Visual walkthrough — Diffraction — single slit intensity pattern derivation

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Step 1 — A slit is not one source, it is a beach of sources

WHAT. We shine light of one pure colour on an opening (a "slit") of width $a$ cut in an opaque screen. Instead of thinking of the light at the slit as a single thing, we chop the opening into many thin vertical strips.

WHY. A famous idea — Huygens Principle — says every point a wave reaches becomes a new tiny source of ripples ("wavelets") that spread outward. So an opening of real width is really a continuous row of little sources sitting side by side. That is the whole reason a single slit can make a pattern at all: many sources can interfere.

PICTURE. Look at the figure. The slit (height $a$) is sliced into strips. Each red dot is one wavelet source. We measure how far down a strip sits from the top edge with the letter $y$ — so $y=0$ at the top, $y=a$ at the bottom.

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Step 2 — Rays that leave at an angle travel unequal distances

WHAT. We pick a viewing direction: an angle $\theta$ tilted off the straight-ahead line. Every strip sends a wavelet toward that same far-off direction. Because the strips sit at different heights, their rays start from different places — so they cover different total distances to the screen.

WHY. The screen is very far away (this "far screen" assumption is the Fraunhofer condition), so all rays heading to one point are effectively parallel. For parallel rays we can measure the extra distance one ray travels with a clean right triangle — pure geometry.

PICTURE. Drop a perpendicular from the top strip onto the ray leaving a strip at depth $y$. The little leg it cuts off — the extra path — is the short side of a right triangle whose long side is $y$ and whose angle is $\theta$.

Reading the triangle: the side opposite the angle $\theta$, divided by the hypotenuse $y$, is what $\sin \theta$ means. So

$\underset{\text{extra path of strip at }y}{\underbrace{\Delta (y)}}=\underset{\text{how deep the strip is}}{\underbrace{y}}\times \underset{\text{tilt of the viewing direction}}{\underbrace{\sin \theta }}.$

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Step 3 — Extra distance becomes extra phase

WHAT. Two waves that travel different distances arrive "out of step." We turn the extra distance $\Delta (y)$ into an amount of lag, called phase.

WHY. One whole wavelength $\lambda$ of extra path pushes a wave exactly one full cycle out of step. We measure one full cycle as an angle of $2\pi$ radians (once around a circle). So extra path and phase are just two units for the same "how far out of step" idea, linked by "$\lambda$ of path = $2\pi$ of phase."

PICTURE. The wheel picture: as a wave travels, imagine a dot going round a circle. Moving one wavelength = one full lap = $2\pi$. Half a wavelength = half a lap = $\pi$ = a crest lands where a trough was.

$\varphi (y)=\underset{\begin{array}{c}\text{phase per}\\ \text{metre of path}\end{array}}{\underbrace{\frac{2\pi }{\lambda }}}\times \underset{\text{extra path }\Delta (y)}{\underbrace{y\sin \theta }}.$

The biggest phase gap is between the very top ($y=0$) and very bottom ($y=a$) strips:

$\Phi =\frac{2\pi }{\lambda }a\sin \theta =2\beta ,\boxed{\beta \equiv \frac{\pi a\sin \theta }{\lambda }}.$

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Step 4 — Adding waves = adding little arrows (phasors)

WHAT. Each strip sends a wave of the same small strength but a different phase. We must add all these waves. We do it by drawing each wave as a little arrow ("phasor") and laying them tip-to-tail.

WHY. A wave has a size and a phase — that is precisely what an arrow has: a length and a direction. Adding waves is then just adding arrows. This is the phasor trick, and it makes the whole thing visual instead of algebraic.

PICTURE. Every arrow has the same length (each strip is equally bright) but each is turned a little more than the one before (phase grows steadily with depth $y$). Laid tip-to-tail, equal arrows with a steadily-turning direction bend into an arc of a circle. The resultant wave is the single arrow from the start of the first to the tip of the last — the chord of that arc.

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Step 5 — The chord length gives $\sin \beta /\beta$

WHAT. We compute the chord length of an arc that bends by a total angle $2\beta$, keeping the arc's length fixed (that length is the total amplitude $A_0$ you'd get if all arrows pointed the same way).

WHY. The chord is the resultant wave's amplitude. Getting its length as a formula in $\beta$ is the entire derivation in one geometric move.

PICTURE. The arc has length $A_0$ and bends through $2\beta$, so it lives on a circle of radius $R=A_0/(2\beta )$ (arc length = radius × angle). The chord subtends the same angle $2\beta$ at the centre, and the chord of an arc is $2R\sin (\text{half the bend})$.

$E=\underset{=A_0/\beta }{\underbrace{2R}}\underset{\text{half-bend}}{\underbrace{\sin \beta }}=\frac{A_0}{\beta }\sin \beta =A_0\frac{\sin \beta }{\beta }.$

Term by term: $R=\frac{A_0}{2\beta }$ comes from "arc $=$ radius $\times$ angle" with arc $=A_0$ and angle $=2\beta$; the chord formula $2R\sin \beta$ uses half the bend, which is why $\beta$ (not $2\beta$) appears inside the $\sin$.

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Step 6 — Square it to get brightness

WHAT. Convert amplitude (arrow length) into intensity — what an eye or sensor actually measures.

WHY. Brightness is proportional to amplitude squared (energy carried by a wave grows like the square of its size). Squaring also kills the direction-only factor $e^{i\beta }$, since its length is $1$.

PICTURE. Same chord as before, now with its length squared — the tall central spike appears because at $\theta =0$ the arc is dead straight and the chord is longest.

$\boxed{I(\theta )=I_0{\left(\frac{\sin \beta }{\beta }\right)}^2,\beta =\frac{\pi a\sin \theta }{\lambda },I_0=A_0^2}$

• $I(\theta )$ ::: brightness at viewing angle $\theta$.
• $I_0$ ::: the brightness dead centre ($\theta =0$).
• ${\left(\sin \beta /\beta \right)}^2$ ::: the "$\mathrm{sinc}$-squared" shape — the chord length squared, in units of $A_0$.

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Step 7 — Edge case: straight ahead ($\theta =0$)

WHAT. Check what the formula does at the centre, where $\theta =0$ so $\beta =0$ — and $\sin \beta /\beta$ looks like the forbidden $0/0$.

WHY. A formula must survive its degenerate input. Physically $\theta =0$ means every strip is in perfect step, so we expect the brightest point, not a hole.

PICTURE. The arc unbends into a straight line: all arrows point the same way, the chord equals the full arc, resultant is longest. As $\beta \to 0$ the ratio $\sin \beta /\beta \to 1$.

${\lim }_{\beta \to 0}\frac{\sin \beta }{\beta }=1\Rightarrow I(0)=I_0\cdot 1^2=I_0.$

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Step 8 — Edge case: the dark fringes ($\beta =m\pi$)

WHAT. Find where the brightness drops to exactly zero. That needs $\sin \beta =0$ with $\beta \ne 0$, i.e. $\beta =m\pi$ for $m=\pm 1,\pm 2,\dots$

WHY. Zeros of $\sin$ are where the arc has bent all the way round into a closed loop — the chord shrinks to nothing. This is the signature "cancel in pairs" of a continuous slit.

PICTURE. When $\beta =\pi$ the total bend is $2\beta =2\pi$: the arrows curl into one full circle, tip meets tail, chord $=0$. Split the slit into a top half and a bottom half — each strip up top has a partner half a slit down that is exactly $\lambda /2$ out of step, so they annihilate.

$\beta =m\pi \Rightarrow \frac{\pi a\sin \theta }{\lambda }=m\pi \Rightarrow \boxed{a\sin \theta =m\lambda },m=\pm 1,\pm 2,\dots$

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Step 9 — Edge case: the dim secondary peaks

WHAT. Between two dark fringes the light does not vanish — it rises to a small "secondary" peak. These sit where the chord is locally longest, not exactly halfway.

WHY. The peaks of $I=I_0(\sin \beta /\beta {)}^2$ are found by asking where the curve stops climbing. Doing that (setting the slope to zero) gives the condition $\tan \beta =\beta$, whose first solution is $\beta \approx 1.43\pi$ — pulled toward the centre by the shrinking $1/\beta$ envelope, so not at the tidy $1.5\pi$.

PICTURE. The full ${\mathrm{sinc}}^2$ curve: a giant central hump, then bumps that fall off fast. The first side bump reaches only about $4.5\%$ of the centre.

$\frac{I}{I_0}{|}_{\beta \approx 1.43\pi }={\left(\frac{\sin (1.43\pi )}{1.43\pi }\right)}^2\approx 0.0472\approx 4.5\%.$

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Worked checks (numbers you can trust)

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The one-picture summary

This figure stacks the whole logic: slit → path difference $y\sin \theta$ → phase → arc of phasors → chord $\propto \sin \beta /\beta$ → square it → the ${\mathrm{sinc}}^2$ pattern with its centre, dark fringes at $a\sin \theta =m\lambda$, and dim side peaks.

Recall Feynman retelling — say it back in plain words

A slit is really a row of tiny light sources (Huygens). Tilt your gaze by an angle $\theta$: each source's ray travels a slightly different distance, the extra bit being $y\sin \theta$ for a source $y$ below the top. Extra distance means extra lag, and one wavelength of lag is one full turn ($2\pi$). So each source's wave points a little more turned than the one above it. Draw them as equal arrows, each rotated a bit more — they curl into an arc, and the sum is the straight chord across it. Do the geometry and the chord's length is $A_0\sin \beta /\beta$, where $\beta$ is half the total twist across the slit. Square it (brightness $=$ amplitude squared) and you get $I_0(\sin \beta /\beta {)}^2$. Straight ahead the arc is flat → longest chord → brightest spot. When the arc curls into a full circle ($\beta =m\pi$, i.e. $a\sin \theta =m\lambda$) the chord vanishes → dark. Between the loops the arc bends a bit and the chord is short but not zero → faint side peaks. Thinner slit, more curling per angle, wider fan — light spread out.

Recall

A single slit gives DARK fringes at ::: $a\sin \theta =m\lambda$ with $m=\pm 1,\pm 2,\dots$ ($m=0$ is the central bright peak). The variable $\beta$ physically is ::: half the total phase spread across the slit, $\beta =\pi a\sin \theta /\lambda$. Why is the resultant amplitude $\propto \sin \beta /\beta$ ::: it is the chord of a circular arc of length $A_0$ bent through total angle $2\beta$. First secondary peak brightness ::: about $4.5\%$ of $I_0$, near $\beta \approx 1.43\pi$.