Wave Optics Class 12 Notes | CBSE Physics Chapter 10 (Free PDF)

Chapter summary

Wave Optics treats light as a wave and builds everything from Huygens’ principle, explaining reflection, refraction, interference, diffraction and polarisation. It covers Young’s double-slit experiment, single-slit diffraction and Malus’s law, which are reliable sources of NEET questions every year. Mastering the fringe-width and intensity formulas here makes a cluster of one and two mark questions almost automatic.

Chapter notes

Key Concepts

1. Huygens’ Principle

Every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront is the forward envelope of all these secondary wavelets.

Used to derive the laws of reflection and refraction.

2. Interference - Young’s Double Slit Experiment (YDSE)

When light from two coherent sources (slits S₁ and S₂) overlaps, it produces a pattern of bright and dark fringes.

Condition for bright fringe: Path difference = nλ (n = 0, 1, 2, …)

Condition for dark fringe: Path difference = (n + ½)λ

Fringe width: β = λD/d

  • λ = wavelength of light
  • D = distance from slits to screen
  • d = separation between slits

Central fringe is always bright and widest.

3. Diffraction - Single Slit

When light passes through a narrow slit, it spreads out and forms a pattern of bright and dark bands.

Central maximum: Width = 2λD/a (where a = slit width)

Minima condition: a sin θ = nλ (n = ±1, ±2, …)

Secondary maxima: a sin θ = (n + ½)λ

Central maximum is twice as wide as secondary maxima.

4. Polarisation

Light is a transverse wave. Polarisation is the restriction of vibrations of the electric field to a single plane.

Malus’s Law: I = I₀ cos² θ (intensity of polarised light through an analyser at angle θ)

Brewster’s Law: tan ip = n (ip = polarising angle; reflected and refracted rays are perpendicular)

Polaroid uses: Sunglasses, LCD screens, photography, 3D movies


Solved Examples

Example 1

In YDSE, slit separation d = 0.5 mm, screen distance D = 1 m, wavelength λ = 600 nm. Find the fringe width.

Answer: β = λD/d = (600 × 10⁻⁹ × 1)/(0.5 × 10⁻³) = 1.2 mm

Example 2

Unpolarised light of intensity I₀ passes through two polaroids with axes at 60°. Find final intensity.

Answer: After first polaroid: I₁ = I₀/2. After second: I₂ = (I₀/2)cos²60° = (I₀/2)(1/4) = I₀/8


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Important Questions for Board Exams

3-Mark

  1. Derive the expression for fringe width in YDSE.
  2. State and prove Malus’s law.
  3. What is diffraction? Compare the diffraction pattern with interference pattern.

5-Mark

  1. Describe Young’s double slit experiment. Derive expressions for bright, dark fringes and fringe width.
  2. What is polarisation? State Brewster’s law and Malus’s law with derivations.

Quick Revision Points

  • Huygens: each point on wavefront → source of secondary wavelets
  • YDSE: β = λD/d; bright: Δ = nλ; dark: Δ = (n+½)λ
  • Diffraction: central max width = 2λD/a; minima: a sin θ = nλ
  • Polarisation: transverse wave property; Malus: I = I₀cos²θ
  • Brewster: tan ip = n; reflected light is fully polarised

Previous: Ch 9 - Ray Optics
Next: Ch 11 - Dual Nature

🃏 Flash Cards: Wave Optics

Class 12 Physics · Chapter 10 – swipe through all 11 cards to understand the whole chapter.

🌊Start here1/11

Wavefronts & Huygens’ Principle

A wavefront joins all points vibrating in the same phase, and every point on it acts as a new source of wavelets.

Each point → secondary wavelet; forward envelope = new wavefront

Rays are perpendicular to the wavefront and point the way light travels.

  • Point source → spherical, line source → cylindrical, source at infinity → plane wavefronts.
  • Forward envelope of the wavelets after time t gives the next wavefront.
  • Explains reflection and refraction purely from wave geometry.
🪟Core idea2/11

Refraction from Huygens

In a denser medium light slows down, so the wavefront bends toward the normal and Snell’s law drops out.

n1 sin i = n2 sin r ; λ_medium = λ_vacuum / n

Frequency f stays constant across a boundary; speed and λ decrease in a denser medium.

  • Speed drops in a denser medium, so wavelets travel less and the front tilts.
  • Only λ and v change at a boundary; f is fixed.
  • Recovers Snell’s law without Newton’s particle picture.
Core law3/11

Superposition & Path Difference

When coherent waves overlap their displacements add, giving steady bright and dark patches.

Bright: Δx = nλ ; Dark: Δx = (2n−1) λ/2 ; φ = (2π/λ)·Δx

Path difference Δx and phase difference φ are two languages for the same thing.

  • Whole-wavelength path difference → crest meets crest → bright.
  • Half-wavelength (odd) path difference → crest meets trough → dark.
  • One wavelength of path = 2π radians of phase.
🔆Key formula4/11

Interference Intensity

Two equal-amplitude coherent waves give an intensity that swings between zero and four times a single wave.

I = 4 I0 cos2(φ/2) ; I_max = (√I1 + √I2)2 , I_min = (√I1 − √I2)2

Energy is only redistributed from dark to bright regions, never lost.

  • Constructive max = 4 I0 at φ = 0; destructive min = 0 at φ = π.
  • Unequal sources: brightness uses the sum/difference of √I.
  • Interference conserves total energy.
🔗Condition5/11

Coherence

A steady fringe pattern needs sources locked in a constant phase relationship.

Coherent → same f + constant phase difference

That is why one source is split into two, not two separate bulbs.

  • Sources must keep a constant phase difference in time.
  • They must have the same frequency.
  • Two independent bulbs are incoherent, so no fringes form.
🎯Key experiment6/11

Young’s Double-Slit (YDSE)

Two coherent slits a distance d apart throw equally spaced fringes onto a screen at distance D.

Δx = y d / D ; y_n = n λ D / d ; β = λ D / d

Central fringe is bright (zero path difference); fringes are equally spaced and equally bright.

  • β grows with λ and D, shrinks with d.
  • Angular fringe width θ = λ/d is independent of D.
  • Immerse in liquid of index n → β becomes β/n.
🌈Exam fact7/11

YDSE with White Light

White light gives a white centre flanked by coloured fringes.

Central fringe white ; violet nearest centre, red farthest

Because β ∝ λ, longer-wavelength colours spread out more.

  • Zero path difference is white for all colours → white central fringe.
  • Violet (small λ) sits closest to the centre.
  • Higher orders overlap and wash out into white.
🚪Key process8/11

Single-Slit Diffraction

One slit of width a spreads light into a wide central band with fainter side bands.

Minima: a sinθ = nλ ; Central width w = 2 λ D / a ; θ_half = λ/a

Here nλ marks DARK, the reverse of double-slit; central max is twice as wide as side maxima.

  • Pairing trick: top-half source cancels its partner a/2 below.
  • Side maxima are weaker and fall off sharply (unlike equal YDSE fringes).
  • Narrower slit → wider central peak (more spreading).
🔭Application9/11

Resolving Limit

Diffraction sets the finest detail an instrument can separate.

θ_min ≈ 1.22 λ / D (circular aperture)

Smaller θ_min = better resolution; bigger aperture D helps, shorter λ helps.

  • Larger aperture D → smaller θ_min → sharper resolution.
  • Shorter wavelength → finer detail resolved.
  • Why telescopes and microscopes chase large apertures / short λ.
🧭Core law10/11

Polarisation & Malus’s Law

Light is transverse, so its vibration can be confined to one plane by a polaroid.

Unpolarised → I = I0/2 ; Malus: I = I0 cos2θ

Sound is longitudinal and CANNOT be polarised (a classic NEET one-marker).

  • A polaroid passes exactly half of unpolarised intensity.
  • θ = 0 → full transmission; θ = 90° → crossed polaroids → zero.
  • Transmitted light is plane-polarised along the polaroid’s axis.
🪞Advanced11/11

Brewster’s Angle

At one special incidence angle the reflected ray is completely plane-polarised.

tan θ_B = n ; θ_B + r = 90°

At θ_B the reflected and refracted rays are perpendicular; for glass (n ≈ 1.5), θ_B ≈ 57°.

  • Reflected ray at θ_B is fully plane-polarised.
  • Reflected ⟂ refracted at this angle.
  • Brewster’s angle increases with refractive index n.
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📝 Practice Wave Optics — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2020
Two coherent sources of light interfere and produce a fringe pattern on a screen. For the central maximum, the phase difference between the two waves will be:
Correct answer: A. At the central maximum the path difference is zero (the point is equidistant from both sources). Phase difference φ = (2π)/(λ)×Δ x = (2π)/(λ)× 0 = 0. Zero phase difference gives full constructive interference.
🔎 See the full step-by-step solution in the app →
Q2NEET 2020
In Young’s double-slit experiment, if the separation between the coherent sources is halved and the distance of the screen from them is doubled, then the fringe width becomes:
Correct answer: B. Fringe width β = (λ D)/(d). New d’ = d/2 and D’ = 2D, so β’ = (λ (2D))/(d/2) = 4 (λ D)/(d) = 4β. The fringe width becomes four times.
🔎 See the full step-by-step solution in the app →
Q3NEET 2020
Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of a telescope whose objective has a diameter of 2 m is:
Correct answer: D. Limit of resolution Δθ = (1.22λ)/(D) = (1.22×600×10⁻⁹)/(2) = (7.32×10⁻⁷)/(2) = 3.66×10⁻⁷ rad.
🔎 See the full step-by-step solution in the app →
Q4NEET 2020
The Brewster angle i_b for an interface (light going from a rarer to a denser medium) should be:
Correct answer: B. Brewster’s law: tan i_b = n. For any interface from a rarer to a denser medium, n > 1, so tan i_b > 1, which means i_b > 45°. Since the index is finite, i_b < 90°. Hence 45° < i_b < 90°.
🔎 See the full step-by-step solution in the app →
Q5NEET 2020
In a Young’s double-slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference:
Correct answer: C. For the nth minimum the path difference is Δ x = (2n-1)(λ)/(2). For the fifth minimum, n = 5: Δ x = (2×5 – 1)(λ)/(2) = 9(λ)/(2).
🔎 See the full step-by-step solution in the app →
Q6NEET 2019
In a double-slit experiment, when light of wavelength 400 nm was used, the angular width of the first minimum formed on a screen placed 1 m away was found to be 0.2°. What will be the angular width of the first minimum if the entire experimental apparatus is immersed in water? (n_water = 4/3)
Correct answer: A. Angular width of a diffraction minimum ∝ λ. In water the wavelength becomes λ/n, so the angular width scales by 1/n: new width = 0.2° × (3)/(4) = 0.15°.
🔎 See the full step-by-step solution in the app →
Q7NEET 2018
Unpolarised light is incident from air on a plane surface of a material of refractive index μ. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following is correct for this situation?
Correct answer: B. When the reflected and refracted rays are perpendicular, the incidence is at Brewster’s angle, where tan i = μ so i = tan⁻¹(μ) (not tan⁻¹(1/μ), ruling out D and A). At this angle the reflected ray is completely plane-polarised with its electric vector perpendicular to the plane of incidence.
🔎 See the full step-by-step solution in the app →
Q8NEET 2018
In Young’s double-slit experiment, the separation d between the slits is 2 mm, the wavelength λ is 5896 angstrom and distance D between the screen and slits is 100 cm. The angular fringe width is found to be 0.20°. To increase the angular fringe width to 0.21° (keeping λ and D the same), the slit separation d needs to be changed to:
Correct answer: B. Angular fringe width θ = (λ)/(d), so θ ∝ (1)/(d), giving (θ₁)/(θ₂) = (d₂)/(d₁). Thus d₂ = d₁(θ₁)/(θ₂) = 2 mm×(0.20)/(0.21) ≈ 1.9 mm. To widen the angle the slits must be brought closer.
🔎 See the full step-by-step solution in the app →
Q9NEET 2016
The intensity at the maximum in a Young’s double-slit experiment is I₀. When one of the two slits is covered, the intensity at a point on the screen where the path difference is λ/4 from that slit’s contribution becomes (treat the geometry so that the path difference there is λ/4):
Correct answer: C. A path difference of λ/4 corresponds to a phase difference φ = (2π)/(λ)×(λ)/(4) = (π)/(2). Using I = I₀cos²(φ/2) = I₀cos²(π/4) = I₀×(1)/(2) = (I₀)/(2).
🔎 See the full step-by-step solution in the app →
Q10NEET 2016
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio (Iₘₐₓ – Iₘᵢₙ)/(Iₘₐₓ + Iₘᵢₙ) is:
Correct answer: B. With I₁/I₂ = n, Iₘₐₓ = (√(I₁)+√(I₂))² and Iₘᵢₙ = (√(I₁)-√(I₂))². Then (Iₘₐₓ-Iₘᵢₙ)/(Iₘₐₓ+Iₘᵢₙ) = (2√(I₁ I₂)× 2)/(2(I₁+I₂)) = (2√(I₁ I₂))/(I₁+I₂). Dividing top and bottom by I₂ and using I₁/I₂=n gives (2√(n))/(n+1).
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Frequently Asked Questions

What is Huygens’ principle in wave optics?

Huygens’ principle states that every point on a wavefront acts as a source of secondary wavelets that spread out at the wave speed, and the forward envelope (tangent surface) of these wavelets gives the new wavefront a moment later. It is used to derive the laws of reflection and refraction from a pure wave picture.

What is the formula for fringe width in Young’s double-slit experiment?

The fringe width, the spacing between two consecutive bright or dark fringes, is beta = lambda D / d, where lambda is the wavelength, D is the slit-to-screen distance and d is the slit separation. So fringe width increases with wavelength and screen distance, and decreases as the slits move farther apart.

Is Wave Optics important for NEET?

Yes, Wave Optics is part of the Class 12 NEET physics syllabus and usually contributes one to two questions every year. Young’s double-slit experiment (fringe width), single-slit diffraction and Malus’s law are the most frequently tested ideas.

What is the difference between interference and diffraction?

Interference is the superposition of waves from two or more separate coherent sources, giving equally spaced and equally bright fringes, while diffraction is the bending and spreading of light from a single slit or edge, giving a wide central maximum with weaker, unequal side maxima. A handy contrast is that in a single slit a sin theta = n lambda marks the dark minima, the reverse of the double-slit bright condition.

Why can light be polarised but sound cannot?

Light is a transverse wave, so its electric field vibration can be restricted to a single plane, which is what polarisation means. Sound is a longitudinal wave with vibrations along the direction of travel, so it has no transverse plane to restrict and cannot be polarised, a point NEET often tests directly.

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