Physics Optics and Modern - How to Solve: Polarisation (Malus’ Law, Brewster Angle) – IIT JEE Guide

How to Solve: Polarisation (Malus’ Law, Brewster Angle) – IIT JEE Guide

(For Students & Teachers – Ready-to-Record Script Included)

Introduction

Mastering polarisation unlocks 2-3 marks in IIT JEE—every year. From sunglasses to fiber optics, polarisation explains why glare disappears when you tilt your head. In exams, it’s a guaranteed question in JEE Main and a high-weightage topic in JEE Advanced. Miss it, and you lose easy marks. Nail it, and you gain an edge in optics.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand: 1. Transverse waves – Light is a transverse electromagnetic wave (oscillations perpendicular to direction of travel). 2. Intensity of light – Defined as power per unit area (I = P/A). For waves, intensity ∝ (amplitude)². 3. Snell’s Law – Relates angles of incidence and refraction at an interface (n₁ sin θ₁ = n₂ sin θ₂).

KEY TERMS & FORMULAS

Key Terms

1. Polarisation – Restriction of light vibrations to a single plane.
2. Plane-polarised light – Light whose electric field oscillates in only one direction.
3. Unpolarised light – Light whose electric field oscillates in all directions perpendicular to propagation.
4. Polariser – A device (e.g., Polaroid sheet) that transmits only one component of the electric field.
5. Analyser – A second polariser used to detect polarisation.
6. Brewster’s Angle (θ_B) – The angle of incidence at which reflected light is completely plane-polarised.

Formulas

1. Malus’ Law

Formula: I = I₀ cos² θ - I = Intensity of light after passing through the analyser (W/m²) - I₀ = Intensity of polarised light incident on the analyser (W/m²) - θ = Angle between the polariser and analyser transmission axes (° or rad)

MEMORISE THIS – This is the most frequently tested formula in polarisation.

2. Brewster’s Angle

Formula: tan θ_B = (n₂ / n₁) - θ_B = Brewster’s angle (angle of incidence for complete polarisation) - n₁ = Refractive index of the incident medium (e.g., air, n₁ ≈ 1) - n₂ = Refractive index of the refracting medium (e.g., glass, n₂ ≈ 1.5)

MEMORISE THIS – Examiners love asking for θ_B when given n₁ and n₂.

Bonus Relation (Given on exam sheet, but useful to know): At Brewster’s angle, the reflected and refracted rays are perpendicular to each other. θ_B + θ_r = 90° (where θ_r = angle of refraction)

3. Intensity After Polariser (Unpolarised → Polarised)

Formula: I = I₀ / 2 - I = Intensity after passing through a polariser (W/m²) - I₀ = Intensity of unpolarised light incident on the polariser (W/m²)

MEMORISE THIS – This is the starting point for most Malus’ Law problems.

STEP-BY-STEP METHOD

How to Solve Polarisation Problems (Malus’ Law & Brewster Angle)

Step 1: Identify the Type of Light

• Unpolarised light? → Use I = I₀ / 2 after first polariser.
• Plane-polarised light? → Use Malus’ Law (I = I₀ cos² θ).

Step 2: Determine the Angle Between Polariser & Analyser

• If two polarisers are used, find the angle θ between their transmission axes.
• If only one polariser is used, θ = angle between polariser axis and initial polarisation direction.

Step 3: Apply Malus’ Law (If Applicable)

• For polarised light incident on an analyser: I = I₀ cos² θ
• For unpolarised light first passing through a polariser: I₁ = I₀ / 2 (now polarised) Then, if it passes through a second polariser (analyser): I₂ = I₁ cos² θ = (I₀ / 2) cos² θ

Step 4: For Brewster’s Angle Problems

• Use tan θ_B = n₂ / n₁ to find θ_B.
• If asked for reflected light intensity, remember:
• At θ_B, reflected light is 100% polarised (parallel to the surface).
• Refracted light is partially polarised.

Step 5: Check for Hidden Conditions

• Is the light initially unpolarised? (Divide I₀ by 2 first.)
• Are there multiple polarisers? (Apply Malus’ Law sequentially.)
• Is the question about reflection/refraction? (Use Brewster’s angle.)

Worked Example (Using Steps Above)

Problem: Unpolarised light of intensity 8 W/m² passes through two polarisers. The first polariser is vertical, and the second is rotated 60° from the vertical. What is the final intensity?

Solution (Step-by-Step):

1. Identify light type: Unpolarised → Use I = I₀ / 2 after first polariser.

2. After first polariser: I₁ = 8 / 2 = 4 W/m² (now polarised vertically).

3. Angle between polarisers: θ = 60° (given).

4. Apply Malus’ Law:

5. I₂ = I₁ cos² θ = 4 × cos² 60°
6. cos 60° = 0.5 → cos² 60° = 0.25
7. I₂ = 4 × 0.25 = 1 W/m²

Final Answer: 1 W/m²

What we did and why: - Unpolarised light loses half its intensity when polarised. - Malus’ Law reduces intensity based on the angle between polarisers. - Always start with unpolarised light if given.

WORKED EXAMPLES

Example 1 – Basic (Malus’ Law)

Problem: Plane-polarised light of intensity 9 W/m² is incident on a polariser. If the polariser’s axis makes an angle of 30° with the light’s polarisation direction, what is the transmitted intensity?

Solution: 1. Light is already polarised → Directly apply Malus’ Law. 2. I = I₀ cos² θ = 9 × cos² 30° 3. cos 30° = √3/2 → cos² 30° = 3/4 4. I = 9 × (3/4) = 6.75 W/m²

Answer: 6.75 W/m²

What we did and why: - Since light was already polarised, we skipped the I₀/2 step. - Malus’ Law gives the exact reduction based on angle.

Example 2 – Medium (Multiple Polarisers)

Problem: Unpolarised light of intensity 16 W/m² passes through three polarisers. The first is vertical, the second is at 45°, and the third is at 90° to the first. What is the final intensity?

Solution: 1. After first polariser (unpolarised → polarised):
I₁ = 16 / 2 = 8 W/m² (vertical polarisation).

1. After second polariser (45° from first):
   I₂ = I₁ cos² 45° = 8 × (1/√2)² = 8 × 0.5 = 4 W/m²

2. After third polariser (90° from first, so 45° from second):
   I₃ = I₂ cos² 45° = 4 × 0.5 = 2 W/m²

Answer: 2 W/m²

What we did and why: - Unpolarised light → First polariser halves intensity. - Multiple polarisers → Apply Malus’ Law sequentially. - 90° from first ≠ 90° from second → Calculate relative angle.

Example 3 – Exam-Style (Brewster’s Angle + Malus’ Law)

Problem (JEE Advanced 2018-Style): Light is incident on a glass slab (n = 1.5) at Brewster’s angle. The reflected light passes through a polariser whose axis is parallel to the plane of incidence. If the incident light has intensity 10 W/m², what is the intensity after the polariser?

Solution: 1. Find Brewster’s angle:
- tan θ_B = n₂ / n₁ = 1.5 / 1 = 1.5
- θ_B = tan⁻¹(1.5) ≈ 56.3°

1. Reflected light at θ_B is 100% polarised (parallel to the surface).
2. Intensity of reflected light = I_ref = (I₀ / 2) × (reflection coefficient)

3. For simplicity, assume I_ref = I₀ / 2 = 5 W/m² (exact value depends on reflection coefficient, but JEE often assumes this).

4. Polariser axis is parallel to plane of incidence → It transmits the entire polarised component.

5. No reduction in intensity (θ = 0° → cos² 0° = 1).
6. Final intensity = 5 W/m²

Answer: 5 W/m²

What we did and why: - Brewster’s angle → Reflected light is fully polarised. - Polariser parallel to polarisation direction → No intensity loss. - Assumed I_ref = I₀ / 2 (common in JEE problems unless reflection coefficient is given).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—polarisation is easy marks if you follow these rules: 1. Unpolarised light? First polariser halves intensity (I = I₀ / 2). 2. Polarised light? Use Malus’ Law (I = I₀ cos² θ)—angle is between polarisers. 3. Brewster’s angle? tan θ_B = n₂ / n₁—reflected light is fully polarised. 4. Multiple polarisers? Apply Malus’ Law one by one, not all at once. 5. Watch for hidden angles—draw a diagram if confused.

Most mistakes happen when you forget to halve intensity for unpolarised light or mix up the angle. Stick to the steps, and you’ll ace every polarisation question in JEE."