Two-source interference & double-slit intensity

Put the slits at ±d/2 and a screen point P at height y, a distance L away. Exactly, Δ = √(L² + (y + d/2)²) − √(L² + (y − d/2)²). Nobody computes with that. Expand for L ≫ y and L ≫ d: the leading term is Δ ≈ yd/L, and the next two are −(yd/L)(y²/2L²), which is only the gap between tanθ and sinθ, and −(yd/L)(d²/8L²), the residual from the two rays not being quite parallel. Writing Δ = d sinθ, the projection of the slit separation onto the ray direction, absorbs the first correction exactly and leaves the second, which for two point slits placed symmetrically is tiny: with d = 0.250 mm and L = 2.00 m it is 2 parts in 10⁹. At y = 5.06 mm on that screen the exact Δ, d sinθ and yd/L agree to 0.002 nm on a 632 nm difference. The stricter condition this section quotes for the far field, L ≫ d²/λ (here d²/λ = 9.9 cm), is not what two point sources need; it is what the finite width of each real slit will demand once its own diffraction enters, three topics on.

One extra wavelength of travel is one full cycle, so a path difference Δ is worth a phase φ = 2πΔ/λ. The field from slit 1 at P is E₀ cos(ωt) and from slit 2 it is E₀ cos(ωt + φ); both oscillate at the same ω, so represent each by a phasor of length E₀ and add them tip to tail. Two equal phasors separated by φ form an isosceles triangle whose closing side is the resultant, of length 2E₀ cos(φ/2) and pointing at φ/2, halfway between them: E₀ cos(ωt) + E₀ cos(ωt + φ) = 2E₀ cos(φ/2) cos(ωt + φ/2). Check the limits: φ = 0 gives 2E₀, φ = π gives zero, and the result repeats every 2π, as a fringe pattern must. With numbers: at Δ = λ/4, φ = π/2 and the resultant is 2E₀ cos(π/4) = 1.41E₀; at Δ = λ/3, φ = 2π/3 and it is exactly E₀, the same as one slit alone. Adding the fields as plain numbers assumes both are polarised the same way; that scalar approximation runs through this whole section. Any extra phase, from a plate over one slit or a different medium in one path, is added to φ at this step, not bolted onto the answer.

Intensity is proportional to the time average of E², and ⟨cos²(ωt + φ/2)⟩ over a cycle is ½, so I ∝ ½(2E₀)² cos²(φ/2). Calling the single-slit level I₀ ∝ ½E₀², the pattern is I = 4I₀ cos²(φ/2). Three numbers to keep: the peaks are 4I₀, not 2I₀; the minima are exactly zero, because the two phasors have equal length; and the average over one fringe is 4I₀ × ½ = 2I₀, exactly I₀ + I₀. Interference moves energy from the dark fringes into the bright ones and creates none. The intensity halves at φ = π/2, that is at Δ = λ/4; for the 0.250 mm slits and the 632.8 nm laser that point sits 1.27 mm from the central peak, a quarter of the 5.06 mm spacing. The identity cos²(φ/2) = (1 + cos φ)/2 gives the equivalent form I = 2I₀(1 + cos φ), the one that generalises when the two sources are unequal.

The peaks fall where the phasors are parallel, φ = 2πm, so d sinθ = mλ; the zeros where they are antiparallel, d sinθ = (m + ½)λ. On a screen at distance L with y ≪ L, sinθ ≈ tanθ = y/L, so yₘ = mλL/d and successive bright fringes are Δy = λL/d apart, whatever m is. That constancy is what makes the pattern a measuring instrument: count N spacings across a length and divide. With d = 0.250 mm and L = 2.00 m, ten spacings of the helium–neon line cover 50.6 mm, and inverting, λ = (50.6 mm/10)(0.250 mm)/(2.00 m) = 633 nm: a wavelength of 0.6 μm from three lengths read off a rule and a travelling microscope. The angular spacing λ/d = 2.53 mrad is the quantity the screen distance cannot change. Watch the small-angle step: at m = 20, y = 101 mm and tanθ exceeds sinθ by 0.13%, so the twentieth fringe already sits 0.13 mm outside the linear formula. The last order that exists at all is the largest integer below d/λ = 395.

The section's habit is to build the total phase before choosing a formula. Cover one slit with a transparent plate of thickness t and index n and that path gains optical length (n − 1)t, because the light inside it has wavelength λ/n. The phasor angle becomes φ = 2π[d sinθ − (n − 1)t]/λ, with θ counted positive towards the covered slit. Every fringe keeps its spacing, since the plate adds the same constant at every angle, but the whole pattern slides: the zero-order fringe moves to where d sinθ = (n − 1)t, on the side of the covered slit, where the shorter geometric path pays for the plate. For t = 10.0 μm and n = 1.50 with the 632.8 nm laser, (n − 1)t = 5.00 μm, which is 7.90 wavelengths, so the pattern shifts by 7.90 fringe spacings: 40.0 mm on the 2.00 m screen. Immersing the whole apparatus in water instead shortens every wavelength to λ/1.33, so Δy shrinks by 1.33 and nothing shifts. The rule is the same both times: convert each path to optical length, take the difference, divide by the vacuum wavelength, and only then write cos².

Each assumption that built the model marks a boundary. Narrow slits: a real slit of width a diffracts, and its single-slit pattern multiplies the cos² fringes as an envelope of half-width λL/a, so outer fringes dim and some orders vanish; that is the business of the diffraction topics. Equal illumination: with intensities I₁ and I₂ the sum is I = I₁ + I₂ + 2√(I₁I₂) cos φ, whose minima no longer reach zero. With I₂ = I₁/4 the peaks are 2.25I₁ and the troughs 0.25I₁, a visibility (Iₘₐₓ − Iₘᵢₙ)/(Iₘₐₓ + Iₘᵢₙ) = 0.80. Mutual coherence: a broad or broadband source lets the phase difference wander, and the fringes wash out, as the previous topic quantified. Small angles: beyond y ≈ 0.1L use d sinθ, not yd/L. And a phase difference is defined for one wavelength only: white light gives a white central fringe and coloured, overlapping orders beyond it, because each λ has its own spacing λL/d.