Coherent & Incoherent Addition of Waves

Superposition and Coherence

The superposition principle: at any point, the resultant displacement is the (vector) sum of the displacements of the individual waves. All of interference is this one sentence, exploited.

Picture two needles S$_1$, S$_2$ bobbing identically in water. At any point P, each wave arrives with some phase; if the phase difference at P stays constant in time, the sources are coherent. Identical bobbing guarantees it.

The bright case: if $S_1P = S_2P$ (or the path difference $S_2P - S_1P = n\lambda$), the waves arrive in phase: $y_1 = a\cos\omega t$, $y_2 = a\cos\omega t$, resultant $y = 2a\cos\omega t$. Intensity goes as amplitude squared:

$$\boxed{I = 4I_0 \quad\text{(constructive: path difference } n\lambda)}$$

The dark case: path difference $(n + \tfrac12)\lambda$ — waves arrive exactly out of phase, $y_2 = -y_1$, resultant zero:

$$\boxed{I = 0 \quad\text{(destructive)}}$$

The General Point — and the Incoherent Collapse

For an arbitrary phase difference $\phi$ between the two arriving waves ($y_1 = a\cos\omega t$, $y_2 = a\cos(\omega t + \phi)$), the resultant amplitude is $2a\cos(\phi/2)$, so

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

— smoothly interpolating between $4I_0$ ($\phi = 0, 2\pi,...$) and 0 ($\phi = \pi, 3\pi,...$). A path difference $\Delta$ corresponds to $\phi = \frac{2\pi}{\lambda}\Delta$.

Incoherent sources: if the two sources' relative phase drifts randomly (as for two separate lamps), $\phi$ at P changes erratically and rapidly; the eye/detector sees the time average $\langle\cos^2(\phi/2)\rangle = \tfrac12$:

$$\boxed{I = 2I_0 \quad\text{(incoherent: intensities simply add, no fringes)}}$$

Key Point: coherent addition averages amplitudes (then squares); incoherent addition averages intensities. The whole drama of fringes lives in that difference: $4I_0$ vs 0 vs the flat $2I_0$.

[JEE Tip] Unequal amplitudes $a_1, a_2$: $I_{max,min} = (\sqrt{I_1}\pm\sqrt{I_2})^2$, so $\frac{I_{max}}{I_{min}} = \left(\frac{a_1+a_2}{a_1-a_2}\right)^2$. Given an amplitude ratio 2:1, the ratio is 9:1 — a JEE perennial.

Solved Examples

Example 1: Conditions, stated cleanly

Write the path-difference conditions for constructive and destructive interference of two coherent sources.

Solution:

1. Constructive (bright): $S_2P - S_1P = n\lambda$, $n = 0, \pm1, \pm2,...$ — arrival in phase, $I = 4I_0$.
2. Destructive (dark): $S_2P - S_1P = (n+\tfrac12)\lambda$ — arrival out of phase, $I = 0$.
3. Valid wherever the two sources maintain a fixed phase relationship — i.e. are coherent.

Example 2: A point off the bisector [NEET Numerical]

Two coherent sources emit 600 nm light in phase. At P, the path difference is 1.8 $\mu$m. Bright or dark?

Solution:

1. $\frac{\Delta}{\lambda} = \frac{1.8\times10^{-6}}{600\times10^{-9}} = 3$ — an integer.
2. Bright (third-order constructive), $I = 4I_0$.
3. Had it been 1.5 $\mu$m ($2.5\lambda$), P would be perfectly dark.

Example 3: Phase from path [JEE Numerical]

For 500 nm light, what phase difference corresponds to a 125 nm path difference, and what is the resulting intensity (each source alone gives $I_0$)?

Solution:

1. $\phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi \times 125}{500} = \frac{\pi}{2}$.
2. $I = 4I_0\cos^2\frac{\pi}{4} = 4I_0 \times \frac12 = 2I_0$.
3. Quarter-wavelength shifts give exactly the average intensity — neither bright nor dark.

Example 4: The incoherent pair [NEET Numerical]

Two identical but independent lamps each light a screen point with $I_0$. What intensity results, and why are no fringes seen?

Solution:

1. Independent lamps have randomly drifting relative phase: incoherent.
2. $\langle 4I_0\cos^2(\phi/2)\rangle = 2I_0$ — intensities add, everywhere the same.
3. No fixed bright/dark geography = no fringes. (Section 5 shows Young's cure.)

Example 5: Unequal amplitudes [JEE Numerical]

Two coherent waves have amplitude ratio 2:1. Find $I_{max}:I_{min}$.

Solution:

1. $I_{max} \propto (a_1+a_2)^2 = 9a^2$; $I_{min} \propto (a_1-a_2)^2 = a^2$.
2. $I_{max}:I_{min} = 9:1$.
3. General rule: $\left(\frac{a_1+a_2}{a_1-a_2}\right)^2$ — amplitude ratios in, intensity ratios out.

Example 6: From intensity ratio to amplitude ratio [JEE Numerical]

Two coherent beams give $I_{max}/I_{min} = 25$. Find the ratio of their intensities.

Solution:

1. $\frac{a_1+a_2}{a_1-a_2} = 5 \Rightarrow a_1 = \frac{3}{2}a_2$.
2. $\frac{I_1}{I_2} = \left(\frac{a_1}{a_2}\right)^2 = \frac{9}{4}$.
3. Unwind the squares carefully — the most common slip in this family.

Example 7: Where does the dark energy go?

In destructive interference the intensity vanishes. Is energy destroyed?

Solution:

1. No. Interference only redistributes energy: dark fringes' missing energy appears in the bright fringes ($4I_0$ — twice the plain sum).
2. Averaged over the pattern, $\langle I\rangle = 2I_0$ — exactly the energy the two sources supply.
3. Conservation holds fringe-by-fringe in the average; nature merely rearranges the lighting.

Example 8: Coherence, defined for marks

Define coherent sources and give the practical requirement behind the definition.

Solution:

1. Two sources are coherent if the phase difference between their waves at a point does not change with time.
2. Practically they must have the same frequency (else the phase difference drifts continuously) and a locked initial phase.
3. NCERT's needles moving 'in identical fashion' — or Young's two slits fed by one source — are the canonical examples.

Example 9: The bisector is always bright

Why is every point on the perpendicular bisector of $S_1S_2$ bright (for in-phase coherent sources)?

Solution:

1. By symmetry, $S_1P = S_2P$ on the bisector: path difference zero everywhere on it.
2. Waves arrive in phase: $I = 4I_0$ along the whole line.
3. This is Young's central fringe in waiting — and why it stays put when wavelength changes.

Example 10: Three-intensity drill [JEE Numerical]

For two coherent sources of equal $I_0$ each, give the intensity at points where the path difference is (a) $\lambda$, (b) $\lambda/2$, (c) $\lambda/3$.

Solution:

1. (a) $\phi = 2\pi$: $I = 4I_0$.
2. (b) $\phi = \pi$: $I = 0$.
3. (c) $\phi = \frac{2\pi}{3}$: $I = 4I_0\cos^2\frac{\pi}{3} = 4I_0\times\frac14 = I_0$. The cosine-squared dial covers every case.