- Decide from the path difference whether a point is bright or dark, and use Δx = Lλ/d to find the fringe spacing, the wavelength or the effect of a change.
- Explain thin-film colours and find the minimum thickness of an anti-reflection coating.
- Tell interference patterns from diffraction patterns and use d · sin θ = kλ for a grating.
- Describe polarisation (and what it proves) and order the electromagnetic spectrum by wavelength, frequency and use.
A soap bubble is made of colourless soap and water, yet it shimmers with every colour of the rainbow. Two car headlights shine on the same wall, yet they never draw dark and bright stripes on it. Both puzzles have one answer: light is a wave, and waves can add up or cancel — but only when they keep in step. In the lesson «Geometrical optics: reflection and refraction» light travelled as straight rays; here we look at the effects that rays cannot explain — interference, diffraction and polarisation — and finish with the whole electromagnetic spectrum.
Interference: coherent sources and Young's double slit
When two waves meet, their displacements add up (the superposition principle from the lesson «Mechanical waves»). Where a crest meets a crest the light is bright; where a crest meets a trough it is dark. A pattern that stays still on a screen appears only if the two sources are coherent. Two separate lamps or headlights are not: every atom emits short, random wave trains, so the phase difference between the lamps changes millions of times a second and the stripes wash out. The way out is to split one wave into two — with two slits or with the two surfaces of a thin film.
Sources of the same frequency with a constant phase difference (and the same direction of vibration). Only coherent waves give a steady interference pattern.
- Δrpath difference: the difference between the distances from the point to the two sources
- λwavelength of the light
- korder of the fringe (0 = the central one)
For two sources vibrating in phase: a whole number of wavelengths gives constructive interference, an odd number of half-wavelengths gives destructive interference.
Two coherent sources vibrate in phase.
1) λ = 600 nm, Δr = 1.8 μm. What is seen at the point?
2) λ = 500 nm, Δr = 1.25 μm. What is seen at the point?
3) Point P is the second dark fringe counted from the central bright fringe, λ = 400 nm. Find the path difference at P.
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2) Δr/λ = 1.25/0.5 = 2.5, i.e. Δr = 5 · (λ/2) — an odd number of half-waves: dark.
3) Dark fringes lie where Δr = λ/2, 3λ/2, 5λ/2, … The second one: Δr = 3λ/2 = 3 · 400/2 = 600 nm.
In Young's experiment (early 19th century) light from one source falls on two narrow slits a distance d apart, and a screen stands at a distance L much larger than d. For a point at a distance x from the centre of the screen the path difference is Δr ≈ d·x/L. The condition Δr = kλ gives the bright fringes at x = kLλ/d, so neighbouring fringes are equally spaced. The centre of the screen (Δr = 0) is always bright.
- Δxfringe spacing: the distance between neighbouring bright (or dark) fringes
- Ldistance from the slits to the screen
- ddistance between the two slits
- λwavelength
The fringes are equally spaced and almost equally bright. Chinese textbooks write Δx = (l/d)λ. Red light (λ ≈ 700 nm) gives wider fringes than violet (λ ≈ 400 nm); in white light the central fringe is white and the others are coloured, violet on the inside and red on the outside.
1) d = 0.25 mm, L = 1.0 m, λ = 500 nm. Find Δx.
2) With d = 0.40 mm and L = 1.0 m, the distance from the 1st to the 6th bright fringe is 7.5 mm. Find λ.
3) The apparatus of 1) is placed in water (n = 4/3). What is the new Δx?
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2) From the 1st to the 6th fringe there are 5 spacings: Δx = 7.5/5 = 1.5 mm. λ = Δx·d/L = 1.5 × 10⁻³ · 4 × 10⁻⁴ / 1.0 = 6 × 10⁻⁷ m = 600 nm.
3) In water the frequency stays the same but the wavelength becomes λ/n, so Δx = 2 mm · 3/4 = 1.5 mm.
In a double-slit experiment with green light, which single change makes the bright fringes on the screen farther apart?
A) Moving the screen closer to the slits
B) Using two slits that are farther apart
C) Replacing the green light with red light
D) Replacing the green light with violet light
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A is tempting because L is in the formula — but a smaller L gives a smaller Δx. B increases d, which is in the denominator. D changes the colour in the wrong direction: violet has the shortest visible wavelength.
A double-slit calculation always goes the same way:
- 1Units first
Write λ, d, L and Δx in metres (1 nm = 10⁻⁹ m, 1 μm = 10⁻⁶ m, 1 mm = 10⁻³ m).
- 2Is there a medium?
If the apparatus is in a liquid or in glass, use λ/n instead of λ; the frequency does not change.
- 3Fringes or spacings?
From the 1st to the nth bright fringe there are n − 1 spacings; from the central fringe to the kth bright fringe there are k.
- 4Formula or ratio
For numbers use Δx = Lλ/d; for «what happens if…» questions write the ratio Δx₂/Δx₁ = (L₂/L₁) · (λ₂/λ₁) · (d₁/d₂).
Thin films: colours and anti-reflection coatings
A thin transparent layer — a soap film, oil on a wet road, the air between two glass plates — reflects light twice: at its front and at its back surface. Both reflected waves come from the same incoming wave, so they are coherent and interfere. The wave from the back surface travels about 2d farther (d is the thickness). Where this extra path strengthens one colour, that colour dominates the reflected light. A vertical soap film drains downwards and becomes a wedge, thin at the top and thick at the bottom, so it shows horizontal coloured bands in white light and bright and dark bands in light of one colour.
- Air wedge: two flat glass plates touch at one edge and a thin paper strip separates them at the other. The fringes are straight, equally spaced and parallel to the edge where the plates touch; a thicker strip (a larger angle) moves them closer together. A bump or a dip on a plate bends the fringes — this is how optical workshops check flatness.
- Newton's rings: a convex lens lying on flat glass gives concentric rings that crowd together towards the edge, because the air gap grows faster and faster.
A thin transparent layer on a lens or a solar cell. Its thickness is chosen so that the waves reflected from its two surfaces cancel for the middle of the spectrum (green–yellow light); less light is reflected and more passes through.
- dₘᵢₙsmallest thickness of the coating
- λwavelength in vacuum (air)
- nrefractive index of the coating (smaller than that of the glass)
The two reflected waves must differ by half a wavelength: 2d = λ/(2n). In Chinese-textbook form: the coating is a quarter of the wavelength inside the film, d = λfilm/4 with λfilm = λ/n. No energy is destroyed: what is not reflected is transmitted.
1) A coating with n = 1.25 must cancel the reflection of light with λ = 550 nm. Find its minimum thickness.
2) A coating with n = 1.5 is 90 nm thick. For which wavelength (in air) does it cancel the reflection best?
3) Why does a coated camera lens look purple?
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2) λ = 4nd = 4 · 1.5 · 90 = 540 nm (green light).
3) The reflection of green is cancelled, while red and violet at the two ends of the spectrum are still partly reflected; a mixture of red and violet looks purple.
Diffraction: single slit, obstacles and gratings
The bending of waves around obstacles and their spreading after passing through an opening. It is noticeable when the opening or the obstacle is about as large as the wavelength or smaller. The wavelength of light is only 0.4–0.76 μm, so everyday objects cast sharp shadows, while sound (λ ≈ 1 m) easily bends around a door.
- The central bright fringe is the widest and the brightest — about twice as wide as each of the others.
- The side fringes are much dimmer and get fainter away from the centre.
- A narrower slit or a longer wavelength gives a wider (and dimmer) pattern; a very wide slit gives just a sharp image of the slit.
- White light gives a white centre with coloured edges (red farther out).
- Behind a small opaque disc a bright spot appears in the middle of the shadow (the Poisson, or Arago, spot) — strong evidence for the wave theory of light.
| Feature | Two slits (interference) | One slit (diffraction) |
|---|---|---|
| Width of the fringes | equal | central fringe twice as wide |
| Brightness | almost the same | falls quickly from the centre |
| λ increases | fringes get wider | pattern gets wider |
| slit separation / slit width increases | fringes get closer | central fringe gets narrower |
A diffraction grating has a very large number of equal, equally spaced slits (hundreds per millimetre). The waves from all the slits add up, so the bright lines are much sharper and brighter than with two slits; more slits make them sharper without moving them. Gratings split white light into spectra in spectrometers.
- dgrating period: the distance between neighbouring slits, d = 1/N for N slits per unit length
- θangle between the k-th order line and the straight-through direction
- korder (0, 1, 2, …)
Since sin θ ≤ 1, only orders with k < d/λ can be seen. A longer wavelength is deviated more: in a grating spectrum red lies farthest out — the opposite of a prism, which bends violet most.
A grating has 500 lines per millimetre.
1) Find its period.
2) Light with λ = 500 nm falls on it perpendicularly. At what angle is the second-order line seen?
3) What is the highest order that can be seen with λ = 600 nm?
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2) sin θ = kλ/d = 2 · 0.5 μm / 2 μm = 0.5, so θ = 30°.
3) k < d/λ = 2/0.6 ≈ 3.3, so the highest order is 3 (on each side of the central line).
Which of the following is caused by the diffraction of light?
A) A bright spot at the centre of the shadow of a small round disc
B) The colours of a soap bubble
C) A rainbow after rain
D) Coloured bands in a thin oil film on a wet road
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B and D are tempting because they are also «wave» colours, but they are thin-film interference. C is dispersion by refraction in raindrops (see «Geometrical optics: reflection and refraction»).
Polarisation and the electromagnetic spectrum
Light is an electromagnetic wave: its electric field vibrates perpendicular to the direction of travel, so light is a transverse wave. Sunlight and lamp light are unpolarised — they contain vibrations in all transverse directions. A polariser lets through only the vibrations along its transmission axis, so after it the light is polarised and half as bright. A second polariser (the analyser) passes all of this light when the axes are parallel and none of it when they are crossed at 90°. Only transverse waves can be polarised: interference and diffraction prove that light is a wave, polarisation proves that it is a transverse wave.
- I₀intensity of the polarised light falling on the analyser
- θangle between the axes of the polariser and the analyser
- Itransmitted intensity
Malus's law. Unpolarised light loses half of its intensity in the first polariser, whatever the direction of its axis.
Unpolarised light of intensity I₀ passes through two polarisers. Find the transmitted intensity when the angle between their axes is 1) 0°, 2) 60°, 3) 90°.
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1) cos²0° = 1: I₀/2.
2) cos²60° = 1/4: I₀/2 · 1/4 = I₀/8.
3) cos²90° = 0: darkness. Turning the analyser through a full circle gives two maxima and two zeros.
Polarisation is used in sunglasses and camera filters that cut glare (light reflected from water, roads and glass is partly polarised), in 3D cinema glasses, which send a differently polarised picture to each eye, and in LCD screens. Maxwell predicted, and Hertz confirmed, that changing electric and magnetic fields travel as electromagnetic waves. All of them move at c = 3.0 × 10⁸ m/s in vacuum and differ only in wavelength — and the wavelength decides how they are produced and used.
- cspeed of light in vacuum, 3.0 × 10⁸ m/s
- λwavelength
- νfrequency
A shorter wavelength means a higher frequency — and, as the lesson «The photoelectric effect» shows, more energetic photons.
1) An FM radio station broadcasts at 100 MHz. Find λ.
2) Find the frequency of orange light with λ = 600 nm.
3) A microwave oven works at 2.5 GHz. Find λ.
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2) ν = c/λ = 3 × 10⁸ / (6 × 10⁻⁷) = 5 × 10¹⁴ Hz.
3) λ = 3 × 10⁸ / (2.5 × 10⁹) = 0.12 m = 12 cm.
| Band | Wavelength (approx.) | Source or use |
|---|---|---|
| Radio waves | > 1 m | radio and TV broadcasting, communication |
| Microwaves | ≈ 1 mm – 1 m | radar, mobile phones, Wi-Fi, microwave ovens |
| Infrared | ≈ 760 nm – 1 mm | warm bodies; remote controls, night vision, heaters |
| Visible light | ≈ 400–760 nm | the Sun, lamps; sight (red to violet) |
| Ultraviolet | ≈ 10–400 nm | the Sun; sterilising, fluorescence (banknote checks), vitamin D |
| X-rays | ≈ 0.01–10 nm | X-ray tubes; medical images, airport security |
| γ-rays | < 0.01 nm | atomic nuclei; cancer therapy, sterilising |
Which pairing of a type of radiation with its use is correct?
A) X-rays — TV remote controls
B) Infrared — images of broken bones
C) γ-rays — FM radio broadcasting
D) Ultraviolet — checking banknotes
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A and B swap two uses: remote controls work with infrared, and bone images are made with X-rays. C is wrong because broadcasting uses radio waves; γ-rays come from atomic nuclei and are used in cancer therapy.
How CSCA asks about this
- Fringe changes: «which change widens the fringes?» — answer with Δx = Lλ/d without computing; in a liquid λ becomes λ/n.
- Δx or λ from data: one line of arithmetic with powers of ten; count spacings, not fringes.
- Bright or dark: divide Δr by λ/2 — an even number means bright, an odd number dark.
- Name the phenomenon: soap bubble, oil film, coated lens, air wedge → interference; a narrow slit, the Poisson spot → diffraction; sunglasses, 3D glasses → polarisation; rainbow, prism → dispersion.
- Statements about light: polarisation → transverse wave; two independent sources → no fringes; every electromagnetic wave travels at c in vacuum.
- The spectrum: sort the bands by λ, ν or photon energy and match each band to its use.
| Term | 中文 | Pinyin |
|---|---|---|
| interference | 干涉 | gānshè |
| diffraction | 衍射 | yǎnshè |
| coherent light source | 相干光源 | xiānggān guāngyuán |
| double-slit interference | 双缝干涉 | shuāngfèng gānshè |
| fringe spacing | 条纹间距 | tiáowén jiānjù |
| path difference | 路程差 | lùchéngchā |
| wavelength | 波长 | bōcháng |
| thin-film interference | 薄膜干涉 | bómó gānshè |
| anti-reflection coating | 增透膜 | zēngtòumó |
| single-slit diffraction | 单缝衍射 | dānfèng yǎnshè |
| diffraction grating | 衍射光栅 | yǎnshè guāngshān |
| polarisation (polariser) | 偏振（偏振片） | piānzhèn (piānzhènpiàn) |
| transverse wave | 横波 | héngbō |
| electromagnetic spectrum | 电磁波谱 | diàncí bōpǔ |
| Poisson spot | 泊松亮斑 | Bósōng liàngbān |
Key points
- Coherent sources (same frequency, constant phase difference) give steady fringes: bright where Δr = kλ, dark where Δr = (2k + 1)λ/2.
- Young's double slit: Δx = Lλ/d — equally spaced, equally bright fringes with a bright centre; in a medium λ becomes λ/n.
- Thin films interfere by reflecting light from both surfaces; an anti-reflection coating is λ/(4n) thick.
- Diffraction is noticeable when the opening is about λ or smaller; one slit gives a central fringe twice as wide; a grating obeys d · sin θ = kλ and deviates red most.
- Polarisation proves that light is transverse; unpolarised light becomes I₀/2 after a polariser, then I = I₀cos²θ.
- By rising frequency: radio, microwaves, infrared, visible, ultraviolet, X-rays, γ-rays; all travel at c = λν.
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