Last Updated: April 2026
Wave Optics is a consistent high-yield chapter in JEE Main 2027 Physics, contributing 2-4 questions every year. The chapter covers the wave nature of light — interference, diffraction, and polarisation — and is treated in NCERT Class 12 Physics Chapter 10. Unlike geometric optics (ray optics), wave optics requires understanding phase relationships and wave mathematics. This guide covers every concept with theory, derivations, and 30 JEE-level practice problems.
JEE Main Wave Optics — Weightage Analysis
| Topic | JEE Questions/Year (Avg.) | Difficulty Level |
|---|---|---|
| Young’s Double Slit Experiment (YDSE) | 1-2 | Medium |
| Interference (Path difference, fringe width) | 1 | Medium |
| Diffraction at single slit | 0-1 | Medium-Hard |
| Polarisation (Brewster’s angle, Malus’s law) | 1 | Easy-Medium |
| Huygens’ Principle | 0-1 | Easy |
1. Huygens’ Wave Theory
Christian Huygens (1678) proposed that light propagates as waves. Key postulates:
- Every point on a wavefront acts as a source of secondary wavelets
- Secondary wavelets spread in all directions with the same speed as primary wave
- The new wavefront is the common tangent to all secondary wavelets (envelope)
- Huygens’ principle explains reflection, refraction, and diffraction of light
Wavefront: Locus of all points of the medium vibrating in the same phase. Types: Spherical (point source), Cylindrical (line source), Plane (distant source).
2. Interference of Light
When two coherent light waves superpose, they produce a pattern of alternating bright and dark fringes — interference pattern.
Conditions for Interference
- Coherent sources: constant phase difference between sources (same frequency, fixed phase relation)
- Sources must be monochromatic (single wavelength) for distinct fringes
- Sources must be close to each other (compared to distance to screen)
Constructive vs Destructive Interference
| Type | Path Difference (Δx) | Phase Difference | Result |
|---|---|---|---|
| Constructive (Bright fringe) | nλ (n = 0, 1, 2, …) | 2nπ | Maximum intensity (I_max) |
| Destructive (Dark fringe) | (2n-1)λ/2 | (2n-1)π | Minimum intensity (I_min = 0 if equal amplitudes) |
Intensity formula: I = I₁ + I₂ + 2√(I₁I₂)cosφ
When I₁ = I₂ = I₀: I_max = 4I₀, I_min = 0
3. Young’s Double Slit Experiment (YDSE)
YDSE demonstrated the wave nature of light by producing interference fringes. Thomas Young (1801) used sunlight and two pinholes close together.
Setup and Key Formulae
- d = slit separation, D = distance from slits to screen, λ = wavelength of light
- Fringe width (β) = λD/d — distance between consecutive bright (or dark) fringes
- Position of nth bright fringe: y_n = nλD/d
- Position of nth dark fringe: y_n = (2n-1)λD/2d
- Angular fringe width: θ = λ/d
YDSE — Effect of Variables on Fringe Width
| Change | Effect on Fringe Width β |
|---|---|
| Increase λ (use red light instead of blue) | β increases (wider fringes) |
| Increase D (move screen farther) | β increases |
| Increase d (wider slit separation) | β decreases |
| Immerse in medium of refractive index μ | β decreases (λ becomes λ/μ, so β = λD/μd) |
| Replace one slit with opaque sheet | Only diffraction pattern, no interference |
YDSE — Shift Due to Slab Insertion
If a glass slab of thickness t and refractive index μ is placed over one slit, the central bright fringe shifts towards that slit by:
Shift = (μ-1)t × D/d
The path difference introduced by the slab = (μ-1)t (extra optical path length).
4. Diffraction
Diffraction is the bending of light around obstacles or through narrow slits. Explained by Huygens’ secondary wavelets.
Single Slit Diffraction
When light passes through a single slit of width ‘a’:
- Central maximum: wide, at θ = 0
- Minima (dark fringes): at a sinθ = nλ (n = ±1, ±2, …)
- Secondary maxima: at a sinθ = (2n+1)λ/2
- Width of central maximum = 2λD/a (twice the width of secondary maxima)
Key difference from YDSE: In single slit, positions of minima use nλ; in YDSE, positions of maxima use nλ. Don’t confuse these!
Resolving Power and Rayleigh’s Criterion
Two sources can just be resolved when the central maximum of one falls on the first minimum of the other. Limiting angle of resolution: θ = 1.22λ/D (for circular aperture of diameter D).
5. Polarisation
Polarisation proves the TRANSVERSE nature of light waves. Sound waves (longitudinal) cannot be polarised.
Brewster’s Law
When light is incident at the Brewster angle (i_B), the reflected light is completely plane-polarised.
tan(i_B) = μ (refractive index of the denser medium)
At Brewster’s angle, reflected and refracted rays are perpendicular to each other: i_B + r = 90°
Malus’s Law
When plane-polarised light of intensity I₀ passes through an analyser (polaroid) at angle θ to the polarisation direction:
I = I₀cos²θ
- θ = 0°: I = I₀ (maximum transmission)
- θ = 90°: I = 0 (complete extinction)
Intensity After Passing Through Polaroid
Unpolarised light of intensity I₀ → passes through polaroid → intensity = I₀/2 (half the incident intensity, regardless of polaroid orientation).
JEE Main Wave Optics — Solved Problems
Problem 1 (YDSE — Fringe Width)
In a YDSE, slits are separated by 0.2 mm, screen is 1 m away, and light of wavelength 600 nm is used. Find fringe width.
Solution: β = λD/d = (600 × 10⁻⁹ × 1) / (0.2 × 10⁻³) = 3 × 10⁻³ m = 3 mm
Problem 2 (Malus’s Law)
Polarised light of intensity I₀ passes through an analyser making 60° with the polariser. What is the transmitted intensity?
Solution: I = I₀cos²60° = I₀ × (1/2)² = I₀/4
Problem 3 (Brewster’s Angle)
Light is reflected from a glass surface (μ = 1.732). Find Brewster’s angle.
Solution: tan(i_B) = μ = 1.732 = tan60° → i_B = 60°
Problem 4 (Slab Shift in YDSE)
In YDSE (d = 1mm, D = 1m, λ = 500nm), a glass slab (t = 1mm, μ = 1.5) is placed over one slit. Find the shift of central fringe.
Solution: Shift = (μ-1)tD/d = (0.5)(1×10⁻³)(1)/(1×10⁻³) = 0.5 m — 500 mm
Number of fringes shifted = shift/β = 500mm / (0.5mm) = 1000 fringes
Frequently Asked Questions (FAQ)
What is the key difference between interference and diffraction?
Interference occurs due to superposition of waves from two or more coherent sources (e.g., YDSE). Diffraction occurs due to superposition of secondary wavelets from different parts of the SAME wavefront (single slit). In interference, all bright fringes have equal intensity; in diffraction, intensity decreases for higher-order fringes. The central maximum in single slit diffraction is twice as wide as secondary maxima.
Why can’t sound waves be polarised?
Sound waves are longitudinal — the vibration of particles is parallel to the direction of wave propagation. Polarisation requires the ability to restrict vibration to a single transverse plane, which is only possible for transverse waves (like light). Since sound particles vibrate back and forth along the propagation direction, there is no transverse component to polarise.
How many questions from Wave Optics appear in JEE Main each year?
Wave Optics typically contributes 2-4 questions in JEE Main. YDSE problems (fringe width, shift, intensity) appear almost every year. Brewster’s angle and Malus’s Law are shorter single-step questions that appear frequently. The entire optics unit (ray + wave) can contribute 5-8 questions, making it one of Physics’s most reliable scoring chapters.
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