Polarization - Proving Light is Transverse

The Movie Hook 🎬

Ever worn polarized sunglasses and noticed they reduce glare from water or roads? Or noticed 3D movie glasses create depth perception? Or seen how LCD screens go dark when you tilt your phone? That’s all polarization - the property that proves light is a transverse wave!

The Big Picture

Polarization: The phenomenon of restricting the vibrations of light (electric field) to a particular direction perpendicular to the direction of wave propagation.

Key applications:

Sunglasses (reducing glare)
3D movies (stereoscopic vision)
LCD screens (controlling light)
Photography (reducing reflections)
Stress analysis (photoelasticity)

Connection Alert: Proves light is transverse, relates to Wave Optics and EM Waves.

Core Concepts

1. Unpolarized vs Polarized Light

Unpolarized light:

Electric field vibrates in all directions perpendicular to propagation
Example: Sunlight, bulb light
Random orientation of E-field vectors

Polarized light:

Electric field vibrates in one particular direction only
Obtained by passing unpolarized light through polarizer
Fixed orientation of E-field

Memory Trick: “Unpolarized = Universe of directions”, “Polarized = Particular direction”

2. Plane of Vibration and Plane of Polarization

Plane of vibration: Plane containing the direction of vibration (E-field) and direction of propagation

Plane of polarization: Plane perpendicular to plane of vibration

Note

Important: Plane of polarization is perpendicular to the direction of E-field vibrations!

In JEE: Usually we just refer to “polarization direction” = direction of E-field vibrations.

3. Transverse Nature of Light

Only transverse waves can be polarized!

Longitudinal waves (like sound) cannot be polarized
The fact that light can be polarized proves it’s transverse
Electric field perpendicular to propagation direction

Why sound cannot be polarized? Sound is longitudinal → particles vibrate parallel to propagation → no perpendicular direction to restrict!

4. Polaroid (Polarizer)

Polaroid: Material that produces polarized light by selective absorption

Working:

Contains long-chain molecules aligned in one direction
Absorbs light with E-field parallel to molecular chains
Transmits light with E-field perpendicular to molecular chains

Transmission axis: Direction perpendicular to aligned molecules (allowed E-field direction)

When unpolarized light passes through polaroid:

Intensity becomes half
Light becomes plane polarized

$$\boxed{I = \frac{I_0}{2}}$$

Where I₀ = intensity of unpolarized light

5. Malus Law

Setup: Polarized light of intensity I₀ passes through an analyzer (second polaroid)

Angle θ: Between transmission axes of polarizer and analyzer

Malus Law:

$$\boxed{I = I_0 \cos^2 \theta}$$

Where:

I = transmitted intensity
I₀ = incident polarized light intensity
θ = angle between polarizer and analyzer axes

Special cases:

θ	cos²θ	I	Condition
0°	1	I₀	Maximum (parallel)
45°	1/2	I₀/2	Half intensity
60°	1/4	I₀/4	Quarter intensity
90°	0	0	Minimum (crossed, dark)

Memory Trick: “Malus with Cosine Squared” (Malus = cos²θ)

6. Two Polaroids Setup

Case 1: Unpolarized light through two polaroids

After first polaroid (polarizer):

$$I_1 = \frac{I_0}{2}$$

After second polaroid (analyzer) at angle θ:

$$I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta$$

Complete formula:

$$\boxed{I = \frac{I_0}{2} \cos^2 \theta}$$

For crossed polaroids (θ = 90°):

$$I = 0$$

(complete darkness!)

7. Three Polaroids Setup

Common JEE question!

Setup: P₁ at 0°, P₂ at θ, P₃ at 90°

Without middle polaroid (P₂):

P₁ and P₃ are crossed (90°)
No light passes → I = 0

With middle polaroid (P₂) at angle θ:

After P₁:

$$I_1 = \frac{I_0}{2}$$

After P₂ (angle θ from P₁):

$$I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta$$

After P₃ (angle 90°-θ from P₂):

$$I_3 = I_2 \cos^2(90° - \theta) = I_2 \sin^2 \theta$$ $$I_3 = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta$$

Using

$$\sin 2\theta = 2\sin\theta \cos\theta$$

$$\sin^2\theta \cos^2\theta = \frac{1}{4}\sin^2 2\theta$$ $$\boxed{I_3 = \frac{I_0}{8} \sin^2 2\theta}$$

Maximum transmission: When θ = 45°

$$I_{max} = \frac{I_0}{8}$$

Key insight: Inserting middle polaroid actually allows some light through crossed polaroids!

Polarization by Reflection

1. Brewster’s Law

When unpolarized light is incident on a transparent surface at a certain angle, the reflected light is completely plane polarized.

Brewster’s angle (i_p):

$$\boxed{\tan i_p = n}$$

Or:

$$\boxed{i_p = \tan^{-1}(n)}$$

Where:

i_p = polarizing angle (Brewster’s angle)
n = refractive index of medium (relative to air)

At Brewster’s angle:

Reflected light is completely polarized (⊥ to plane of incidence)
Refracted light is partially polarized
Reflected and refracted rays are perpendicular to each other

Relation:

$$\boxed{i_p + r = 90°}$$

Where r = angle of refraction

Memory Trick: “Brewster needs Tan” (Brewster’s angle uses tan, not sin!)

2. Derivation of Brewster’s Law

At Brewster’s angle: i_p + r = 90°

So: r = 90° - i_p

Using Snell’s law:

$$\sin i_p = n \sin r = n \sin(90° - i_p) = n \cos i_p$$ $$\frac{\sin i_p}{\cos i_p} = n$$ $$\boxed{\tan i_p = n}$$

3. Polarization by Scattering

Rayleigh scattering: Scattering of light by particles much smaller than wavelength

Sky appears blue:

Shorter wavelengths (blue) scatter more
Intensity ∝ 1/λ⁴

Scattered light is partially polarized:

Maximum polarization perpendicular to incident beam
At 90° scattering angle, light is completely polarized

This is why:

Sky is blue (blue scatters most)
Sunset is red (blue scattered away, red transmitted)
Polarized sunglasses reduce glare from sky

Methods of Polarization

Summary Table

Method	Principle	Polarization	Example
Polaroid	Selective absorption	Complete	Sunglasses
Reflection	Brewster’s law	Complete (reflected)	Water surface
Refraction	Double refraction	Partial/Complete	Calcite crystal
Scattering	Rayleigh scattering	Partial	Blue sky

The Formula Cheat Sheet

POLARIZATION ESSENTIALS:
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
1. Through polaroid:     I = I₀/2
2. Malus law:            I = I₀cos²θ
3. Two polaroids:        I = (I₀/2)cos²θ
4. Three polaroids:      I = (I₀/8)sin²2θ
5. Brewster's law:       tan i_p = n
6. At Brewster:          i_p + r = 90°
7. Crossed polaroids:    θ = 90° → I = 0
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Common Traps & Mistakes

Trap #1: Forgetting Factor of 1/2

❌ Wrong: I = I₀ cos²θ for unpolarized light through two polaroids ✅ Right: I = (I₀/2) cos²θ (factor 1/2 from first polaroid!)

Trap #2: Brewster’s Angle Formula

❌ Wrong: sin i_p = n (confusing with Snell’s law) ✅ Right: tan i_p = n (Brewster uses tangent!)

Trap #3: Three Polaroids Maximum

❌ Wrong: Maximum at θ = 0° or 90° ✅ Right: Maximum at θ = 45° → I_max = I₀/8

Trap #4: Longitudinal Wave Polarization

❌ Wrong: Sound can be polarized ✅ Right: Only transverse waves can be polarized!

Trap #5: Angle in Malus Law

❌ Wrong: Using angle with horizontal or vertical ✅ Right: Use angle between transmission axes of polarizer and analyzer

Practice Problems

Level 1: JEE Main Warmup

Problem 1.1: Unpolarized light of intensity I₀ passes through a polaroid. What is the intensity of emerging light?

Solution

When unpolarized light passes through single polaroid:

$$I = \frac{I_0}{2}$$

Answer: I₀/2

Reason: Polaroid blocks half the randomly oriented vibrations.

Problem 1.2: Polarized light of intensity I₀ is incident on an analyzer. The angle between polarizer and analyzer is 60°. Find transmitted intensity.

Solution

Using Malus law:

$$I = I_0 \cos^2 \theta = I_0 \cos^2 60°$$ $$I = I_0 \times \left(\frac{1}{2}\right)^2 = I_0 \times \frac{1}{4} = \frac{I_0}{4}$$

Answer: I₀/4

Problem 1.3: Find Brewster’s angle for water (n = 4/3).

Solution

$$\tan i_p = n = \frac{4}{3}$$ $$i_p = \tan^{-1}\left(\frac{4}{3}\right) = \tan^{-1}(1.33) \approx 53.1°$$

Answer: 53.1°

At this angle, reflected light from water surface is completely polarized!

Level 2: JEE Main/Advanced

Problem 2.1: Unpolarized light passes through two polaroids with transmission axes at 30° to each other. What fraction of incident intensity is transmitted?

Solution

After first polaroid:

$$I_1 = \frac{I_0}{2}$$

After second polaroid (at 30°):

$$I_2 = I_1 \cos^2 30° = \frac{I_0}{2} \times \left(\frac{\sqrt{3}}{2}\right)^2$$ $$I_2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8}$$

Fraction transmitted:

$$\frac{I_2}{I_0} = \frac{3}{8}$$

Answer: 3/8 or 37.5%

Problem 2.2: At what angle of incidence will light reflected from glass (n = 1.5) be completely polarized?

Solution

Using Brewster’s law:

$$\tan i_p = n = 1.5$$ $$i_p = \tan^{-1}(1.5) \approx 56.3°$$

Angle of refraction:

$$r = 90° - i_p = 90° - 56.3° = 33.7°$$

Verification using Snell’s law:

$$\sin 56.3° = 1.5 \times \sin 33.7°$$ $$0.832 = 1.5 \times 0.555 = 0.833$$

✓

Answer: i_p = 56.3°

Level 3: JEE Advanced

Problem 3.1: Two polaroids P₁ and P₂ are crossed (90°). A third polaroid P₃ is placed between them at 45° to P₁. If unpolarized light of intensity I₀ is incident on P₁, find the final transmitted intensity.

Solution

After P₁:

$$I_1 = \frac{I_0}{2}$$

After P₃ (at 45° from P₁):

$$I_2 = I_1 \cos^2 45° = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$

After P₂ (at 45° from P₃):

$$I_3 = I_2 \cos^2 45° = \frac{I_0}{4} \times \frac{1}{2} = \frac{I_0}{8}$$

Using formula directly:

$$I_3 = \frac{I_0}{8} \sin^2(2 \times 45°) = \frac{I_0}{8} \times 1 = \frac{I_0}{8}$$

Answer: I₀/8

Key: Without P₃, intensity is zero. With P₃ at 45°, we get maximum transmission = I₀/8!

Quick Revision Cards

╔══════════════════════════════════════╗
║  POLARIZATION QUICK FACTS           ║
╚══════════════════════════════════════╝

BASICS:
🎯 Only transverse waves polarize
🎯 Light is transverse EM wave
🎯 Through 1 polaroid: I = I₀/2
🎯 Malus law: I = I₀cos²θ

TWO POLAROIDS:
🎯 Parallel (0°): I = I₀/2
🎯 At 45°: I = I₀/4
🎯 At 60°: I = I₀/8
🎯 Crossed (90°): I = 0

THREE POLAROIDS:
🎯 I = (I₀/8)sin²2θ
🎯 Maximum at θ = 45°
🎯 I_max = I₀/8

BREWSTER:
🎯 tan i_p = n
🎯 i_p + r = 90°
🎯 Reflected light 100% polarized

Cross-Topic Connections

Wave Optics: Transverse nature of waves
EM Waves: Electric field oscillations
Refraction: Brewster’s angle uses Snell’s law
Interference: Polarized light interference

Final Tips

Malus law needs polarized incident light!
First polaroid always gives I₀/2 (for unpolarized)
Brewster’s angle uses tan i_p = n (easy to confuse with sin!)
Three polaroids - remember sin²2θ formula
Only transverse waves can be polarized

Pro Tip: In three polaroid problems, middle polaroid at 45° gives maximum transmission = I₀/8!

Congratulations! You’ve completed the comprehensive Optics chapter covering all major JEE topics!

Polarization - Proving Light is Transverse

The Movie Hook 🎬

The Big Picture

Core Concepts

1. Unpolarized vs Polarized Light

2. Plane of Vibration and Plane of Polarization

3. Transverse Nature of Light

4. Polaroid (Polarizer)

5. Malus Law

6. Two Polaroids Setup

7. Three Polaroids Setup

Polarization by Reflection

1. Brewster’s Law

2. Derivation of Brewster’s Law

3. Polarization by Scattering

Methods of Polarization

Summary Table

The Formula Cheat Sheet

Common Traps & Mistakes

Trap #1: Forgetting Factor of 1/2

Trap #2: Brewster’s Angle Formula

Trap #3: Three Polaroids Maximum

Trap #4: Longitudinal Wave Polarization

Trap #5: Angle in Malus Law

Practice Problems

Level 1: JEE Main Warmup

Level 2: JEE Main/Advanced

Level 3: JEE Advanced

Quick Revision Cards

Cross-Topic Connections

Final Tips

Stay Updated with JEE Notes