Aperture Integrals, Fresnel Number & Combined Patterns

Write A(x) for the amplitude an opening transmits at position x: 1 inside a clear slit, 0 outside, a smaller number where a filter dims the light, a complex number where a plate shifts its phase. Every element A(x) dx radiates a wavelet that, far away in direction θ, leads the centre by the phase kx sin θ, so E(θ) ∝ ∫ A(x) e^{(ikx sin θ)} dx over the whole opening. That is the Fourier transform of A at the spatial frequency k sin θ, in rad m⁻¹, and the exponent is a length times an inverse length, as the argument of e must be. The rectangle of the last topic transforms to sin β/β. Change the function and the transform changes with it: a slit whose transmission falls linearly from 1 at the centre to 0 at the edges transforms to (sin(β/2)/(β/2))², because a triangle is a rectangle of width a/2 convolved with itself. Its intensity is the fourth power of a sinc: the first side lobe drops from 4.72% to 0.22% of the peak, but the central lobe doubles in width, since a sin θ = 2λ is now the first zero. That trade, side lobes against resolution, is apodisation, read straight off the integral.

Move the whole aperture sideways by x₀, so the transmission becomes A(x − x₀). Substitute x' = x − x₀: the limits move with it, and the factor e^{(ikx₀ sin θ)} comes outside untouched. A shifted aperture therefore has exactly the same intensity pattern; only the phase of the far field records where it was. Now put two identical slits at x₀ = +d/2 and x₀ = −d/2. Linearity adds their fields: E = E₁(θ)[e^(iα) + e^(−iα)] = 2E₁(θ) cos α with α = πd sin θ/λ = ½kd sin θ, where E₁ is one slit's own field. Squaring gives I = 4I₁ cos²α, so with I₁ = Iₛ (sin β/β)² the whole pattern is I₀ cos²α (sin β/β)² and I₀ = 4Iₛ: the two-slit centre is four times as bright as one slit's, not twice, because amplitudes add before squaring. Try a = 25 μm, d = 75 μm and λ = 633 nm at sin θ = 0.010. Then α = 3.72 rad and β = 1.24 rad; cos²α = 0.699 and (sin β/β)² = 0.581, so I/I₀ = 0.406. The point sits near the m = 1 fringe (d sin θ = 1.18λ), and the envelope has already taken 42% of what that fringe would have had.

Fringe maxima sit where d sin θ = mλ; envelope zeros where a sin θ = nλ with n ≠ 0. They coincide when m/n = d/a, and there the bright direction of the pair is a direction in which each slit alone sends nothing, so the order is missing. With d/a = 3 the orders ±3, ±6, ±9 vanish. Every other fringe is scaled by the envelope at its own angle, βₘ = mπa/d: for d = 3a the first order keeps (sin(π/3)/(π/3))² = 0.684 of the central intensity, the second 0.171, and the third is gone. Inside the central lobe |m| < d/a, so an integer ratio leaves 2(d/a) − 1 fringes, five for d = 3a and seven for d = 4a. A non-integer ratio hides no order until ma/d first reaches an integer: at d/a = 2.5 the third-order fringe survives at a dim 2.4%, the first casualty is m = 5, and the central lobe still holds five fringes. Two refinements a first course skips: the product's maxima lie a little inside mλ/d, pulled by the envelope's slope, and narrowing a with d fixed broadens the envelope and admits more orders without moving any of them.

Every step so far took the phase across the aperture to be linear in x. From an element at x to a screen point y at distance L the exact path is √(L² + (y − x)²) ≈ L + (y − x)²/(2L), so k times it contains −kxy/L, the Fraunhofer term with sin θ ≈ y/L, and kx²/(2L), which the far-field integral throws away. At the edge x = a/2 the discarded phase is πa²/(4λL) = (π/4)N_F with N_F = a²/(λL). Read N_F as (a/L)/(λ/a): the angle the aperture subtends over its own diffraction angle. For N_F ≲ 0.1 the dropped phase is under 0.08 rad and the pattern is Fraunhofer; near N_F = 1 it is 0.8 rad and neither the sinc² nor a shadow is right; for N_F ≫ 1 the shadow returns, fringed at its edges. A 0.50 mm slit at 500 nm has a²/λ = 0.50 m, so 5.0 m gives N_F = 0.10, 0.50 m gives 1.0 and 5.0 cm gives 10. For two slits the width that matters is the illuminated extent d + a, not a. A lens changes the rule: its own phase −kx²/(2f) cancels the dropped term exactly at L = f, so its focal plane shows the true far field on a short bench, with y = f tan θ.

When N_F is not small the integrand keeps the whole paraxial phase: E(y) ∝ ∫ A(x) e^{(iπ(x − y)²/(λL))} dx. Substitute t = (x − y)√(2/(λL)) and the exponent becomes iπt²/2, so the field is the difference of C(v) = ∫₀ᵛ cos(πt²/2) dt and S(v) = ∫₀ᵛ sin(πt²/2) dt between the two edges. These Fresnel integrals have no elementary antiderivative; they are tabulated, and plotted against each other they trace the Cornu spiral, a phasor arc whose curvature grows along its length instead of staying fixed. On the axis of a slit each edge sits at v = √(N_F/2), and I/I_inc = 2[C(v)² + S(v)²], with I_inc the intensity of the unobstructed beam. Check the limits. As N_F → 0, C ≈ v and S ≈ πv³/6, so I/I_inc → 2v² = N_F: exactly the on-axis Fraunhofer prediction, an amplitude proportional to a. As N_F → ∞, C and S → ½ and I/I_inc → 1, the geometric shadow, where the axis is as bright as the open beam. Between them the axis oscillates: N_F = 1 gives 0.946, N_F = 2 gives 1.60, brighter than with no slit at all, N_F = 4 gives 1.58 and N_F = 6 gives 0.74.

A Fresnel integral, a measured A(x) or an unsymmetrical aperture all end the same way: the phasor sum is done by machine. Cut the aperture into M strips of equal width, evaluate the phase at each midpoint and add the phasors, E ≈ (1/M) Σ A(uⱼ) e^(iφⱼ), with uⱼ = −½ + (j − ½)/M the fractional position and, on axis, φⱼ = πN_F uⱼ². For N_F = 2 the edge phase is π/2, and M = 4 gives a mean phasor of squared modulus 0.854, so I/I_inc = N_F × 0.854 = 1.71; M = 8 gives 1.63 and M = 16 gives 1.61, converging on the exact 1.60 from above with an error falling as 1/M². The choice of M is set where the phase varies fastest, at the edge: neighbouring strips should differ by less than about π/8, and since dφ/du = 2πN_F u reaches 2π at u = ½, a strip width 1/M gives a step 2π/M and M ≥ 16. For a two-dimensional aperture the same sum on a grid is the fast Fourier transform. Numbers from a machine still owe the three checks: every phase must be dimensionless, N_F → 0 must return the sinc² curve, and integrating I across the screen must return the power that entered the aperture.