By the end of this chapter you'll be able to…

  • 1State Huygens' principle and derive reflection and refraction
  • 2Analyse Young's double-slit experiment and fringe width
  • 3Describe single-slit diffraction and the central maximum
  • 4Distinguish interference from diffraction
  • 5Explain polarisation, Brewster's law, and Malus' law
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Why this chapter matters
Wave optics reveals the true wave nature of light, explaining interference, diffraction, and polarisation that ray optics cannot. Young's double-slit experiment, diffraction, and polarisation underlie holography, optical instruments' resolving power, and LCD/3D display technology.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Wave Optics

'When light passes through a narrow slit, it SPREADS — not because it is obstructed, but because it is a WAVE.'

1. Chapter Overview

Wave optics treats light as a WAVE (specifically, an ELECTROMAGNETIC wave). This chapter covers: HUYGENS' PRINCIPLE (every point on a wavefront is a source of secondary wavelets — the foundation of wave optics), the WAVEFRONT (the locus of points in the same phase), COHERENT SOURCES (essential for observing interference), YOUNG'S DOUBLE SLIT EXPERIMENT (YDSE — the classic demonstration of interference), DIFFRACTION (the bending of light around obstacles), and POLARISATION (restricting the vibration of light to one plane).


2. Huygens' Principle

Statement

  • 'Every point on a WAVEFRONT acts as a source of SECONDARY SPHERICAL WAVELETS. The new wavefront is the ENVELOPE of these wavelets.'

Reflection and Refraction Using Huygens' Principle

  • Reflection: Angle of incidence = Angle of reflection (proved using wave theory).
  • Refraction: Snell's law n₁ sin θ₁ = n₂ sin θ₂ — derived from Huygens' principle by comparing speeds in the two media.
  • 'Huygens' principle explains BOTH reflection and refraction — the change in wave speed at the boundary causes the bending.'

3. Interference — Young's Double Slit Experiment

Conditions for Interference

  • Sources must be COHERENT (constant phase difference). Obtained by splitting a SINGLE wavefront (YDSE).

Path Difference and Fringe Pattern

  • Path difference: Δx = d sin θ ≈ dy/D for small angles.
  • Constructive interference (bright fringe): Δx = nλ, n = 0, 1, 2, ...
  • Destructive interference (dark fringe): Δx = (2n+1)λ/2.
  • Fringe width: β = λD/d. 'The distance between two consecutive bright or dark fringes.'

Intensity Distribution

  • I = I₁ + I₂ + 2√(I₁I₂) cos Δφ.
  • For I₁ = I₂ = I₀: I_max = 4I₀, I_min = 0.
  • 'Constructive interference → waves ADD. Destructive → waves CANCEL.'

Worked Example 1

Problem: In YDSE, slits are 0.5 mm apart and the screen is 1 m away. The third bright fringe is at 3 mm from centre. Find λ. Solution: For bright fringe: y_n = nλD/d. y₃ = 3λD/d ⇒ λ = y₃d/(3D) = (3×10⁻³×0.5×10⁻³)/(3×1) = (1.5×10⁻⁶)/3 = 5×10⁻⁷ m = 500 nm.


4. Diffraction

Single Slit Diffraction

  • 'When light passes through a slit comparable to its wavelength, it SPREADS — each point in the slit acts as a source.'

Conditions

  • Minima: a sin θ = nλ (n = 1, 2, 3, ...). a = slit width.
  • Angular width of central maximum: 2θ = 2λ/a.
  • Width of central maximum: w = 2λD/a.

Difference Between Interference and Diffraction

AspectInterference (YDSE)Diffraction (Single Slit)
SourceTWO coherent sourcesSINGLE slit (many point sources)
Fringe widthβ = λD/d — ALL fringes EQUALCentral max is TWICE as wide
IntensityAll maxima have SAME intensityCentral max is BRIGHTEST — decreases away
Minima conditiond sin θ = (2n+1)λ/2a sin θ = nλ

5. Polarisation

What is Polarisation?

  • Unpolarised light: Electric field vectors vibrate in ALL directions perpendicular to propagation.
  • Polarised light: Electric field vibrates in ONLY ONE plane.
  • 'Polarisation proves that light is a TRANSVERSE wave — longitudinal waves (sound) cannot be polarised.'

Methods of Polarising Light

  1. Polaroid: A sheet that transmits light vibrating in only one plane.
  2. Reflection: Light reflected at BREWSTER'S ANGLE is completely polarised perpendicular to the plane of incidence.
  3. Double refraction: Some crystals (calcite) split light into two polarised rays.

Brewster's Law

  • tan θ_B = μ (where θ_B = Brewster's angle).
  • 'At Brewster's angle, the REFLECTED and REFRACTED rays are PERPENDICULAR to each other.'

Law of Malus

  • I = I₀ cos² θ (where θ = angle between the polariser and analyser transmission axes).
  • 'When two polarisers are CROSSED (θ = 90°), NO light passes through.'

6. Resolving Power

  • Rayleigh's criterion: Two point sources are JUST RESOLVED when the central maximum of one falls on the FIRST MINIMUM of the other.
  • Resolving power of a microscope: RP = 1/d_min = (2NA)/λ (NA = numerical aperture).
  • Resolving power of a telescope: RP = 1/θ_min = D/(1.22λ) — depends on the DIAMETER of the objective.

7. Common Mistakes

  1. Coherent sources requirement: Two INDEPENDENT sources (two different bulbs) do NOT produce interference — they are INCOHERENT. Both slits must be illuminated by the SAME source.
  2. Interference vs diffraction: Interference uses TWO sources. Diffraction uses ONE extended source (the slit). The patterns look similar but have DIFFERENT intensity distributions.
  3. Brewster's angle: The reflected ray is 100% polarised, but the refracted ray is ONLY PARTIALLY polarised.
  4. Polarisation direction: A polaroid DOES NOT affect the DIRECTION of propagation — it only selects one plane of VIBRATION.

8. CBSE Exam Focus

  1. Huygens' principle — reflection and refraction using wave theory
  2. Interference — YDSE (β = λD/d), constructive/destructive conditions
  3. Diffraction — single slit, minima, width of central maximum
  4. Difference between interference and diffraction
  5. Polarisation — Brewster's law (tan θ_B = μ), Malus' law (I = I₀ cos² θ)
  6. Resolving power of microscopes and telescopes

9. Self-Test

Q1: In YDSE, λ = 600 nm, d = 0.2 mm, D = 1.5 m. Find the fringe width. A1: β = λD/d = (600×10⁻⁹×1.5)/(0.2×10⁻³) = (9×10⁻⁷)/(2×10⁻⁴) = 4.5×10⁻³ m = 4.5 mm.

Q2: A single slit of width 0.1 mm is illuminated by λ = 500 nm. The screen is at 1 m. Find width of central maximum. A2: w = 2λD/a = 2×500×10⁻⁹×1/(0.1×10⁻³) = (10⁻⁶)/(10⁻⁴) = 10⁻² m = 1 cm.

Q3: Find Brewster's angle for glass (μ = 1.5). A3: tan θ_B = μ = 1.5 ⇒ θ_B = tan⁻¹(1.5) ≈ 56.3°.

Q4: Two polarisers are at 60° to each other. If I₀ is incident on the first, find the transmitted intensity. A4: After first polariser: I₁ = I₀/2. After second (analyser): I₂ = I₁ cos² 60° = (I₀/2)(1/2)² = (I₀/2)(1/4) = I₀/8.

Q5: In YDSE, what happens to the fringe width if the whole apparatus is immersed in water (μ = 4/3)? A5: In water, λ' = λ/μ = λ/(4/3) = 3λ/4. β' = λ'D/d = (3/4)(λD/d) = (3/4)β. Fringe width DECREASES by factor of 3/4.


10. Conclusion

Wave optics reveals the TRUE NATURE of light:

  • HUYGENS: 'The wavefront model — every point is a source of secondary waves. Reflection, refraction, and diffraction all explained.'
  • INTERFERENCE: 'Light + Light can give DARKNESS — when the waves are exactly out of phase.'
  • DIFFRACTION: 'Light BENDS around corners when the obstacle is small enough. The narrower the slit, the MORE the spreading.'
  • POLARISATION: 'Light waves vibrate in all directions — or in just one. The ultimate proof that light is TRANSVERSE.'

'Wave optics does not replace ray optics — it SUPERSEDES it, explaining phenomena that the ray model cannot.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Fringe width (YDSE)
beta = lambda D / d
Bright: path difference = n lambda; dark: (2n+1)lambda/2.
Single-slit diffraction
Minima: a sin(theta) = n lambda; central max width = 2 lambda D / a
Central maximum is twice as wide as others.
Brewster's law
tan(theta_B) = mu
Reflected light is fully polarised at Brewster's angle.
Malus' law
I = I0 cos^2(theta)
theta is angle between polariser and analyser axes.
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Using two independent bulbs as coherent sources
Interference needs coherent sources; both slits must be lit by the same single source.
WATCH OUT
Treating interference and diffraction as identical
Interference uses two sources with equal-intensity fringes; diffraction uses one slit with a bright, wide central maximum.
WATCH OUT
Saying the refracted ray at Brewster's angle is fully polarised
Only the reflected ray is fully polarised at Brewster's angle; the refracted ray is partially polarised.
WATCH OUT
Forgetting fringe width changes in a medium
In a medium the wavelength shrinks to lambda/mu, so the fringe width decreases by the same factor.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· YDSE
In YDSE, lambda = 600 nm, d = 0.2 mm, D = 1.5 m. Find the fringe width.
Show solution
beta = lambda D/d = (600e-9 x 1.5)/(0.2e-3) = 4.5e-3 m = 4.5 mm.
Q2MEDIUM· Diffraction
A single slit of width 0.1 mm with lambda = 500 nm and screen at 1 m. Find the central maximum width.
Show solution
w = 2 lambda D / a = 2 x 500e-9 x 1 / 0.1e-3 = 1e-2 m = 1 cm.
Q3EASY· Polarisation
Find Brewster's angle for glass (mu = 1.5).
Show solution
tan(theta_B) = 1.5, so theta_B = arctan(1.5) = 56.3 degrees.
Q4MEDIUM· Malus
Two polarisers are at 60 degrees. If I0 falls on the first, find the transmitted intensity.
Show solution
After the first polariser I1 = I0/2. After the second I2 = I1 cos^2(60) = (I0/2)(1/4) = I0/8.
Q5MEDIUM· Medium Effect
What happens to YDSE fringe width if the apparatus is immersed in water (mu = 4/3)?
Show solution
Wavelength becomes lambda/mu = 3 lambda/4, so the fringe width decreases to 3/4 of its original value.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Huygens' principle: each wavefront point is a source of secondary wavelets.
  • Interference needs coherent sources (single wavefront split).
  • YDSE fringe width beta = lambda D/d; bright at n lambda, dark at (2n+1)lambda/2.
  • Single-slit diffraction minima at a sin(theta) = n lambda; central maximum twice as wide.
  • Interference: equal fringes; diffraction: bright central maximum decreasing outward.
  • Polarisation shows light is transverse; Brewster's law tan(theta_B) = mu.
  • Malus' law I = I0 cos^2(theta); resolving power set by Rayleigh's criterion.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks across the chapter

Question typeMarks eachTypical countWhat it tests
Interference (YDSE)3-51Fringe width and conditions
Diffraction31Single-slit minima and central maximum
Polarisation2-31Brewster's and Malus' laws
Prep strategy
  • Memorise fringe-width and diffraction formulas
  • Tabulate interference vs diffraction differences
  • Learn Brewster's and Malus' laws
  • Note how a medium changes wavelength and fringe width

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Displays and glare reduction

Polarisation is used in LCD screens, polarised sunglasses, and 3D cinema glasses.

Precision measurement

Interference is used in interferometers to measure tiny distances and test optical surfaces.

Imaging resolution

Diffraction sets the resolving power limit of microscopes and telescopes.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Use beta = lambda D/d for fringe problems
  2. Apply a sin(theta) = n lambda for diffraction minima
  3. Account for the I0/2 drop at the first polariser
  4. Adjust wavelength by mu when the medium changes

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Derive the intensity distribution of single-slit diffraction.
  • Analyse thin-film interference and Newton's rings.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Physics examHigh
JEE Main and Advanced (Wave Optics)High
NEET PhysicsMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Stable interference requires the two sources to be coherent, meaning they maintain a constant phase difference over time. Two independent bulbs emit light in random, rapidly changing phases because their atoms radiate independently. The phase difference between them fluctuates millions of times per second, so any interference pattern shifts too quickly to be seen and averages out. Young's experiment solves this by using a single source and splitting its wavefront with two slits, ensuring the two beams stay coherent and produce steady fringes.

Polarisation is the restriction of a wave's vibrations to a single plane. Only transverse waves, whose oscillations are perpendicular to the direction of travel, can be polarised, because there is a meaningful plane of vibration to restrict. Longitudinal waves like sound oscillate along the direction of travel and cannot be polarised. Since light can be polarised by polaroids, reflection at Brewster's angle, and double-refracting crystals, this demonstrates conclusively that light is a transverse (electromagnetic) wave.
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Last reviewed on 30 May 2026. Written and reviewed by subject-matter experts — read about our process.
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