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Wave Optics

Interference of light

Solve Young's double-slit problems.

When two waves meet, they add. Crest on crest gives a bigger wave (constructive interference: bright). Crest on trough cancels (destructive: dark). This is superposition.

A steady pattern needs coherent sources: they must keep a constant phase difference. Two separate bulbs are not coherent: each sends out light in random bursts, so their pattern shifts millions of times a second and just looks evenly lit. Young's trick: shine one source through two slits. The two slits then act as coherent sources.

The result depends on the path difference between the two waves reaching a point:

  • Bright fringe: path difference = nλ (0, λ, 2λ, ...).
  • Dark fringe: path difference = (n + ½)λ (λ/2, 3λ/2, ...).

With slits d apart and the screen D away (D much larger than d), the bright fringes sit at x_n = nλD/d. The spacing between neighbouring fringes, the fringe width, is:

β = λD / d

Example: d = 0.5 mm, D = 1 m, λ = 500 nm. β = 500 × 10⁻⁹ × 1 / (0.5 × 10⁻³) = 1 mm.

So fringes get wider with longer wavelength (red more than blue), a farther screen, or slits closer together. Doing the experiment under water shrinks λ and the fringes.

Intensity: if each slit alone gives I₀, the pattern goes from 4I₀ at the bright fringes to 0 at the dark ones: I = 4I₀ cos²(φ/2), where φ = 2π × path difference / λ. Energy is not lost, only moved from dark places to bright ones.

With white light, the central fringe is white (path difference 0 for every colour) and the others are coloured.

x_n = nλD/d; β = λD/d.
β = λD/d.

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