Wave Optics - Light as a Wave
The Movie Hook 🎬
Remember the water ripples in Inception that represent reality layers? Or the circular waves when Thor’s hammer hits water? Or the sound waves visualization in A Quiet Place? These wave patterns are exactly how light behaves - and Huygens showed us how!
The Big Picture
Wave optics treats light as a wave phenomenon, explaining:
- Interference: Bright and dark fringes
- Diffraction: Bending around obstacles
- Polarization: Transverse nature of light
- Wavefronts: How light propagates
Connection Alert: Foundation for Interference, Diffraction, and Polarization. Links to EM Waves.
Core Concepts
1. Wave Nature of Light
Light is an electromagnetic wave:
- Electric field E oscillates
- Magnetic field B oscillates (perpendicular to E)
- Both perpendicular to direction of propagation
- Transverse wave
Wave equation:
$$\boxed{y = A \sin(\omega t - kx + \phi)}$$Where:
- A = amplitude
- ω = angular frequency = 2πf
- k = wave number = 2π/λ
- φ = initial phase
Wave speed:
$$\boxed{c = f\lambda = \frac{\omega}{k}}$$2. Wavefront
Wavefront: Locus of all points vibrating in same phase
Types:
Spherical wavefront: From point source
- Expands as sphere
- Radius increases with time
Cylindrical wavefront: From line source
- Expands as cylinder
Plane wavefront: At large distance from source
- OR from source at infinity
- Parallel planes
Key property: Wavefront is perpendicular to direction of propagation (rays)
Memory Trick: “Waves Propagate Perpendicular” (wavefront ⊥ rays)
3. Huygens Principle
Statement: Every point on a wavefront acts as a source of secondary wavelets. The new wavefront is the tangent envelope to these wavelets.
Applications:
- Explains propagation of waves
- Derives laws of reflection
- Derives laws of refraction
- Explains diffraction
Construction:
- Draw primary wavefront
- From each point, draw secondary wavelets (circles/spheres)
- Draw common tangent → new wavefront
4. Huygens Derivation of Refraction (Snell’s Law)
Consider plane wavefront traveling from medium 1 (speed v₁) to medium 2 (speed v₂).
Time for wavelet to travel in medium 1:
$$t = \frac{AC}{v_1}$$Distance traveled in medium 2:
$$BD = v_2 t = v_2 \frac{AC}{v_1}$$From geometry:
$$\frac{AC}{\sin i} = AB = \frac{BD}{\sin r}$$ $$\frac{AC}{\sin i} = \frac{v_2 AC/(v_1)}{\sin r}$$ $$\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1}$$Therefore:
$$\boxed{n_1 \sin i = n_2 \sin r}$$Snell’s law derived from wave theory!
5. Coherent Sources
Coherent sources: Two sources having:
- Same frequency
- Constant phase difference
Why needed?
- For stable interference pattern
- Random phase changes → no fixed pattern
How to obtain:
- Division of wavefront: Single source split (Young’s slits, Fresnel biprism)
- Division of amplitude: Partial reflection (thin films, Newton’s rings)
Incoherent sources:
- Independent sources (two bulbs)
- Random phase changes
- No stable interference
6. Path Difference and Phase Difference
Relation:
$$\boxed{\phi = \frac{2\pi}{\lambda} \times \Delta x}$$Or:
$$\boxed{\Delta x = \frac{\lambda}{2\pi} \times \phi}$$Where:
- φ = phase difference (radians)
- Δx = path difference
- λ = wavelength
For constructive interference:
- Path difference = nλ (n = 0, 1, 2, …)
- Phase difference = 2nπ
For destructive interference:
- Path difference = (n + 1/2)λ = (2n+1)λ/2
- Phase difference = (2n+1)π
Memory Trick: “Complete Wavelengths Construct” (nλ → constructive)
7. Superposition Principle
When two waves meet:
$$\boxed{y = y_1 + y_2}$$If:
$$y_1 = A_1 \sin(\omega t)$$and
$$y_2 = A_2 \sin(\omega t + \phi)$$Resultant amplitude:
$$\boxed{A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi}}$$Special cases:
Constructive (φ = 0, 2π, 4π, …):
$$A_{max} = A_1 + A_2$$ $$I_{max} = (I_1^{1/2} + I_2^{1/2})^2$$Destructive (φ = π, 3π, 5π, …):
$$A_{min} = |A_1 - A_2|$$ $$I_{min} = (I_1^{1/2} - I_2^{1/2})^2$$For equal amplitudes (A₁ = A₂ = A₀):
$$A = 2A_0 \cos(\phi/2)$$ $$I = 4I_0 \cos^2(\phi/2)$$
8. Intensity Relations
Intensity ∝ (Amplitude)²:
$$\boxed{I = kA^2}$$For two coherent sources:
$$\boxed{I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi}$$Maximum intensity:
$$\boxed{I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2}$$Minimum intensity:
$$\boxed{I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = (\sqrt{I_1} - \sqrt{I_2})^2}$$For equal intensities (I₁ = I₂ = I₀):
- I_max = 4I₀
- I_min = 0
- Average intensity = 2I₀ (energy conserved!)
The Formula Cheat Sheet
WAVE OPTICS ESSENTIALS:
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1. Wave equation: y = A sin(ωt - kx + φ)
2. Wave speed: c = fλ = ω/k
3. Phase-path: φ = (2π/λ)Δx
4. Constructive: Δx = nλ, φ = 2nπ
5. Destructive: Δx = (n+½)λ, φ = (2n+1)π
6. Resultant amp: A² = A₁² + A₂² + 2A₁A₂cosφ
7. Intensity: I = I₁ + I₂ + 2√(I₁I₂)cosφ
8. Equal sources: I_max = 4I₀, I_min = 0
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Common Traps & Mistakes
Trap #1: Confusing Path and Phase Difference
❌ Wrong: Using path difference directly in cosφ ✅ Right: Convert: φ = (2π/λ) × Δx first!
Trap #2: Intensity Addition
❌ Wrong: I_total = I₁ + I₂ (ignoring interference term) ✅ Right: I = I₁ + I₂ + 2√(I₁I₂)cosφ
Trap #3: Wavelength in Medium
❌ Wrong: Using vacuum wavelength in medium ✅ Right: λ_medium = λ_vacuum / n
Trap #4: Coherence Requirement
❌ Wrong: Expecting interference from any two sources ✅ Right: Sources must be coherent (same frequency, constant phase)
Practice Problems
Level 1: JEE Main Warmup
Problem 1.1: Two waves have path difference of 3λ/2. What is the phase difference? Is interference constructive or destructive?
Solution
Given: Δx = 3λ/2
Phase difference:
$$\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{3\lambda}{2} = 3\pi$$Since φ = 3π = (2×1+1)π → odd multiple of π
Answer: Phase difference = 3π, Destructive interference
Problem 1.2: Two coherent sources have intensities in ratio 4:1. Find the ratio of maximum to minimum intensity.
Solution
Given: I₁/I₂ = 4/1
Let I₂ = I, then I₁ = 4I
$$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{4I} + \sqrt{I})^2 = (2\sqrt{I} + \sqrt{I})^2 = (3\sqrt{I})^2 = 9I$$ $$I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I$$ $$\frac{I_{max}}{I_{min}} = \frac{9I}{I} = 9$$Answer: 9:1
Formula trick:
$$\frac{I_{max}}{I_{min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \left(\frac{\sqrt{4} + \sqrt{1}}{\sqrt{4} - \sqrt{1}}\right)^2 = \left(\frac{3}{1}\right)^2 = 9$$Level 2: JEE Main/Advanced
Problem 2.1: Two coherent sources of equal intensity I₀ produce interference. Find: (a) Maximum and minimum intensities (b) Intensity at a point where phase difference is π/3 (c) Intensity at a point where path difference is λ/4
Solution
Given: I₁ = I₂ = I₀
(a) Maximum and minimum:
$$I_{max} = 4I_0$$ $$I_{min} = 0$$(b) At φ = π/3:
$$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$$ $$I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos(\pi/3)$$ $$I = 2I_0 + 2I_0 \times \frac{1}{2} = 2I_0 + I_0 = 3I_0$$(c) At Δx = λ/4:
$$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$$ $$I = 2I_0 + 2I_0 \cos(\pi/2) = 2I_0 + 0 = 2I_0$$Answers:
- (a) I_max = 4I₀, I_min = 0
- (b) I = 3I₀
- (c) I = 2I₀
Problem 2.2: Two waves are represented by: y₁ = 5 sin(ωt) y₂ = 5 sin(ωt + π/2)
Find the amplitude of resultant wave.
Solution
Given: A₁ = A₂ = 5, φ = π/2
$$A = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi}$$ $$A = \sqrt{25 + 25 + 2 \times 5 \times 5 \times \cos(\pi/2)}$$ $$A = \sqrt{50 + 50 \times 0} = \sqrt{50} = 5\sqrt{2}$$Answer: A = 5√2
Alternative (for π/2): For equal amplitudes and φ = 90°:
$$A = A_0\sqrt{2} = 5\sqrt{2}$$Level 3: JEE Advanced
Problem 3.1: Three coherent sources of equal intensity I₀ are placed at vertices of an equilateral triangle. Find the intensity at the centroid. (Assume all have same phase)
Solution
All three sources have:
- Same amplitude A₀
- Same phase (φ = 0)
- Equal distance from centroid
The three waves arriving at centroid:
$$y_1 = A_0 \sin(\omega t)$$ $$y_2 = A_0 \sin(\omega t)$$ $$y_3 = A_0 \sin(\omega t)$$Resultant:
$$y = 3A_0 \sin(\omega t)$$Resultant amplitude:
$$A = 3A_0$$Intensity:
$$I = kA^2 = k(3A_0)^2 = 9kA_0^2 = 9I_0$$Answer: I = 9I₀
General formula for n coherent sources in phase:
$$I = n^2 I_0$$Problem 3.2: A point source S emits light of wavelength λ. Two points A and B are at distances 10λ and 10.5λ from S. What is the phase difference between waves reaching A and B?
Solution
Path from S to A: r_A = 10λ Path from S to B: r_B = 10.5λ
Path difference:
$$\Delta x = r_B - r_A = 10.5\lambda - 10\lambda = 0.5\lambda = \frac{\lambda}{2}$$Phase difference:
$$\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi$$Answer: φ = π radians (180°)
This is destructive interference condition.
Advanced Concepts
1. Doppler Effect for Light
$$\boxed{f' = f\sqrt{\frac{1 \pm v/c}{1 \mp v/c}}}$$Approaching: Use upper signs Receding: Use lower signs
Red shift: Source receding → frequency decreases, wavelength increases Blue shift: Source approaching → frequency increases
2. Electromagnetic Spectrum
| Type | Wavelength | Frequency |
|---|---|---|
| Radio | > 1 m | < 300 MHz |
| Microwave | 1 mm - 1 m | 300 MHz - 300 GHz |
| Infrared | 700 nm - 1 mm | 300 GHz - 430 THz |
| Visible | 400-700 nm | 430-750 THz |
| UV | 10-400 nm | 750 THz - 30 PHz |
| X-ray | 0.01-10 nm | 30 PHz - 30 EHz |
| Gamma | < 0.01 nm | > 30 EHz |
Visible spectrum: VIBGYOR (400-700 nm)
Quick Revision Cards
╔══════════════════════════════════════╗
║ WAVE OPTICS QUICK FACTS ║
╚══════════════════════════════════════╝
🎯 Light is EM wave (transverse)
🎯 Wavefront ⊥ rays
🎯 Huygens: Every point → secondary source
🎯 Coherence: same f, constant φ
🎯 Constructive: Δx = nλ
🎯 Destructive: Δx = (n+½)λ
🎯 φ = (2π/λ)Δx
🎯 I_max/I_min = (√I₁+√I₂)²/(√I₁-√I₂)²
🎯 Equal sources: I_max = 4I₀
Cross-Topic Connections
- Interference: Young’s double slit uses wave optics
- Diffraction: Explained by Huygens principle
- Polarization: Shows transverse nature
- EM Waves: Light is EM radiation
- Dual Nature: Wave-particle duality
Exam Strategy
Time Allocation:
- Wavefront/Huygens questions: 2 min
- Coherence/superposition: 3 min
- Intensity calculations: 3 min
Common JEE Patterns:
- Intensity ratio calculations
- Path/phase difference conversions
- Huygens construction
- Coherent source requirements
- Wavelength in different media
Final Tips
- Always convert path difference to phase difference for intensity
- Coherence is must for stable interference
- Energy is conserved: Average intensity = I₁ + I₂
- In medium: λ_medium = λ_vacuum / n
- Wavefront ⊥ rays: Use this for constructions
Pro Tip: For equal intensities, I_max = 4I₀ (not 2I₀!) - very common JEE trap!
Next up: Interference - Young’s double slit experiment!