Brewster's Angle: The Magic Angle for Perfect Polarization by Reflection
Master Brewster's Law derivation and understand why reflected light becomes perfectly polarized at this special angle.
Why Brewster's Angle Matters in JEE
Brewster's angle is a fundamental concept in wave optics that explains polarization by reflection. It's crucial for:
- Direct formula applications in JEE Main
- Conceptual understanding of polarization
- Derivation-based questions in JEE Advanced
- Real-world applications like anti-glare sunglasses
Brewster's Law Formula
where $\theta_B$ is Brewster's angle and $\mu$ is the refractive index of medium 2 relative to medium 1
Brewster's Angle Setup
Incident light | Reflected light (polarized) | Refracted light (partially polarized)
Incident ↗
/
/ θ_B
Interface ------------
\
Refracted \
Reflected ← (Polarized)
Derivation of Brewster's Law
📐 Method 1: Using Snell's Law and Reflection Condition
Step 1: At Brewster's angle, reflected and refracted rays are perpendicular
$\theta_B + \theta_r = 90^\circ$
Step 2: Apply Snell's Law at interface
$n_1 \sin \theta_B = n_2 \sin \theta_r$
Step 3: Substitute $\theta_r = 90^\circ - \theta_B$
$n_1 \sin \theta_B = n_2 \sin(90^\circ - \theta_B) = n_2 \cos \theta_B$
Step 4: Rearrange to get Brewster's Law
$\frac{\sin \theta_B}{\cos \theta_B} = \frac{n_2}{n_1}$
$\tan \theta_B = \mu = \frac{n_2}{n_1}$
📐 Method 2: Using Fresnel's Equations
Step 1: Fresnel's reflection coefficient for parallel component:
$r_\parallel = \frac{n_2 \cos \theta_i - n_1 \cos \theta_t}{n_2 \cos \theta_i + n_1 \cos \theta_t}$
Step 2: Set $r_\parallel = 0$ for no reflection of parallel component
Step 3: Solve to get the same result: $\tan \theta_B = \frac{n_2}{n_1}$
Why Perfect Polarization Occurs
💡 The Molecular Dipole Explanation
Key Insight: At Brewster's angle, the reflected light contains only the component perpendicular to the plane of incidence.
Molecular Oscillations: When light hits the interface, molecules in the medium oscillate
• Parallel component drives oscillations along reflection direction
• Perpendicular component drives oscillations perpendicular to reflection direction
No Radiation Condition: An oscillating dipole doesn't radiate along its axis
• At Brewster's angle, parallel oscillations point along reflection direction
• Therefore, no parallel component in reflected light
🎯 JEE Application Example:
Problem: Calculate Brewster's angle for light going from air (n=1) to glass (n=1.5)
Solution: Using Brewster's Law:
$\tan \theta_B = \frac{n_2}{n_1} = \frac{1.5}{1} = 1.5$
$\theta_B = \tan^{-1}(1.5) \approx 56.3^\circ$
🚀 Problem-Solving Strategies
Key Points to Remember:
- Reflected ray is always perpendicular to refracted ray at θ_B
- Only perpendicular component reflects at Brewster's angle
- Refracted light is partially polarized
- Works for both internal and external reflection
JEE Exam Tips:
- Remember the perpendicular ray condition
- Practice numerical problems with different media
- Understand the physical reason behind polarization
- Relate to real-world applications
Advanced Applications Available
Includes polarization by scattering, Malus' Law applications, and JEE Advanced level problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Find Brewster's angle for water (n=1.33) to air interface
2. If Brewster's angle is 60°, find the refractive index of the medium
3. Explain why photographers use polarizing filters at Brewster's angle
Ready to Master Wave Optics?
Get complete access to polarization techniques and JEE practice problems