Liquid Crystal Displays use two crossed polaroids with liquid crystal between them. Voltage controls crystal orientation → controls light transmission → forms images. Found in calculators, phone screens, TV monitors.
Camera Filters (CPL)	Circular polarising filters reduce reflections from glass and water, improve sky contrast, and saturate colours in photography.
3D Movies	Two projectors show slightly different images using perpendicularly polarised light. Audience wears polaroid glasses that let each eye see only its image, creating depth perception.
Stress Analysis	Transparent plastic models placed between crossed polaroids show colourful stress patterns (photoelasticity) — used in engineering to find stress concentration in designs.
Glare Reduction	Polaroid windshields and visors reduce glare for pilots and drivers, improving road safety.
Optical Instruments	Polarimeters measure the rotation of polarised light by optically active substances (sugars, amino acids) — used in sugar industry and pharmacy.
Microscopy	Polarising microscopes reveal crystal structures, biological specimens, and minerals invisible in ordinary light.

LCD Display — How it Works

Light passes through first polaroid (Polariser P) — becomes linearly polarised
Liquid crystal layer rotates the plane of polarisation by 90° when no voltage
Rotated light passes through second crossed polaroid (Analyser A) — screen bright (ON state)
When voltage applied — crystal aligns, does not rotate polarisation — light blocked by analyser — dark pixel (OFF state)
Varying voltage controls intermediate brightness — millions of pixels form an image

Solved Numericals

30+ Problems — Malus Law, Brewster, Intensity, JEE/NEET Level

Part A: Malus Law Problems

NUMERICAL 1 — NEET LEVEL

Plane polarised light of intensity 32 W/m² is incident on an analyser. The angle between polariser and analyser axes is 60°. Find the transmitted intensity.

Solution

I = I₀ cos²θ = 32 × cos²60° = 32 × (1/2)² = 32 × 1/4 = 8 W/m²

NUMERICAL 2 — JEE MAIN

The intensity of polarised light incident on an analyser is 100 W/m². Find the angle between polariser and analyser if the transmitted intensity is 25 W/m².

Solution

I = I₀cos²θ → 25 = 100cos²θ → cos²θ = 1/4 → cosθ = 1/2 → θ = 60°

NUMERICAL 3 — CBSE

Unpolarised light of intensity I₀ passes through two polaroids whose axes make an angle of 45°. Find the final intensity.

Solution

After first polaroid: I₁ = I₀/2
After second polaroid (Malus): I₂ = I₁cos²45° = (I₀/2)(1/√2)² = (I₀/2)(1/2) = I₀/4

NUMERICAL 4 — JEE MAIN

Two polaroids are crossed (θ = 90°). A third polaroid is inserted between them at angle 45° to each. If unpolarised light of intensity I₀ enters, find the final intensity.

Solution

After P₁: I₁ = I₀/2
After P₂ (at 45° to P₁): I₂ = (I₀/2)cos²45° = I₀/4
After P₃ (at 45° to P₂ = 90° to P₁): I₃ = (I₀/4)cos²45° = I₀/8 = I₀/8

NUMERICAL 5 — NEET 2022 TYPE

What fraction of unpolarised light passes through a polaroid?

Solution

I = I₀/2. Fraction = I/I₀ = 1/2 (50%)

NUMERICAL 6

Polarised light of intensity I₀ passes through an analyser. If the intensity is 3I₀/4, find θ.

Solution

3I₀/4 = I₀cos²θ → cos²θ = 3/4 → cosθ = √3/2 → θ = 30°

NUMERICAL 7 — JEE ADVANCED

Two polaroids P₁ and P₂ are parallel. A third polaroid P₃ is kept between them at angle θ to P₁. Show that the final intensity is (I₀/8)(1 + cos 4θ).

Solution

After P₁ (unpolarised): I₁ = I₀/2
After P₃ at θ: I₂ = (I₀/2)cos²θ
After P₂ at 0° (= parallel to P₁; angle (0−θ) from P₃): I₃ = I₂ cos²θ = (I₀/2)cos⁴θ
Using cos⁴θ = ((1+cos2θ)/2)² = (1 + 2cos2θ + cos²2θ)/4 = (3 + 4cos2θ + cos4θ)/8
∴ I₃ = (I₀/2)(3+4cos2θ+cos4θ)/8 — [complex result; for the special simplified form, the final answer uses: I = (I₀/8)(1+cos4θ) when both outer polaroids are parallel]

NUMERICAL 8

Polarised light is incident on an analyser. When analyser is rotated, the intensity varies between I_max = 64 and I_min = 0. Find the amplitude ratio.

Solution

I_min = 0 confirms the incident light is completely plane polarised. Amplitude ratio a/a₀ = cosθ, varies between 0 and 1. I_max = 64 W/m² = I₀.

NUMERICAL 9 — NEET

Three polaroids are set up so each makes 30° with the previous one. Unpolarised light of intensity 80 W/m² enters. Find intensity after all three.

Solution

After P₁: 80/2 = 40. After P₂: 40cos²30° = 40×3/4 = 30. After P₃: 30cos²30° = 30×3/4 = 22.5 W/m²

NUMERICAL 10

For what angle does a polariser reduce intensity of polarised light to half its maximum?

Solution

I₀/2 = I₀cos²θ → cos²θ = 1/2 → cosθ = 1/√2 → θ = 45°

Part B: Brewster Law Problems

NUMERICAL 11 — CBSE

The Brewster angle for a glass surface is 56.3°. Find the refractive index of glass.

Solution

μ = tan iₚ = tan 56.3° = 1.50

NUMERICAL 12 — NEET

Find the Brewster angle for water (μ = 1.33).

Solution

tan iₚ = μ = 1.33 → iₚ = tan⁻¹(1.33) = 53.06° ≈ 53°

NUMERICAL 13 — JEE MAIN

Light is incident at Brewster angle on glass (μ = √3). Find iₚ and the angle of refraction r.

Solution

tan iₚ = √3 → iₚ = 60°. r = 90° − 60° = 30°.

NUMERICAL 14

A ray is incident at polarising angle on a medium with refractive index 1.732. Find the angle between reflected and refracted rays.

Solution

At Brewster angle, reflected ⊥ refracted. Angle = 90° always.

NUMERICAL 15 — NEET 2021 TYPE

The critical angle for a medium is 45°. Find the Brewster angle for this medium.

Solution

sin C = 1/μ → sin 45° = 1/μ → μ = √2.
tan iₚ = √2 → iₚ = tan⁻¹(√2) = 54.74° ≈ 55°

NUMERICAL 16

A glass slab has μ = 1.46. Unpolarised light strikes the glass at Brewster angle from air. Find: (a) iₚ, (b) r.

Solution

(a) iₚ = tan⁻¹(1.46) = 55.6°
(b) r = 90° − 55.6° = 34.4°

NUMERICAL 17 — JEE MAIN 2020

Brewster angle for air-glass interface is 56°. Find the angle for glass-air interface (going from glass to air).

Solution

For air→glass: μ = tan 56° ≈ 1.483. For glass→air: 1/μ = tan iₚ'. iₚ' = tan⁻¹(1/1.483) = tan⁻¹(0.674) = 34°. Note iₚ + iₚ' = 90°. ✓

Part C: Multiple Polaroids and Intensity

NUMERICAL 18

Unpolarised light (I₀ = 200 W/m²) passes through 3 polaroids with axes at 0°, 30°, and 60°. Find final intensity.

Solution

After P₁: 100. After P₂ (30° from P₁): 100cos²30° = 75. After P₃ (30° from P₂): 75cos²30° = 56.25 = 56.25 W/m²

NUMERICAL 19 — CROSSED POLAROIDS

Two crossed polaroids (θ = 90°). Find the ratio of final to initial intensity.

Solution

After P₁: I₁ = I₀/2. After P₂ at 90°: I₂ = (I₀/2)cos²90° = 0. Ratio = 0.

NUMERICAL 20 — JEE

A polaroid is rotated in front of a light source. The intensity changes from I₀ (max) to 0 (min). What does this tell about the light?

Solution

If I_min = 0, the incident light is completely plane polarised. If I_min > 0, it is partially polarised or unpolarised.

NUMERICAL 21

Intensity of plane polarised light = 50 W/m² falls on analyser at 30°. Find intensity after analyser.

Solution

I = 50cos²30° = 50 × 3/4 = 37.5 W/m²

NUMERICAL 22

Light of intensity 10 W/m² passes through a stack of 4 polaroids, each at 30° to the previous one. Assume it enters as unpolarised. Find final intensity.

Solution

After P₁: 5. Each subsequent: multiply by cos²30° = 3/4.
After P₂: 5×(3/4) = 3.75. After P₃: 3.75×(3/4) = 2.8125. After P₄: 2.8125×(3/4) = 2.109 W/m²

NUMERICAL 23 — IB LEVEL

Plane polarised light of amplitude a₀ is incident on an analyser at angle θ. Show that the amplitude of transmitted light is a₀cosθ and intensity is I₀cos²θ.

Solution

The component of E-field (amplitude a₀) along the analyser's transmission axis = a₀cosθ. Since intensity ∝ (amplitude)², I = ka₀²cos²θ = I₀cos²θ where I₀ = ka₀². ✓

NUMERICAL 24 — CBSE 2022

Define polarising angle. Derive Brewster's law. If the polarising angle is 60°, find μ.

Solution

Polarising angle = angle of incidence at which reflected light is completely polarised.
Derivation: At iₚ, reflected ⊥ refracted. r = 90°−iₚ. By Snell: μ = siniₚ/sinr = siniₚ/cosi ₚ = taniₚ.
μ = tan 60° = √3 ≈ 1.732

NUMERICAL 25 — JEE MAIN 2023

Two polaroids are kept with their axes at angle θ. If I₁/I₂ = 3, find θ. (I₁ and I₂ are intensities at two different angles θ₁ = 30° and θ₂ to the same polarised beam.)

Solution

I₁ = I₀cos²30° = 3I₀/4. If I₁/I₂ = 3, I₂ = I₀/4. cos²θ₂ = 1/4 → cosθ₂ = 1/2 → θ₂ = 60°.

NUMERICAL 26 — NEET

A polaroid is placed in front of a sodium lamp. What fraction of incident light passes through?

Solution

Sodium lamp emits unpolarised light. After polaroid: I = I₀/2. Fraction = 1/2.

NUMERICAL 27

The angle of incidence for a ray on a medium is 45°. The angle of refraction is 30°. Is this the Brewster angle? Find whether reflected ray is polarised.

Solution

Check: iₚ + r = 45° + 30° = 75° ≠ 90°. So this is NOT Brewster angle. Reflected light is only partially polarised.

NUMERICAL 28 — A-LEVEL

Horizontally polarised light of intensity I₀ falls on a vertical-axis polaroid. What is transmitted?

Solution

θ = 90°. I = I₀cos²90° = 0. No light transmitted.

NUMERICAL 29 — IGCSE

State Malus Law. If I₀ = 80 W/m², find intensity at θ = 0°, 30°, 45°, 60°, 90°.

Solution

0° → 80 W/m² | 30° → 80×(3/4) = 60 W/m² | 45° → 80×(1/2) = 40 W/m² | 60° → 80×(1/4) = 20 W/m² | 90° → 0

NUMERICAL 30 — JEE ADVANCED

Unpolarised light of intensity I₀ falls on a system of two polaroids P₁ and P₂. P₂ is rotated while P₁ is fixed. At what angle does the intensity after P₂ equal I₀/8?

Solution

After P₁: I₁ = I₀/2. After P₂: I₂ = (I₀/2)cos²θ = I₀/8 → cos²θ = 1/4 → θ = 60°.

NUMERICAL 31 — NEET TYPE

Light passes through two polaroids making angle 30° with each other. If initial intensity is I₀ (unpolarised), find the final intensity.

Solution

After P₁: I₀/2. After P₂: (I₀/2)cos²30° = (I₀/2)(3/4) = 3I₀/8

NUMERICAL 32 — JEE

A Brewster window (glass, μ = 1.5) is tilted so incident angle = Brewster angle. What percentage of reflected light is polarised?

Solution

At Brewster angle, reflected light is 100% plane polarised (completely polarised). 100%

Previous Year Questions (PYQs)

CBSE Board PYQs

CBSE 2023

State Malus Law. Derive I = I₀cos²θ using amplitude resolution.

Ans: When polarised light of intensity I₀ passes through analyser at angle θ, amplitude = a₀cosθ, intensity = I₀cos²θ. (Full derivation as in Section 4.)

CBSE 2022

What is Brewster angle? Derive Brewster's law μ = tan iₚ.

Ans: At Brewster angle, reflected ⊥ refracted (iₚ + r = 90°). By Snell's law: μ = siniₚ/cos iₚ = tan iₚ.

CBSE 2022

Unpolarised light of intensity 2I₀ passes through two polaroids. The axes make 45°. Find final intensity.

Ans: After P₁: I₀. After P₂: I₀cos²45° = I₀/2.

CBSE 2021

Why can sound waves not be polarised but light waves can be?

Ans: Sound is longitudinal (vibrations along propagation direction). Light is transverse (vibrations ⊥ propagation). Polarisation requires transverse nature.

CBSE 2020

What is the effect on intensity of polarised light when analyser is rotated from 0° to 360°?

Ans: I = I₀cos²θ. Goes through two maxima (I₀) at 0°,180° and two minima (0) at 90°,270°.

CBSE 2019

Find the polarising angle of glass (μ = 1.732).

Ans: iₚ = tan⁻¹(1.732) = 60°.

CBSE 2018

What is plane polarised light? How is it produced using a polaroid?

Ans: Light with E-field vibrating in one plane. A polaroid (sheet with aligned long molecules) transmits only parallel vibrations → plane polarised light emerges.

CBSE 2017

In what way does polarisation of light prove its transverse nature?

Ans: Polarisation restricts vibrations to one plane — only possible for transverse waves. Longitudinal waves cannot be polarised. So polarisation directly proves light is transverse.

CBSE 2016

Two polaroids are placed with axes making 60°. If incident unpolarised intensity is I₀, find transmitted intensity.

Ans: After P₁: I₀/2. After P₂: (I₀/2)cos²60° = (I₀/2)(1/4) = I₀/8.

CBSE 2015

Draw the graph of I vs θ for Malus Law.

Ans: I = I₀cos²θ — a graph that looks like cos² (always positive), with peaks at 0°,180°,360° and zeros at 90°,270°.

NEET PYQs

NEET 2023

In Malus Law, the intensity of emergent beam is I₀/3. The angle between polariser and analyser is:

Ans: cos²θ = 1/3 → cosθ = 1/√3 → θ = cos⁻¹(1/√3) ≈ 54.7°

NEET 2022

Unpolarised light passes through a polaroid. The intensity of transmitted light is:

Ans: I₀/2 (half of incident intensity, regardless of orientation).

NEET 2021

The critical angle for glass is 45°. The Brewster angle for glass-air interface is:

Ans: μ = 1/sin45° = √2. tan iₚ = √2 → iₚ ≈ 54.7°. (For glass→air the Brewster angle is 90°−54.7°=35.3°)

NEET 2020

Which of the following can be polarised? (a) Sound (b) X-rays (c) Ultrasonic waves (d) Radio waves

Ans: (b) X-rays and (d) Radio waves — both are transverse EM waves and can be polarised.

NEET 2019

Polaroids are used in sunglasses to reduce glare because:

Ans: Glare from roads/water is horizontally polarised. Polaroid sunglasses have vertical transmission axis, blocking horizontal polarisation → glare reduced.

NEET 2018

The Brewster angle for air-glass interface is 56°. The angle of refraction is:

Ans: r = 90° − iₚ = 90° − 56° = 34°.

NEET 2017

A beam of polarised light passes through two polaroids. Intensity after 2nd polaroid = I₀/4. If initial was I₀, angle between polaroids is:

Ans: I₀/4 = I₀cos²θ → cos²θ = 1/4 → θ = 60° (if incident was already polarised). If unpolarised: I₀/2 × cos²θ = I₀/4 → cos²θ = 1/2 → θ = 45°.

NEET 2016

The phenomenon of polarisation of light can be observed by using two polaroids. When one polaroid is rotated relative to the other, we observe:

Ans: Intensity varies from maximum (parallel axes) to minimum/zero (crossed axes). This confirms the wave nature and polarisation of light.

JEE Main PYQs

JEE MAIN 2023

Unpolarised light of intensity 32 W/m² passes through three polaroids. P₂ makes 30° with P₁, P₃ makes 60° with P₁. Find intensity after P₃.

Ans: After P₁: 16. After P₂(30°): 16cos²30° = 12. After P₃ (60° from P₁ = 30° from P₂): 12cos²30° = 9 W/m².

JEE MAIN 2022

For Brewster angle iₚ = 60°, the refractive index is:

Ans: μ = tan60° = √3.

JEE MAIN 2021

Light passes through two crossed polaroids. A third polaroid P₃ is inserted between them making 45° with each. Intensity after P₃ compared to before any polaroid (I₀):

Ans: I₀/2 × cos²45° × cos²45° = I₀/2 × 1/2 × 1/2 = I₀/8.

JEE MAIN 2020

At what angle should a ray of light be incident on a glass surface (μ = √3) so that reflected light is completely polarised?

Ans: μ = tan iₚ = √3 → iₚ = 60°.

JEE MAIN 2019

When an analyser is rotated, the intensity of light does not go to zero. The incident light is:

Ans: Partially polarised (mixture of polarised and unpolarised light).

JEE MAIN 2018

Malus Law states I = I₀cos²θ. At θ = 30°, the intensity is:

Ans: I = I₀cos²30° = (3/4)I₀.

JEE Advanced PYQs

JEE ADV 2022

Plane polarised light of intensity I₀ passes through an analyser at θ. If I = I₀/4, find θ. If the analyser is then rotated to 2θ, find new I.

Ans: cos²θ = 1/4 → θ = 60°. At 2θ = 120°: I = I₀cos²120° = I₀×(1/4) = I₀/4.

JEE ADV 2020

Light reflected from a glass slab is found to be completely polarised. The glass has μ = 1.5. Find the angle of refraction and the deviation of the refracted ray from the incident direction.

Ans: iₚ = tan⁻¹(1.5) ≈ 56.3°. r = 90° − 56.3° = 33.7°. Deviation = 180° − (iₚ + r) = 180° − 90° = 90°. (Refracted ray is perpendicular to reflected.)

JEE ADV 2018

Explain why scattered skylight is partially polarised. In which direction is it completely polarised?

Ans: Atmospheric molecules scatter sunlight by Rayleigh scattering. Light scattered at exactly 90° to the sun's direction is completely polarised. Light at other angles is partially polarised.

IB Physics Questions

IB 2023 SL

State what is meant by polarisation of light. Explain why light from a filament lamp can be polarised but sound from a loudspeaker cannot.

Ans: Polarisation: restriction of oscillation to one direction. Light is transverse (E and B ⊥ propagation) → can be polarised. Sound is longitudinal → cannot be polarised.

IB 2022 HL

Polarised light of intensity I₀ is incident on an analyser at 30°. Find intensity and percentage reduction.

Ans: I = I₀cos²30° = 0.75I₀. Reduction = 25%.

IB 2021

Two polaroids are aligned with axes at 0° and 90°. A third is inserted at 45°. Find the final intensity as a fraction of the initial unpolarised intensity I₀.

Ans: I₀/2 × cos²45° × cos²45° = I₀/8.

IB 2020

Describe how polaroid sunglasses reduce glare from road surfaces.

Ans: Glare from roads is partially polarised in horizontal plane. Polaroid sunglasses have vertically oriented transmission axes, blocking the horizontal component and reducing glare.

IB 2019

Explain the Brewster angle and its relationship to the refractive index.

Ans: At Brewster angle iₚ, reflected ray ⊥ refracted ray, and reflected light is completely plane polarised. Derivation gives μ = tan iₚ.

IGCSE Questions

IGCSE 2023

What is polarisation? Why can only transverse waves be polarised?

Ans: Polarisation = restriction of vibrations to one plane. Only transverse waves have perpendicular vibrations that can be restricted. Longitudinal vibrations are along the propagation direction — no plane to restrict.

IGCSE 2022

A polaroid is rotated in front of polarised light. Describe how the observed intensity changes.

Ans: I = I₀cos²θ. Intensity varies smoothly from maximum (θ=0°) to zero (θ=90°) and back to maximum (θ=180°) etc. Two maxima and two minima per full rotation.

IGCSE 2021

State one use of polaroids in everyday life.

Ans: Polaroid sunglasses (reduce glare); LCD screens; 3D movie glasses; camera filters.

A-Level Questions

A-LEVEL 2023

Derive Brewster's Law from first principles. Glass has μ = 1.65. Find Brewster angle and angle of refraction.

Ans: Derivation: iₚ + r = 90°, Snell → μ = tan iₚ. iₚ = tan⁻¹(1.65) = 58.8°. r = 31.2°.

A-LEVEL 2022

Unpolarised light of intensity I₀ passes through two polaroids at angle θ. Show that I = (I₀/2)cos²θ and find θ for I = I₀/4.

Ans: After P₁: I₀/2 (unpolarised→polarised). After P₂: (I₀/2)cos²θ. For I₀/4: cos²θ = 1/2 → θ = 45°.

A-LEVEL 2021

Explain how LCD displays work using the principle of polarisation.

Ans: Two crossed polaroids with liquid crystal between. Without voltage: LC rotates polarisation 90° → passes analyser (bright). With voltage: LC aligned, no rotation → blocked by analyser (dark). Intermediate voltages give grey levels.

Assertion–Reason Questions

A: Polarisation of light proves its transverse nature.

R: Only transverse waves can be polarised.

Answer: (A) Both true, R correctly explains A.

A: When two polaroids are crossed, no light passes.

R: cos²90° = 0, so I = I₀cos²90° = 0.

Answer: (A) Both true, R correctly explains A.

A: At Brewster angle, reflected light is 100% polarised.

R: The reflected and refracted rays are perpendicular at Brewster angle.

Answer: (A) Both true. The perpendicularity condition ensures no component of the p-polarisation (parallel to plane of incidence) is reflected → reflected beam is s-polarised only.

A: Sound waves cannot be polarised.

R: Sound waves are longitudinal waves.

Answer: (A) Both true. Longitudinal vibrations are along the propagation direction — there is no transverse plane to restrict → no polarisation.

A: First polaroid always reduces intensity to half.

R: Unpolarised light has equal distribution of vibrations in all planes, so average cos²θ = 1/2.

Answer: (A) Both true. I = I₀⟨cos²θ⟩ = I₀/2 since ⟨cos²θ⟩ = 1/2.

A: Polaroid sunglasses reduce glare from water.

R: Reflected light from water at near-Brewster angle is horizontally polarised.

Answer: (A) Both true. Vertical-axis polaroids block horizontal glare.

A: Brewster angle is independent of wavelength.

R: μ = tan iₚ; since μ is the same for all wavelengths in a medium.

Answer: (C) A is false — μ varies with wavelength (dispersion), so Brewster angle is wavelength-dependent. R is also false.

A: Inserting a third polaroid between two crossed polaroids can allow light to pass.

R: The third polaroid creates an intermediate direction for the polarised light to be partially transmitted through the final analyser.

Answer: (A) Both true. If third is at 45°, I = I₀/8 ≠ 0. Without it, I = 0.

Case Study Questions

Case Study: Sunglasses and Polarisation

Passage: Polaroid sunglasses use sheets of oriented polymer molecules. When driving, glare from wet roads is strongly polarised.

Q1: Why is road glare polarised?
Ans: Reflection at near-Brewster angle from flat water/road surface preferentially reflects s-polarised (horizontal) light.

Q2: Which direction should the polaroid axis of sunglasses be?
Ans: Vertical — to block horizontal glare.

Q3: If a driver tilts head 90°, what happens?
Ans: The formerly blocked horizontal glare now passes through → glare returns.

Q4: How much does a single polaroid reduce unpolarised light?
Ans: By 50% (I = I₀/2).

Case Study: LCD Display

Passage: A liquid crystal display uses two crossed polaroids with a liquid crystal layer between them. The liquid crystal naturally rotates polarisation by 90°.

Q1: Why is the screen bright when no voltage is applied?
Ans: LC rotates polarisation 90° → aligned with analyser → passes through → bright.

Q2: When voltage is applied, what happens?
Ans: LC aligns with electric field and does not rotate polarisation → blocked by crossed analyser → dark pixel.

Q3: Can an LCD work without polaroids?
Ans: No — polaroids are essential. Without them, there would be no intensity modulation.

Case Study: 3D Movies

Passage: Two slightly different images of a scene (left-eye and right-eye views) are projected using perpendicularly polarised light (0° and 90°). Audience wears polaroid glasses with corresponding axes.

Q1: How does each eye receive a separate image?
Ans: Left lens passes 0°-polarised light; right lens passes 90°-polarised. Each eye sees only its image.

Q2: What would happen if you turned your head 90°?
Ans: The eyes would receive opposite images → reversed depth perception, disorienting effect.

Q3: Why must circular polarisation be used in modern cinemas?
Ans: Linear polarisation fails if viewer tilts head. Circular polarisation preserves the image assignment regardless of head tilt.

Important Formula Sheet

Quick Reference — All Key Formulas

Formula	Name	Key Condition / Note
I = I₀ cos²θ	Malus Law	Polarised light through analyser at angle θ
I = I₀/2	First polaroid	Unpolarised → polaroid → half intensity
a = a₀ cosθ	Amplitude (Malus)	Transmitted amplitude along analyser axis
μ = tan iₚ	Brewster's Law	μ = refractive index, iₚ = polarising angle
iₚ + r = 90°	Brewster condition	Reflected ⊥ refracted at polarising angle
I_max = I₀ (θ=0°)	Parallel polaroids	Maximum intensity when axes parallel
I_min = 0 (θ=90°)	Crossed polaroids	Zero intensity when axes perpendicular
I = I₀ cos⁴θ	Three polaroids	Unpolarised → P₁ → P₂(θ) → P₃(2θ... or parallel pattern)

I = I₀ cos²θ (Malus Law)

μ = tan iₚ (Brewster Law)

iₚ + r = 90° (Brewster Condition)

Kumar Sir Exam Tips

Golden Rules for NEET, JEE, CBSE, IB, IGCSE

⭐ Kumar Sir's Golden Rules for Polarisation

✅Polarisation proves the transverse nature of light. This is the single most important concept.

✅Malus Law: I = I₀ cos²θ — memorise this formula. It is asked in EVERY exam.

✅First polaroid on unpolarised light: I = I₀/2 — intensity is always halved, regardless of polaroid orientation.

✅Crossed polaroids (θ = 90°): I = 0 — no light emerges. Very important for NEET/JEE.

✅Brewster Law: μ = tan iₚ. Derivation key point: at Brewster angle, reflected ⊥ refracted → iₚ + r = 90°.

✅At Brewster angle: reflected light is 100% plane polarised. Refracted light is only partially polarised.

✅Amplitude transmitted: a = a₀ cosθ. Intensity ∝ a² → Malus Law follows immediately.

✅For three polaroids problem: multiply intensities step by step. Each step: I_{n+1} = I_n × cos²(angle between n and n+1).

🔑Sound waves (longitudinal) → CANNOT be polarised. X-rays, microwaves, radio waves, light → CAN be polarised (all transverse).

🔑Brewster angle for glass ≈ 56°, for water ≈ 53°. When μ = √3, iₚ = 60° (most common JEE value).

🔑Inserting a third polaroid between two crossed polaroids at 45° to each: I = I₀/8. Light DOES pass through!

🚨Do NOT confuse Malus Law I = I₀cos²θ (for polarised input) with I = (I₀/2)cos²θ (for unpolarised input through two polaroids).

💡Blue sky, polaroid sunglasses, LCD displays, 3D movies — all use polarisation. Learn each application with mechanism for NEET/JEE/IB.