Coherent Sources and the Intensity of Two Waves

Path difference and phase difference

The chapters on waves found the rule for two waves that start in step. If one wave travels a distance \(\Delta x\) farther than the other, it arrives behind in its cycle. A wave repeats after one wavelength, and its phase grows by \(2\pi\) over that distance. So a path difference \(\Delta x\) is \(\Delta x/\lambda\) of a cycle, and the phase difference is \(2\pi\Delta x/\lambda\).

For light the distances are tiny. With \(\lambda = 600\,\text{nm}\), a path difference of \(150\,\text{nm}\) is a quarter of a wavelength, and \(\Delta\phi = 2\pi \times 150/600 = \pi/2\). A path difference of \(300\,\text{nm}\) gives \(\pi\), and \(600\,\text{nm}\) gives \(2\pi\). A millionth of a metre of extra path already makes a large change of phase.

Two cases matter most. When the path difference is a whole number of wavelengths, the waves arrive in phase, crest on crest, and the light is brightest. This is constructive interference. When it is an odd number of half wavelengths, the waves arrive in opposite phase, crest on trough. For equal waves the light cancels completely. This is destructive interference.

The dark condition \((2n - 1)\lambda/2\) is the one the NCERT uses, with \(n\) counting the dark places from 1. The chapter on sound wrote the same thing as \((n + \tfrac12)\lambda\) with \(n\) from 0. Both give \(\lambda/2, 3\lambda/2, 5\lambda/2\), and so on. Path differences of \(\lambda/2\), \(\lambda\) and \(3\lambda/2\) therefore give phase differences \(\pi\), \(2\pi\) and \(3\pi\): dark, bright and dark again.

Why two torches never make fringes

A bright place needs the waves to arrive in phase, and a dark place needs them half a cycle apart. That pattern stays put only if the phase difference of the two sources stays fixed. Sources whose waves have the same frequency and a phase difference constant in time are coherent sources.

The light of a bulb or a torch comes from a huge number of atoms. Each atom sends out a short burst, a wave train that lasts about \(10^{-8}\,\text{s}\) or less. Its phase has nothing to do with the next burst. So the phase of a bulb's light jumps at random a hundred million times a second or faster. A second bulb jumps on its own, quite independently.

Two such sources do make bright and dark bands, but only for an instant. When the phase difference jumps, the bands jump to new places. The eye holds a picture for about a tenth of a second, so it averages about ten million different patterns. The average of all of them is an even brightness. Fig 3.1 shows the two cases side by side.

How, then, can two coherent sources be made? The answer is to take them from one source. If the light of one narrow slit is divided between two slits, any jump of phase at the source reaches both slits together. The two slits then keep a fixed phase difference; this is Young's method, the subject of the next lesson. Splitting one wavefront in this way is called division of wavefront. A thin film splits one wave into two by partial reflection instead, a division of amplitude; the next chapter treats it.

A laser is a far better source. Its atoms are made to emit in step, so a laser's wave train can be many centimetres long, even for a cheap laser pointer. That is why a laser pointer through two slits shows bands at once, while two torches never do.

How bright is a place in between?

At a point on the screen, the electric field of each wave swings up and down as a simple harmonic motion. Write the two as \(y_1 = a_1\sin\omega t\) and \(y_2 = a_2\sin(\omega t + \phi)\). Their sum is again a simple harmonic motion of the same frequency. Its amplitude \(A\) comes from adding the two as arrows, as the chapter on the superposition of oscillations did.

For two equal waves, \(I_1 = I_2 = I_0\), the formula becomes \(I\) \(= 2I_0(1 + \cos\phi)\) \(= 4I_0\cos^2(\phi/2)\). In phase, \(\phi = 0\), it gives \(4I_0\). At \(\phi = \pi/2\) it gives \(2I_0\), and at \(\phi = \pi\) it gives zero. At \(\phi = 60°\), \(I = 4I_0\cos^2 30° = 3I_0\), and the amplitude is \(2a\cos 30° = \sqrt3\,a\). At \(\phi = 2\pi/3\), \(I = 4I_0\cos^2 60° = I_0\).

Try the sliders of the figure above. With equal waves the curve runs from \(4\) at \(0°\) through \(2\) at \(90°\) to \(0\) at \(180°\). Now make the second wave half as strong. The curve swings only between \(2.25\) and \(0.25\): the dark places are no longer dark. Whatever the amplitudes, the curve is centred on the dashed mean.

The ratio can also be run backwards. Suppose a pattern has \(I_{\max}/I_{\min} = 25\). Then \((a_1 + a_2)/(a_1 - a_2) = 5\), so \(a_1 + a_2 = 5a_1 - 5a_2\), and \(4a_1 = 6a_2\). The amplitudes are in the ratio \(3 : 2\), and the intensities in the ratio \(9 : 4\). In the same way, intensities \(4I_0\) and \(I_0\) have amplitudes in the ratio \(2 : 1\). They give \(I_{\max} = (2 + 1)^2 I_0 = 9I_0\) and \(I_{\min} = (2 - 1)^2 I_0 = I_0\).

Energy is moved, not made

Interference makes some places four times as bright as one wave, and others dark. It does not create or destroy light. Across the pattern the phase difference takes every value equally often, and \(\cos\phi\) averages to zero. So the average intensity is \(\langle I\rangle = I_1 + I_2\), the sum of the two waves alone. The dark places lose exactly what the bright places gain.

Now return to the two torches. Each torch is a crowd of atoms firing short, independent bursts. At any instant their light does make bright and dark bands on the wall. The phase difference jumps many millions of times a second, and the bands jump with it. The eye averages them into an even patch, of brightness \(I_1 + I_2\). One laser split by two slits keeps a fixed phase difference, so its bands stand still.