Chapter 02 · Wave Optics · Full book

Wave Optics

The previous chapter told a clean and powerful lie. It said that light travels in straight lines called rays, that a ray bends only at a surface, and that the whole behaviour of mirrors, lenses, prisms, and instruments follows from tracking those rays with a ruler and the law of refraction. Every formula we built there—the mirror equation, the lens-maker's formula, the deviation of a prism—is correct, and you should keep all of it. But it was built on a deliberate omission. We never asked what light is. We only asked where its rays go.

The omission was harmless as long as the obstacles light met—lenses, slits, apertures—were enormous compared with something we never named. Shrink the obstacle, and the lie breaks. Send light through a slit narrower than a hair and it refuses to cast a sharp shadow: it fans out, spreading into the geometric darkness. Overlap two beams from the same source and the screen does not simply get brighter everywhere—it breaks into stripes of light and dark, as though light added to light could make blackness. Tilt a piece of plastic in front of a glare off water and the glare vanishes, though the plastic is perfectly clear. None of these can be drawn with rays. They are the fingerprints of the thing ray optics hid: light is a wave.

This chapter restores what Chapter 10 set aside. The hidden length scale is the wavelength \(\lambda\)—for visible light, around half a micrometre, which is why a wavelength-sized obstacle had to be unusually small before the wave nature showed. When the obstacle is far larger than \(\lambda\), the wave marches forward in lockstep and its leading edge is a ray; ray optics is the large-obstacle limit of wave optics, exactly as Newtonian mechanics is the slow limit of relativity. When the obstacle approaches \(\lambda\), the marching breaks up and we see interference and diffraction. Our task is to find the single principle—Huygens'—from which both the old ray laws and the new wave phenomena fall out, and then to follow it into the four great experiments of wave optics: interference (Young), thin films, diffraction (the single slit), and polarisation (Malus and Brewster).

As before, two audiences travel together. The geometry of fringes and the algebra of \(\beta=\lambda D/d\) are for everyone (NEET, AP Physics 2); the intensity integrals and the diffraction envelope are for those going further (JEE Advanced, the Olympiads). Both describe the same striped screen.

Core intuition: Chapter Roadmap

Huygens' principle — Every point of a wavefront is itself a source of secondary wavelets; the envelope of those wavelets is the wavefront an instant later. A ray is just a line drawn perpendicular to the wavefronts. This one construction is the engine of the whole chapter.
Reflection and refraction, re-derived — Run Huygens' construction at a surface and the law of reflection (\(\theta_i=\theta_r\)) and Snell's law (\(n_1\sin\theta_1=n_2\sin\theta_2\)) drop out—together with the reason light slows in glass: its wavelength shortens while its frequency holds fixed.
Superposition and coherence — Two waves at a point add amplitude to amplitude, so intensity is not simply additive: \(I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta\). Sustained fringes demand a steady phase relation—coherence—which is why two light bulbs never interfere but two slits fed by one source do.
Young's double slit — The keystone experiment. Path difference \(d\sin\theta\) sets bright and dark fringes; the fringe width is \(\beta=\lambda D/d\), and the screen intensity is \(I=4I_0\cos^2(\pi d y/\lambda D)\). A thin sheet over one slit shifts the whole pattern by \((\mu-1)tD/d\).
Thin films — Why a colourless soap bubble blazes with colour: the two reflections off the front and back of the film interfere, with a compulsory \(\lambda/2\) jump at the harder surface. Condition \(2\mu t\cos r=(n-\tfrac12)\lambda\) for a bright reflection—the basis of anti-reflection coatings.
Diffraction at a single slit — A slit is not two sources but a continuum; summing them gives minima at \(a\sin\theta=n\lambda\) and the envelope \(I=I_0(\sin\beta/\beta)^2\). The Fresnel distance \(z_F=a^2/\lambda\) marks where ray optics finally fails.
Resolving power — Two stars, or two cells, blur into one when their diffraction discs overlap. Rayleigh's criterion fixes the limit: \(\theta_{\min}=1.22\lambda/D\) for a telescope, sharpening with aperture, not with magnification.
Polarisation — The phenomenon that settles the deepest question: light is a transverse wave. Malus's law \(I=I_0\cos^2\theta\) governs a polaroid; Brewster's angle \(\tan\theta_B=n\) produces perfectly polarised reflection and explains polaroid sunglasses.

Perplexing Questions

Perplexing questions

Six puzzles, each impossible to draw with a ray. Each is resolved at the point in the chapter where the relevant idea is built—watch for the [Resolution: PQ N] callout.

Light plus light makes darkness. Shine one narrow beam on a screen and you get a bright patch. Add a second, identical beam beside it and—at certain places—the screen goes darker than it was with one beam alone. Energy was poured in from two sources, yet some spots receive less than before. Where did the light go, and does this not break the conservation of energy?
Two lamps never stripe a wall. The striped pattern of light-and-dark is easy to make with two narrow slits cut in a card and lit by a single lamp behind them. But place two separate lamps side by side, as bright and as close as you like, and the wall is uniformly lit: no stripes, ever. What does one-source-through-two-slits possess that two-sources-side-by-side can never have?
The colour with no pigment. A soap film and a slick of oil on a wet road are both made of colourless, transparent stuff. Yet both flare with shifting bands of colour, and the colours move as you change your viewing angle or as the film thins. Where does the colour come from, if there is no coloured substance anywhere in the film?
The shadow that fans out. Make a slit narrower and narrower and shine light through it onto a far screen. Common sense says the bright band on the screen should get narrower as the slit closes. Instead, past a point, it starts to get wider, fanning into the shadow. How can closing the gate make the beam spread?
Why a bigger telescope, not a stronger eyepiece. Two stars sit very close together in the sky. A small telescope shows them as one smeared blob, and no amount of extra magnification in the eyepiece splits them—it only enlarges the blur. Swap in a telescope with a wider mirror and suddenly they separate into two clean points. What does the width of the mirror do that magnification cannot?
The glare that one tilt erases. Glare bouncing off a wet road or a lake is fierce. Hold up a clear plastic filter and rotate it: at one orientation the glare nearly vanishes, while the rest of the scene dims only slightly. The filter is not coloured and not a mirror. How does a mere rotation kill the reflection from the water but spare the view?

Every one of these is answered explicitly before the chapter ends.
Look for the [Resolution: PQ N] callout as you read.

Huygens' Principle: Light as a Marching Wavefront

◈ Huygens' wavefronts

live bench

Watch secondary wavelets build the next front — and Snell's law fall out of it. Toggle reflection vs refraction.

◈ loading the interactive bench…

↗ open in the Simulations page

◇ Figure 1

Fig 1. Huygens' construction. Each point of the wavefront (cyan) emits a secondary wavelet of radius \(v\,\Delta t\) (grey); their forward envelope is the next wavefront. The ray (blue) is the normal to the fronts.

To treat light as a wave we need a way to track where the wave is as it advances—and a wave, unlike a particle, is not at one place. It is spread along a front. The natural bookkeeping object is the wavefront: a surface joining all points of the wave that are in the same phase—all the crests at one instant, say. A point source radiates spherical wavefronts, like the expanding shells of a struck bell; far from the source a small patch of that sphere is effectively flat, and we speak of plane wavefronts, the optical equivalent of straight, parallel ocean swells rolling toward a beach.

The relationship between this new object and the ray of Chapter 10 is the first thing to fix, because it is the bridge between the two chapters.

Core intuition: A Ray Is the Normal to a Wavefront

A ray is the line along which energy travels; a wavefront is the surface of constant phase. They are perpendicular everywhere: the ray points the way the front advances. In a uniform medium the front moves straight ahead and the rays are straight parallel lines—this is precisely the “light travels in straight lines” of ray optics, now seen as a consequence of an undistorted wavefront rather than a starting assumption. Everything that bends a ray (a surface, a slit) does so by reshaping the wavefront. Hold this picture: in wave optics we track fronts; the rays are wherever the fronts point.

How does a wavefront know where to be an instant later? Christiaan Huygens (1678) gave the rule that runs this entire chapter:

Huygens' principle. Every point of a wavefront may be regarded as a fresh source of secondary spherical wavelets, spreading forward at the wave speed of the medium. After a time \(t\), the new wavefront is the surface that is tangent to all these wavelets—their forward envelope.

The construction is mechanical. Take the present wavefront; from each of its points draw a small sphere (in two dimensions, a circle) of radius \(vt\), the distance the wave travels in time \(t\); the smooth surface kissing the leading edges of all those spheres is the wavefront at the later instant. A plane front begets a parallel plane front a distance \(vt\) ahead; a spherical front begets a larger sphere. (Huygens left a loose end—why we see only the forward envelope and not a backward wave returning toward the source. The honest repair came later, with Fresnel's obliquity factor and finally the wave equation; for our purposes the forward envelope is the rule, and it never fails us.)

The power of the principle is that it makes no mention of straight lines, of surfaces, or of slits. It is a single local rule—each point spawns a wavelet—and from it the whole of optics, ray and wave alike, must follow. We spend the rest of the chapter cashing that promise.

Worked Examples

Example 1 · How far a wavefront marches

A point source flashes in vacuum. (a) What is the radius of its spherical wavefront \(2.0\,\text{ns}\) later? (b) The same flash enters glass (\(n=1.5\)); how far does each Huygens wavelet spread in \(1.0\,\text{ns}\) inside the glass?

Solution. Given: \(t_a=2.0\,\text{ns}\) (vacuum), \(t_b=1.0\,\text{ns}\) (glass, \(n=1.5\)), \(c=3.0\times10^{8}\,\text{m\,s}^{-1}\). Find: wavefront radius \(r_a\); wavelet radius \(r_b\) in glass.
Setup: a Huygens wavelet spreads a distance (speed)\(\times\)(time); in vacuum the speed is \(c\), in glass it is \(v=c/n\).
Solve: \[ r_a = c\,t_a = (3.0\times10^{8})(2.0\times10^{-9}) = 0.60\,\text{m}, \] \[ v = \frac{c}{n} = \frac{3.0\times10^{8}}{1.5} = 2.0\times10^{8}\,\text{m\,s}^{-1}, \qquad r_b = v\,t_b = (2.0\times10^{8})(1.0\times10^{-9}) = 0.20\,\text{m}. \] Answer: \(\boxed{r_a = 0.60\,\text{m},\ \ r_b = 0.20\,\text{m}}\).
Check: the envelope of these wavelets is the next wavefront; in glass the wavelets are smaller in the same time (slower speed), so the front advances more slowly—the very fact that later makes it pivot at a boundary and bend the ray. ✓

Example 2 · From wavefront tilt to ray angle

A plane wavefront approaches a flat mirror, making an angle of \(30^\circ\) with the mirror surface. Find the ray's angle of incidence and its angle of reflection.

Solution. Given: the wavefront makes \(30^\circ\) with the surface. Find: \(\theta_i\) and \(\theta_r\) (measured from the normal).
Setup: the ray is perpendicular to the wavefront and the normal is perpendicular to the surface. Two lines each rotated \(90^\circ\) keep the angle between them, so the ray-to-normal angle equals the wavefront-to-surface angle.
Solve: \[ \theta_i = 30^\circ, \qquad \theta_r = \theta_i = 30^\circ \ \ (\text{Huygens: both wavelets born in one medium travel at one speed}). \] Answer: \(\boxed{\theta_i = \theta_r = 30^\circ}\).
Check: a front lying flat on the mirror (\(0^\circ\) to the surface) is normal incidence (\(\theta_i=0\)), and a steeply tilted front is grazing—so the wavefront-tilt\(=\)incidence-angle rule behaves correctly at both extremes, and the equal-speed construction returns \(\theta_r=\theta_i\). ✓

Reflection and Refraction from Wavefronts

◇ Figure 2

Fig 2. Refraction by wavefronts. The end of the wavefront that enters the slower medium first falls behind, so the front (cyan) pivots toward the normal and the ray (blue) bends: \(\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{v_1}{v_2}=\dfrac{n_2}{n_1}\).

The first test of any new principle is whether it reproduces the laws we already trust. Huygens' construction must give back the law of reflection and Snell's law—and it does, with a bonus: it tells us why light slows in glass, a question ray optics could not even pose.

Birth of the formula: Birth of the Law of Reflection: \(\theta_i=\theta_r\) from a Pivoting Wavefront

A plane wavefront \(AB\) falls on a flat mirror, its lower edge \(A\) touching the surface while the upper edge \(B\) is still a distance away. Let the incident wavefront make angle \(\theta_i\) with the mirror; the rays (perpendicular to \(AB\)) make the same \(\theta_i\) with the normal. While the wavelet from \(B\) travels forward to strike the mirror at a point \(C\), covering a distance \(BC=v\,t\), the point \(A\)—which struck first—has been re-radiating into the same medium for the same time \(t\), throwing out a secondary wavelet of radius \(AD=v\,t=BC\). The reflected wavefront \(CD\) is the tangent from \(C\) to that wavelet. Now compare the two right triangles \(ABC\) and \(CDA\). They share the hypotenuse \(AC\); they have a right angle each (\(\angle ABC\) and \(\angle CDA\)); and their other matching sides are equal, \(BC=AD\). The triangles are congruent, so the angles at \(A\) and \(C\) match: \[ \sin\theta_i=\frac{BC}{AC}=\frac{AD}{AC}=\sin\theta_r \quad\Longrightarrow\quad \boxed{\;\theta_i=\theta_r\;} \] The incident ray, reflected ray, and normal lie in one plane (the plane of the page, the plane of the wavefront's motion), and the angles are equal. The law of reflection is not a separate postulate; it is the statement that both wavelets, born in one medium, travel at one speed.

Refraction is the richer case, because the wave crosses into a medium where its speed changes.

Birth of the formula: Birth of Snell's Law: A Wavefront Pivots Because One End Slows First

A plane wavefront \(AB\) in medium 1 (speed \(v_1\)) reaches the flat boundary with medium 2 (speed \(v_2\lt v_1\)). Edge \(A\) enters the second medium first; edge \(B\) is still in the first medium and continues at \(v_1\). In a time \(t\), the still-outside edge \(B\) advances to the surface point \(C\), covering \(BC=v_1 t\) in medium 1. Meanwhile the wavelet launched by \(A\) into medium 2 has spread only a distance \(AE=v_2 t\), because it travels at the slower speed. The refracted wavefront is the tangent \(CE\) from \(C\) to that smaller wavelet. Because \(A\)'s wavelet is smaller, the front pivots: it tilts toward the normal. Both triangles share the hypotenuse \(AC\). With \(\theta_1\) the angle of incidence and \(\theta_2\) the angle of refraction (each measured from the normal, and equal to the angle each wavefront makes with the surface), \[ \sin\theta_1=\frac{BC}{AC}=\frac{v_1 t}{AC}, \qquad \sin\theta_2=\frac{AE}{AC}=\frac{v_2 t}{AC}. \] Divide: \[ \frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}. \] Define the refractive index of a medium by \(n=c/v\) (the factor by which it slows light). Then \(v_1/v_2=n_2/n_1\), and \[ \boxed{\;n_1\sin\theta_1=n_2\sin\theta_2\;} \] Snell's law, derived—not assumed—from a wavefront that pivots because one end of it entered the slow medium before the other. The geometry is the same that tilts a marching band as it walks off pavement onto mud: the file that hits the mud first falls behind, and the whole line swings.

The derivation hands us, for free, the answer to a question ray optics never addressed: what changes when light enters glass?

Core intuition: In a Denser Medium the Wavelength Shrinks, Not the Frequency

When light passes from vacuum into glass, its speed drops from \(c\) to \(v=c/n\). But the wave's frequency \(\nu\) is fixed by the source and cannot change at a passive boundary—the crests arrive at the interface at a certain rate and must leave at the same rate, or they would pile up. Since \(v=\nu\lambda\) and \(\nu\) holds while \(v\) falls, \[ \lambda_{\text{medium}}=\frac{v}{\nu}=\frac{c/n}{\nu}=\frac{\lambda_{\text{vacuum}}}{n}. \] The wavelength shortens by the factor \(n\); the colour (frequency) does not change. This is why a single-slit or double-slit pattern shrinks when the whole apparatus is immersed in water: every fringe spacing carries a \(\lambda\) that has become \(\lambda/n\). Remember it as the rule that resolves half the “what happens in a liquid” traps in this chapter.

Misconception alert: Light Does Not “Choose” the Fastest Path

A popular shortcut says light “decides” to bend so as to reach its destination in the least time (Fermat's principle), as though the beam surveys all routes and selects the quickest. This is a true and beautiful summary of the result, but it is not the mechanism, and taken literally it is teleological nonsense—light cannot know its destination when it sets out. The wavefront does nothing global. Each point obeys one local rule (Huygens: re-radiate at the local speed), and the least-time path emerges from that local obedience, exactly as the marching band's swing emerges from each player slowing on the mud without any player planning the turn. When you are tempted to say light “picks” a path, replace it with: the wavefront pivoted because one end slowed first.

Worked Examples

Example 3 · Snell's law from two wave speeds

Light travels at \(v_1 = 2.4\times10^{8}\,\text{m\,s}^{-1}\) in medium 1 and strikes the boundary with medium 2, where its speed is \(v_2 = 2.0\times10^{8}\,\text{m\,s}^{-1}\), at an angle of incidence \(40^\circ\). Find the angle of refraction and the ratio of the two refractive indices.

Solution. Given: \(v_1 = 2.4\times10^{8}\), \(v_2 = 2.0\times10^{8}\,\text{m\,s}^{-1}\), \(\theta_1 = 40^\circ\). Find: \(\theta_2\) and \(n_2/n_1\).
Setup: the wavefront derivation gives \(\dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}\).
Solve: \[ \sin\theta_2 = \frac{v_2}{v_1}\sin\theta_1 = \frac{2.0}{2.4}\sin 40^\circ = 0.833\times 0.643 = 0.536 \;\Rightarrow\; \theta_2 = 32.4^\circ, \] \[ \frac{n_2}{n_1} = \frac{v_1}{v_2} = \frac{2.4}{2.0} = 1.2 . \] Answer: \(\boxed{\theta_2 = 32.4^\circ,\ \ n_2/n_1 = 1.2}\).
Check: medium 2 is slower, hence optically denser (\(n_2\gt n_1\)), and the ray bends toward the normal (\(\theta_2\lt \theta_1\))—exactly the pivot the slower wavelet forces. ✓

Example 4 · Speed, wavelength, and colour in water

Light of vacuum wavelength \(600\,\text{nm}\) enters water (\(n = 1.33\)). Find its speed, wavelength, and frequency inside the water.

Solution. Given: \(\lambda_0 = 600\,\text{nm}\), \(n = 1.33\). Find: \(v\), \(\lambda_{\text{water}}\), \(\nu\).
Setup: \(v = c/n\); frequency is fixed at the boundary, so \(\lambda = \lambda_0/n\) and \(\nu = c/\lambda_0 = v/\lambda_{\text{water}}\).
Solve: \[ v = \frac{3.0\times10^{8}}{1.33} = 2.26\times10^{8}\,\text{m\,s}^{-1}, \qquad \lambda_{\text{water}} = \frac{600}{1.33} = 451\,\text{nm}, \] \[ \nu = \frac{c}{\lambda_0} = \frac{3.0\times10^{8}}{600\times10^{-9}} = 5.0\times10^{14}\,\text{Hz}. \] Answer: \(\boxed{v = 2.26\times10^{8}\,\text{m\,s}^{-1},\ \lambda = 451\,\text{nm},\ \nu = 5.0\times10^{14}\,\text{Hz}}\).
Check: \(v/\lambda_{\text{water}} = 2.26\times10^{8}/451\times10^{-9} = 5.0\times10^{14}\,\text{Hz}\), the same frequency—the colour is unchanged. It is the shortened \(\lambda\) that shrinks a fringe pattern immersed in water. ✓

✓ Checkpoint · Huygens and the Wavefront Construction

State what a wavefront is and how a ray is related to it.
A wavefront is a surface of constant phase; a ray is the line of energy flow, everywhere perpendicular to the wavefronts.
A plane wave passes from air into glass (\(n=1.5\)). What happens to its speed, wavelength, and frequency?
Speed \(v=c/n\) falls; wavelength \(\lambda/n\) shortens; frequency is unchanged (fixed by the source).
(Trap) “Huygens says every point radiates wavelets in all directions, so light should also travel backward.” Resolve.
Only the forward envelope is physically realised; the backward-going wavelets cancel (formally, via the obliquity factor). Huygens' bare construction needs this repair.

Superposition, Interference, and Coherence

Two ripples crossing a pond pass through each other and emerge unchanged, but while they overlap the water's displacement is the sum of what each ripple alone would produce. This is the principle of superposition, and for light it is the gateway to everything that follows: where two light waves overlap, their electric fields add, vector to vector, at every instant and every point.

The subtlety—and the source of every fringe—is that what our eyes and instruments record is not the field but the intensity, proportional to the square of the resultant amplitude, averaged over the wave's fast oscillation. Squaring a sum is not summing the squares, and that single fact is interference.

Birth of the formula: Birth of the Interference Term: Why Two Intensities Do Not Simply Add

Let two waves of the same frequency arrive at a point with amplitudes \(a_1,a_2\) and a constant phase difference \(\delta\): \[ y_1=a_1\sin\omega t,\qquad y_2=a_2\sin(\omega t+\delta). \] By superposition the resultant is \(y=y_1+y_2\). Adding two sinusoids of the same frequency gives another sinusoid, \(y=A\sin(\omega t+\phi)\), whose amplitude follows from the phasor (or trigonometric) sum: \[ A^2=a_1^2+a_2^2+2a_1a_2\cos\delta. \] Intensity is proportional to amplitude squared, \(I\propto A^2\), so with \(I_1\propto a_1^2\) and \(I_2\propto a_2^2\), \[ \boxed{\;I=I_1+I_2+2\sqrt{I_1 I_2}\,\cos\delta\;} \] The first two terms are what you would get if the beams merely piled up. The third—the interference term—is the whole story. It rides up and down with \(\cos\delta\):

Constructive (\(\delta=0,2\pi,4\pi,\dots\), i.e. path difference an integer number of wavelengths): \(\cos\delta=+1\), and \(I=(\sqrt{I_1}+\sqrt{I_2})^2\), a maximum.
Destructive (\(\delta=\pi,3\pi,\dots\), half-integer wavelengths): \(\cos\delta=-1\), and \(I=(\sqrt{I_1}-\sqrt{I_2})^2\), a minimum—exactly zero if \(I_1=I_2\).

Equal sources. If \(I_1=I_2=I_0\), set \(a_1=a_2\) and the formula collapses beautifully: \[ I=2I_0(1+\cos\delta)=4I_0\cos^2\!\frac{\delta}{2}. \] This \(\cos^2(\delta/2)\) is the shape of every two-slit fringe pattern. It runs from \(4I_0\) at the bright fringes down to \(0\) at the dark—four times one source's intensity at the peaks, zero in the troughs. Energy is conserved [Resolution: PQ 1]. Average \(4I_0\cos^2(\delta/2)\) over a full cycle of \(\delta\): since \(\langle\cos^2\rangle=\tfrac12\), the mean intensity is \(2I_0=I_1+I_2\). Not a joule is created or destroyed. Interference only redistributes the energy—robbing the dark fringes to pay the bright ones. The light removed from a dark fringe reappears, doubled, at the neighbouring bright fringe.

Misconception alert: Interference Does Not Create or Destroy Energy

The dark fringe tempts a real worry: two beams arrive, the screen is black, where did the energy go? It went next door. The space-average of the pattern equals the plain sum \(I_1+I_2\); the bright fringes carry exactly the surplus that the dark fringes lack. Conservation of energy is never threatened—it is enforced, fringe by fringe. (If you ever compute a pattern whose average is not \(I_1+I_2\), you have made an arithmetic error, not discovered free energy.) “Destructive interference” destroys light here only by sending it there.

There is a condition lurking in the phrase “a constant phase difference \(\delta\).” If \(\delta\) drifts randomly in time, \(\cos\delta\) averages to zero before your eye can register a single pattern, the interference term washes out, and you are left with \(I=I_1+I_2\)—plain, fringe-free brightness. A steady, sustained pattern therefore demands that the two sources keep a fixed phase relationship. This property is called coherence, and it is the reason interference is rare in everyday life.

Core intuition: Coherence: A Steady Phase Relationship, Not Just “Same Colour”

Two sources are coherent if their phase difference at the overlap point stays constant in time. An ordinary light source emits in independent bursts a few nanoseconds long, each burst starting with a random phase; two separate lamps—even of the identical colour—have phase differences that scramble billions of times a second, and their interference term averages instantly to zero [Resolution: PQ 2]. The cure, before lasers, was Young's trick: take one source and split its light into two paths (two slits). Whatever random phase the source throws out, both slits inherit it together, so their difference stays locked. One source, split, is automatically coherent with itself; two sources never are. (A laser is special precisely because it emits one long coherent wavetrain, so two lasers can—barely, and only briefly—interfere.) Coherence is about the constancy of the phase difference, not the sameness of the colour.

Misconception alert: Same Colour Is Not the Same as Coherent

The tempting shortcut is “two sources of the same wavelength will interfere.” They will not, unless their phase difference is also steady. Two sodium lamps emit the identical yellow line, yet each radiates in independent nanosecond bursts with random starting phase, so their phase difference scrambles billions of times a second and the fringes average away to uniform brightness. Equal frequency is necessary (different frequencies give a moving, time-averaged-flat pattern) but not sufficient—coherence is the missing ingredient. This is exactly why Young split one beam into two slits rather than using two lamps, and why even two separate lasers interfere only briefly and with difficulty. Check for a locked phase relationship, not merely a matched colour.

Making Two Coherent Sources: Biprism and Lloyd's Mirror

If no two independent lamps will ever interfere, every interference experiment must do the same thing: take one beam and split it. There are only two ways to split it, and the classical experiments are all one or the other. Division of wavefront takes different parts of the same wavefront — Young's two slits, Fresnel's biprism, Lloyd's mirror. Division of amplitude takes the same part of the wavefront and splits its energy at a surface — a thin film, a Michelson interferometer. Either way the phase relation is inherited from the parent beam, and is therefore locked.

Fresnel's biprism. Two very thin prisms joined base to base, each of refracting angle \(\alpha\) of the order of half a degree, act on a slit source \(S\) placed a distance \(a\) in front of them. Each half deviates the light through \[ \delta = (\mu - 1)\alpha, \] one half upward and one half downward, so an observer on the far side sees two virtual images of the slit, \(S_1\) and \(S_2\), separated by \[ d = 2a(\mu - 1)\alpha. \] From there nothing is new: the two virtual sources are coherent, and with the screen a distance \(D\) from the source the fringes have the usual width \(\beta = \lambda D/d\). Because \(\alpha\) is small, \(d\) is small and the fringes are comfortably wide.

Its historical point is worth more than its formula. Young's fringes could be — and were — dismissed as an artefact of light bending round the edges of two slits. A biprism has no edges in the beam at all: the light is refracted, never obstructed, and the fringes appear anyway. That killed the objection.

Lloyd's mirror. Cruder still: let a source at height \(h\) above a plane mirror send light both directly to the screen and by grazing reflection. The mirror produces a virtual source at depth \(h\) below it, so the separation is simply \[ d = 2h, \] and again \(\beta = \lambda D/d\). This one carries a sting that no other two-source arrangement does. The reflected beam grazes a denser medium and picks up an extra \(\lambda/2\), so the two conditions swap: \[ \Delta x = n\lambda \;\Rightarrow\; \text{dark}, \qquad \Delta x = \left(n - \tfrac{1}{2}\right)\lambda \;\Rightarrow\; \text{bright}. \] The fringe at mirror level, where the path difference is zero and every other experiment shows its brightest maximum, is dark. It is the cleanest visible proof that reflection off a denser medium flips the phase.

Worked Examples

Example 5 · Resultant intensity at a given path difference

Two equal coherent sources (\(I_1=I_2=I_0\)) overlap at a point where the path difference is \(\lambda/3\). Find the resulting intensity in units of \(I_0\).

Solution. Given: \(I_1=I_2=I_0\), path difference \(\Delta=\lambda/3\). Find: \(I\).
Setup: phase difference \(\delta = \dfrac{2\pi}{\lambda}\Delta\); equal sources give \(I = 4I_0\cos^2(\delta/2)\).
Solve: \[ \delta = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{3} = \frac{2\pi}{3}, \qquad I = 4I_0\cos^2\!\frac{\delta}{2} = 4I_0\cos^2\!\frac{\pi}{3} = 4I_0\left(\tfrac12\right)^2 = I_0 . \] Answer: \(\boxed{I = I_0}\).
Check: the limits bracket the answer—\(\Delta=0\) gives \(4I_0\) (full constructive), \(\Delta=\lambda/2\) gives \(0\) (full destructive)—so a path difference of \(\lambda/3\) landing at \(I_0\) sits sensibly between them, well inside the \(0\)-to-\(4I_0\) range. ✓

Example 6 · Coherence length of a spectral line

A source emits light of mean wavelength \(600\,\text{nm}\) with a spectral bandwidth \(\Delta\lambda = 1.0\,\text{nm}\). Estimate its coherence length and coherence time.

Solution. Given: \(\lambda = 600\,\text{nm}\), \(\Delta\lambda = 1.0\,\text{nm}\). Find: coherence length \(L_c\) and coherence time \(\tau_c\).
Setup: a wavetrain stays in step only over \(L_c \approx \lambda^2/\Delta\lambda\); the coherence time is \(\tau_c = L_c/c\).
Solve: \[ L_c = \frac{\lambda^2}{\Delta\lambda} = \frac{(600\times10^{-9})^2}{1.0\times10^{-9}} = 3.6\times10^{-4}\,\text{m} = 0.36\,\text{mm}, \] \[ \tau_c = \frac{L_c}{c} = \frac{3.6\times10^{-4}}{3.0\times10^{8}} = 1.2\times10^{-12}\,\text{s} = 1.2\,\text{ps}. \] Answer: \(\boxed{L_c \approx 0.36\,\text{mm},\ \ \tau_c \approx 1.2\,\text{ps}}\).
Check: interference washes out once the path difference between the two beams exceeds \(L_c\); a narrower spectral line (smaller \(\Delta\lambda\)) gives a longer \(L_c\), which is why a laser's near-single wavelength coheres over metres while a white-light source coheres over barely a wavelength. ✓

Example 7 · Fringe contrast from unequal sources

Two coherent sources have intensities in the ratio \(1:9\). Find the ratio of maximum to minimum intensity in their interference pattern.

Solution. Given: \(I_1:I_2 = 1:9\). Find: \(I_{\max}:I_{\min}\).
Setup: intensities combine through amplitudes (\(a\propto\sqrt{I}\)); \(I_{\max}=(a_1+a_2)^2\), \(I_{\min}=(a_1-a_2)^2\).
Solve: amplitudes are in the ratio \(\sqrt1:\sqrt9 = 1:3\), so \[ \frac{I_{\max}}{I_{\min}} = \frac{(3+1)^2}{(3-1)^2} = \frac{16}{4} = 4 . \] Answer: \(\boxed{I_{\max}:I_{\min} = 4:1}\).
Check: the minimum is not zero because the amplitudes are unequal (\(1\ne3\)); only equal sources give a perfectly black fringe. Add amplitudes, not intensities—the whole point of interference. ✓

✓ Checkpoint · Coherence and Superposition

Two equal coherent sources meet with phase difference \(\delta=\pi/2\). Find \(I/I_{\max}\).
\(I/I_{\max}=\cos^2(\delta/2)=\cos^2(\pi/4)=1/2\).
Why do the two headlights of a car never produce interference fringes on a wall?
They are independent sources with a randomly fluctuating phase difference (incoherent); the interference term time-averages to zero.
Two coherent sources have intensities \(I\) and \(4I\). Find \(I_{\max}:I_{\min}\).
Amplitudes \(1:2\), so \(I_{\max}:I_{\min}=(2+1)^2:(2-1)^2=9:1\).

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Fringe width from the apparatus

In a double-slit experiment the slits are \(0.5\,\text{mm}\) apart and the screen is \(1.0\,\text{m}\) away. Light of wavelength \(600\,\text{nm}\) is used. Find the fringe width.

Solution. Given: \(d=0.5\,\text{mm}=5\times10^{-4}\,\text{m}\), \(D=1.0\,\text{m}\), \(\lambda=600\,\text{nm}=6\times10^{-7}\,\text{m}\). Find: fringe width \(\beta\). Setup: the spacing of consecutive bright (or dark) fringes is \(\beta=\lambda D/d\). Solve: \[ \beta=\frac{\lambda D}{d}=\frac{(6\times10^{-7})(1.0)}{5\times10^{-4}}=1.2\times10^{-3}\,\text{m}. \] Answer: \(\boxed{\beta=1.2\,\text{mm}}\). Check. \(\beta=\lambda D/d\) scales with \(\lambda\) and \(D\), inversely with \(d\); the equal angular spacing \(\lambda/d=1.2\times10^{-3}\,\)rad confirms it, and halving \(d\) would double \(\beta\). ✓

Example 2 · Wavelength from a measured pattern

In a Young's experiment with slits \(1.0\,\text{mm}\) apart and a screen \(1.0\,\text{m}\) away, the centres of the first and the eleventh bright fringes are \(5.0\,\text{mm}\) apart. Find the wavelength of the light.

Solution. Given: \(d=1.0\,\text{mm}\), \(D=1.0\,\text{m}\); ten fringe widths span \(5.0\,\text{mm}\). Find: \(\lambda\). Setup: between fringe \(1\) and fringe \(11\) there are \(10\) gaps, so \(10\beta=5.0\,\text{mm}\Rightarrow\beta=0.5\,\text{mm}\); then \(\lambda=\beta d/D\). Solve: \[ \lambda=\frac{\beta d}{D}=\frac{(5\times10^{-4})(1\times10^{-3})}{1.0}=5\times10^{-7}\,\text{m}. \] Answer: \(\boxed{\lambda=500\,\text{nm}}\) (green). Check. Forward: \(\beta=\lambda D/d=(5\times10^{-7})(1.0)/(1\times10^{-3})=0.5\,\text{mm}\), so ten gaps span \(5.0\,\text{mm}\) ✓—counting many fringe widths, not one, is what makes the measurement precise. ✓

Example 3 · Positions of bright and dark fringes

Light of wavelength \(500\,\text{nm}\) falls on two slits \(1.0\,\text{mm}\) apart; the screen is \(1.0\,\text{m}\) away. Find the distance from the centre to (a) the third bright fringe and (b) the third dark fringe.

Solution. Given: \(\lambda=500\,\text{nm}\), \(d=1.0\,\text{mm}\), \(D=1.0\,\text{m}\); so \(\lambda D/d=5\times10^{-4}\,\text{m}=0.5\,\text{mm}\). Find: \(y_3^{\text{bright}}\), \(y_3^{\text{dark}}\). Solve: \[ y_3^{\text{bright}}=\frac{3\lambda D}{d}=3(0.5\,\text{mm})=1.5\,\text{mm}, \qquad y_3^{\text{dark}}=\frac{(3-\tfrac12)\lambda D}{d}=2.5(0.5\,\text{mm})=1.25\,\text{mm}. \] Answer: \(\boxed{y_3^{\text{bright}}=1.5\,\text{mm},\quad y_3^{\text{dark}}=1.25\,\text{mm}}\). Check. The third dark fringe (\(1.25\,\text{mm}\)) sits exactly midway between the second bright (\(1.0\,\text{mm}\)) and third bright (\(1.5\,\text{mm}\))—dark fringes always bisect adjacent bright ones. ✓

Example 4 · Pattern immersed in water

A double-slit pattern has fringe width \(1.2\,\text{mm}\) in air. The entire apparatus is now submerged in water (\(\mu=4/3\)). Find the new fringe width.

Solution. Given: \(\beta_{\text{air}}=1.2\,\text{mm}\), \(\mu=4/3\). Setup: in a medium the wavelength shrinks to \(\lambda/\mu\), and \(\beta\propto\lambda\), so \(\beta_{\text{water}}=\beta_{\text{air}}/\mu\). Solve: \[ \beta_{\text{water}}=\frac{1.2\,\text{mm}}{4/3}=0.9\,\text{mm}. \] Answer: \(\boxed{\beta_{\text{water}}=0.9\,\text{mm}}\) (the pattern contracts). Check. \(\beta\propto\lambda\propto1/\mu\), so \(1.2/(4/3)=0.9\,\text{mm}\) ✓; the wavelength shrinks by \(\mu\) inside the water and the whole pattern shrinks with it. ✓

Example 5 · Maximum-to-minimum intensity ratio

Two coherent sources have intensities in the ratio \(4:1\). Find the ratio of maximum to minimum intensity in their interference pattern.

Solution. Given: \(I_1:I_2=4:1\), so amplitudes \(a_1:a_2=2:1\). Setup: \(I_{\max}=(a_1+a_2)^2\), \(I_{\min}=(a_1-a_2)^2\). Solve: \[ \frac{I_{\max}}{I_{\min}}=\frac{(a_1+a_2)^2}{(a_1-a_2)^2}=\frac{(2+1)^2}{(2-1)^2}=\frac{9}{1}. \] Answer: \(\boxed{I_{\max}:I_{\min}=9:1}\). Check. Amplitudes \(2:1\) give \((2+1)^2:(2-1)^2=9:1\) ✓; the minimum is non-zero precisely because the sources are unequal—equal sources (\(a_1=a_2\)) would give a perfectly black fringe. ✓

Wave Optics

Perplexing Questions

Huygens' Principle: Light as a Marching Wavefront

Worked Examples

Reflection and Refraction from Wavefronts

Worked Examples

Superposition, Interference, and Coherence

Making Two Coherent Sources: Biprism and Lloyd's Mirror

Worked Examples

Solved examples

Problem bank

Chapter test

The chapter continues.