Coherence & Interference: Path Difference, Phase & Intensity

The amplitude-addition of this article is the working principle of an entire measurement technology. Anti-reflection coatings on camera lenses and spectacles are engineered so the reflections from their two surfaces meet with $\Delta = \lambda/2$ and cancel by equation (3), killing glare. Noise-cancelling headphones do the same trick with sound, generating a wave phase-shifted by $\pi$ to reach your ear $\lambda/2$ out of step with the noise. Radio astronomers link dishes across continents into an interferometer whose resolution comes from combining coherent signals with tracked path differences, and the LIGO detectors that caught gravitational waves are giant interferometers reading a path difference smaller than a proton's width as a shift in their fringes. Every one of these turns geometry into $\Delta$, $\Delta$ into $\varphi$, and $\varphi$ into a measured intensity — the exact chain built here. The next two articles apply it to the double slit and to thin films.

Interference is superposition wearing a spotlight. In the earlier part of this chapter you met the rule that when two waves overlap, the displacements add point by point — linear superposition. Interference is what that rule looks like when the two waves keep a steady relationship: instead of a shifting blur, you get a fixed pattern of places that are permanently bright and places that are permanently dark. The condition for a fixed pattern is coherence. Two sources are coherent if they share one frequency and hold a constant phase difference over the time you observe them. Light from ordinary sources fails this badly: a lamp emits in bursts a few nanoseconds long as its countless atoms flash independently, so the phase relationship between any two points jitters at random and averages the fringes to a uniform grey. The cure, discovered by Young, is never to use two lamps but to split one wavefront into two — with two slits, a biprism, a thin film, or a half-silvered mirror — so that any flicker in the source appears identically in both copies and their phase difference stays locked. That is why every real interference experiment traces both beams back to a single source. Once the sources are coherent, the accounting is done by the path difference $\Delta$, the difference in the distances the two waves travelled to reach a given point. If one wave is behind the other by exactly one wavelength, it is behind by one full cycle and they still line up crest-to-crest; if it is behind by half a wavelength, its crests fall on the other's troughs and they annihilate. Every point in the pattern gets its brightness from its own value of $\Delta$, and the whole art of interference is turning geometry into $\Delta$.

Two coherent waves that started in step arrive at a point after travelling different distances; that extra distance is the path difference $\Delta$. Phase counts cycles, and one whole wavelength of extra path is exactly one whole cycle — a phase of $2\pi$. Scaling linearly, a path difference $\Delta$ corresponds to a phase difference

where: $\Delta$ is the extra distance one wave travelled, $\lambda$ the wavelength, and $\varphi$ the phase difference in radians. The physical reading is a dictionary between distance and angle: feed in $\Delta = \lambda$ and you get $\varphi = 2\pi$ (a full turn, back in step); feed in $\Delta = \lambda/2$ and you get $\varphi = \pi$ (a half turn, exactly opposed); $\Delta = \lambda/4$ gives $\varphi = \pi/2$; and $\Delta = \lambda/3$ gives $\varphi = 2\pi/3$. This one line is the hinge of the whole chapter — geometry supplies $\Delta$, equation (1) turns it into $\varphi$, and the intensity formula turns $\varphi$ into brightness. The two headline conditions follow immediately. The waves reinforce (constructive interference, a bright fringe) when they are an integer number of full turns apart, $\varphi = 2m\pi$, i.e.

and they cancel (destructive interference, a dark fringe) when they are an odd number of half-turns apart, $\varphi = (2m+1)\pi$, i.e.

where: $m$ is the order of the fringe. Constructive means "a whole number of wavelengths of path difference"; destructive means "an odd number of half-wavelengths." Confusing these two lines is the single most common interference error, so pin them down now: whole $\lambda$ is bright, half-odd $\lambda$ is dark.

Conditions tell you where the fringes are; intensity tells you how bright each point is between them. Add two coherent waves of equal amplitude $a$ with phase difference $\varphi$, and the sum-to-product identity (worked below) gives a resultant of amplitude $2a\cos(\varphi/2)$. Since intensity goes as amplitude squared, $I \propto (2a\cos(\varphi/2))^2$, and writing $I_0$ for the intensity of one beam alone ($I_0 \propto a^2$),

where: $I_0$ is the intensity each beam would deliver on its own, $\varphi$ the phase difference from equation (1), and $I$ the resultant intensity. Read it at the special points. At a bright fringe $\varphi = 0$: $\cos^2 0 = 1$, so $I = 4I_0$ — four times one beam, not two. At a dark fringe $\varphi = \pi$: $\cos^2(\pi/2) = 0$, so $I = 0$ — total darkness. Halfway, at $\varphi = \pi/2$ ($\Delta = \lambda/4$): $\cos^2(\pi/4) = 0.5$, so $I = 2I_0$. And at $\varphi = 2\pi/3$ ($\Delta = \lambda/3$): $\cos^2(\pi/3) = (0.5)^2 = 0.25$, so $I = I_0$ — the two beams together deliver exactly what one beam alone would. These four values — $4I_0$, $2I_0$, $I_0$, $0$ — are worth memorising as sanity checks. Two things about equation (4) matter more than any single number. First, coherent light adds amplitudes, not intensities: the maximum is $4I_0$, whereas two incoherent beams would just add powers to give $2I_0$ everywhere. The factor of four at the peaks, paid for by zeros in the troughs, is the fingerprint of coherence. Second, energy is conserved. The average of $\cos^2$ over a full pattern is $\tfrac12$, so the average intensity is $\tfrac12 \times 4I_0 = 2I_0$ — precisely the incoherent sum. Interference does not manufacture light; it takes the energy that would have been spread as a flat $2I_0$ and heaps it into bright fringes of $4I_0$, borrowing from the dark fringes of $0$.

Real beams are rarely a perfect match — one slit may be wider, or a beam may be partly absorbed. If the amplitudes are $a_1$ and $a_2$, the resultant amplitude ranges from $a_1 + a_2$ (in step) down to $|a_1 - a_2|$ (opposed), so the brightest and darkest points are

where: $a_1, a_2$ are the two amplitudes and intensity goes as amplitude squared. The physical reading is that the dark fringes are only truly black when the amplitudes are equal; any mismatch leaves a residual glow in the minima and lowers the contrast. Take $a_1 = 2a_2$ (so the first beam carries $I_1 = 4I_2$, four times the power): the ratio is $(2a_2 + a_2)^2/(2a_2 - a_2)^2 = (3a_2)^2/(a_2)^2 = 9$. The maxima are nine times the minima — bright, but the minima never reach zero. Only when $a_1 = a_2$ does $(a_1 - a_2)^2 = 0$, giving $I_{\min} = 0$ and the perfect black fringes of equation (4). This is why interference demonstrations work hardest to make the two beams equally strong.

Board level — intensity at a chosen path difference (CBSE and NEET pattern)

Two coherent beams of equal intensity $I_0$ each overlap at a point where the path difference is $\Delta = \lambda/3$. Find the phase difference and the resultant intensity, and state the maximum and minimum intensities the pattern can reach.

Step 1. Convert path to phase with equation (1): $\varphi = (2\pi/\lambda)\times(\lambda/3) = 2\pi/3$ radians (that is $120^\circ$).

Step 2. Resultant intensity from equation (4): $I = 4I_0\cos^2(\varphi/2) = 4I_0\cos^2(\pi/3)$. Since $\cos(\pi/3) = 0.5$, $\cos^2(\pi/3) = 0.25$, so $I = 4I_0 \times 0.25 = \boxed{I_0}$. Two beams here deliver exactly what one would.

Step 3. Extremes of the pattern: the maximum occurs where the waves are in step, $I_{\max} = 4I_0$, and the minimum where they are exactly opposed, $I_{\min} = 0$. So this point, at $I_0$, sits a quarter of the way up from dark to bright.

Step 4. Sanity check on energy: the pattern averages $\tfrac12(4I_0) = 2I_0$, the same as if the two beams simply added their powers with no interference — the bright and dark fringes balance.

Competitive level — contrast from unequal slits (JEE Advanced pattern)

In a two-source interference experiment the amplitude from one slit is twice that from the other, $a_1 = 2a_2$. Find the ratio of the intensities from the two slits, and the ratio of maximum to minimum intensity in the pattern. Contrast this with the equal-amplitude case.

Step 1. Intensity goes as amplitude squared, so $I_1/I_2 = (a_1/a_2)^2 = 2^2 = 4$: the first slit carries four times the power of the second.

Step 2. Maximum-to-minimum ratio from equation (5): $\dfrac{I_{\max}}{I_{\min}} = \dfrac{(a_1+a_2)^2}{(a_1-a_2)^2} = \dfrac{(2a_2+a_2)^2}{(2a_2-a_2)^2} = \dfrac{(3a_2)^2}{(a_2)^2} = \boxed{9}$.

Step 3. Interpret it: the bright fringes are nine times the dark ones, but the dark fringes are not black — because $a_1 \ne a_2$, the minimum amplitude $|a_1 - a_2| = a_2$ is nonzero, leaving a residual intensity $I_{\min} = a_2^2 \ne 0$. The fringe contrast is high but imperfect.

Step 4. Compare the equal case: if instead $a_1 = a_2$, then $(a_1 - a_2)^2 = 0$, so $I_{\min} = 0$ and the ratio $I_{\max}/I_{\min}$ is infinite — the perfect black fringes and full contrast of equation (4). Equal amplitudes give the cleanest pattern. Timing note: about 75 seconds. The trap is squaring at the wrong step — take the ratio of amplitudes first, then square once at the end; do not square $a_1$ and $a_2$ before forming $(a_1 \pm a_2)$.

Whole $\lambda$ is bright, odd half-$\lambda$ is dark. Constructive interference is $\Delta = m\lambda$; destructive is $\Delta = (m+\tfrac12)\lambda$. Swapping these two lines is the commonest interference error. Write both at the top of your working and label which is which.

Add amplitudes, not intensities. For coherent light the maximum is $4I_0$ (from amplitude $2a$ squared), not $2I_0$. If your bright fringe came out at twice one beam, you added intensities and treated the light as incoherent.

The average is $2I_0$, always. Interference cannot raise the total brightness of the field; it only moves energy from dark fringes into bright ones. A question that implies "the screen is now brighter everywhere" is testing this misconception.

Two lamps do not interfere. Independent sources are incoherent, so no steady fringes form. Any real fringe experiment splits a single wave. Do not describe two separate bulbs, lasers, or headlights as producing an interference pattern.

Dark fringes are black only for equal amplitudes. If the two beams differ in strength, $I_{\min} = (a_1 - a_2)^2 \ne 0$; the minima glow and contrast drops. Use $I_{\max}/I_{\min} = (a_1+a_2)^2/(a_1-a_2)^2$, never assume the minima vanish.

Board scope. Coherence, the path–phase relation, and the constructive and destructive conditions are assessed on CBSE, ICSE, JEE, NEET, AP Physics 2, IB (SL and HL), and A-Level (CIE 9702). The explicit intensity formula $I = 4I_0\cos^2(\varphi/2)$ is CBSE, JEE, and NEET depth; AP, IB, and A-Level expect the path-difference conditions and a qualitative description of the intensity swing, not this equation. IGCSE has no wave optics at all.

Texts. NCERT Physics Class 12, Chapter 10 ("Wave Optics"), the sections on coherent sources, interference, and the intensity distribution · H. C. Verma, Concepts of Physics Vol. 1, "Light Waves" · Halliday, Resnick and Walker, the wave-optics chapter on coherence and intensity.

Provenance. Standard results. Recompute: $\varphi = 2\pi\Delta/\lambda$ gives $\varphi(\lambda)=2\pi$, $\varphi(\lambda/2)=\pi$, $\varphi(\lambda/4)=\pi/2$, $\varphi(\lambda/3)=2\pi/3$; $I = 4I_0\cos^2(\varphi/2)$ gives $I(0)=4I_0$, $I(\pi)=0$, $I(\pi/2)=2I_0$, $I(2\pi/3)=I_0$ (using $\cos^2(\pi/3)=0.25$); average intensity $2I_0$; unequal amplitudes $a_1=2a_2$ give $I_1/I_2=4$ and $I_{\max}/I_{\min}=(3)^2/(1)^2=9$; equal amplitudes give $I_{\min}=0$.

Prerequisites. Huygens' Principle & Wavefronts — where the coherent secondary sources come from · Superposition & Interference of Waves — the amplitude-addition rule. Leads to. Young's Double-Slit Experiment — this intensity engine on two-slit geometry · Thin-Film Interference & Newton's Rings — the same conditions with a reflection twist · and, for enrichment, the university seams Optical Path Difference and Phase Retardation and Temporal Coherence and Coherence Length (go-deeper only, never an exam dependency here).