Physical-Optics Synthesis

A slit is not two edges radiating; every point of the open aperture is a Huygens source, and all of them are driven in step by the same incoming wavefront. Divide the opening into strips of width dx. Each strip sends the same tiny amplitude, proportional to dx, towards a chosen point on the screen. What distinguishes them is arrival time. If the screen sits far enough away that the rays reaching that point are effectively parallel at angle θ, the strip at x saves exactly x sinθ of path against the one at the centre, so it arrives with a phase lead of (2π/λ) x sinθ. That single line is the whole physical input: equal amplitudes, phases linear in position. For a 0.050 mm slit in 633 nm light at θ = 12.66 mrad, the strip at the far edge leads the one at the near edge by (2π/633 nm)(50 μm)(0.01266) = 2π rad — a full turn across the aperture, which is the condition you will meet again in a moment.

Summing thousands of phasors is an integral. Write each strip as a unit phasor e^{(i(2π/λ) x sinθ)} dx and integrate across the slit from −a/2 to +a/2. Put k = 2π/λ and let u = k sinθ. The integral of e^(iux) is e^(iux)/(iu), so evaluating between the limits gives (e^(iua/2) − e^(−iua/2))/(iu) = 2 sin(ua/2)/u, which is a sin β / β with β = ua/2 = πa sinθ/λ. Intensity is the squared modulus, so I(θ) = I₀ (sin β / β)². Check it at the centre: as β → 0, sin β / β → 1, and every strip is in phase, which is exactly the largest amplitude the aperture can produce. Zeros arrive whenever β = pπ with p = 1, 2, 3, … — never p = 0, which is the central maximum — that is, whenever a sinθ = pλ. Between the zeros the side-lobe peaks do not sit at the halfway points: the first true maximum falls at β = 4.4934, short of 3π/2, and the worked example turns that root into a height.

Draw the strips as arrows laid nose to tail. Each is the same length and each is turned from its neighbour by the same small angle, so the chain bends into a circular arc whose total arc length is fixed by the aperture width and whose total turning is 2β. The resultant is the straight chord across that arc. At β = 0 the arc is straight and the chord equals the full arc: maximum. As β grows the arc curls, the chord shortens, and at 2β = 2π the arc has closed into a complete circle with zero chord — the first dark fringe, reached when a sinθ = λ. Curl further and the chain wraps more than once, leaving only a short leftover chord, which is why the side lobes are so faint. The picture also explains the awkward 4.4934: the chord peaks slightly before the arc has turned one and a half circles.

Now open a second identical slit with its centre a distance d from the first. The integral splits into two pieces that differ only by the constant phase the centres carry, so each returns the same sin β / β and the pair contributes a common factor times (e^(iγ) + e^(−iγ)) = 2 cos γ, with γ = πd sinθ/λ. Squaring gives I(θ) = I₀ (sin β / β)² cos²γ. The result is a product, and the two factors watch different lengths: the slit width a fixes the slowly varying envelope, the separation d fixes the rapid fringes. For N identical slits the same argument stacks N phase factors into a geometric sum and the bracket becomes [sin(Nγ)/(N sin γ)]², which is cos²γ at N = 2 and narrows each principal maximum until its first zero lies only λ/(Nd) away in sinθ.

Because the factors multiply, an interference maximum lands on an envelope zero whenever m λ/d equals p λ/a, that is when m = p(d/a). Take a = 0.060 mm and d = 0.18 mm, so d/a = 3: the third order is dark on both sides and the central lobe carries only five fringes, m = 0 out to ±2. The product also grades the ones that survive. Order m has β = mπa/d, so at d/a = 3 the m = 1 fringe keeps 68.4% of the central height and m = 2 only 17.1% — the envelope is already falling steeply while the fringes are still evenly spaced. If you plot the profile numerically, sample it finely: with N slits each principal maximum runs only λ/(Nd) in sinθ from its peak to the zero beside it, so a step coarser than about a tenth of that walks straight over the peaks and returns a curve that is smooth and wrong.

Three assumptions are doing work, and each has a number. First, parallel rays: they are legitimate only when the Fresnel number F = W²/(λL) is far below 1 for the full illuminated width W. A 2.00 mm aperture in 633 nm light gives F = 31.6 at a screen 0.200 m away, which is a Fresnel shadow with structure of its own rather than this integral; the same aperture gives F = 0.316 at 20.0 m and only 0.0316 at 200 m. Far-field geometry is expensive in length as soon as the aperture is wide, which is why a wide aperture is normally brought into the far field with a lens instead: the Fraunhofer pattern then forms in the lens's focal plane at any convenient distance. Second, small angles: sinθ ≈ tanθ ≈ y/L costs 1.5% at 10° but 15% at 30°, so read a grating order as an angle and never as y/L. Third, coherence: the path difference must stay inside the coherence length, and a sodium lamp filtered to Δλ = 0.6 nm at 589 nm offers only λ²/Δλ ≈ 0.58 mm, while a source of finite width smears each fringe by imaging itself across the pattern.