Superposition and interference of light
When two coherent light waves overlap, their displacements add. A path difference of nλ gives a bright point; (n + ½)λ gives a dark point.
Superposition and interference
The principle of superposition says: when two or more waves overlap, the resultant displacement at any point is the sum of the individual displacements. Interference is the change in intensity produced when two coherent light waves superpose. At some points the waves arrive in phase (crest on crest), and they reinforce each other. This is constructive interference, and the point is bright. At other points they arrive out of phase (crest on trough), and they cancel. This is destructive interference, and the point is dark. Energy is not destroyed at the dark points. It is only redistributed to the bright points.
Conditions for bright and dark points
| Quantity | Constructive (bright) | Destructive (dark) |
|---|---|---|
| Path difference S₂P − S₁P | nλ, that is 0, λ, 2λ, ... | (n + ½)λ, that is λ/2, 3λ/2, ... |
| Phase difference φ | 0, 2π, 4π, ... (2nπ) | π, 3π, 5π, ... ((2n + 1)π) |
| Resultant amplitude (each wave a) | 2a | 0 |
| Intensity (each wave I₀) | 4I₀ | 0 |
Resultant intensity of two coherent waves
Two waves of equal amplitude a reach a point P with a constant phase difference φ.
- 1Write the waves: y₁ = a cos ωt and y₂ = a cos(ωt + φ).
- 2By superposition: y = y₁ + y₂ = a[cos ωt + cos(ωt + φ)].
- 3Use cos A + cos B = 2 cos((A + B)/2) cos((A − B)/2): y = 2a cos(φ/2) cos(ωt + φ/2).
- 4So the resultant amplitude is A = 2a cos(φ/2).
- 5★Intensity is proportional to amplitude squared, and I₀ ∝ a². So I = 4I₀ cos²(φ/2).
- 6So, I = 4I₀ when φ = 0, ±2π, ... (bright) and I = 0 when φ = ±π, ±3π, ... (dark).
★ key step
Worked example: intensity at a point
In an interference pattern, the intensity at the centre of a bright fringe is I. Find the intensity at a point where the path difference is (a) λ/4, (b) λ/3.
- At a bright fringe φ = 0, so I = 4I₀. This gives I₀ = I/4.
- (a) Path difference λ/4 gives φ = (2π/λ)(λ/4) = π/2.
- Intensity = 4I₀ cos²(π/4) = 4I₀ × ½ = 2I₀ = I/2.
- (b) Path difference λ/3 gives φ = (2π/λ)(λ/3) = 2π/3.
- Intensity = 4I₀ cos²(π/3) = 4I₀ × ¼ = I₀ = I/4.
(a) I/2; (b) I/4
Worked example: bright or dark?
Two coherent sources S₁ and S₂ emit light of wavelength 500 nm. A point P is 2.5 μm farther from S₂ than from S₁. Another point Q is 1.75 μm farther from S₂ than from S₁. Is each point bright or dark?
- Path difference at P = 2.5 μm = 2500 nm.
- 2500 nm ÷ 500 nm = 5, a whole number, so the path difference is 5λ.
- P is bright (constructive interference).
- Path difference at Q = 1750 nm; 1750 nm ÷ 500 nm = 3.5.
- This is (3 + ½)λ, an odd multiple of λ/2, so Q is dark (destructive interference).
P is bright; Q is dark.
Think of the shining colours on a soap bubble, or on a film of oil on a wet road after monsoon rain. They come from interference between light reflected from the top and the bottom of the thin film.
2 marks: “Write the conditions, in terms of path difference and phase difference, for constructive and destructive interference.” Constructive: path difference nλ, phase difference 2nπ. Destructive: path difference (n + ½)λ, phase difference (2n + 1)π, where n = 0, 1, 2, ...
3 marks: “Two coherent waves of equal intensity I₀ meet at a point with phase difference φ. Show that the resultant intensity is 4I₀ cos²(φ/2). Where is it maximum and where is it zero?” y₁ = a cos ωt, y₂ = a cos(ωt + φ); add using cos A + cos B; amplitude 2a cos(φ/2); I ∝ amplitude², so I = 4I₀ cos²(φ/2); maximum 4I₀ at φ = 2nπ; zero at φ = (2n + 1)π.
1 marks: “Is energy lost at the dark fringes of an interference pattern?” No. Energy is only redistributed from the dark fringes to the bright fringes; total energy is conserved.
Whole λ, bright as day; half λ, fades away. And I = 4I₀ cos²(φ/2).