Wavefronts, reflection and refraction
A wavefront is a line joining points of a wave that move together, for example all the crests. A stone in a pond makes circular wavefronts. Far from the source they look almost straight (plane waves).
Reflection: when a wave hits a hard wall it bounces back. The angle it comes in at equals the angle it goes out at. An echo is reflected sound.
Refraction: when a wave moves into a place where its speed is different, it changes direction. The frequency f stays the same (the source decides it). So from v = fλ, if v falls, λ falls too. Water waves slow down in shallow water; light slows down in glass.
Diffraction: waves spread round corners
Diffraction is the spreading of a wave after it passes through a gap or round an edge. It is strongest when the gap is about the same size as the wavelength. That is why you can hear a friend round a corner (sound λ is about 1 m) but cannot see them (light λ is less than a millionth of a metre).
For light through one narrow slit of width a, the first dark place is at an angle θ where a sin θ = λ. A narrower slit gives a wider central bright band.
Superposition and interference
Principle of superposition: when two waves meet, the total displacement at a point is the sum of the two displacements. After they pass, each wave carries on unchanged.
Constructive and destructive interference
If two waves arrive in step (crest with crest), they reinforce: constructive interference, amplitude adds up. If they arrive exactly out of step (crest with trough), they cancel: destructive interference.
The path difference rule
For two sources vibrating in step, at a point P find the path difference Δ = |S₁P − S₂P|.
- Δ = nλ (0, λ, 2λ …) → constructive (bright / loud)
- Δ = (n + ½)λ (λ/2, 3λ/2 …) → destructive (dark / quiet)
A steady pattern needs coherent sources: same frequency and a fixed phase difference.
Double-slit fringes
For two slits a distance d apart and a screen at distance D, the bright fringes are equally spaced: Δx = λD/d. Bigger λ or D spreads the fringes; bigger d squeezes them.
Thin-film interference
Light reflects from the top and from the bottom of a thin film (soap, oil). The two reflected waves interfere. The extra path is about 2nt (n = refractive index, t = thickness). Reflection from a denser medium adds half a wavelength shift. Different colours are reinforced at different thicknesses, so we see bands of colour. Anti-reflection coatings on spectacles use a film about λ/4n thick so the reflections cancel.
Try it at home
Fill a wide tray with 2 cm of water. Tap the surface with one finger: watch circular ripples reflect off the side. Put two blocks with a 2 cm gap in the middle and push a ruler to make straight waves: see them spread after the gap. Then tap with two fingers at once, 5 cm apart: look for calm lines between the ripples. Predict first: will the calm lines get closer if you move your fingers apart? Then check.
Key formulas and definitions
- v = f λ
- Path difference Δ = |S₁P − S₂P|
- Constructive: Δ = nλ
- Destructive: Δ = (n + ½)λ
- Fringe spacing Δx = λD / d
- Single slit first minimum: a sin θ = λ
- Thin film (one phase flip): 2nt = (m + ½)λ for bright reflection
Worked examples
1. A water wave has frequency 4 Hz and wavelength 0.5 m. Find its speed.
v = fλ = 4 × 0.5 = 2 m/s.
2. The same wave enters shallow water where its speed falls to 1.2 m/s. Find the new wavelength.
Frequency stays 4 Hz. λ = v/f = 1.2/4 = 0.3 m. The wave gets shorter.
3. Two speakers in step give sound of λ = 0.8 m. A point is 5.0 m from one and 6.2 m from the other. Loud or quiet?
Δ = 6.2 − 5.0 = 1.2 m. Δ/λ = 1.2/0.8 = 1.5 = 1 + ½. So destructive: quiet.
4. Two slits 0.5 mm apart, screen 1.5 m away, light λ = 600 nm. Find the fringe spacing.
Δx = λD/d = (600 × 10⁻⁹ × 1.5)/(0.5 × 10⁻³) = 1.8 × 10⁻³ m = 1.8 mm.
5. Light of 500 nm falls on a slit 0.01 mm wide. Find the angle of the first dark band.
sin θ = λ/a = (500 × 10⁻⁹)/(1 × 10⁻⁵) = 0.05, so θ ≈ 2.9°.
6. A soap film (n = 1.33) looks bright for 532 nm light. Find the thinnest possible thickness.
One phase flip (top surface), so bright when 2nt = ½λ for m = 0. t = λ/(4n) = 532/(4 × 1.33) = 100 nm.
Common mistakes
- Thinking frequency changes in refraction. It is the speed and wavelength that change; frequency stays the same.
- Saying energy is destroyed in destructive interference. It is only moved: the bright/loud places get the extra energy.
- Using any two bulbs for interference. They are not coherent, so no steady pattern forms.
- Forgetting to convert mm and nm to metres before using Δx = λD/d.