Visual walkthrough — Interference — constructive, destructive conditions

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Step 1 — A single wave is a moving up-and-down number

WHAT. Pick one point in space — call it $P$. A wave passing through $P$ makes the medium there (air, water, a string) move up and down over time. We write that up-or-down amount as $y$.

WHY a sine. The simplest, smoothest possible repeating motion — the motion of a mass on a spring, of a swing — traces out a sine curve when you plot displacement against time. So we model one wave as

$y_1=a\sin (\omega t)$

• $y_1$ — the displacement: how far up ($+$) or down ($-$) point $P$ is, right now.
• $a$ — the amplitude: the biggest displacement it ever reaches. The height of a crest.
• $t$ — time, ticking forward.
• $\omega$ — the angular frequency: how fast the wave cycles. Bigger $\omega$ = quicker wiggle. It turns clock-time $t$ into an "angle" $\omega t$ that sweeps through a full circle ($2\pi$) each cycle.

PICTURE. The red curve is $y_1$ over time. Notice it starts at zero, rises to $+a$, falls to $-a$, and repeats.

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Step 2 — A second wave, running behind

WHAT. Now a second wave of the same kind reaches $P$, but it left its source a little later (or travelled farther). It is the same shape, same amplitude, same speed — just shifted along the time axis.

WHY $\varphi$. We measure that shift as an angle $\varphi$, the phase difference. "Phase" is where in its cycle a wave is. Two waves in the same place in their cycle have $\varphi =0$; one a full cycle behind has $\varphi =2\pi$.

$y_2=a\sin (\omega t+\varphi )$

• $\varphi$ — the phase difference: how many radians of its own cycle wave 2 is ahead of wave 1. Adding $\varphi$ inside the sine slides the curve leftward (earlier) in the picture.

PICTURE. Red is $y_1$, teal is $y_2$. They have the same height and rhythm; teal is just nudged sideways by $\varphi$.

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Step 3 — Superposition: just add the two heights

WHAT. At every instant, the medium at $P$ can only be in one place. So its actual displacement is the sum of what each wave alone would demand:

$y=y_1+y_2=a\sin (\omega t)+a\sin (\omega t+\varphi )$

WHY add, not multiply or average. This is the Principle of Superposition — a physical fact for small waves. Push a swing twice at once, the pushes literally add. Nothing fancier is going on.

PICTURE. The plum curve is the point-by-point sum. Where red and teal are both up, plum is tall. Where one is up and the other down, they partly cancel and plum is small.

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Step 4 — Two sines of equal frequency add to ONE sine

WHAT. We use a trig identity to collapse the sum into a single sine:

$\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}$

Put $A=\omega t$ and $B=\omega t+\varphi$:

$y=\underset{A_{\text{res}}}{\underbrace{2a\cos \frac{\varphi }{2}}}\sin \left(\omega t+\frac{\varphi }{2}\right)$

Term by term:

• $\frac{A+B}{2}=\omega t+\frac{\varphi }{2}$ — the new sine still cycles at frequency $\omega$; it's just shifted by half the phase gap. Same rhythm.
• $\frac{A-B}{2}=-\frac{\varphi }{2}$, and $\cos (-x)=\cos x$, so the front factor is $2a\cos \frac{\varphi }{2}$.
• That whole front factor is a constant in time (it has no $t$ in it) — it is the new amplitude $A_{\text{res}}$.

WHY this matters. The sum of two equal-frequency sines is still a sine of that frequency — only its height and starting point change. So all the physics of "how big is the combined wave" lives in the single number $A_{\text{res}}=2a\cos (\varphi /2)$.

PICTURE. The plum sum from Step 3 is redrawn, and a dashed envelope marks its true amplitude $2a\cos (\varphi /2)$ — one clean sine.

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Step 5 — Read the amplitude as a phasor triangle

WHAT. There's a picture that makes $2a\cos (\varphi /2)$ obvious. Draw each wave as an arrow of length $a$ (a "phasor"). Wave 1 points along a direction; wave 2 is turned by angle $\varphi$ from it. The resultant is the arrow you get by placing them tip-to-tail.

WHY arrows. A sine wave's amplitude-and-phase behaves exactly like a 2-D arrow's length-and-direction. Adding waves = adding arrows. This turns algebra into geometry.

The geometry. Two arrows of equal length $a$ with angle $\varphi$ between them form a rhombus. The resultant is its long diagonal. That diagonal bisects the angle, splitting the rhombus into two right triangles, each with the resultant's half-length as the adjacent side to angle $\varphi /2$:

$\frac{A_{\text{res}}/2}{a}=\cos \frac{\varphi }{2}⟹A_{\text{res}}=2a\cos \frac{\varphi }{2}$

• $\cos \frac{\varphi }{2}$ — "adjacent over hypotenuse" in that little right triangle; it measures how much the two arrows point the same way.

PICTURE. Red arrow $a$, teal arrow $a$ at angle $\varphi$, plum resultant along the diagonal, with the $\varphi /2$ right triangle shaded.

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Step 6 — From amplitude to intensity (why we square)

WHAT. Brightness (light) or loudness (sound) is the intensity $I$, and intensity is proportional to amplitude squared:

$I\propto A_{\text{res}}^2={\left(2a\cos \frac{\varphi }{2}\right)}^2=4a^2{\cos }^2\frac{\varphi }{2}$

Let one wave alone carry intensity $I_0\propto a^2$. Then:

$\boxed{I=4I_0{\cos }^2\frac{\varphi }{2}}$

WHY squared. A wave's energy grows with amplitude squared — doubling a wave's height quadruples its energy (see Energy in Waves). So we must finish with amplitude, then square. Never add intensities first.

• ${\cos }^2(\varphi /2)$ — always between $0$ and $1$: it dials the intensity between fully dark and $4\times$ bright.

PICTURE. Plot of $I/I_0=4{\cos }^2(\varphi /2)$ against $\varphi$: peaks of $4$, valleys of $0$, and the dashed average line at $2$.

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Step 7 — Every case: bright, dark, and the neutral in-between

WHAT. Now sweep $\varphi$ through all its values and read the boxed law.

$\varphi$	$\cos (\varphi /2)$	$I$	meaning
$0,2\pi ,4\pi \dots$	$\pm 1$	$4I_0$	constructive — brightest
$\pi /2$	$\frac{1}{\sqrt{2}}$	$2I_0$	neutral — equals the average
$\pi ,3\pi \dots$	$0$	$0$	destructive — dark
$2\pi /3$	$1/2$	$I_0$	dim, one-source-worth

WHY it repeats. ${\cos }^2$ has period $2\pi$ in $\varphi /2$, i.e. period... the pattern of bright and dark simply repeats every $2\pi$ of phase — which, via the master link, means every one wavelength of path.

The master link (path ↔ phase). A path difference $\Delta x$ of one whole wavelength $\lambda$ is one whole cycle $2\pi$ of phase:

$\varphi =\frac{2\pi }{\lambda }\Delta x$

Feed the bright/dark values of $\varphi$ back through this to get the path conditions (see Coherence and Path Difference):

$\text{Constructive: }\Delta x=n\lambda \text{Destructive: }\Delta x=\left(n+\frac{1}{2}\right)\lambda$

PICTURE. The intensity curve again, but with the three regions flagged: WOW peaks, HUSH zeros, and the neutral crossing at $\varphi =\pi /2$.

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Step 8 — Degenerate case: unequal amplitudes can't fully cancel

WHAT. Drop the "equal $a$" assumption. Two arrows of lengths $a_1$ and $a_2$ at angle $\varphi$ combine by the law of cosines:

$A_{\text{res}}=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \varphi }$

• $2a_1{a}_2\cos \varphi$ — the "interference" cross-term; it swings positive (add) or negative (subtract) as $\varphi$ turns.

WHY it can't reach zero. At $\varphi =\pi$ (fully opposed) this gives $A_{\text{res}}=|a_1-a_2|$. Only when $a_1=a_2$ does that become $0$. Unequal waves leave a leftover — destructive interference is only total for equal amplitudes.

PICTURE. Arrows $3a$ and $4a$: at $\varphi =0$ they stack to $7a$; at $\varphi =\pi$ they oppose to leftover $1a$.

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The one-picture summary

Everything above, compressed: two arrows of length $a$ at angle $\varphi$ → resultant $2a\cos (\varphi /2)$ → intensity $4I_0{\cos }^2(\varphi /2)$ → bright when $\varphi =2\pi n$, dark when $\varphi =(2n+1)\pi$.

Recall Feynman: tell the whole walkthrough as a story

Picture one point in a pond where two ripples arrive. Each ripple, on its own, would bob that spot up and down — a smooth sine wiggle (Step 1). The second ripple is the same size but arrives a beat late; that lateness is an angle $\varphi$ (Step 2). The water can only be in one place, so its motion is just the two wiggles added (Step 3). Add two equal wiggles and — magic of trig — you get one wiggle of the same rhythm but a new height, $2a\cos (\varphi /2)$ (Step 4). You can see that height without algebra: draw each wave as an arrow of length $a$, put them at angle $\varphi$, and the diagonal is the answer — a right triangle hands you the $\cos (\varphi /2)$ (Step 5). Loudness/brightness is height squared, so $I=4I_0{\cos }^2(\varphi /2)$ (Step 6). Turn the $\varphi$ knob: in step → four times as bright; opposite → total silence; quarter-turn → exactly the plain average (Step 7). And if the two waves aren't equal, the silence isn't perfect — there's always leftover $|a_1-a_2|$ (Step 8). That's interference, start to finish.

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Connections

• Interference — constructive, destructive conditions — the parent result we derived here.
• Principle of Superposition — Step 3's "just add" rule.
• Coherence and Path Difference — Step 7's path ↔ phase master link.
• Energy in Waves — Step 6's "intensity is amplitude squared".
• Young's Double Slit Experiment — where these $\varphi$ conditions paint fringes.
• Standing Waves — interference of oppositely-moving waves.
• Beats — what happens when the two frequencies differ.