Young's Double-Slit Interference

Light reaching a point P on the screen leaves slit S₁ having travelled r₁ and slit S₂ having travelled r₂, and the entire pattern depends on r₂ − r₁. Computing that exactly means subtracting two square roots, which buries the physics. Instead notice how nearly parallel the two rays are: with slit separation d = 0.25 mm and a screen L = 2.0 m away, the two routes to the same point differ in direction by only about d/L = 1.25 × 10⁻⁴ rad, or 0.0072°. Treat them as parallel, both leaving at angle θ to the axis, and drop a perpendicular from S₁ onto the ray from S₂. The small right triangle it makes has hypotenuse d and opposite side d sin θ, so Δ = d sin θ. The price is tiny: for that geometry, at a point 5.0 mm off the axis, subtracting the exact square roots gives 6.24998 × 10⁻⁷ m while the far-field form d y/L gives 6.25000 × 10⁻⁷ m — three parts in a million, far below anything you could measure.

A path difference only means something measured against the wavelength, so convert it. One whole wavelength of extra path is one full turn of phase, 2π rad, so φ = 2π Δ/λ = 2π d sin θ/λ. Crest meets crest when φ is a whole number of turns, which is d sin θ = mλ, with the integer m called the order. Crest meets trough when the extra path is a half-integer number of wavelengths, d sin θ = (m + ½)λ. Put numbers on it: a helium-neon laser at λ = 633 nm through slits d = 0.25 mm apart gives sin θ₁ = (633 × 10⁻⁹ m) ÷ (0.25 × 10⁻³ m) = 2.53 × 10⁻³, so θ₁ = 0.145°, and even the tenth order is only 1.45° off-axis. That is why the screen must stand metres away before the fringes separate enough to see. Notice also what m = 0 means: on the axis the two routes are equal, so every wavelength arrives in step there whatever its value, and in white light the central fringe is the only white one.

Interference is a statement about fields, not about intensities. Let each slit alone deliver amplitude E₁ at P, with the far slit's contribution lagging by φ. Then E = E₁cos(ωt) + E₁cos(ωt − φ) = 2E₁cos(φ/2) cos(ωt − φ/2): two phasors of equal length with angle φ between them, closing to a resultant of length 2E₁cos(φ/2). Intensity goes as amplitude squared, so I = 4I₁cos²(φ/2), where I₁ ∝ E₁² is one slit's intensity at that point. In the geometry of the experiment, I(θ) = 4I₁cos²(π d sin θ/λ). Three things fall out. The maxima are four times as bright as one slit, not twice. The minima are exactly zero, and only because the two amplitudes are equal — unequal slits give grey minima instead. And since cos² averages to ½, the mean intensity across the pattern is 2I₁, exactly what two slits deliver.

Angles are not what you measure; positions on a screen are. A fringe at angle θ lands at y = L tan θ, while the interference condition gives sin θₘ = mλ/d. Those are different functions, and only while θ is small does sin θ ≈ tan θ let you write yₘ = mλL/d. Adjacent maxima are then Δy = λL/d apart, the same gap everywhere — and that even spacing is a prediction of the small-angle step, not of the interference condition itself. With λ = 633 nm, d = 0.25 mm and L = 2.0 m, Δy = 5.06 mm. The approximation is generous here: tan θ/sin θ = 1/cos θ, still within 1% at θ = 8.0°, and for this slit spacing 8.0° is already order m = 55. Push beyond that — a coarse grating rather than a double slit — and the fringes on a flat screen spread apart toward the edges, because equal steps in sin θ become ever larger steps in tan θ: the gap from m = 55 to m = 56 is 5.22 mm, about 3% wider than the 5.06 mm near the axis.

The cos² formula predicts fringes of equal brightness all the way out to ±90°, and no experiment shows that. The culprit is an assumption hidden in the word narrow — each slit alone was taken to spread light smoothly across the whole screen. A slit of finite width a does no such thing — it makes its own diffraction pattern, and the two-slit fringes appear multiplied by it. So the observed fringes fade away from the centre, dying out where one slit's own light does. There is a sharp consequence: when d/a is a whole number, order m = d/a falls exactly on a zero of that envelope and the bright fringe there is simply missing. For d = 0.25 mm with slits a = 50 μm wide, orders 5, 10 and 15 are absent. Keeping a ≪ d is what makes the ideal two-slit model usable; the envelope itself has a topic of its own.

Turn the spacing formula around: λ = d Δy/L, and every quantity on the right is a bench measurement. Δy is the awkward one, because a single spacing of a few millimetres cannot be read to better than a few per cent. Measure across ten spacings instead — the distance from the m = −5 fringe to the m = +5 fringe is 10Δy — and dividing by ten divides the reading uncertainty by ten with it. The relation runs the other way too. With λ known, the fringes measure d, which is how slit separations too small for a ruler are found. And an unknown medium declares itself: immerse the whole apparatus in water, where the frequency is unchanged but the wavelength becomes λ/n with n = 1.33, and a 4.8 mm spacing in air contracts to 3.6 mm. Nothing geometric moved — only the ruler the light is using.