Superposition of Waves
The entire field of interference rests on the superposition principle: at a particular point in a medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each wave individually. NCERT introduces this through a concrete picture — two needles $S_1$ and $S_2$ moving periodically up and down in an identical fashion in a trough of water, each launching its own circular wave.
Consider a point $P$ equidistant from the two sources, so that $S_1P = S_2P$. Waves from $S_1$ and $S_2$ take the same time to reach $P$, and waves that leave the sources in phase therefore arrive at $P$ in phase. If the displacement produced by $S_1$ at $P$ is $y_1 = a\cos\omega t$, then the displacement from $S_2$ is also $y_2 = a\cos\omega t$, so the resultant is
$$y = y_1 + y_2 = 2a\cos\omega t$$
Two coherent sources $S_1$ and $S_2$ emit identical wavefronts. The path lengths $S_1P$ and $S_2P$ decide the phase difference at $P$, and hence whether the waves reinforce or cancel.
Because intensity is proportional to the square of amplitude, doubling the amplitude quadruples the intensity. Writing $I_0$ for the intensity due to one source ($I_0 \propto a^2$), the resultant intensity at $P$ is $I = 4I_0$. Every point on the perpendicular bisector of $S_1S_2$ shares this maximum — the two sources are said to interfere constructively.
Phase Difference and Resultant Intensity
For an arbitrary point $G$ the two waves arrive with a phase difference $\phi$. Taking $y_1 = a\cos\omega t$ and $y_2 = a\cos(\omega t + \phi)$, the resultant displacement follows from a standard trigonometric identity:
$$y = a\big[\cos\omega t + \cos(\omega t + \phi)\big] = 2a\cos\!\left(\tfrac{\phi}{2}\right)\cos\!\left(\omega t + \tfrac{\phi}{2}\right)$$
The amplitude of the resultant is $2a\cos(\phi/2)$. For two sources of unequal intensities $I_1$ and $I_2$, the more general result for the resultant intensity is
$$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\phi$$
The first two terms are what the sources would deliver on their own; the third — the interference term $2\sqrt{I_1 I_2}\cos\phi$ — carries all the physics. It redistributes energy: where $\cos\phi$ is positive, intensity is enhanced; where it is negative, intensity is reduced. The total energy is conserved, only its spatial distribution changes.
| Quantity | Symbol / relation | Note |
|---|---|---|
| Single-source intensity | I0 ∝ a² | Intensity scales as amplitude squared |
| Resultant amplitude (equal sources) | 2a cos(φ/2) | Amplitudes add, then square |
| Resultant intensity (general) | I = I1 + I2 + 2√(I1I2) cosφ | Interference term carries φ |
| Path difference ↔ phase | φ = (2π/λ) × Δx | Δx = path difference |
Constructive and Destructive Interference
The outcome at any point is fixed by the path difference $S_1P \sim S_2P$ (the tilde denotes the magnitude of the difference). NCERT works two cases explicitly. At a point $Q$ with $S_2Q - S_1Q = 2\lambda$, the wave from $S_1$ arrives exactly two cycles earlier; a path difference of $2\lambda$ corresponds to a phase difference of $4\pi$, the waves are back in phase, and the intensity is again $4I_0$. At a point $R$ with path difference $2.5\lambda$ (phase difference $5\pi$), the two displacements are exactly out of phase and cancel, giving zero intensity.
| Condition | Phase difference φ | Path difference Δx | Resultant intensity (equal sources) |
|---|---|---|---|
| Constructive (maxima) | 2nπ | nλ | 4I0 |
| Destructive (minima) | (2n+1)π | (n + ½)λ | 0 |
Summarising NCERT Eqs. (10.9) and (10.10): for two coherent sources vibrating in phase, constructive interference (intensity $4I_0$) occurs whenever the path difference is $S_1P \sim S_2P = n\lambda$ with $n = 0, 1, 2, \ldots$, and destructive interference (zero intensity) occurs whenever $S_1P \sim S_2P = \left(n + \tfrac{1}{2}\right)\lambda$.
Top: two in-phase waves (teal and dashed purple) sum to a wave of doubled amplitude. Bottom: two waves a phase $\pi$ apart cancel to a flat resultant. Displacements add first; intensity is the square of the sum.
Intensity does not add — amplitude does
A common error is to add the two intensities directly for coherent waves and conclude that two $I_0$ sources give $2I_0$. For coherent sources you must first add the displacements (amplitudes), then square. Two in-phase $I_0$ sources give amplitude $2a$, hence intensity $4I_0$ — not $2I_0$. Direct intensity addition is reserved for incoherent sources.
Coherent: add amplitudes, then square → up to $4I_0$. Incoherent: add intensities → $2I_0$.
Intensity for Equal Amplitudes
When the two sources have equal amplitude $a$ (equal intensity $I_0$), squaring the resultant amplitude $2a\cos(\phi/2)$ gives NCERT Eq. (10.11):
$$I = 4I_0\cos^2\!\left(\tfrac{\phi}{2}\right)$$
This single expression encodes both extremes. When $\phi = 0, \pm 2\pi, \pm 4\pi, \ldots$, $\cos^2(\phi/2) = 1$ and the intensity is maximum at $4I_0$ (constructive). When $\phi = \pm\pi, \pm 3\pi, \pm 5\pi, \ldots$, $\cos^2(\phi/2) = 0$ and the intensity vanishes (destructive). The intensity varies smoothly between these limits, tracing the $\cos^2$ curve below.
$I = 4I_0\cos^2(\phi/2)$. Maxima of $4I_0$ (green) at $\phi = 2n\pi$; minima of zero (red) at $\phi = (2n+1)\pi$. The pattern is stable in $\phi$ only for coherent sources.
These conditions feed directly into the fringe analysis. See Young's Double-Slit Experiment to turn path difference into fringe width.
Coherent versus Incoherent Sources
The stability of the pattern hinges on whether the phase difference $\phi$ stays constant. If the two needles oscillate up and down regularly, $\phi$ at any point does not change with time and the positions of maxima and minima are fixed — a stable interference pattern. Such sources are coherent. If instead $\phi$ changes rapidly with time, the maxima and minima shift too fast to be seen and the eye registers only a time-averaged intensity. Such sources are incoherent.
For incoherent sources, averaging $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$ over the rapidly varying $\phi$ kills the interference term, because the time-average of $\cos\phi$ is zero. What survives is the plain sum:
$$\langle I \rangle = I_1 + I_2 \quad\Rightarrow\quad I = 2I_0 \ \text{(equal sources)}$$
This is exactly what happens when two separate light sources illuminate a wall: the intensities just add up and no fringes appear. NIOS §22.2 frames interference in the same way — a redistribution of energy due to the superposition of light waves from two coherent sources — underscoring that coherence is the prerequisite for any observable pattern.
| Feature | Coherent sources | Incoherent sources |
|---|---|---|
| Phase difference φ | Constant in time | Varies rapidly in time |
| Interference term | Retained: $2\sqrt{I_1I_2}\cos\phi$ | Averages to zero |
| Resultant (equal sources) | $4I_0\cos^2(\phi/2)$, $0$ to $4I_0$ | $2I_0$ everywhere |
| Pattern observed | Stable fringes | Uniform illumination |
Coherence needs a constant phase difference, not equal phase
Sources need not be exactly in phase to be coherent — they only need a phase difference that does not change with time. Two independent light bulbs, even of the same colour, are incoherent because their phase relationship fluctuates randomly, so they cannot produce sustained fringes. This is why Young split light from a single source rather than using two lamps.
Coherent = constant (steady) φ. Same frequency alone is not enough.
Intensity Ratios and Traps
For unequal sources the general formula gives the bright and dark extremes directly. The maximum occurs at $\cos\phi = +1$ and the minimum at $\cos\phi = -1$:
$$I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2, \qquad I_{\min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2$$
Their ratio is a frequently tested result:
$$\frac{I_{\max}}{I_{\min}} = \frac{\left(\sqrt{I_1} + \sqrt{I_2}\right)^2}{\left(\sqrt{I_1} - \sqrt{I_2}\right)^2}$$
Minimum is zero only when the intensities are equal
Complete cancellation ($I_{\min} = 0$) requires $I_1 = I_2$, i.e. equal amplitudes. If the two beams differ in intensity, the dark fringes are not perfectly dark; $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 > 0$. Watch out for problems that quote an amplitude ratio $a_1 : a_2$ — convert to intensities using $I \propto a^2$ before forming the ratio.
Amplitude ratio $a_1/a_2$ ⇒ $I_1/I_2 = (a_1/a_2)^2$ before using $I_{\max}/I_{\min}$.
Two coherent sources have intensities in the ratio $I_1 : I_2 = 4 : 1$. Find the ratio of maximum to minimum intensity in the pattern.
With $\sqrt{I_1} : \sqrt{I_2} = 2 : 1$, $I_{\max} = (2+1)^2 = 9$ and $I_{\min} = (2-1)^2 = 1$. Hence $I_{\max}/I_{\min} = 9 : 1$. The dark fringes are not fully dark because the intensities are unequal.
Coherent and Incoherent Addition in one screen
- Superposition: resultant displacement is the vector sum of individual displacements.
- General intensity: $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$; the third term is the interference term.
- Equal sources: $I = 4I_0\cos^2(\phi/2)$, ranging from $0$ to $4I_0$.
- Constructive: $\phi = 2n\pi$, path difference $n\lambda$, intensity $4I_0$.
- Destructive: $\phi = (2n+1)\pi$, path difference $(n+\tfrac{1}{2})\lambda$, intensity $0$.
- Coherent = constant φ → stable fringes; incoherent = rapidly varying φ → $I = I_1 + I_2 = 2I_0$.
- $I_{\max}/I_{\min} = (\sqrt{I_1}+\sqrt{I_2})^2 / (\sqrt{I_1}-\sqrt{I_2})^2$; zero minimum only if $I_1 = I_2$.