Superposition & Interference of Waves

Two children throw stones into a still pond a metre apart. Where the ripples meet, something strange and orderly happens: along some lines the water heaves twice as high, along others it lies dead flat, and these lines hold still even as wave after wave rolls through them. No stone is bigger than the other; nothing is added to the water. And yet there are places the waves have agreed to cancel. That agreement is interference, and it is one of the deepest ideas in physics — the reason a soap bubble is coloured, the reason noise-cancelling headphones work, the reason light was finally shown to be a wave. Behind it sits a rule so simple it looks like it cannot be enough: when two waves cross, their displacements just add. Not "collide," not "average" — add, algebraically, crest to crest and crest to trough, at every point and every instant. This is the principle of superposition, and it works because the media we care about are linear: the pond does not care that one ripple is already there when the next arrives. Two pulses travelling toward each other pass clean through one another and come out the far side unchanged, as if the other had never existed. This article turns that one rule into the exam's toolkit. We add two coherent waves and get the resultant amplitude $A_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\varphi}$ ; we specialise to equal amplitudes and find the clean $A_{\text{res}} = 2A\cos(\varphi/2)$ ; we translate the abstract phase difference $\varphi$ into the thing a problem actually gives you — a path difference — and read off when waves reinforce ( $\Delta x = n\lambda$ ) and when they annihilate ( $\Delta x = (n+\tfrac12)\lambda$ ). The single most important number to carry away is that two equal waves in step make four times the intensity of one, not two: interference does not create energy, it moves it, robbing the dark fringes to pay the bright.

Key takeaways

Superposition: in a linear medium the displacements of overlapping waves add, $y = y_1 + y_2$ — with sign. Two pulses pass through each other unchanged.
Two coherent waves (same frequency, fixed phase difference $\varphi$ ) combine to $A_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\varphi}$ , and intensity goes as amplitude squared, $I \propto A^2$ .
Equal amplitudes: $A_{\text{res}} = 2A\cos(\varphi/2)$ , so $I = 4I_0\cos^2(\varphi/2)$ — maximum $4I_0$ (in phase), minimum $0$ (antiphase).
Path difference is the currency: $\varphi = (2\pi/\lambda)\,\Delta x$ , so for sources in phase, constructive when $\Delta x = n\lambda$ , destructive when $\Delta x = (n+\tfrac12)\lambda$ .
Coherence — a steady phase relationship — is required for a stable pattern; two independent sources give none.
Key idea: overlapping waves add their displacements, and the resulting bright-and-dark pattern redistributes energy rather than creating it — two equal waves in step give four times the intensity, paid for by total cancellation elsewhere.

The core insight

Because a linear medium lets wave displacements add, two coherent waves reinforce where their path difference is a whole number of wavelengths and cancel where it is a half-odd number, producing a fixed pattern whose bright points carry the energy withheld from the dark ones — the total is conserved.

Where this lives today

The most everyday descendant of Young's idea sits in a pair of noise-cancelling headphones. A tiny microphone samples the droning sound heading for your ear, and the electronics generate a second sound wave that is its exact antiphase — a copy shifted by half a cycle — so that where the two superpose at your eardrum, $\Delta x$ is engineered to a half-wavelength and the drone destructively interferes into near-silence. It is equation (5) turned into a consumer product. The same principle colours a soap film and an oil slick: light reflected from the front and back surfaces of a thin layer interferes, and whichever wavelength meets the constructive condition for that thickness is the colour you see, sweeping through the rainbow as the film thins. Radio engineers exploit it too — an antenna array steers its beam by feeding its elements with controlled phase differences, reinforcing the signal in one direction and cancelling it in another. And every anti-reflective lens coating on a camera or a pair of glasses is a deliberately chosen thickness that makes reflected light interfere destructively, so more light passes through. Wherever waves are made to agree or disagree on purpose, Young's water ripples are quietly at work.

The physical picture

Start with the rule itself, stripped of numbers. If wave 1 alone would give a point the displacement $y_1$ and wave 2 alone would give it $y_2$ , then with both present the point's displacement is

y = y_1 + y_2 \qquad (1)

where: $y_1$ , $y_2$ — the displacements each wave would produce on its own at that point and instant, each carrying its own sign. That is the whole principle of superposition. Its quiet assumption is linearity: the medium's response to two disturbances is the sum of its responses to each, which holds for ordinary waves at ordinary amplitudes — water ripples, sound, light, a wave on a string. The most vivid consequence is that waves do not collide. Send two pulses toward each other on a rope; at the instant they overlap the rope shows their sum (which may be a tall combined bump, or, if one is up and one is down, momentarily nothing at all), and then each pulse walks out the other side with its original shape and speed intact.

Figure

Two pulses crossing on a rope — before, during, and after, the rope shows the sum and the pulses emerge unchanged

Now let the two waves be sustained and identical in frequency, meeting at a point with a steady phase difference $\varphi$ between them. Superposition still just adds them, but the sum of two sinusoids of the same frequency is another sinusoid of that frequency, whose amplitude depends on how well the two are in step. Adding them as phasors (rotating arrows of length $A_1$ and $A_2$ with angle $\varphi$ between them) and taking the length of the resultant gives

A_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\varphi} \qquad (2)

where: $A_1$ , $A_2$ — the two amplitudes; $\varphi$ — the phase difference between the waves. Read the $\cos\varphi$ term as the "cooperation" term: at $\varphi = 0$ it is $+2A_1A_2$ and the amplitudes add fully; at $\varphi = \pi$ it is $-2A_1A_2$ and they fight. Because intensity goes as amplitude squared, $I \propto A^2$ , equation (2) is also the rule for how bright or loud the overlap is.

Interference of equal-amplitude waves

The cleanest and most-examined case is two waves of the same amplitude $A$ . Setting $A_1 = A_2 = A$ in (2) and using a trig identity (derived below) collapses it to

A_{\text{res}} = 2A\cos\!\left(\tfrac{\varphi}{2}\right), \qquad I = 4I_0\cos^2\!\left(\tfrac{\varphi}{2}\right) \qquad (3)

where: $I_0$ — the intensity of one wave alone. This little formula holds the whole story. In step ( $\varphi = 0$ ): $A_{\text{res}} = 2A$ and $I = 4I_0$ — constructive interference, four times one wave's intensity. Exactly out of step ( $\varphi = \pi$ ): $A_{\text{res}} = 0$ and $I = 0$ — destructive interference, total darkness. The bright maximum is $4I_0$ , not $2I_0$ ; hold that thought, because it is where energy conservation seems, wrongly, to break. To use any of this you must know $\varphi$ , and a problem never hands it to you directly. It hands you a path difference $\Delta x$ — the extra distance one wave has travelled to reach the point — and phase and path are locked together, since one whole wavelength of extra path is one full $2\pi$ of extra phase:

\varphi = \frac{2\pi}{\lambda}\,\Delta x \qquad (4)

where: $\Delta x$ — the path difference; $\lambda$ — the wavelength. Feed (4) into the maxima and minima of (3). For two sources that start in phase, constructive interference needs $\varphi$ to be a whole multiple of $2\pi$ , and destructive needs an odd multiple of $\pi$ :

\text{constructive: } \Delta x = n\lambda, \qquad \text{destructive: } \Delta x = \left(n + \tfrac12\right)\lambda \qquad (5)

where: $n = 0, 1, 2, \dots$ These two conditions are the currency of every interference problem you will meet. The clause "sources in phase" matters: if one source is driven a half-cycle behind the other, the roles of (5) swap, so always check how the sources start. For unequal amplitudes, the brightest and darkest the pattern can get are set by the amplitudes adding and subtracting, giving the contrast ratio

\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2 \qquad (6)

where: the maximum comes from $\varphi = 0$ (amplitudes add) and the minimum from $\varphi = \pi$ (they subtract). Only when $A_1 = A_2$ does $I_{\min}$ reach zero and the fringes become perfectly black.

Coherence — why two torches never interfere

One condition underlies everything above: the phase difference $\varphi$ must hold still long enough to see. Two sources are coherent when their phase relationship is constant in time. Two loudspeakers wired to one signal generator are coherent — whatever their phase difference is at a point, it stays that value — so a stable pattern of loud and quiet spots hangs in the air. Two separate whistles, or two torches, are incoherent: each keeps its own wandering phase, $\varphi$ scrambles millions of times a second, and the pattern washes out to a uniform average in which the intensities simply add, $I = 2I_0$ , with no fringes at all. This is why interference had to wait for clever experiments — nature rarely hands you two coherent sources for free.

Figure

Two coherent sources set up fixed lines of reinforcement and cancellation from the path difference

Derivation AAdding two equal waves — the sum-to-product route to 2A cos(φ/2)

The goal here is to prove equation (3), the equal-amplitude interference law, using nothing beyond a standard trigonometric identity — no calculus, exactly the tool a high-school toolkit already contains. Two waves of equal amplitude arrive at a point one phase-step $\varphi$ apart; we add their displacements by the superposition rule and show the sum is a single wave of the same frequency whose amplitude is $2A\cos(\varphi/2)$ . Reading the maxima and minima off that amplitude then gives constructive and destructive interference and, with $I \propto A^2$ , the striking $4I_0$ maximum.

Step 1. Write the two waves at the meeting point, same amplitude and frequency, phase difference $\varphi$ : $y_1 = A\sin(kx - \omega t)$ and $y_2 = A\sin(kx - \omega t + \varphi)$ .

Step 2. Add them by superposition, $y = y_1 + y_2$ , and apply the identity $\sin P + \sin Q = 2\sin\!\frac{P+Q}{2}\cos\!\frac{P-Q}{2}$ with $P = kx - \omega t$ and $Q = kx - \omega t + \varphi$ . The half-sum is $kx - \omega t + \varphi/2$ ; the half-difference is $-\varphi/2$ .

Step 3. Since cosine is even, $\cos(-\varphi/2) = \cos(\varphi/2)$ , and the sum becomes $y = 2A\cos\!\left(\tfrac{\varphi}{2}\right)\sin\!\left(kx - \omega t + \tfrac{\varphi}{2}\right)$ . This is one wave of the original frequency with amplitude $A_{\text{res}} = 2A\cos(\varphi/2)$ — equation (3).

Step 4. Read off the extremes. In phase, $\varphi = 0$ : $A_{\text{res}} = 2A$ , and since $I \propto A^2$ , $I = 4I_0$ . Antiphase, $\varphi = \pi$ : $\cos(\pi/2) = 0$ , so $A_{\text{res}} = 0$ and $I = 0$ . The bright fringe is four times one wave, the dark fringe is nothing. Verified: for $A = 2.0$ mm and $\varphi = 60°$ , $A_{\text{res}} = 2(2.0)\cos 30° = 3.464$ mm; and $\cos^2(0) = 1$ gives the $4I_0$ maximum — recomputed in verify_hs2-superposition-of-waves.py.

Go deeper · Derivation B · 1 minFrom phase difference to path difference — and the energy audit that saves conservation

Derivation BFrom phase difference to path difference — and the energy audit that saves conservation

The goal here is to turn the abstract phase difference $\varphi$ into the measurable path difference $\Delta x$ that problems actually supply, deriving the constructive and destructive conditions of equation (5), and then to answer the objection that interference seems to destroy energy at the dark fringes. The tool is the direct proportionality between extra path and extra phase — one wavelength of path is one full cycle of phase — followed by an average of the intensity over the whole pattern. The payoff is a clean rule for maxima and minima and a reassurance that energy is only moved, never lost.

Step 1. Relate path to phase. A wave advances its phase by $2\pi$ for every wavelength $\lambda$ it travels, so an extra path $\Delta x$ contributes an extra phase $\varphi = (2\pi/\lambda)\,\Delta x$ — equation (4).

Step 2. Impose constructive interference for sources that start in phase: the waves must arrive a whole number of cycles apart, $\varphi = 2\pi n$ . Substituting gives $(2\pi/\lambda)\Delta x = 2\pi n$ , i.e. $\Delta x = n\lambda$ .

Step 3. Impose destructive interference: the waves must arrive exactly out of step, $\varphi = (2n+1)\pi$ . Then $(2\pi/\lambda)\Delta x = (2n+1)\pi$ gives $\Delta x = (n + \tfrac12)\lambda$ — equation (5).

Step 4. Audit the energy. For two equal waves the pattern's intensity is $I = 4I_0\cos^2(\varphi/2)$ , which swings between $4I_0$ and $0$ . Averaged over all phase differences, $\cos^2$ averages to $\tfrac12$ , so the mean intensity is $2I_0$ — exactly the sum $I_0 + I_0$ of the two waves acting separately. Interference has not created the $4I_0$ of a bright fringe from nothing; it has borrowed it from the dark fringes, where $I = 0$ . Energy is conserved, merely rearranged in space. Verified: two speakers with $\lambda = 1.00$ m give destructive interference at $\Delta x = 0.50$ m and constructive at $\Delta x = 1.00$ m; the average of $4I_0\cos^2(\varphi/2)$ is $2I_0$ — recomputed in verify_hs2-superposition-of-waves.py.

Limiting cases

case	resultant	meaning
$\varphi = 0$ (in phase)	$A_{\text{res}} = A_1 + A_2$ , $I = (A_1+A_2)^2/A^2 \cdot I_0$	fully constructive — amplitudes add
$\varphi = \pi$ (antiphase)	$A_{\text{res}} = \lvert A_1 - A_2\rvert$	fully destructive — amplitudes subtract
equal amplitudes, $\varphi = 0$	$A_{\text{res}} = 2A$ , $I = 4I_0$	four times one wave, not two
equal amplitudes, $\varphi = \pi$	$A_{\text{res}} = 0$ , $I = 0$	perfect cancellation, black fringe
$\Delta x = n\lambda$ (sources in phase)	constructive	bright / loud line
$\Delta x = (n+\tfrac12)\lambda$	destructive	dark / quiet line
incoherent sources	no steady pattern; $I = 2I_0$	intensities simply add — no fringes

Common trapInterference destroys energy at the dark fringes

It does not, and believing it is the deepest misunderstanding of this topic. At a point of destructive interference two waves cancel and the displacement is zero — but you must look at the whole pattern, not one point. The energy that is missing from every dark fringe is exactly the energy piled up in the bright fringes, where the intensity is $4I_0$ , twice the $2I_0$ the two waves would deliver if they never interfered. Interference is a redistribution: it takes energy from where the waves disagree and hands it to where they agree, and when you average over the pattern the mean intensity is $2I_0$ , precisely conserved. The partner error is quantitative: writing the bright maximum as $2I_0$ instead of $4I_0$ . Intensity is not amplitude — it is amplitude squared. Two equal waves in step make an amplitude of $2A$ , and $(2A)^2 = 4A^2$ , so four times the intensity of a single wave. Adding the two intensities directly ( $I_0 + I_0 = 2I_0$ ) is what incoherent sources do, precisely because they never hold a steady phase; coherent sources reach $4I_0$ at the bright fringes and $0$ at the dark, and the two only average back to $2I_0$ .

Worked examples

1 · Board level — the resultant of two coherent waves (CBSE/ICSE pattern, 3 marks)

Two coherent waves of amplitudes $A_1 = 3.0$ and $A_2 = 4.0$ (arbitrary units) meet at a point. Find the resultant amplitude when the phase difference is (a) $0$ , (b) $60°$ , (c) $90°$ , (d) $180°$ .

Step 1. Use equation (2), $A_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\varphi}$ , with $A_1^2 + A_2^2 = 9 + 16 = 25$ and $2A_1A_2 = 24$ .

Step 2. In phase, $\varphi = 0$ , $\cos 0 = 1$ : $A_{\text{res}} = \sqrt{25 + 24} = \sqrt{49} = \boxed{7.0}$ — the amplitudes simply add.

Step 3. At $\varphi = 60°$ , $\cos 60° = 0.5$ : $A_{\text{res}} = \sqrt{25 + 12} = \sqrt{37} = \boxed{6.083}$ ; and at $\varphi = 90°$ , $\cos 90° = 0$ : $A_{\text{res}} = \sqrt{25} = \boxed{5.000}$ — the familiar $3$ – $4$ – $5$ , because at $90°$ the phasors are perpendicular.

Step 4. Antiphase, $\varphi = 180°$ , $\cos 180° = -1$ : $A_{\text{res}} = \sqrt{25 - 24} = \sqrt{1} = \boxed{1.0}$ — the amplitudes subtract. Notice the resultant slides smoothly from $7$ down to $1$ as the phase difference opens from $0$ to $180°$ .

2 · Board level — equal amplitudes and the fourfold maximum (CBSE/NEET pattern, 3 marks)

Two coherent waves, each of amplitude $2.0\ \mathrm{mm}$ , meet with a phase difference of $60°$ . (a) Find the resultant amplitude. (b) If the intensity of one wave alone is $I_0$ , what is the resultant intensity in the bright fringe where they are in phase?

Step 1. Equal amplitudes, so use equation (3): $A_{\text{res}} = 2A\cos(\varphi/2) = 2(2.0)\cos 30°$ .

Step 2. With $\cos 30° = 0.8660$ : $A_{\text{res}} = 4.0 \times 0.8660 = \boxed{3.464\ \mathrm{mm}}$ .

Step 3. Part (b): in the bright fringe $\varphi = 0$ , so $I = 4I_0\cos^2 0 = \boxed{4I_0}$ — four times one wave, not two.

Step 4. Sanity on energy: the dark fringes have $I = 0$ , and averaging $4I_0\cos^2(\varphi/2)$ over the pattern gives $2I_0$ , the plain sum of the two waves. The bright fringe's extra brightness is borrowed, not created.

3 · Competitive level — two loudspeakers and the contrast ratio (JEE-Main/NEET pattern)

Two loudspeakers driven by one signal generator (so they are coherent and in phase) emit sound of frequency $340\ \mathrm{Hz}$ ; the speed of sound is $340\ \mathrm{m/s}$ . (a) A listener stands where the path difference from the two speakers is $0.50\ \mathrm{m}$ : loud or quiet? What about at $1.00\ \mathrm{m}$ ? (b) If instead the two amplitudes were in the ratio $3:1$ , find the ratio of the loudest to the quietest intensity.

Step 1. Wavelength first: $\lambda = v/f = 340/340 = 1.00\ \mathrm{m}$ .

Step 2. Part (a). At $\Delta x = 0.50\ \mathrm{m} = \lambda/2 = (0 + \tfrac12)\lambda$ , equation (5) says destructive — a quiet spot. At $\Delta x = 1.00\ \mathrm{m} = 1\lambda$ , it says constructive — a loud spot. Walking from one to the other, the listener passes through silence into full loudness.

Step 3. Part (b). With amplitudes in ratio $3:1$ , the loudest fringe has amplitude $3 + 1 = 4$ and the quietest $3 - 1 = 2$ , so by equation (6), $I_{\max}/I_{\min} = (4/2)^2 = \boxed{4.0}$ .

Step 4. Read the physics: because the amplitudes are unequal, the dark fringes are no longer perfectly silent — the stronger wave is only partly cancelled — so the contrast is a modest $4$ , not infinite. Perfect blackness needs equal amplitudes. Timing note: about two minutes. The whole of part (a) hinges on computing $\lambda = 1.00\ \mathrm{m}$ in the first five seconds and then comparing $\Delta x$ to it — a half-wavelength is dark, a whole one is bright. Part (b) is a one-line ratio; do not compute any actual intensity. The examiner's bait is the phrase "in phase," which you must use to pick $\Delta x = n\lambda$ for constructive rather than the antiphase rule.

Exam traps

Adding amplitudes when you should add intensities, or the reverse. Displacements (and amplitudes, as phasors) add for coherent waves; raw intensities add only for incoherent ones. Two coherent equal waves in step give amplitude $2A$ and intensity $4I_0$ ; two incoherent ones give $2I_0$ . Decide coherence first, then choose.

Writing $I_{\max} = 2I_0$ . The single most common slip. Intensity is amplitude squared, so two equal waves in step give $(2A)^2 = 4A^2$ , that is $4I_0$ . The $2I_0$ answer is the incoherent average, not the bright fringe.

Forgetting "sources in phase" before using $\Delta x = n\lambda$ . The conditions in equation (5) assume the two sources start in step. If one is driven a half-cycle behind, constructive and destructive swap: $\Delta x = n\lambda$ becomes dark. Always read how the sources are driven.

Assuming any two same-frequency sources interfere. They do not — a stable pattern needs coherence, a constant phase relationship. Two independent tuning forks of the same pitch produce no fringes; their phase difference drifts and the pattern averages away. Coherence, not merely equal frequency, is the requirement.

Thinking energy vanishes at a dark fringe. It is relocated to the bright fringes. Board schemes reward the sentence "energy is redistributed, not destroyed"; write it.

Board scope. IGCSE candidates: you need the idea — two coherent sources in a ripple tank give reinforcement where crests meet crests, cancellation where crests meet troughs, and the path-difference rule $\Delta x = n\lambda$ — but the amplitude formula $A_{\text{res}} = \sqrt{A_1^2+A_2^2+2A_1A_2\cos\varphi}$ and the $4I_0$ intensity algebra are not on your syllabus. A-Level, IB, AP and the Indian boards need the full toolkit.

Origins · 1 minThomas Young and how this was discovered

Thomas Young

1773–1829 · England

Thomas Young was a physician, linguist and physicist who could read at two and later helped decipher the Rosetta Stone — and who, around 1801, gave physics the word and the idea of interference. Arguing against Newton's authority that light was a stream of particles, Young pointed first not to light but to water: where two sets of ripples cross, he observed, there are lines of doubled swell and lines of dead calm, and only a wave picture could explain them. In his Bakerian Lecture to the Royal Society he proposed that light does the same, and by 1803 he was demonstrating the bright and dark bands of two-source light interference — the experiment that, a century later, would sit at the heart of quantum mechanics. His insistence that overlapping waves simply add, cancelling here and reinforcing there, is exactly the principle this article is built on.

History

1665

— Francesco Grimaldi's posthumously published work describes light spreading into coloured fringes at a narrow opening and coins the word "diffraction" — an early hint that light behaves like a wave.

1801

— Thomas Young states the principle of interference in his Bakerian Lecture and argues it first from ripples on water, not from light.

1803

— Young demonstrates the interference of two coherent light sources, the bright and dark bands appearing exactly where his wave theory predicts.

1807

— Young's Lectures on Natural Philosophy publishes the full account and fixes the word "interference" in the language of physics.

1878

— Lord Rayleigh's The Theory of Sound gives the interference of sound waves its quantitative treatment, the acoustic twin of Young's optics.

Going further

Texts. NCERT Physics Class 11, Ch. 15 "Waves", §15.4 (the principle of superposition) · H. C. Verma, Concepts of Physics Vol. 1, "Superposition and Interference" · Halliday, Resnick and Walker, "Waves — II" · Cambridge A-Level 9702 topic 8 (superposition); IGCSE 0625 §3.3 covers the ripple-tank picture only.

Provenance. Every number is recomputed in `verify_hs2-superposition-of-waves.py`: the $3$ – $4$ resultants $7.0$ , $6.083$ , $5.000$ , $1.0$ at $\varphi = 0, 60°, 90°, 180°$ ; the equal-amplitude $A_{\text{res}} = 3.464$ mm at $60°$ ; the intensities $4I_0$ (in phase) and $0$ (antiphase), with the $2I_0$ average; the speakers' $\lambda = 1.00$ m with destructive at $0.50$ m and constructive at $1.00$ m; and the $3:1$ contrast ratio $I_{\max}/I_{\min} = 4.0$ .

Prerequisites. Describing a Wave: y(x,t), Amplitude, Wavelength & Phase — where phase and phase difference are defined and $\varphi = 2\pi\Delta x/\lambda$ first appears. Leads to. Reflection of Waves & Phase Change at Boundaries — where a wave interferes with its own reflection · Standing Waves & Stationary-Wave Formation — the interference of two waves travelling opposite ways · and, in optics, Conditions for Observable Interference, the university account of coherence. Go deeper. Conditions for Observable Interference sets out exactly when a stable pattern forms — enrichment, not a dependency.

What is the difference between superposition and interference?

Superposition is the underlying rule; interference is what you see when you apply it to sustained waves. Superposition says that at any point the displacements of all the waves present simply add, with sign — it holds for any waves in a linear medium, a passing pair of pulses included. Interference is the specific, stable pattern of reinforcement and cancellation that appears when two coherent waves of the same frequency superpose: bright and dark fringes, loud and quiet regions. Every interference result on this page is just superposition, equation (1), worked out for two steady sinusoids.

Why is the maximum intensity 4I0 and not 2I0? It feels like cheating.

Because intensity depends on the square of amplitude, not on amplitude itself. Two equal waves in step reinforce to an amplitude of $2A$ , and squaring gives $(2A)^2 = 4A^2$ , four times the $A^2$ of a single wave. It is not cheating and it does not create energy: the bright fringes are $4I_0$ , but the dark fringes are $0$ , and averaged across the whole pattern the intensity is $2I_0$ — exactly what the two waves carry between them. The interference simply moves energy from the dark places to the bright ones. Adding the intensities to get $2I_0$ is right only for incoherent sources, which never build a fringe.

What exactly does coherent mean, and why do I need it?

Coherent means the two sources keep a constant phase difference over time. You need it because the interference pattern is fixed in space only if $\varphi$ at each point stays put; if $\varphi$ wanders, the bright and dark fringes at any point flicker between reinforcement and cancellation faster than you can see, and everything averages to a dull uniform glow. Two speakers wired to one generator are coherent and give a stable pattern; two separate whistles are incoherent, their phases drift independently, and no pattern forms. Equal frequency alone is not enough — the phase relationship must hold still.

The formula assumes the sources are in phase. What if they are not?

Then the roles of the path-difference conditions swap. Equation (5) — constructive at $\Delta x = n\lambda$ , destructive at $(n+\tfrac12)\lambda$ — is written for sources that start in step. If one source is driven exactly a half-cycle behind the other, that built-in half-wavelength head start adds to every path difference, so now $\Delta x = n\lambda$ gives destructive interference and $(n+\tfrac12)\lambda$ gives constructive. The safe habit is to compute the total phase difference at the point, including any built-in phase between the sources, and only then decide bright or dark.

Do the two waves have to have the same amplitude to interfere?

No. Waves of any two amplitudes interfere; equation (2) handles the general case, and the bright and dark fringes still fall at the same path differences. What unequal amplitudes change is the contrast: the dark fringes are no longer perfectly black, because the stronger wave is only partly cancelled by the weaker one. The contrast ratio is $I_{\max}/I_{\min} = ((A_1+A_2)/(A_1-A_2))^2$ , which is finite unless the amplitudes are equal. Perfectly dark fringes — total cancellation — require $A_1 = A_2$ .

How can two pulses pass through each other and come out unchanged?

Because superposition only adds the displacements while the pulses overlap; it does not alter the pulses themselves. Each pulse carries its own shape and velocity, and the medium, being linear, lets both instructions run at once. At the instant of overlap the rope shows the sum — perhaps a tall combined bump, perhaps, if one is up and one down, a momentarily flat rope — but each pulse's "information" is still travelling in its own direction at its own speed. A moment later they separate, each exactly as it was. The flat instant is the eeriest: the rope is momentarily still, yet the energy is all in its motion, and the pulses reappear.