Coherent and incoherent addition of waves

Shine two ordinary torches on the same patch of wall and the wall simply gets brighter — a smooth wash of light with no pattern at all. Yet let a single narrow beam of light pass through two closely spaced slits, and on a screen behind them a neat ladder of bright and dark bands appears, spaced with almost machine-like regularity. Same physics — two light waves arriving at a point and adding together — but wildly different outcomes. What separates the two situations is not the colour of the light, not its intensity, and not even its frequency. It is a single, subtle property called coherence: whether the phase relationship between the two waves stays fixed in time or drifts randomly, many times a second, in a way no eye or detector can follow. This article is about that one property and its consequence. We will build the general formula for the intensity produced when two waves of the same frequency are superposed at a point, carrying a phase difference $\phi$ between them. When $\phi$ is locked — the sources are coherent — that formula predicts a real, standing pattern of bright and dark fringes, the signature of interference. When $\phi$ jitters randomly — the sources are incoherent — the very same formula, time-averaged, collapses to a boring sum of intensities, and the pattern washes out to uniform brightness. Everything that follows in the Wave Optics chapter, especially Young's double-slit experiment, depends on understanding exactly why splitting one wavefront into two coherent parts is the only reliable trick for keeping $\phi$ fixed, and why two independent lamps — however identical they look — can never manage it.

Key takeaways

Superposition first. When two waves of the same frequency overlap at a point, the resultant displacement is the algebraic sum of the two, and this leads to a resultant intensity $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\phi$ , where $\phi$ is the phase difference between them.
Coherent sources keep $\phi$ fixed. A constant (non-random) phase difference makes $\cos\phi$ hold a steady value, so the interference term survives — a genuine, sustained pattern of bright and dark fringes appears.
Incoherent sources let $\phi$ drift randomly. The phase difference changes many times within the eye's response time, $\cos\phi$ averages to zero, and only the plain sum $I = I_1 + I_2$ is observed — no fringes, just a uniform glow.
Equal-intensity coherent sources give a clean cosine-squared law. With $I_1 = I_2 = I_0$ , the resultant becomes $I = 4I_0\cos^2(\phi/2)$ , running smoothly from $I_{\max} = 4I_0$ at $\phi = 0, 2\pi,\dots$ down to $I_{\min} = 0$ at $\phi = \pi, 3\pi,\dots$ .
Coherence is engineered, not found. Two independent sources — even two lasers of the same nominal frequency — cannot be relied on to hold a fixed $\phi$ ; the standard trick, used in Young's experiment, is to split a single wavefront into two so that any drift in the original source affects both parts identically and cancels out of the difference $\phi$ .
The instantaneous cross-term never vanishes; only its time average does. At any single instant, $2\sqrt{I_1I_2}\cos\phi$ is a real, nonzero number for incoherent sources too — it is only the average recorded over an observation time (far longer than the phase's random-flip timescale) that goes to zero.
Key idea: two waves add to a resultant intensity $I = I_1+I_2+2\sqrt{I_1I_2}\cos\phi$ , and whether the cross-term survives as a real, sustained bright-dark fringe pattern (coherent, fixed $\phi$ ) or averages away to a plain sum $I_1+I_2$ (incoherent, randomly drifting $\phi$ ) decides whether interference is visible at all.

The core insight

Two waves of the same frequency superpose to give $I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi$ , and this interference term survives as a sustained, visible fringe pattern only when the phase difference $\phi$ between the sources is held constant — the defining property of coherent sources — while for incoherent sources the randomly drifting $\phi$ averages the cross-term to zero, leaving the plain sum $I = I_1 + I_2$ .

Where it fits in the syllabus

Coherent and incoherent addition of waves sits between wavefronts and Huygens' principle (how a wave propagates and how reflection and refraction follow from it) and interference proper — Young's double-slit experiment and its fringe-width formula $\beta = \lambda D/d$ — in the CBSE Class 12 Wave Optics chapter. This section supplies the vocabulary and the one formula, resultant intensity for two superposed waves, that Young's experiment then applies to a specific geometry. The CBSE 2025-26 syllabus asks only for the final fringe-width expression in the next node, with no geometric derivation required there; here, by contrast, building the superposition/intensity result from two sinusoidal waves is fair game and is the natural algebra-and-trigonometry derivation this topic supports. Polarisation of light (Malus's law, Brewster's angle) is explicitly out of the 2025-26 syllabus and is not referenced here.

The physical picture

Picture two waves of the same frequency $\omega$ arriving at the same point in space, one of amplitude $a_1$ and one of amplitude $a_2$ , but shifted from each other by a constant phase $\phi$ . The principle of superposition says the actual disturbance at that point at any instant is simply the sum of what each wave would produce alone. Because both waves oscillate at the same frequency, their sum is itself a sinusoid at that frequency, but with a new, combined amplitude that depends on $a_1$ , $a_2$ , and, crucially, on $\phi$ . Since intensity of a wave is proportional to the square of its amplitude, this combination rule translates directly into a rule for how the intensities combine, and that rule is the single formula this whole topic is built around:

I = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\phi

where: $I_1 \propto a_1^2$ and $I_2 \propto a_2^2$ are the intensities the two waves would separately produce at that point, $\phi$ is the (instantaneous) phase difference between them at that point, and $I$ is the actual resultant intensity observed there. Read physically, the first two terms, $I_1+I_2$ , are what you would get if the waves simply added their energies with no interplay at all; the third term, $2\sqrt{I_1I_2}\cos\phi$ , is the genuinely wave-like correction — it can be positive (adding extra brightness, constructive interference) or negative (subtracting brightness, even down to darkness, destructive interference), depending only on $\phi$ . This cross-term is the entire content of "interference": without it, optics would just be additive photometry, with no fringes anywhere. Two extreme, named cases of $\phi$ 's behaviour in time decide whether this cross-term is something you can ever see on a screen.

Coherent sources hold $\phi$ fixed — the same value, instant after instant, for as long as you care to watch. Then $\cos\phi$ is a fixed number (somewhere between $-1$ and $+1$ ) at every point in space, so the intensity $I$ at that point is a fixed number too, different from the intensity at a neighbouring point where $\phi$ happens to take a different fixed value. Walk across the screen and $\phi$ changes smoothly from point to point (because the two path lengths from the sources change), so $I$ traces out a smooth, sustained pattern of bright fringes (where $\cos\phi\to+1$ ) and dark fringes (where $\cos\phi\to-1$ ) that does not wash out or flicker with time. This is precisely the fringe pattern Young's experiment relies on.

Incoherent sources let $\phi$ wander randomly and extremely rapidly — far faster than the eye (or any ordinary detector) can respond. Over any realistic observation time, $\cos\phi$ takes every value between $-1$ and $+1$ with roughly equal likelihood, and its time average is zero:

\langle\cos\phi\rangle_{\text{time}} \approx 0 \quad\Longrightarrow\quad I = I_1 + I_2.

There is no dependence left on position, so no fringe pattern is visible anywhere — just a flat, uniform intensity everywhere the two beams overlap. This is exactly what happens with two separate light bulbs, or with two different points of one ordinary extended source: each atom in the source emits light in short, random bursts with an unpredictable phase, so the phase difference between light reaching a screen from two independent points changes millions of times a second. A particularly clean and exam-friendly special case is when the two coherent sources have equal intensity, $I_1 = I_2 = I_0$ . Substituting into the general formula and using the double-angle identity $1+\cos\phi = 2\cos^2(\phi/2)$ :

I = I_0 + I_0 + 2\sqrt{I_0^2}\cos\phi = 2I_0(1+\cos\phi) = 4I_0\cos^2\!\left(\frac{\phi}{2}\right).

where: the maximum value $I_{\max} = 4I_0$ occurs whenever $\phi = 0, 2\pi, 4\pi,\dots$ (the two waves crest together, constructive interference), and the minimum $I_{\min} = 0$ occurs whenever $\phi = \pi, 3\pi,\dots$ (crest meets trough exactly, destructive interference, total cancellation because the amplitudes here are equal). Notice that $4I_0$ is four times a single source's intensity, not merely twice — that "extra" doubling above the naive $2I_0$ is the visible fingerprint of coherent addition, and it is the quantity a bright fringe in Young's experiment actually reaches.

Figure

A graph of resultant intensity against phase difference showing a coherent cosine squared curve oscillating between zero and four times the single source intensity, alongside a flat dashed line at twice the single source intensity for the incoherent case

Derivation ASuperposition of two coherent waves

The goal of this derivation is to obtain the resultant-intensity formula $I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$ directly from adding two sinusoidal displacements of the same frequency but different amplitude and phase, using nothing beyond the trigonometric expansion of a sine of a sum and the basic fact that intensity is proportional to the square of amplitude — exactly the algebra a Class 12 student already has from the mechanical-waves chapter, applied here to light.

Step 1. Write the two waves arriving at the chosen point as $y_1 = a_1\sin(\omega t)$ and $y_2 = a_2\sin(\omega t + \phi)$ , where $a_1,a_2$ are their amplitudes, $\omega$ is the common angular frequency, and $\phi$ is the fixed phase by which the second wave leads the first. Both waves are evaluated at the same point in space, so only the time dependence and the phase offset need to be tracked.

Step 2. Add the two displacements using the principle of superposition, $y = y_1+y_2$ , and expand the second term with the compound-angle identity $\sin(\omega t+\phi)=\sin\omega t\cos\phi+\cos\omega t\sin\phi$ . This gives $y = a_1\sin\omega t + a_2\cos\phi\sin\omega t + a_2\sin\phi\cos\omega t = (a_1+a_2\cos\phi)\sin\omega t + (a_2\sin\phi)\cos\omega t$ , which is now a sum of a sine and a cosine term at the same frequency $\omega$ , with coefficients that depend only on the amplitudes and $\phi$ , not on time.

Step 3. Recognise that any expression of the form $P\sin\omega t + Q\cos\omega t$ is itself a single sinusoid of amplitude $A=\sqrt{P^2+Q^2}$ at the same frequency (write $P=A\cos\theta$ , $Q=A\sin\theta$ and use the compound-angle identity in reverse). Here $P=a_1+a_2\cos\phi$ and $Q=a_2\sin\phi$ , so the resultant wave is $y=A\sin(\omega t+\theta)$ with $A^2 = (a_1+a_2\cos\phi)^2+(a_2\sin\phi)^2$ .

Step 4. Expand $A^2$ and simplify using $\sin^2\phi+\cos^2\phi=1$ : $A^2 = a_1^2 + 2a_1a_2\cos\phi + a_2^2\cos^2\phi + a_2^2\sin^2\phi = a_1^2+a_2^2+2a_1a_2\cos\phi$ . This is the resultant amplitude-squared, and it already carries the whole physics: it is not simply $a_1^2+a_2^2$ (which is what independent, non-interfering sources would give) but has the extra $2a_1a_2\cos\phi$ term that depends on the phase relationship.

Step 5. Convert to intensities using $I\propto a^2$ (write $I_1=ka_1^2$ , $I_2=ka_2^2$ for the same proportionality constant $k$ , since both waves are of the same kind of disturbance at the same point). Then $I = kA^2 = ka_1^2+ka_2^2+2k a_1 a_2\cos\phi = I_1+I_2+2\sqrt{I_1I_2}\cos\phi$ , using $k a_1 a_2 = \sqrt{(ka_1^2)(ka_2^2)}=\sqrt{I_1I_2}$ . This is exactly the formula stated above, now derived rather than merely asserted, and it holds at every instant for any two waves of the same frequency, whatever the origin of $\phi$ .

Worked examples

Board level - resultant intensity at a fixed phase difference

Two coherent sources of light, each of intensity $I_0 = 4$ units, superpose at a point where the phase difference is $\phi = 60^\circ$ . Find the resultant intensity at that point, and check it against the equal-intensity cosine-squared form.

Step 1. Apply the general formula with $I_1=I_2=4$ and $\phi=60^\circ$ , so $\cos\phi=0.5$ : $I = I_1+I_2+2\sqrt{I_1I_2}\cos\phi = 4+4+2\sqrt{4\times4}\,(0.5)$ .

Step 2. Evaluate: $\sqrt{4\times4}=4$ , so $I = 8 + 2(4)(0.5) = 8+4=12$ units.

Step 3. Cross-check with the equal-intensity form $I=4I_0\cos^2(\phi/2)$ , using $\phi/2=30^\circ$ , $\cos30^\circ=0.866$ : $I=4(4)(0.866)^2 = 16\times0.75 = 12$ units. Answer: the resultant intensity is $I=12$ units, comfortably between the extremes $I_{\min}=0$ and $I_{\max}=16$ units, exactly as expected since $\phi=60^\circ$ lies between the perfectly constructive ( $\phi=0$ ) and perfectly destructive ( $\phi=180^\circ$ ) cases.

Competitive level - comparing coherent maxima with the incoherent baseline

Two incoherent sources, each of intensity $3$ units, illuminate a screen. (a) Find the resultant intensity at any point on the screen. (b) If the same two sources were instead made coherent, find the maximum (brightest-fringe) intensity they could produce. (c) Express the coherent maximum as a multiple of the incoherent (uniform) intensity, and comment on what this ratio signifies physically.

Step 1. For incoherent sources, the phase difference $\phi$ drifts randomly and $\langle\cos\phi\rangle\to 0$ , so only the plain sum survives: $I_{\text{incoherent}} = I_1+I_2 = 3+3 = 6$ units, and — the key point — this value is the same at every point on the screen, since it no longer depends on $\phi$ or therefore on position.

Step 2. For coherent sources of the same individual intensities, the resultant is maximised when $\cos\phi=+1$ (waves arriving in step): $I_{\max} = I_1+I_2+2\sqrt{I_1I_2} = 3+3+2\sqrt{9} = 6+6=12$ units.

Step 3. Form the ratio: $I_{\max}/I_{\text{incoherent}} = 12/6 = 2$ . Answer: the incoherent intensity is $6$ units everywhere; the coherent maximum is $12$ units, exactly double. Physically, this factor of two (which is general for equal-intensity sources: $I_{\max}=4I_0$ against an incoherent baseline of $2I_0$ ) is the entire visible signature of interference — bright fringes rise to twice the "flat" level while dark fringes fall to zero, and it is this contrast, not any change in total light energy delivered, that your eye reads as a fringe pattern. Timing note: under ninety seconds. The whole problem reduces to two applications of one formula (once with $\cos\phi\to0$ on average, once with $\cos\phi=1$ ) plus a ratio — no geometry, no small-angle approximations are needed here at all, since $\phi$ itself is given or set to its extreme value directly.

Common traps

Believing that two waves of the same frequency automatically interfere to give a fringe pattern — the frequency match is necessary but not sufficient; without a constant phase difference, the pattern averages away to nothing, however identical the two frequencies are.
Writing the incoherent result as though the cross-term $2\sqrt{I_1I_2}\cos\phi$ simply "does not exist" for incoherent sources. It exists at every instant, exactly as for coherent sources — it is only the time average over many random phase flips that vanishes, which is a subtly different (and more accurate) statement.
Assuming the coherent maximum intensity is just $I_1+I_2$ — forgetting the interference term entirely. The true maximum is $I_1+I_2+2\sqrt{I_1I_2}$ , strictly greater than $I_1+I_2$ unless one of the sources has zero intensity.
Treating "same laser, same frequency" as automatically coherent. Two independent emitters, however identical in specification, do not hold a fixed relative phase in practice; true coherence in the syllabus sense is obtained by deriving both waves from a single original wavefront (split, as in Young's experiment), not from two separate sources however alike.
Forgetting that the equal-intensity peak is $4I_0$ , not $2I_0$ . Students who correctly recall $I=I_1+I_2$ for the incoherent case sometimes carry the habit of simple addition into the coherent maximum and undercount it by a factor of two.

Common trapCoherence means a constant phase difference, not merely equal frequency or equal amplitude

The single biggest scoring error on this topic is conflating "same frequency" (or "same colour," or "same amplitude") with "coherent." Frequency matching is necessary for a stable interference pattern to even make sense (otherwise $\phi$ itself would sweep continuously in time even for two perfectly steady sources), but it is not sufficient. Coherence specifically requires that the phase difference $\phi$ between the two waves stay constant, not that the waves individually be "steady." Two separate sodium lamps emit the same characteristic frequency, yet they are never coherent, because each lamp's atoms emit independently, in short random bursts, so the phase relationship between light from the two lamps changes utterly unpredictably many times each second — far too fast for any detector to track, so only the time-averaged, fringe-free sum $I_1+I_2$ is ever seen. The only reliable route to a fixed $\phi$ used in this syllabus is to derive both interfering waves from a single wavefront (splitting it into two, as Young's slits do), so that any phase wobble in the original source is common to both derived waves and cancels out of their difference.

You should be able to…

State precisely what makes two sources coherent (a constant, not merely equal-frequency, phase difference) versus incoherent (a randomly fluctuating phase difference).
Write and correctly apply the general resultant-intensity formula $I = I_1+I_2+2\sqrt{I_1I_2}\cos\phi$ for any given $I_1$ , $I_2$ , and $\phi$ .
Derive the resultant amplitude and intensity for two superposed sinusoidal waves of the same frequency using the compound-angle expansion, without needing calculus.
Reduce the general formula to the equal-intensity case $I=4I_0\cos^2(\phi/2)$ and identify its maxima ( $4I_0$ ) and minima ( $0$ ).
Explain physically why the incoherent-addition result $I=I_1+I_2$ holds as a time average, even though the interference term is nonzero at any single instant.
Explain why splitting a single wavefront (as in Young's double-slit experiment), rather than using two independent sources, is the standard method for achieving coherence.

Ask the tutor: Want to see how this same resultant-intensity formula turns into the actual fringe-width expression $\beta=\lambda D/d$ once the path-difference geometry of Young's double slits is added in?

Why can't two identical torches or two identical lasers ever show interference fringes, even if their frequencies match exactly?

Frequency matching alone is not enough — what matters is whether the phase difference between the two sources stays fixed. Inside an ordinary lamp or even a laser, light is emitted by enormous numbers of atoms independently, each firing off a short wave train with essentially random phase, so the overall phase of the beam keeps jumping around unpredictably many times every second. Two separate lamps, however identically manufactured, have no way of synchronising these random jumps with each other, so the phase difference between the two beams at any point drifts randomly too. Averaged over any real observation time, the interference cross-term washes out to zero, leaving only the plain sum of intensities — no stable bright-dark pattern, just extra brightness.

If the interference term 2 root I1 I2 cos phi is genuinely there at every instant for incoherent sources too, why do we never see any flicker or trace of a pattern at all?

The response time of your eye (and of most ordinary light detectors) is far, far longer than the timescale over which an incoherent source's phase difference randomly flips — typically many orders of magnitude longer. So what you perceive is not the instantaneous intensity but its average over that comparatively huge response window. Because $\phi$ takes essentially every value between $0$ and $2\pi$ many times within that window, positive and negative excursions of $\cos\phi$ cancel almost perfectly, and the observed, time-averaged intensity settles to the plain sum $I_1+I_2$ with no detectable residual pattern. A detector fast enough to outrun the source's phase jitter would, in principle, catch fleeting flickers of interference — but nothing in an ordinary classroom or exam setting can.

In Young's double-slit experiment, the two slits are lit by the same lamp, so how is that different from using two separate lamps?

The crucial difference is where the randomness in phase originates and whether it is shared. In Young's setup, a single wavefront from one source first passes through (or reaches) both slits together; each slit then re-radiates a spherical wavelet, but both wavelets are derived from the very same patch of the original wavefront at the very same instant. If the original source's phase jumps randomly, that jump reaches both slits simultaneously and shifts both derived waves by the same amount — so their difference in phase, which is all that determines $\phi$ at the screen, stays essentially unchanged. With two separate lamps, by contrast, each lamp's random phase jumps are completely independent of the other's, so their difference drifts just as randomly as each one alone. Splitting one wavefront, not merely using "the same kind of light," is what manufactures coherence.

Does coherent addition mean more total light energy is created compared to incoherent addition?

No — total energy is conserved either way; interference only redistributes it in space. Compare the equal-intensity cases: incoherent addition gives a flat $2I_0$ everywhere, while coherent addition gives $4I_0\cos^2(\phi/2)$ , which swings between $0$ and $4I_0$ but averages back to $2I_0$ over a full cycle of $\phi$ (since the average value of $\cos^2$ over a period is exactly one half). So interference does not add or remove energy overall — it simply concentrates the same total energy into bright fringes at the expense of the dark fringes in between, which is exactly why a fringe pattern looks striking even though nothing about the light sources themselves changed in total output.