Coherent sources and interference

Two stones in a pond

Drop two stones into still water at the same moment, close together. The rings spread and cross. In some places the water moves up and down with a large height. In other places it hardly moves. The two sets of ripples have added up in the first case and cancelled in the second. The idea is the principle of superposition: when two waves meet, the displacement at any point is the sum of the displacements that each wave would give alone.

The picture shows the result for two sources that vibrate together. The bright bands are places where the waves arrive in step. The dark bands are places where they arrive half a cycle out of step.

Adding two waves

At a point, the first wave gives the displacement $y_1 = a\cos\omega t$. The second wave has the same frequency and the same amplitude $a$, but it is ahead by a phase $\varphi$: $y_2 = a\cos(\omega t + \varphi)$. The sum is

This is a wave of the same frequency with amplitude $2a\cos(\varphi/2)$. The intensity of a wave goes as the square of its amplitude. If one wave alone has intensity $I_0$, then

• $\varphi = 0, 2\pi, 4\pi, \dots$ gives $I = 4I_0$. This is constructive interference.
• $\varphi = \pi, 3\pi, \dots$ gives $I = 0$. This is destructive interference.

Two waves of different amplitudes $a_1$ and $a_2$ give the resultant intensity

The greatest value is $I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and the least is $I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$. The ratio $\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{a_1 + a_2}{a_1 - a_2}\right)^2$.

Set the amplitude ratio to $1.0$. How dark is the darkest place? It is zero. Now bring the ratio down to $0.5$. The dark places lose their zero, since a weaker wave cannot cancel a stronger one.

Phase difference and path difference

The phase of a wave goes round by $2\pi$ when the path grows by one wavelength. Two waves from two sources reach a point $P$ by paths that differ by $\Delta$. Then the phase difference is

• Bright (constructive): $\Delta = n\lambda$, with $n = 0, 1, 2, \dots$
• Dark (destructive): $\Delta = \left(n + \tfrac{1}{2}\right)\lambda$

Energy is not lost. The average intensity over a whole pattern is $I_1 + I_2$. The waves only move the energy from the dark places to the bright ones.

Coherent sources

Why do two torches not make fringes on a wall? A light bulb sends light from a huge number of atoms. Each atom sends a short train of waves for about $10^{-8}$ s and then starts a new train with a new phase. So the phase difference between two separate bulbs changes at random, many millions of times in each second. At any point, $\cos\varphi$ then averages to zero, and the intensities simply add: $I = I_1 + I_2$. We see an even brightness.

Two sources are coherent when the phase difference between their waves stays fixed in time. Only then is there a steady pattern. Both waves must also have the same frequency.

There are two ways to get coherent light from an ordinary source:

• Divide the wavefront. Let one wavefront fall on two narrow slits (Young's double slit). The two slits act as two sources that always have the same phase.
• Divide the amplitude. Let a wave reflect partly and pass partly at a thin film. The two parts come from the same wave (course three of this chapter).

A laser gives light that is coherent over a long distance, so two laser beams can interfere with no special trick.

A worked example. Two coherent waves have amplitudes $3$ and $1$ in arbitrary units. Find $I_{\max}/I_{\min}$ and the intensity ratio of the two waves.

1. The intensities. They go as the squares of the amplitudes: $I_1 : I_2 = 9 : 1$.
2. The brightest and the darkest. $I_{\max} \propto (3 + 1)^2 = 16$ and $I_{\min} \propto (3 - 1)^2 = 4$.
3. The ratio. $\dfrac{I_{\max}}{I_{\min}} = \dfrac{16}{4} = 4$.
4. Check. The fringes are clear, but not perfect. The visibility $\dfrac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \dfrac{16 - 4}{16 + 4} = 0.60$.

Practise with new numbers

Try it in Python

The program adds two waves for one whole cycle of the phase and looks at the brightest, the darkest and the average intensity.

Common mistakes

• Adding intensities when the waves are coherent. For coherent waves, add the amplitudes first (with the phase), then square. For incoherent light, add the intensities.
• Using $\cos\varphi$ for equal waves. For equal waves $I = 4I_0\cos^2(\varphi/2)$, with the half angle.
• Thinking that dark places "destroy" energy. The energy is moved to the bright places. The average stays $I_1 + I_2$.
• Thinking that two separate lamps are coherent. They are not, since the atoms of each lamp send out waves with random phases.

Remember these

• Superposition: the displacements add.
• $\varphi = \dfrac{2\pi}{\lambda}\Delta$. Bright: $\Delta = n\lambda$. Dark: $\Delta = (n + \tfrac12)\lambda$.
• Equal waves: $I = 4I_0\cos^2(\varphi/2)$. In general: $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\varphi$.
• Coherent sources have a fixed phase difference. Get them by dividing the wavefront or the amplitude of one wave.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Two coherent waves of intensity $I_0$ each meet at a point. What is the intensity if they arrive (a) in phase, (b) half a cycle apart?

2. Why do the beams of two torches not make fringes on a wall?

3. Two waves have amplitudes in the ratio $1 : 2$. What is $I_{\max}/I_{\min}$?

4. A path difference of one and a half wavelengths is a bright place or a dark place?

References

• OpenStax, University Physics Volume 3, Chapter 3 (Interference): the sum of two waves and coherence.
• NCERT Physics Class 12, Chapter 10 (Wave Optics): coherent and incoherent addition of waves.