Wavefronts and Huygens' principle
Ripples spread in circles from a pebble dropped in a pond; each ring is a wavefront. Link to earlier classes : waves from Class 11 and refraction from Chapter 9.
A wavefront is the surface joining all the points that are in the same phase at a given moment. A point source gives spherical wavefronts, and a very distant source gives plane wavefronts. Huygens' principle says that every point on a wavefront acts as a source of secondary wavelets, and the new wavefront a little later is the forward envelope of these wavelets.
Rule 1: Reflection and refraction by wave theory — Using Huygens' construction one proves that i = r for reflection, and sin i ÷ sin r = v₁ ÷ v₂ = n₂ ÷ n₁ for refraction (Snell's law). On refraction the frequency stays the same and the wavelength changes as λ′ = λ ÷ n.
- Easy : Light of 600 nm in air has λ′ = 600 ÷ 1.5 = 400 nm in glass.
- Medium : Light of 589 nm entering water (n = 1.33) has λ′ = 589 ÷ 1.33 = 443 nm.
- Exam : The speed in water is 3 × 10⁸ ÷ 1.33 = 2.26 × 10⁸ m s⁻¹, less than in air, as the wave theory predicts (the older particle theory predicted a larger speed).
Rule 2: Wavefront shapes — The ray is always perpendicular to the wavefront. A convex lens turns a plane wavefront into a spherical one converging on the focus.
- Easy : Light from the Sun reaching the Earth has nearly plane wavefronts.
- Medium : A point source gives spherical wavefronts that become flatter as they spread.
- Exam : Huygens' principle explains how the wave bends around corners (diffraction).
Wavefront ⟂ ray; frequency unchanged on refraction; λ′ = λ/n.
Saying that the frequency of light changes on entering glass. Only speed and wavelength change.
Memory clue : "Frequency is the colour's ID: it never changes."
Spot the difference
| Source | Wavefront |
|---|---|
| Point source (near) | spherical |
| Very distant source | plane |
| Line source | cylindrical |
Practice questions
- 1Define a wavefront.
- 2What is the shape of the wavefront from a very distant source?
- 3Which quantity does not change when light is refracted?
- 4Find the wavelength in glass (n = 1.5) of light that has 600 nm in air.
- 5Find the wavelength in water (n = 1.33) of 589 nm light.
- 6Find the speed of light in water.
- 7Is the speed of light in water less or more than in air according to the wave theory?
- 8State Huygens' principle.
Show answers (try them yourself first)
- 1A surface on which all points are in the same phase.
- 2Plane.
- 3Frequency.
- 4400 nm.
- 5≈ 443 nm.
- 6≈ 2.26 × 10⁸ m s⁻¹.
- 7Less.
- 8Every point of a wavefront is a source of secondary wavelets; the new wavefront is the forward envelope of these wavelets.
Interference and Young's double-slit experiment
The coloured bands on a soap bubble or on a thin oil film on water come from the interference of light. Link to earlier classes : superposition of waves from Class 11.
Interference is the redistribution of energy when two coherent waves overlap, giving bright and dark fringes. Two sources are coherent if they keep a constant phase difference. In Young's double-slit experiment, a single source illuminates two close slits S₁ and S₂, which act as coherent sources.
Rule 1: Conditions and fringe width — Bright fringe (constructive): path difference Δ = n λ. Dark fringe (destructive): Δ = (n + ½) λ. The distance of the n-th bright fringe from the centre is y = n λ D ÷ d, and the fringe width is β = λ D ÷ d (D = distance to the screen, d = slit separation).
- Easy : λ = 600 nm, D = 1.0 m, d = 0.50 mm ⇒ β = 600 × 10⁻⁹ × 1.0 ÷ 0.5 × 10⁻³ = 1.2 mm.
- Medium : The third bright fringe is at 3β = 3.6 mm; the first dark fringe is at β ÷ 2 = 0.6 mm.
- Exam : The whole set-up in water (n = 1.33) has β = 1.2 ÷ 1.33 = 0.90 mm, since λ becomes λ ÷ n.
Rule 2: Intensity — I = 4 I₀ cos²(φ ÷ 2) for equal slits (φ = phase difference). For unequal intensities, I_max ÷ I_min = (√I₁ + √I₂)² ÷ (√I₁ − √I₂)².
- Easy : For φ = 0, I = 4I₀, the maximum; for φ = π, I = 0.
- Medium : Intensities in the ratio 9 : 1 (amplitudes 3 : 1) give I_max ÷ I_min = (3 + 1)² ÷ (3 − 1)² = 4.
- Exam : The path difference at y = 3.6 mm for d = 0.5 mm and D = 1 m is Δ = yd ÷ D = 1.8 μm = 3λ, so the point is bright.
In the figure : Young's arrangement: two slits S₁, S₂ a distance d apart and a screen at a distance D showing equally spaced bright and dark fringes.
β = λD/d; bright at nλ, dark at (n + ½)λ; fringes are equally spaced.
Treating two separate bulbs as coherent sources.
Memory clue : "Fringe width: lambda D over d."
Spot the difference
| Path difference | Result |
|---|---|
| nλ | bright |
| (n + ½)λ | dark |
Practice questions
- 1Write the condition for a bright fringe.
- 2What are coherent sources?
- 3Write the formula for the fringe width.
- 4Find β for λ = 600 nm, D = 1.0 m and d = 0.50 mm.
- 5Find the position of the third bright fringe.
- 6Find β if the whole set-up is immersed in water (n = 1.33).
- 7Two coherent sources have intensities in the ratio 9 : 1. Find I_max ÷ I_min.
- 8Find the position of the first dark fringe in the set-up of question 4.
Show answers (try them yourself first)
- 1Path difference Δ = nλ.
- 2Sources that keep a constant phase difference.
- 3β = λD ÷ d.
- 4β = 1.2 mm.
- 53β = 3.6 mm.
- 6β ≈ 0.90 mm.
- 74.
- 8β ÷ 2 = 0.6 mm.
Diffraction at a single slit
We hear a person talking around a corner but cannot see around it, because sound waves bend more than light waves. Link to earlier classes : Huygens' principle from the first section.
Diffraction is the bending of waves around the edges of an obstacle or an opening. It is noticeable when the size of the opening is comparable to the wavelength. Light passing through a narrow slit gives a bright central maximum with weaker maxima on both sides.
Rule 1: Minima of a single slit — The dark fringes (minima) occur at a sin θ = n λ (n = ±1, ±2, …), where a is the slit width. The first minimum is at sin θ = λ ÷ a.
- Easy : λ = 600 nm and a = 0.20 mm ⇒ sin θ = 3 × 10⁻³ ≈ 3 × 10⁻³ rad.
- Medium : On a screen at D = 1.5 m, the first minimum is at x = λD ÷ a = 4.5 mm from the centre.
- Exam : The central maximum is twice as wide as each other maximum; there are no minima within it.
Rule 2: Width of the central maximum — 2 λ D ÷ a. A narrower slit gives a wider pattern.
- Easy : For the data above, the width is 2 × 4.5 = 9 mm.
- Medium : If the slit is made 0.10 mm wide, the width doubles to 18 mm.
- Exam : We do not see light diffracting around a door because the wavelength (about 5 × 10⁻⁷ m) is far smaller than the door, while sound has wavelengths of about a metre.
In the figure : Intensity pattern of single-slit diffraction: a broad central peak with small side peaks.
Minima at a sin θ = nλ; central width 2λD/a; narrow slit, wide pattern.
Using the double-slit formula nλ for the maxima of the single slit. In the single slit this condition gives minima.
Memory clue : "Narrow slit spreads wide: diffraction likes small openings."
Spot the difference
| Feature | Interference (YDSE) | Diffraction (single slit) |
|---|---|---|
| Fringe width | equal | central one is twice the others |
| Intensity of bright fringes | nearly equal | falls away from the centre |
Practice questions
- 1Write the condition for minima in single-slit diffraction.
- 2Which maximum is the brightest and widest?
- 3What happens to the pattern when the slit is made narrower?
- 4Find sin θ for the first minimum for λ = 600 nm and a = 0.20 mm.
- 5Find the width of the central maximum on a screen 1.5 m away.
- 6Find the width if the slit width is halved.
- 7Find the distance of the first minimum from the centre in the set-up of question 5.
- 8Why does light not visibly bend around a door while sound does?
Show answers (try them yourself first)
- 1a sin θ = nλ.
- 2The central maximum.
- 3It becomes wider.
- 4sin θ = λ ÷ a = 3 × 10⁻³.
- 52λD ÷ a = 9 mm.
- 618 mm.
- 7x = λD ÷ a = 4.5 mm.
- 8The wavelength of light is much smaller than the size of the door, whereas for sound it is comparable.
Polarisation
Polaroid sunglasses cut the glare from a road or from water. Link to earlier classes : transverse waves from Class 11 and electromagnetic waves from Chapter 8.
In unpolarised light the electric field vibrates in all directions at right angles to the ray. Polarisation restricts the vibrations to one direction. Only transverse waves can be polarised, which proves that light is a transverse wave. A polaroid passes only the component along its pass axis.
Rule 1: Malus' law — I = I₀ cos² θ (I₀ = intensity of the plane-polarised light falling on the analyser, θ = angle between the pass axes). Unpolarised light of intensity I₀ falling on a polaroid emerges with I₀ ÷ 2.
- Easy : Unpolarised light of intensity I₀ through one polaroid becomes I₀ ÷ 2.
- Medium : A second polaroid at 60° to the first gives I = (I₀ ÷ 2) cos² 60° = I₀ ÷ 8.
- Exam : Crossed polaroids (θ = 90°) pass no light.
Rule 2: Brewster's law — Light reflected from a transparent surface is completely polarised at the polarising angle i_B given by tan i_B = n. At this angle the reflected and refracted rays are at right angles, i_B + r = 90°.
- Easy : For glass (n = 1.5), i_B = tan⁻¹ 1.5 = 56.3°.
- Medium : The angle of refraction is then 90° − 56.3° = 33.7°.
- Exam : For water (n = 1.33), i_B = 53.1°. Polaroid glasses block the horizontally polarised glare reflected from water.
I = I₀cos²θ; unpolarised → I₀/2; tan i_B = n; only transverse waves polarise.
Using Malus' law directly on unpolarised light. First apply the factor ½.
Memory clue : "Brewster: tan of the angle equals n."
Spot the difference
| Light | After one polaroid |
|---|---|
| Unpolarised I₀ | I₀ ÷ 2 |
| Polarised I₀ at θ | I₀ cos²θ |
Practice questions
- 1State Malus' law.
- 2State Brewster's law.
- 3Can longitudinal waves be polarised?
- 4Unpolarised light of intensity I₀ falls on a polaroid. Find the transmitted intensity.
- 5A second polaroid at 60° follows it. Find the final intensity.
- 6Find the polarising angle for glass of n = 1.5.
- 7Find the angle of refraction at the polarising angle for glass.
- 8Find the polarising angle for water (n = 1.33).
Show answers (try them yourself first)
- 1The intensity of light passing the analyser is I₀cos²θ.
- 2tan i_B = n.
- 3No, only transverse waves.
- 4I₀ ÷ 2.
- 5(I₀ ÷ 2) × cos² 60° = I₀ ÷ 8.
- 6i_B = 56.3°.
- 7r = 90° − 56.3° = 33.7°.
- 8i_B ≈ 53.1°.