10.2 Coherent Sources, Interference and Young's Double-Slit Experiment

A soap bubble shimmers with bands of colour, a thin film of oil on a wet road shows rainbow swirls, and the blue of a peacock's feather or a morpho butterfly's wing contains no blue pigment at all. Each of these colours is made by light waves adding — reinforcing each other for some colours and cancelling for others. Light added to light can give darkness. That sounds impossible if light is a stream of particles, and it is exactly what the wave picture of Section 10.1 predicts.

In 1801 Thomas Young, a London physician, made the first clean demonstration. He let sunlight through a pinhole, then through two pinholes (later two slits) very close together, and saw on a screen a row of bright and dark bands — interference fringes. From their spacing he measured the wavelength of light, about half a micrometre, for the first time. In this section we find out when two light waves can interfere (they must be coherent), how their intensities combine, and where the fringes of Young's double-slit experiment lie. Today the same idea measures the thickness of thin films to a fraction of a wavelength, tests the flatness of telescope mirrors, and — in kilometre-long interferometers — detects gravitational waves from colliding black holes.

Coherent and Incoherent Addition of Waves

Let two sources send light waves of the same frequency to a point $P$. By the principle of superposition (Class 11 Physics, Chapter 14 — Waves) the electric field at $P$ is the sum of the two fields. If the two oscillations at $P$ are $$y_1=a_1\cos\omega t,\qquad y_2=a_2\cos(\omega t+\phi),$$ the sum is another oscillation of the same frequency. Its amplitude $A$ follows from adding the two amplitudes as phasors — arrows of lengths $a_1$ and $a_2$ with an angle $\phi$ between them, exactly as in the AC phasor diagrams of Chapter 7. By the cosine rule, $$A^2=a_1^2+a_2^2+2a_1a_2\cos\phi.$$ The intensity of light is proportional to the square of its amplitude, $I\propto A^2$, so with $I_1\propto a_1^2$ and $I_2\propto a_2^2$ we have the central result of this section.

The term $2\sqrt{I_1I_2}\cos\phi$ is the interference term. Whether we can see it depends on how $\phi$ behaves in time.

Coherent sources. If the phase difference $\phi$ at each point stays constant, the intensity at each point stays fixed: some places are permanently bright, others permanently dark. The pattern can be seen and photographed.

Incoherent sources. An ordinary lamp emits light from billions of atoms, each sending out a short burst of waves lasting roughly $10^{-9}$–$10^{-8}$ s, with a phase that has nothing to do with the bursts before it. Light from two separate lamps (or two different parts of one lamp) therefore has a phase difference that jumps about at random millions of times a second. The eye — or any detector — sees only the average, and the average of $\cos\phi$ over all angles is zero: $$\overline{I}=I_1+I_2.$$ The intensities simply add; no fringes appear. That is why two car headlights lighting the same wall never produce dark bands.

Two independent light sources are never coherent. The trick — Young's trick — is to take one source and split its light into two parts, so that whatever random phase jumps the source makes, both parts make them together. A laser, whose atoms emit in step, makes the job much easier, but the principle is the same.

Where does the energy go? At a dark fringe the intensity is zero, and at a bright fringe of two equal waves it is $4I_0$, twice the $2I_0$ we would get without interference. Averaged over a whole fringe, $\cos^2(\phi/2)$ averages to $\tfrac12$, so the average intensity is $4I_0\times\tfrac12=2I_0$. Interference does not create or destroy energy; it only redistributes it, taking it from the dark fringes and putting it into the bright ones.

Path difference and phase difference. Suppose the two sources oscillate in phase. A wave that travels a distance $x$ falls behind in phase by $\dfrac{2\pi}{\lambda}x$ (Section 10.1), so if the paths from the two sources to $P$ differ by $\Delta=S_2P-S_1P$, the phase difference at $P$ is $$\phi=\frac{2\pi}{\lambda}\,\Delta .$$

Young's Double-Slit Experiment

In the modern version of Young's experiment, light from a narrow source slit $S$ (or a laser) falls on two very narrow, parallel slits $S_1$ and $S_2$, a distance $d$ apart (typically a fraction of a millimetre), placed symmetrically about the axis. Both slits are lit by the same wavefront from $S$, so they act as two coherent sources oscillating in phase. A screen stands a distance $D$ away (typically a metre or more), so $D\gg d$. On the screen appears a row of equally spaced bright and dark fringes parallel to the slits.

The path difference. Let $P$ be a point on the screen at height $x$ above the centre $O$. By Pythagoras, $$(S_2P)^2=D^2+\left(x+\tfrac d2\right)^2,\qquad (S_1P)^2=D^2+\left(x-\tfrac d2\right)^2,$$ so $(S_2P)^2-(S_1P)^2=2xd$. Factorising the left side, $$S_2P-S_1P=\frac{2xd}{S_2P+S_1P}.$$ For points near the centre ($x$ and $d$ much smaller than $D$), $S_2P+S_1P\approx2D$, so the path difference is $$\Delta=S_2P-S_1P\approx\frac{xd}{D}.$$ Equivalently, the two rays are almost parallel, the path difference is $d\sin\theta$, and for small angles $\sin\theta\approx\tan\theta=x/D$.

The fringes. Bright fringes appear where $\Delta=n\lambda$, that is where $\dfrac{x d}{D}=n\lambda$; dark fringes where $\dfrac{xd}{D}=\left(n+\tfrac12\right)\lambda$.

The formula $\beta=\lambda D/d$ explains why Young needed slits so close together: with $\lambda\approx5\times10^{-7}$ m and $D=1$ m, a separation $d=1$ mm gives $\beta=0.5$ mm, just visible; $d=1$ cm would give fringes $0.05$ mm apart, too fine to see. Reading the formula backwards, $\lambda=\dfrac{\beta d}{D}$: measure three lengths with a ruler and a travelling microscope and you have measured the wavelength of light.

The fringe width grows with $\lambda$ and $D$ and shrinks as the slits move apart. If the whole apparatus is immersed in a liquid of refractive index $n$, the wavelength becomes $\lambda/n$, so $\beta$ shrinks to $\beta/n$.

The Intensity Pattern and White-Light Fringes

Result 10.5 tells us more than where the brightest and darkest points are: it gives the brightness everywhere. At height $x$ the phase difference is $\phi=\dfrac{2\pi}{\lambda}\cdot\dfrac{xd}{D}=\dfrac{2\pi x}{\beta}$, so for two slits giving equal intensities $I_0$:

(The slits have a small but finite width, so in a real experiment the fringes slowly fade far from the centre; Section 10.3 explains why.)

White light. Each colour makes its own fringe system with its own $\beta=\lambda D/d$. At the centre every colour has zero path difference, so all are bright together and the central fringe is white. Next to it each colour's first dark fringe lies at $\tfrac12\beta$, nearest for violet and furthest for red, so the first dark band is bordered with colour — bluish on the inside, reddish on the outside. After a few fringes the colour systems overlap so thoroughly that the screen looks uniformly white again. Only a handful of coloured fringes are visible. The white central fringe is precisely what experimenters use to identify the point of zero path difference.

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