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Physics · Ch 10 — Wave Optics

Coherent and Incoherent Addition of Waves

10.4

Coherent and Incoherent Addition of Waves

Core Idea: Superposition and Interference

When two or more waves overlap in space, the principle of superposition applies: the resultant displacement at any point is the vector sum of the displacements due to each individual wave. This leads to the phenomenon of interference — the redistribution of energy in space, forming a pattern of alternating maxima (bright) and minima (dark) regions.


Coherent vs. Incoherent Sources

  • Coherent sources: Two sources are coherent if the phase difference between the waves they produce at any point does not change with time.

    • Example: Two identical needles oscillating in phase in a water trough (Fig. 10.8a).
    • For coherent sources, a stable interference pattern (fixed positions of maxima and minima) is observed.
  • Incoherent sources: If the phase difference between the two sources changes rapidly with time, the sources are incoherent.

    • In this case, the interference pattern fluctuates so fast that the eye (or detector) sees only a time-averaged intensity.
    • The result: intensities simply add up — no stable pattern.

Constructive and Destructive Interference

Consider two coherent sources S1S_1 and S2S_2 vibrating in phase. Let the displacement at a point PP due to S1S_1 be:

y1=acos⁡ωty_1 = a \cos \omega t

Case 1: Path difference = integer multiple of λ\lambda

If the path difference Δx=S1P∼S2P=nλ\Delta x = S_1P \sim S_2P = n\lambda (where n=0,1,2,…n = 0, 1, 2, \dots), the waves arrive in phase.

The displacement due to S2S_2 is:

y2=acos⁡(ωt−2nπ)=acos⁡ωty_2 = a \cos (\omega t - 2n\pi) = a \cos \omega t

Resultant displacement:

y=y1+y2=2acos⁡ωty = y_1 + y_2 = 2a \cos \omega t

Since intensity I∝(amplitude)2I \propto (\text{amplitude})^2, the resultant intensity is:

I=4I0I = 4 I_0

where I0I_0 is the intensity from each source alone (I0∝a2I_0 \propto a^2). This is constructive interference.

Case 2: Path difference = half-integer multiple of λ\lambda

If Δx=(n+12)λ\Delta x = \left(n + \frac{1}{2}\right)\lambda (where n=0,1,2,…n = 0, 1, 2, \dots), the waves arrive exactly out of phase (phase difference π\pi).

The displacement due to S2S_2 is:

y2=acos⁡(ωt+π)=−acos⁡ωty_2 = a \cos (\omega t + \pi) = -a \cos \omega t

Resultant displacement:

y=y1+y2=0y = y_1 + y_2 = 0

Resultant intensity is zero. This is destructive interference.


General Expression for Intensity

For an arbitrary point GG, let the phase difference between the two waves be ϕ\phi. Then:

y1=acos⁡ωt,y2=acos⁡(ωt+ϕ)y_1 = a \cos \omega t, \quad y_2 = a \cos (\omega t + \phi)

Using the trigonometric identity cos⁡A+cos⁡B=2cos⁡A−B2cos⁡A+B2\cos A + \cos B = 2 \cos\frac{A-B}{2} \cos\frac{A+B}{2}, the resultant displacement is:

y=2acos⁡(ϕ2)cos⁡(ωt+ϕ2)y = 2a \cos\left(\frac{\phi}{2}\right) \cos\left(\omega t + \frac{\phi}{2}\right) …

Figure 10.8(a) Two needles oscillating in phase in water represent two coherent sources. (b) The pattern of displacement of water molecules at an instant on the surface of water showing nodal N (no displacement) and antinodal A (maximum displacement) lines.
Fig. 10.8 — (a) Two needles oscillating in phase in water represent two coherent sources. (b) The pattern of displacement of water molecules at an instant on the surface of water showing nodal N (no displacement) and antinodal A (maximum displacement) lines.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What the Figure Shows

Figure 10.8 has two panels from a water trough experiment.

Panel (a) shows two needles, labelled S₁ and S₂, dipping into water and oscillating up and down in phase — meaning they reach their highest and lowest points together. Each needle acts as a coherent source, sending out circular ripples (crests and troughs) that spread outward.

Panel (b) shows the overlapping ripple pattern at a single instant. Solid circles represent crests (maximum upward displacement) and dashed circles represent troughs (maximum downward displacement) from each source. Where crests from S₁ and S₂ cross, the water displacement is maximum — these are antinodal lines (A). Where a crest from one source meets a trough from the other, the displacement cancels — these are nodal lines (N). The lines fan out radially from the region between S₁ and S₂.

The key physical idea is that two coherent sources produce a stable interference pattern of alternating constructive and destructive interference, because the phase difference at any point does not change with time.


The Key Formulas Developed from This Figure

The textbook uses this figure to derive the conditions for interference:

  1. Constructive interference (maximum intensity, antinodal lines):

S1P∼S2P=nλ(n=0,1,2,3,… )S_1P \sim S_2P = n\lambda \quad (n = 0, 1, 2, 3, \dots)

where S1PS_1P and S2PS_2P are distances from the two sources to point PP, λ\lambda is the wavelength, and ∼\sim means the absolute difference. The resultant intensity is I=4I0I = 4I_0, where I0I_0 is the intensity from one source alone.

  1. Destructive interference (zero intensity, nodal lines):

S1P∼S2P=(n+12)λ(n=0,1,2,3,… )S_1P \sim S_2P = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, 1, 2, 3, \dots)

The resultant intensity is zero.

  1. General intensity formula for any point with phase difference ϕ\phi:

I=4I0cos⁡2(ϕ2)I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)

Here ϕ\phi is the phase difference between the two waves at that point. The phase difference is related to the path difference Δx=S1P∼S2P\Delta x = S_1P \sim S_2P by:

ϕ=2πλΔx\phi = \frac{2\pi}{\lambda} \Delta x


Physical Meaning …

Figure 10.9(a) Constructive interference at a point Q for which the path difference is 2λ. (b) Destructive interference at a point R for which the path difference is 2.5λ.
Fig. 10.9 — (a) Constructive interference at a point Q for which the path difference is 2λ. (b) Destructive interference at a point R for which the path difference is 2.5λ.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Figure 10.9 is a schematic diagram with two separate panels, (a) and (b), each showing two coherent point sources S1S_1 and S2S_2 separated by a distance dd on the left, and a single field point on the right. Straight line segments are drawn from each source to the field point, representing the path lengths S1QS_1Q, S2QS_2Q in panel (a) and S1RS_1R, S2RS_2R in panel (b). The labels QQ and RR mark the field points, and the path differences are explicitly written as 2λ2\lambda and 2.5λ2.5\lambda respectively.

Physical idea: The figure illustrates how the path difference between two coherent waves determines whether they interfere constructively or destructively. When the path difference is an integer multiple of the wavelength λ\lambda, the waves arrive in phase and produce a bright fringe (constructive interference). When the path difference is a half-integer multiple of λ\lambda, the waves arrive exactly out of phase and cancel completely (destructive interference).

Panel (a) — Constructive interference at Q:

Here S2Q−S1Q=2λS_2Q - S_1Q = 2\lambda. Since 2λ2\lambda corresponds to a phase difference of 4π4\pi (because Δϕ=2πλ×path difference\Delta \phi = \frac{2\pi}{\lambda} \times \text{path difference}), the two waves are in phase. The resultant displacement is y=y1+y2=2acos⁡ωty = y_1 + y_2 = 2a \cos \omega t, so the amplitude doubles. The intensity, proportional to the square of the amplitude, becomes I=4I0I = 4I_0, where I0I_0 is the intensity from a single source.

Panel (b) — Destructive interference at R:

Here S2R−S1R=2.5λS_2R - S_1R = 2.5\lambda. This path difference corresponds to a phase difference of 5π5\pi. The waves are exactly out of phase: if y1=acos⁡ωty_1 = a \cos \omega t, then y2=acos⁡(ωt+5π)=−acos⁡ωty_2 = a \cos(\omega t + 5\pi) = -a \cos \omega t. The resultant displacement is y=0y = 0, giving zero intensity.

Key formula developed from this figure:

For two coherent sources vibrating in phase, the condition for constructive interference at a point PP is

S1P∼S2P=nλ(n=0,1,2,3,… )S_1P \sim S_2P = n\lambda \quad (n = 0, 1, 2, 3, \dots)

and for destructive interference …

Figure 10.10Locus of points for which S1P – S2P is equal to zero, ±λ, ±2λ, ±3λ.
Fig. 10.10 — Locus of points for which S1P – S2P is equal to zero, ±λ, ±2λ, ±3λ.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What the Figure Shows

The figure is a schematic diagram of the interference pattern produced by two coherent point sources, S₁ and S₂. The sources are placed close together on the vertical axis, with S₁ above S₂. The diagram does not show wavefronts or intensity directly; instead, it shows the loci of points where the path difference Δ=S1P−S2P\Delta = S_1P - S_2P is constant.

  • The horizontal axis (left–right) represents the direction perpendicular to the line joining the sources.
  • The vertical axis (up–down) is the line containing S₁ and S₂.
  • The curves are a family of confocal hyperbolas, all sharing S₁ and S₂ as foci.
  • The central curve (n=0n = 0) is the straight horizontal line that is the perpendicular bisector of S₁S₂. Here Δ=0\Delta = 0.
  • Above and below this line are hyperbolas labelled n=±1,±2,±3n = \pm 1, \pm 2, \pm 3, corresponding to Δ=±λ,±2λ,±3λ\Delta = \pm \lambda, \pm 2\lambda, \pm 3\lambda.
  • Two specific points are marked: Q (upper region) and G (lower region), representing arbitrary points on the pattern.

Physical Idea Taught

The figure visualises the condition for constructive and destructive interference from two coherent sources. The key idea is that the path difference Δ\Delta determines the phase difference ϕ\phi between the waves arriving at a point P:

ϕ=2πλΔ\phi = \frac{2\pi}{\lambda} \Delta

  • Constructive interference (maximum intensity 4I04I_0) occurs when Δ=nλ\Delta = n\lambda (n=0,±1,±2,…n = 0, \pm 1, \pm 2, \dots). These are the hyperbolas labelled n=0,±1,±2,±3n = 0, \pm 1, \pm 2, \pm 3 in the figure.
  • Destructive interference (zero intensity) occurs when Δ=(n+12)λ\Delta = \left(n + \frac12\right)\lambda (n=0,±1,±2,…n = 0, \pm 1, \pm 2, \dots). These hyperbolas lie between the labelled ones (not explicitly labelled in the figure).

The figure thus maps out the stable interference pattern in space: bright fringes (maxima) lie on the labelled hyperbolas, and dark fringes (minima) lie halfway between them.

Key Formulas Developed with This Figure

The textbook uses the geometry of the figure to derive the general intensity formula. For two coherent sources vibrating in phase, the resultant intensity at any point P is:

I=4I0cos⁡2(ϕ2)I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)

where:

  • I0I_0 = intensity from each source alone (proportional to a2a^2, with aa the amplitude)
  • ϕ\phi = phase difference between the two waves at P, given by ϕ=2πλΔ\phi = \frac{2\pi}{\lambda} \Delta
  • Δ=S1P−S2P\Delta = S_1P - S_2P = path difference

The conditions for maxima and minima follow directly: …