CLASS 12 PHYSICS
CHAPTER 10 — WAVE OPTICS
Conceptual • NCERT-Focused • NEET/JEE-Oriented
Ray optics (Chapter 9) treated light as straight lines and ignored its wavelength. This chapter puts the wavelength back in. Once light is treated as a wave, three phenomena become explainable that ray optics cannot touch at all: interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. (light + light = dark), diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. (light bending round obstacles), and polarisationpolarisationRestriction of a transverse wave's vibrations to a particular direction. (light having a direction of vibration). Everything here traces back to one geometric idea — the Huygens wavefrontwavefrontA surface joining points of a wave that are in the same phase..
WAVEFRONTwavefrontA surface joining points of a wave that are in the same phase. (Huygens) → REFLECTION / REFRACTION (re-derived as waves) → INTERFERENCEinterferenceRedistribution of intensity produced when coherent waves superpose. → DIFFRACTIONdiffractionBending and spreading of waves around an obstacle or through an aperture. → POLARISATIONpolarisationRestriction of a transverse wave's vibrations to a particular direction.
HUYGENS PRINCIPLEHuygens principleEvery point on a wavefront acts as a source of secondary wavelets whose envelope forms the next wavefront.
A wavefrontA wavefrontA wavefront is a surface of constant phase (e.g., all points in same condition of vibration). Huygens' principle states that every point on a wavefront acts as a source of secondary wavelets. is a surface of constant phase — every point on it oscillates in step. Huygens' principle says every point on a wavefrontA wavefrontA wavefront is a surface of constant phase (e.g., all points in same condition of vibration). Huygens' principle states that every point on a wavefront acts as a source of secondary wavelets. is itself a source of secondary spherical wavelets travelling outward at the wave speed; the common tangent (envelope) to all these wavelets gives the new wavefrontwavefrontA surface joining points of a wave that are in the same phase. an instant later.
Point source — produces spherical wavefronts.
Far from a point source — a small patch of the spherical wavefrontwavefrontA surface joining points of a wave that are in the same phase. looks flat — a plane wave.
Energy travels — perpendicular to the wavefrontwavefrontA surface joining points of a wave that are in the same phase., i.e. along what ray optics calls a 'ray'.
📖 NCERT FOCUS
Huygens' construction predicts a spurious 'backwave' travelling backward from the wavefrontwavefrontA surface joining points of a wave that are in the same phase., which is not observed. Huygens patched this with an ad-hoc assumption (secondary wavelets have zero amplitude backward); the absence of the backwave is properly justified only by more rigorous wave theory (Fresnel), not by Huygens' original argument.
Reflection of a plane wave — the same construction, at a mirror
Huygens' construction applies equally to reflection: draw a sphere of radius vτ from the point where the wavefrontwavefrontA surface joining points of a wave that are in the same phase. first touches the reflecting surface, then draw the common tangent from the point the far edge of the wavefrontwavefrontA surface joining points of a wave that are in the same phase. reaches the surface a time τ later. The two triangles formed (incident-side and reflected-side) are congruent, which immediately gives the law of reflection, i = r, without needing a separate postulate.
PLANE WAVEFRONTwavefrontA surface joining points of a wave that are in the same phase. AB → STRIKES MIRROR AT ANGLE i → SECONDARY WAVELET FROM A → CONGRUENT TRIANGLES → ANGLE OF REFLECTION = i
Snell's law and the law of reflection, re-derived
Using Huygens' construction at a boundary, geometry alone (similar triangles formed by the wavefrontwavefrontA surface joining points of a wave that are in the same phase. and the two rays) gives:
sin i / sin r = v₁/v₂ = n₂/n₁ (Snell's law)
🔥 HIGH-YIELD
The wave picture predicts light SLOWS DOWN in a denser medium (v₂ < v₁ when bending toward the normal). This was the opposite of Newton's corpuscular prediction (which said light should speed up) — Foucault's 1850 experiment, measuring light travelling slower in water than air, settled the dispute decisively in favour of the wave theory.
Refractive index n = c/v. When a wave crosses into a denser medium, both its speed AND wavelength decrease by the same factor, but its frequency stays exactly the same (frequency is fixed by the source, not the medium):
v₁/λ₁ = v₂/λ₂ = ν (frequency unchanged on refraction)
✏️ SOLVED EXAMPLE Refraction at a rarer medium — critical angle.
A ray travels from a medium of n = 1.5 toward air (n = 1). Find the critical angle.
sin iC = n₂/n₁ = 1/1.5 = 0.667
iC = sin⁻¹(0.667) ≈ 41.8°.
Answer: iC ≈ 41.8° — for any angle of incidence beyond this, the wave undergoes total internal reflection instead of refracting.
COHERENT AND INCOHERENT ADDITION OF WAVES
Superposition principle: the resultant displacement at any point is the vector sum of displacements from each wave individually. This alone doesn't guarantee a visible interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. pattern — that additionally needs coherence.
| Term | Meaning | Consequence |
|---|---|---|
| Coherent sourcescoherent sourcesSources that maintain a constant phase difference and have the same frequency. | Constant phase difference with time | Stable interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. pattern — fixed bright/dark fringes |
| Incoherent sources | Phase difference changes randomly, rapidly | No stable pattern; intensities simply add: I = I₁ + I₂ |
For two coherent sourcescoherent sourcesSources that maintain a constant phase difference and have the same frequency. of equal amplitude a and phase difference φ at a point, resultant intensity is:
I = 4I₀ cos²(φ/2)
| Condition | Path differencepath differenceThe difference between distances travelled by two waves reaching the same point. | Phase difference | Result |
|---|---|---|---|
| Constructive | nλ | 2nπ | I = 4I₀ (bright) |
| Destructive | (n + 1/2)λ | (2n+1)π | I = 0 (dark) |
⚠️ NEET/JEE TRAP
Two independent bulbs never show interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. — light from an ordinary source undergoes random phase jumps roughly every 10⁻¹⁰ s, so two separate sources are always incoherent. This is exactly why Young had to derive both slits from a single original source.
For a general phase difference φ (not just 0 or π), resultant displacement y = y₁ + y₂ = 2a cos(φ/2) cos(ωt + φ/2) — amplitude 2a cos(φ/2), so intensity I = 4I₀cos²(φ/2) follows directly since intensity ∝ amplitude². When sources are incoherent, φ itself varies randomly and rapidly with time, the time-averaged cos²(φ/2) term settles to 1/2, and I = 2I₀ everywhere — a flat, patternless wash of light, not the 4I₀-to-0 oscillation of a coherent pair.
YOUNG'S DOUBLE-SLIT EXPERIMENTYoung's double-slit experimentAn experiment that produces interference fringes using two coherent light sources. (YDSE)
Young's trick: illuminate a single pinhole S, which in turn illuminates two closely-spaced pinholes S₁ and S₂. Because both derive from the same S, any abrupt phase jump at S affects both identically — S₁ and S₂ become a coherent pair, exactly like two synchronised vibrating needles in a ripple tank.
For slit separation d and screen distance D (with D >> d), the path differencepath differenceThe difference between distances travelled by two waves reaching the same point. at a screen point a distance x from the centre is approximately xd/D. This gives equally spaced bright and dark fringes:
BRIGHT FRINGES: x = nλD/d
DARK FRINGES: x = (n + 1/2) λD/d
FRINGE WIDTHfringe widthThe distance between consecutive bright or consecutive dark fringes in an interference pattern.: β = λD/d
🔥 HIGH-YIELD
Fringe widthfringe widthThe distance between consecutive bright or consecutive dark fringes in an interference pattern. β is the same for every fringe (all bright-dark spacings are equal) — this uniform spacing is the single most exam-tested numerical fact of YDSE. β increases with wavelength and screen distance, decreases as slits are moved apart.
✏️ SOLVED EXAMPLE YDSE: d = 0.28 mm, D = 1.4 m, 4th bright fringe is 1.2 cm from the centre. Find λ.
x = nλD/d ⇒ λ = xd/(nD) = (1.2×10⁻² × 0.28×10⁻³) / (4 × 1.4)
= (3.36×10⁻⁶) / 5.6 = 6.0×10⁻⁷ m
Answer: λ = 600 nm (visible orange-red light).
Intensity at a general point — not just maxima and minima
At any point where the path differencepath differenceThe difference between distances travelled by two waves reaching the same point. is a fraction of λ (not exactly nλ or (n+1/2)λ), intensity is I = 4I₀cos²(πΔ/λ), where Δ is the path differencepath differenceThe difference between distances travelled by two waves reaching the same point. — the same cos² formula as before, just written in terms of path differencepath differenceThe difference between distances travelled by two waves reaching the same point. instead of phase angle (they are linked by φ = 2πΔ/λ).
✏️ SOLVED EXAMPLE Path differencepath differenceThe difference between distances travelled by two waves reaching the same point. = λ gives intensity K units. Find the intensity where path differencepath differenceThe difference between distances travelled by two waves reaching the same point. = λ/3.
At Δ = λ: I = 4I₀cos²(π) = 4I₀ = K ⇒ I₀ = K/4.
At Δ = λ/3: I = 4I₀cos²(π/3) = 4I₀ × (1/2)² = 4I₀ × 1/4 = I₀ = K/4.
Answer: intensity = K/4 at path differencepath differenceThe difference between distances travelled by two waves reaching the same point. λ/3.
DIFFRACTIONdiffractionBending and spreading of waves around an obstacle or through an aperture. — THE SINGLE SLIT
Replace the double slit with one narrow slit of width a. Each point on the wavefrontwavefrontA surface joining points of a wave that are in the same phase. across that single opening now acts as a secondary source (Huygens again), and their mutual interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. produces a broad central bright band flanked by much weaker secondary maxima.
CENTRAL MAXIMUM at θ = 0 → MINIMA at θ = nλ/a (n = ±1, ±2, …) → SECONDARY MAXIMA at θ ≈ (n + 1/2)λ/a, rapidly weakening
📖 NCERT FOCUS
The central maximum is TWICE as wide (angularly, 2λ/a) as every other bright band, and carries far more intensity than all the secondary maxima combined — this asymmetry is what makes a single-slit pattern visually so different from the evenly-spaced YDSE fringes.
InterferenceinterferenceRedistribution of intensity produced when coherent waves superpose. vs diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. — is there even a difference?
Feynman's own answer: there is no fundamental physical distinction. 'InterferenceinterferenceRedistribution of intensity produced when coherent waves superpose.' is the term conventionally used for a few discrete coherent sourcescoherent sourcesSources that maintain a constant phase difference and have the same frequency. (like YDSE's two slits); 'diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture.' is used when a continuum of secondary sources across one aperture is summed. In a real double-slit experiment, the observed pattern is actually a single-slit diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. envelope MULTIPLIED by the double-slit interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. fringes — both effects are present together.
✏️ SOLVED EXAMPLE Single slit of width 0.3 mm illuminated by 600 nm light; screen 2 m away. Find the width of the central maximum.
First minima occur at θ ≈ ±λ/a on either side of centre, so angular width of central max = 2λ/a.
Linear half-width on screen ≈ Dλ/a = (2)(600×10⁻⁹)/(0.3×10⁻³) = 4×10⁻³ m = 4 mm.
Full width of central maximum = 2 × 4 mm = 8 mm.
Answer: central maximum is 8 mm wide — note this uses the SAME small-angle geometry as YDSE, only the physical origin (single aperture vs two sources) differs.
POLARISATIONpolarisationRestriction of a transverse wave's vibrations to a particular direction.
Light is a transverse electromagnetic wave — its electric field oscillates perpendicular to the direction of propagation. In unpolarised light (e.g. from the sun or a bulb), this oscillation direction changes randomly and rapidly. A polaroid contains long-chain molecules aligned in one direction; it absorbs the field component along that direction and transmits only the component along its pass-axis, producing linearly (plane) polarised light.
UNPOLARISED LIGHT → POLAROID P₁ (pass-axis) → LINEARLY POLARISED, I = I₀/2 → POLAROID P₂ at angle θ to P₁ → I = I₀cos²θ (Malus' law)
🔥 HIGH-YIELD
Malus' law: I = I₀cos²θ, where θ is the angle between the pass-axes of the two polaroids. θ = 0° → full transmission; θ = 90° ('crossed polaroids') → zero transmission. This is a direct, frequently-numerical exam favourite.
⚠️ NEET/JEE TRAP
A single polaroid always cuts intensity to exactly half (I₀/2) regardless of its orientation, because unpolarised light has all directions equally represented on average — rotating ONE polaroid alone does nothing to the transmitted intensity. It is only the RELATIVE angle between two polaroids that matters.
✏️ SOLVED EXAMPLE A third polaroid is inserted between two crossed polaroids, at angle θ to the first. Find θ for maximum transmitted intensity.
After P₁: I₁ = I₀/2. After P₂ (at angle θ to P₁): I₂ = I₁cos²θ. Angle between P₂ and P₃ (crossed with P₁) is (90° − θ).
After P₃: I₃ = I₂cos²(90° − θ) = I₁cos²θ sin²θ = (I₁/4)sin²2θ.
sin²2θ is maximum when 2θ = 90°, i.e. θ = 45°.
Answer: θ = 45° gives maximum transmission — inserting a polaroid between two crossed ones INCREASES transmitted light, a counter-intuitive but classic result.
Sound waves are longitudinal, so interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. and diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. occur for sound too — but polarisationpolarisationRestriction of a transverse wave's vibrations to a particular direction. is exclusively a transverse-wave phenomenon and therefore has no sound-wave analogue. This is the cleanest experimental proof that light is transverse, not longitudinal.
A NOTE ON PHOTON ENERGY VS WAVE INTENSITY
A short conceptual bridge to the next chapter: in the wave picture, intensity of light is set by the square of the amplitude of the electric and magnetic fields. In the photon (particle) picture — covered fully in the next chapter — intensity instead reflects how many photons cross a unit area per unit time, while each photon's energy is fixed by frequency alone. Wave optics and photon behaviour are not competing theories; they are two languages describing the same electromagnetic radiation, each best suited to a different class of experiment (interferenceinterferenceRedistribution of intensity produced when coherent waves superpose./diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture./polarisationpolarisationRestriction of a transverse wave's vibrations to a particular direction. call for the wave picture; photoelectric and Compton effects call for the photon picture).
📖 NCERT FOCUS
Reflection and refraction arise from the interaction of light with the atomic constituents of matter — atoms behave like oscillators forced to vibrate at the frequency of the incident light. Since a forced oscillator radiates at its driving frequency, the frequency of reflected and refracted light always equals the frequency of the incident light, even though direction and (for refraction) speed change.
DiffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. sets the resolution limit of every optical instrument
Ray optics (Chapter 9) treats microscopes and telescopes as if they can magnify without limit — in reality, diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. at the objective's finite aperture blurs every point-image into a small diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. pattern, capping how close two objects can be and still be seen as separate (their resolving power). A telescope's ability to distinguish two close stars, or a microscope's ability to resolve two close cell structures, is ultimately set by the wavelength of light and the aperture size — not by how strong the eyepiece is. This is precisely why the largest telescopes chase bigger objective diameters rather than ever more powerful eyepieces (Chapter 9), and it is diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture., not any lens defect, that sets this fundamental ceiling.
🔥 HIGH-YIELD
The everyday absence of visible diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. (why you don't see light 'bend around' a doorway) is simply because the wavelength of visible light (~0.4–0.7 μm) is minuscule compared to everyday obstacles — geometrical (ray) optics is the λ → 0 limiting case of the full wave theory, valid only when apertures and obstacles are large compared to λ.
RAY OPTICS vs WAVE OPTICS — WHEN EACH PICTURE IS USED
Chapters 9 and 10 are not rival theories — ray optics is the practical shortcut you reach for whenever the wavelength of light is negligible compared to the sizes involved, and wave optics is the fuller theory that ray optics approximates.
| Situation | Correct picture to use |
|---|---|
| Designing a camera lens, telescope tube, or eyeglass prescription | Ray optics — apertures and image distances are huge compared to λ |
| Explaining fringes in YDSE, colours on a soap film or CD surface | Wave optics — spacing between features is comparable to λ |
| Predicting the ultimate resolving power of a microscope/telescope | Wave optics (diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture.) sets the limit; ray optics alone wrongly predicts unlimited magnification |
| Sign convention, mirror/lens formulas, image position and size | Ray optics — treating light as straight-line rays is entirely sufficient |
🧠 MEMORY HOOK
Ray optics ≈ wave optics in the limit λ → 0. Whenever an exam question mentions fringes, colours-from-thin-films, bending round an edge, or a resolution LIMIT, reach for wave optics; whenever it is about image position/size/magnification with mirrors, lenses, or prisms, ray optics from Chapter 9 is all that is needed.
QUICK FORMULA SHEET
| Concept | Formula |
|---|---|
| WavefrontwavefrontA surface joining points of a wave that are in the same phase. speed relation | sin i / sin r = v₁/v₂ = n₂/n₁ |
| Refractive index | n = c/v |
| Frequency on refraction | unchanged; v₁/λ₁ = v₂/λ₂ = ν |
| Two-source resultant intensity | I = 4I₀cos²(φ/2) |
| Constructive interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. | path diff. = nλ; phase diff. = 2nπ |
| Destructive interferenceinterferenceRedistribution of intensity produced when coherent waves superpose. | path diff. = (n+½)λ; phase diff. = (2n+1)π |
| Incoherent sources | I = I₁ + I₂ = 2I₀ (equal sources) |
| YDSE bright fringe position | x = nλD/d |
| YDSE dark fringe position | x = (n+½)λD/d |
| YDSE fringe widthfringe widthThe distance between consecutive bright or consecutive dark fringes in an interference pattern. | β = λD/d |
| Single-slit minima | θ = nλ/a |
| Single-slit central max width | angular: 2λ/a |
| Malus' law | I = I₀cos²θ |
| Intensity after one polaroid | I₀/2 (unpolarised input) |
Seeing the diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. pattern with everyday objects
NCERT describes a genuinely doable home experiment: hold two razor blades edge-to-edge to form a very narrow adjustable slit, and view a straight bulb filament through it, right up against the eye. With careful adjustment, alternating bright and dark diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. bands become visible directly — coloured, since only the central band's position is wavelength-independent. This is a favourite 'explain the observation' conceptual question.
✏️ SOLVED EXAMPLE Two wavelengths 650 nm and 520 nm are used together in a YDSE. Find where their bright fringes first coincide (besides the centre).
Bright fringes coincide when n₁λ₁ = n₂λ₂ for integers n₁, n₂ (same x).
650 n₁ = 520 n₂ ⇒ n₁/n₂ = 520/650 = 4/5 ⇒ smallest integers: n₁ = 4, n₂ = 5.
x = n₁λ₁D/d — plug in the given D, d to get the numeric distance; the KEY step examiners test is spotting the 4:5 ratio.
Answer: the two wavelengths' bright fringes first coincide at the position corresponding to the 4th order of 650 nm (= 5th order of 520 nm).
MUST-KNOW DIAGRAMS
⭐ MUST KNOW
Huygens construction for a plane wave refracting at a boundary — Structure → secondary wavelet spheres of radius v₂τ from points on the wavefrontwavefrontA surface joining points of a wave that are in the same phase., common tangent → Function: derive Snell's law geometrically.
YDSE geometry — Structure → S, S₁, S₂, screen at distance D, path differencepath differenceThe difference between distances travelled by two waves reaching the same point. xd/D → Function: source of both fringe formulas.
Single-slit diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. intensity pattern — Structure → tall central peak flanked by small secondary peaks → Function: recall relative peak widths and minima positions instantly.
Malus' law set-up (two polaroids) — Structure → unpolarised light → P₁ → P₂ at angle θ → Function: derive I = I₀cos²θ directly from the geometry of the field component along P₂'s axis.
LAST-MINUTE CONFUSION BUSTERS
| Easily confused pair | Correct distinction |
|---|---|
| Coherent vs Incoherent sources | Coherent: constant phase difference → stable pattern, I up to 4I₀. Incoherent: random phase → no pattern, I = I₁+I₂ everywhere. |
| Refraction into a denser medium: speed | Wave theory: speed DECREASES (correct, Foucault-confirmed). Corpuscular theory wrongly predicted an increase. |
| YDSE fringe widthfringe widthThe distance between consecutive bright or consecutive dark fringes in an interference pattern. vs single-slit central maximum width | YDSE: all fringes equally spaced, β = λD/d. Single slit: central maximum is uniquely TWICE as wide as the side maxima. |
| InterferenceinterferenceRedistribution of intensity produced when coherent waves superpose. vs DiffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. | No sharp physical boundary (Feynman) — 'interferenceinterferenceRedistribution of intensity produced when coherent waves superpose.' for a few discrete coherent sourcescoherent sourcesSources that maintain a constant phase difference and have the same frequency., 'diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture.' for a continuum across one aperture. |
| Which wave phenomena need transverse waves | InterferenceinterferenceRedistribution of intensity produced when coherent waves superpose. & diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. occur for both longitudinal and transverse waves; polarisationpolarisationRestriction of a transverse wave's vibrations to a particular direction. occurs ONLY for transverse waves. |
| Coherent sum vs incoherent sum of two equal-intensity sources | Coherent: intensity varies point to point between 0 and 4I₀. Incoherent: uniformly I₀ + I₀ = 2I₀ everywhere — the coherent case is never simply the 'average' of the incoherent case. |
ACTIVE RECALL
Use Huygens' construction to explain, in words, why a wave bends toward the normal when entering a denser medium.
Why does frequency stay unchanged on refraction while speed and wavelength both change?
State the condition on path differencepath differenceThe difference between distances travelled by two waves reaching the same point. for a bright fringe and a dark fringe in YDSE, and derive the fringe widthfringe widthThe distance between consecutive bright or consecutive dark fringes in an interference pattern. formula from it.
Why is the central maximum of a single-slit diffractiondiffractionBending and spreading of waves around an obstacle or through an aperture. pattern twice as wide as the secondary maxima?
Explain why rotating a single polaroid alone never changes the intensity of transmitted unpolarised light, but rotating a second polaroid does.
Derive the law of reflection from Huygens' construction using congruent triangles.
If two coherent sourcescoherent sourcesSources that maintain a constant phase difference and have the same frequency. are incoherent instead, what replaces I = 4I₀cos²(φ/2), and why?
A polaroid is inserted between two crossed polaroids at 45° to both. Explain, without heavy algebra, why light now gets through when none did before.