Class 12 Physics NCERT REPRINT 2026-27

Chapter 10: Wave Optics

Complete concise study notes, Huygens principle, Snell's law derivation, Young's double-slit experiment, interference, single-slit diffraction, polarization, Malus' law, solved examples 10.1–10.2, and exercises 10.1–10.6.

01 / Exam-Focused Notes

Chapter 10: Wave Optics

Topics covered: Huygens Principle & Wavefront Construction, Derivation of Snell's Law & Law of Reflection, Interference & Coherent Sources, Young's Double-Slit Experiment (YDSE) with Fringe Width derivation ($β = λ D/d$), Single Slit Fraunhofer Diffraction with Central Maxima Width ($2λ D/a$), and Polarisation by Polaroids with Malus's Law ($I = I_0 \cos^2θ$).

100% SYLLABUS
Chapter Navigation • NCERT Sections 10.1 – 10.7
10.1-10.2Huygens Principle 10.3Refraction & Reflection 10.4-10.5Interference & YDSE 10.6Single Slit Diffraction 10.7Polarisation & Malus
10.1 & 10.2
NCERT Sections

Introduction & Huygens Principle

Wavefront & Geometrical Construction

10.2.1 Definition of Wavefront & Huygens Principle

  • Wavefront: The continuous locus of all points in a medium that vibrate with the same phase.
    • Spherical Wavefront: Produced by a point source emitting waves uniformly in 3D space.
    • Cylindrical Wavefront: Produced by a linear slit or line source.
    • Plane Wavefront: A small portion of a spherical or cylindrical wavefront observed at a large/infinite distance from the source.
  • Direction of Ray: Light rays are always perpendicular to the wavefront at every point, indicating the direction of energy propagation.
  • Huygens Principle:
    1. Every point on a primary wavefront acts as a fresh source of secondary spherical disturbances called secondary wavelets, which spread out in all directions with the speed of the wave in that medium ($v$).
    2. The forward envelope (tangential surface) touching all these secondary wavelets at a later time $t + Δ t$ gives the new position and shape of the wavefront.
    3. Absence of Backwave: The amplitude of secondary wavelets is maximum in the forward direction and zero in the backward direction (rigorously proven in wave theory).
10.3
NCERT Section

Refraction and Reflection of Plane Waves using Huygens Principle

Wave Optics Derivations

10.3.1 Proof of Laws of Refraction (Snell's Law) & Reflection

  • Derivation of Snell's Law:
    Incident plane wavefront $AB$ meets refracting surface $PP'$ at angle $i$. Time taken by ray from $B$ to reach $C$ is $τ = BC/v_1$.
    During this time $τ$, secondary wavelet from $A$ travels distance $AE = v_2 τ$ in second medium.
    From right-angled triangles $\triangle ABC$ and $\triangle AEC$: $$\sin i = BCAC = v_1 τAC, \qquad \sin r = AEAC = v_2 τAC$$ $$\implies \sin i\sin r = v_1v_2 = c/n_1c/n_2 = n_2n_1 \implies n_1 \sin i = n_2 \sin r \quad [Snell's Law]$$
  • Wavelength and Frequency Relation: $$λ_1λ_2 = v_1v_2 \implies λ_2 = λ_1n \quad (Wavelength and speed decrease in denser medium, frequency ν remains constant)$$
  • Derivation of Law of Reflection:
    Incident wavefront $AB$ on reflecting surface $MN$: $BC = vτ, AE = vτ$.
    In $\triangle EAC$ and $\triangle BAC$, $AC$ is common, $AE = BC$, $\angle AEC = \angle ABC = 90^° \implies \triangle EAC \cong \triangle BAC$. $$\implies \angle i = \angle r \quad [Law of Reflection]$$
Fig 10.4: Refraction of Plane Wavefront at Rare-Dense Interface (Snell's Law Proof)
Interface PP' Medium 1 (Rare, v₁, n₁) Medium 2 (Dense, v₂, n₂) B A C E BC = v₁ τ AE = v₂ τ sin i / sin r = v₁/v₂ = n₂/n₁ (Snell's Law)
Example 10.1

Frequency Constancy, Energy & Photon Intensity

(a) Why do reflected and refracted light have the same frequency as incident light?
(b) When light slows down in a denser medium, does it carry less energy?
(c) In wave theory, intensity $∝ A^2$. What determines intensity in photon theory?

(a) Reflection/refraction arises from forced oscillations of bound electrons/atoms in matter. The forced oscillators vibrate with the exact frequency of the driving incident light wave and re-radiate waves of the identical frequency $ν$.
(b) No. The energy carried by a light wave depends strictly on its amplitude, not on its propagation speed $v$.
(c) In the photon picture, for a given frequency $ν$, intensity is determined by the number of incident photons crossing unit area per unit time ($I = N hν / A t$).
10.4 & 10.5
NCERT Sections

Interference of Light Waves & Young's Double-Slit Experiment (YDSE)

Superposition & Coherence

10.4.1 Principle of Superposition & Resultant Intensity Formula

  • Coherent Sources: Two sources are coherent if they emit light waves of identical frequency with a constant (zero or fixed) phase difference with time. Two independent sources can never be coherent because atomic emissions undergo random phase jumps every $\sim 10-10 s$.
  • Superposition of Two Coherent Waves ($y_1 = a\cosω t, y_2 = a\cos(ω t + φ)$): $$y = y_1 + y_2 = 2a \cos(φ/2) \cos(ω t + φ/2)$$ $$I = 4 I_0 \cos^2(φ2)$$
$$Constructive (Bright Maxima): Δ x = nλ, \quad φ = 2nπ \implies Imax = 4 I_0$$ $$Destructive (Dark Minima): Δ x = (n + 12)λ, \quad φ = (2n+1)π \implies Imin = 0$$
YDSE Fringe Geometry

10.5.1 Young's Double-Slit Experiment & Fringe Width

  • Fringe Position on Screen (Slit separation $d$, Screen distance $D$):
    Path difference $Δ x = x dD$.
    • $nth$ Bright Fringe Position: $x_n = n λ Dd \quad (n = 0, ± 1, ± 2, \dots)$
    • $nth$ Dark Fringe Position: $x_n = (n - 12)λ Dd \quad (n = 1, 2, \dots)$
  • Fringe Width ($β$): Distance between any two successive bright or dark fringes: $$β = xn+1 - x_n = λ Dd$$
  • Angular Fringe Width: $θ = βD = λd$ (rad).
Fig 10.8: Young's Double-Slit Interference Geometry & Fringe Width β
S₁ S₂ d O (Central Maxima) Screen P(x) Distance D >> d ⟹ Fringe Width β = λD/d
10.6
NCERT Section

Diffraction of Light: Single Slit

Single Slit Diffraction

10.6.1 Fraunhofer Diffraction at a Single Slit (Width $a$)

  • Diffraction: The bending of light around the corners of obstacles or apertures into the region of geometrical shadow. Significant when obstacle/slit size $a \sim λ$.
  • Minima & Maxima Conditions:
    For slit width $a$ and diffraction angle $θ$:
    • Diffraction Minima (Zero Intensity): $$a \sinθ = n λ \implies θ_n ≈ nλa \quad (n = ± 1, ± 2, ± 3, \dots)$$
    • Secondary Maxima: $$θ ≈ (n + 12)λa \quad (n = ± 1, ± 2, \dots)$$
  • Central Maximum Width:
    • Angular Width: $2θ_1 = 2λa$
    • Linear Width on Screen: $β_0 = 2 x_1 = 2λ Da = 2β$
Fig 10.16: Single Slit Diffraction Intensity Distribution Curve
Angle θ → I₀ (Central Max) -λ/a +λ/a -2λ/a +2λ/a Central Width 2θ = 2λ/a
10.7
NCERT Section

Polarisation of Light & Malus' Law

Transverse Nature & Malus' Law

10.7.1 Polaroids, Plane Polarised Light & Law of Malus

  • Unpolarised vs Polarised Light: In unpolarised light, the electric field $\mathbf{E}$ oscillates randomly in all transverse directions. In plane polarised light, $\mathbf{E}$ vibrations are restricted to a single plane.
  • Action of a Polaroid:
    Long polymer chains absorb electric vectors parallel to them and transmit vectors along the perpendicular pass-axis.
    When unpolarised light of intensity $Iunpol$ passes through a polaroid: $$I_1 = 12 Iunpol$$
  • Law of Malus:
    When plane-polarised light of intensity $I_0$ falls on an analyzer whose pass-axis makes angle $θ$ with the polariser: $$I = I_0 \cos^2θ$$
Example 10.2

Polaroid Sheet Rotated Between Two Crossed Polaroids

Discuss the intensity of transmitted light when a polaroid sheet $P_2$ is rotated between two crossed polaroids $P_1$ and $P_3$.

Let $I_0$ be intensity after $P_1$. Angle between pass-axes of $P_1$ and $P_2$ is $θ$.
Intensity emerging from $P_2$: $I_2 = I_0 \cos^2θ$.
Angle between $P_2$ and $P_3$ is $(90^° - θ)$.
Intensity emerging from $P_3$: $$I = I_2 \cos^2(90^° - θ) = I_0 \cos^2θ \sin^2θ = I_04 (2\sinθ\cosθ)^2 = I_04 \sin^2(2θ)$$
Transmitted intensity is maximum when $\sin(2θ) = 1 \implies 2θ = 90^° \implies θ = 45^°$, with $Imax = I_0/4$.
02 / Interactive Assessment

Concept Check & Practice Quiz

15 targeted MCQs covering Huygens principle, Snell's law derivation, Young's double-slit experiment, single-slit diffraction, and polarization/Malus' law with in-depth explanations.

15 MCQS
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03 / Textbook Solutions

NCERT Exercises 10.1 – 10.6

Stepwise solutions for exercise questions in NCERT Class 12 Physics Chapter 10 Wave Optics Reprint 2026-27.

6 QUESTIONS
Ex 10.1Reflected & Refracted Light Wave Parameters

10.1 Monochromatic light of wavelength $589 nm$ is incident from air on a water surface ($n = 1.33$). What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light?

(a) Reflected: $λ = 589 nm, ν = 5.09 × 1014 Hz, v = 3.0 × 10^8 m/s$
(b) Refracted: $λ' = 443 nm, ν = 5.09 × 1014 Hz, v' = 2.26 × 10^8 m/s$.
Frequency of incident light:
$$ν = cλ = 3.0 × 10^8 m s-1589 × 10-9 m = 5.09 × 1014 Hz$$
(a) Reflected light (in air): Speed $v = c = 3.0 × 10^8 m s-1$, Wavelength $λ = 589 nm$, Frequency $ν = 5.09 × 1014 Hz$.
(b) Refracted light (in water):
Frequency remains unchanged: $ν = 5.09 × 1014 Hz$.
Speed $v' = cn = 3.0 × 10^81.33 = 2.26 × 10^8 m s-1$.
Wavelength $λ' = λn = 589 nm1.33 = 442.86 nm ≈ 443 nm$.
Ex 10.2Shapes of Wavefronts

10.2 What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

(a) Spherical wavefront: Light radiates uniformly in all directions in 3D space from a point source.
(b) Plane wavefront: Rays emerging from a convex lens with source at focus become parallel to the principal axis, so the wavefront normal to parallel rays is a plane.
(c) Plane wavefront: Due to astronomical distance, the intercepted small portion of the vast spherical wavefront arriving at Earth approximates an ideal plane wavefront.
Ex 10.3Speed of Light in Glass & Dispersion

10.3 (a) The refractive index of glass is $1.5$. What is the speed of light in glass?
(b) Is the speed of light in glass independent of colour? Which colour travels slower: red or violet?

(a) $v = 2.0 × 10^8 m/s$ • (b) Violet light travels slower than red.
(a) $v = cn = 3.0 × 10^8 m s-11.5 = 2.0 × 10^8 m s-1$.
(b) No, speed depends on colour (wavelength). Since $nviolet > nred$, speed $v = c/n \implies vviolet < vred$. Hence violet light travels slower in glass.
Ex 10.4YDSE Wavelength Determination

10.4 In a YDSE, slits are separated by $d = 0.28 mm$ and screen is placed at $D = 1.4 m$. Distance between central bright fringe and 4th bright fringe is $x_4 = 1.2 cm$. Determine wavelength $λ$.

Wavelength $λ = 6.0 × 10-7 m = 600 nm$.
Position of $nth$ bright fringe: $x_n = n λ Dd \implies x_4 = 4 λ Dd$.
$$λ = x_4 d4 D = (1.2 × 10-2 m) × (0.28 × 10-3 m)4 × 1.4 m = 3.36 × 10-65.6 = 6.0 × 10-7 m = 600 nm$$
Ex 10.5YDSE Intensity Calculation

10.5 In YDSE, intensity where path difference is $λ$ is $K$ units. What is the intensity where path difference is $λ/3$?

Intensity $I' = K/4$.
General intensity: $I = 4 I_0 \cos^2(φ/2)$.
For path difference $Δ x = λ \implies φ = 2π \implies I = 4 I_0 \cos^2(π) = 4 I_0 = K$.
For path difference $Δ x' = λ/3 \implies φ' = 2πλ(λ/3) = 2π3 = 120^°$.
$$I' = 4 I_0 \cos^2(120^°2) = K \cos^2(60^°) = K × (12)^2 = K4$$
Ex 10.6Coinciding Bright Fringes of Two Wavelengths

10.6 A beam of two wavelengths $λ_1 = 650 nm$ and $λ_2 = 520 nm$ is used in YDSE ($D$ and $d$).
(a) Find the distance of 3rd bright fringe for $λ_1 = 650 nm$.
(b) What is the least distance from central maximum where bright fringes of both wavelengths coincide?

(a) $x_3 = 1950 Dd nm = 1.95 Dd mm$ • (b) $xcoincide = 2600 Dd nm = 2.6 Dd mm$ ($n_1 = 4, n_2 = 5$).
(a) 3rd Bright fringe for $λ_1 = 650 nm$:
$$x_3 = 3 λ_1 Dd = 3 × (650 × 10-9) Dd = 1.95 × 10-6Dd m$$
(b) Coincidence condition:
$$x = n_1 λ_1 Dd = n_2 λ_2 Dd \implies n_1 λ_1 = n_2 λ_2$$ $$n_1n_2 = λ_2λ_1 = 520 nm650 nm = 45$$ Smallest non-zero integers: $n_1 = 4, n_2 = 5$.
$$xmin = 4 × (650 × 10-9 m) Dd = 2.6 × 10-6Dd m = 2.6 Dd mm$$ (The 4th bright fringe of $650 nm$ coincides with the 5th bright fringe of $520 nm$).
04 / Rapid Reference

Chapter Summary & Points to Ponder

Core wave optics principles, YDSE fringe formulae, single-slit diffraction criteria, Malus' law, and official NCERT Points to Ponder for Chapter 10.

100% SYLLABUS

Huygens Principle

Secondary Wavelets • Envelope

Every point on a wavefront acts as a source of secondary spherical wavelets; their forward envelope gives the new wavefront.

Snell's Law from Wave Theory

sin i / sin r = v₁ / v₂ = n₂ / n₁

Proves light travels slower in optically denser medium ($v = c/n$), with $λ' = λ/n$ and frequency constant.

YDSE Fringe Width

β = λ D / d

Constructive at $Δ x = nλ$ ($Imax = 4I_0$); Destructive at $Δ x = (n + 1/2)λ$ ($Imin = 0$).

Single Slit Diffraction

Minima: a sinθ = nλ

Central bright maximum has angular width $2θ = 2λ/a$ and linear width $β_0 = 2λ D/a$.

Polarisation & Malus' Law

I = I₀ cos²θ • I₁ = ½ Iunpol

Proves light is a transverse wave. Intensity transmitted by analyzer varies as the square of the cosine of the angle.

Energy Redistribution

Conservation of Energy

Interference and diffraction redistribute light energy without any net loss: energy depleted at dark fringes appears at bright fringes.

NCERT Official

Points to Ponder

  1. Wave vs Ray Picture: Wavefronts spread in 3D space, but when wavelength $λ$ is negligible compared to apparatus dimensions, energy travels along narrow rectilinear rays (geometrical optics limit).
  2. Interference of Amplitudes: The unique characteristic of waves is coherent superposition of complex amplitudes, yielding regions of complete destructive cancellation ($I=0$) and intense reinforcement ($I=4I_0$).
  3. Diffraction as the Resolution Limit: The wave nature and diffraction of light set the ultimate physical limit on the resolving power of microscopes and telescopes (Rayleigh criterion).
  4. Polarisation Confirms Transverse Character: Interference and diffraction occur for all wave types (including longitudinal sound waves). Polarisation occurs exclusively for transverse waves like light.
05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to evaluate your preparation.

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