6.1 Superposition and interference — Learn

Prerequisites: SI prefixes and units from Section 1.1; speed as distance divided by time from Section 3.1; the sine and cosine of an angle from Section 2.2.

By the end of this section you can: describe a progressive wave; tell a transverse wave from a longitudinal wave; use $v = f\lambda$ and $T = \dfrac{1}{f}$ ; find a phase difference and a path difference; decide whether two waves reinforce or cancel; explain what makes two sources coherent; and use the fringe formula for two slits.

1. What a wave is

A progressive wave carries energy from one place to another without carrying matter with it. The particles of the material move about their own positions, and the disturbance travels through them. A wave on a rope moves along the rope, but each piece of rope stays near where it started.

The particles can move in two ways, and this gives two families of wave.

Type	How the particles move	Examples
Transverse	at right angles to the direction of travel	light, a wave on a string, a wave on water
Longitudinal	along the direction of travel	sound, a push along a spring

A longitudinal wave has compressions, where the particles are crowded together, and rarefactions, where they are spread out. A sound wave in air is longitudinal, so it has no direction of vibration that you can see from the side.

2. The quantities that describe a wave

Quantity	Symbol	Meaning	Unit
Displacement	$x$	distance and direction of a point from its rest position	m
Amplitude	$A$	the maximum displacement	m
Wavelength	$\lambda$	the distance between two neighbouring points in step	m
Period	$T$	the time for one complete wave to pass	s
Frequency	$f$	the number of waves passing a point each second	Hz

Frequency and period are reciprocals:

f = \frac{1}{T} \qquad T = \frac{1}{f}

One wavelength passes a point in one period, so the wave travels a distance $\lambda$ in a time $T$ . Its speed is therefore

v = \frac{\lambda}{T} = f\lambda

The symbol $c$ is used for the speed of light and of other electromagnetic waves. The equation is the same: $c = f\lambda$ .

Worked example 1. A sound wave has a frequency of 256 Hz. The speed of sound in air is $340\,\text{m}\,\text{s}^{-1}$ . Find its wavelength.

\begin{aligned} v &= f\lambda \\ \lambda &= \frac{v}{f} = \frac{340}{256} \\ &= 1.33\,\text{m} \end{aligned}

The wavelength is 1.33 m.

Worked example 2. Red light has a wavelength of 600 nm in a vacuum. The speed of light is $3.00 \times 10^{8}\,\text{m}\,\text{s}^{-1}$ . Find its frequency.

\begin{aligned} c &= f\lambda \\ f &= \frac{c}{\lambda} = \frac{3.00 \times 10^{8}}{600 \times 10^{-9}} \\ &= 5.00 \times 10^{14}\,\text{Hz} \end{aligned}

The frequency is $5.00 \times 10^{14}\,\text{Hz}$ . The frequency of a wave does not change when the wave enters a different material, but its speed and wavelength do.

3. Phase and phase difference

Points on a wave are in phase when they are doing the same thing at the same time. Two crests are in phase. Two points one wavelength apart are also in phase.

The phase difference tells you how far apart two points are in their cycle. It can be measured in degrees or in radians. One whole wavelength is $360^\circ$ or $2\pi$ radians, so

\Delta\phi = \frac{\Delta x}{\lambda} \times 360^\circ = \frac{\Delta x}{\lambda} \times 2\pi

where $\Delta x$ is the distance between the points.

Worked example 3. Two points on a wave are 0.150 m apart. The wavelength is 0.600 m. Find the phase difference between them.

\begin{aligned} \Delta\phi &= \frac{\Delta x}{\lambda} \times 360^\circ \\ &= \frac{0.150}{0.600} \times 360^\circ \\ &= 90^\circ \end{aligned}

The phase difference is $90^\circ$ , which is $\dfrac{\pi}{2}$ radians.

4. The principle of superposition

When two waves meet, they pass through each other. At every point the resultant displacement is the sum of the displacements the two waves would have on their own. This is the principle of superposition.

Read the rule carefully. It is the displacements that add, and displacement has a direction. A crest and a trough at the same place add to zero.

Constructive interference happens when the two waves arrive in phase. A crest meets a crest, and the resultant amplitude is larger.
Destructive interference happens when the two waves arrive antiphase. A crest meets a trough, and the resultant amplitude is smaller. If the two amplitudes are equal, the resultant is zero.

For two sources that started in step, the phase at a point depends on how far each wave has travelled. Call the difference in the distances the path difference $\Delta x$ .

Path difference	Phase difference	Result
$n\lambda$	$0^\circ$ , $360^\circ$ , …	constructive
$(n + \tfrac{1}{2})\lambda$	$180^\circ$ , $540^\circ$ , …	destructive

Here $n$ is a whole number: 0, 1, 2, ….

Worked example 4. Two loudspeakers are driven in phase at 500 Hz. The speed of sound is $340\,\text{m}\,\text{s}^{-1}$ . A point P is 2.38 m from one speaker and 1.70 m from the other. Decide whether the sound at P is loud or quiet.

\begin{aligned} \lambda &= \frac{v}{f} = \frac{340}{500} = 0.68\,\text{m} \\ \Delta x &= 2.38 - 1.70 = 0.68\,\text{m} \end{aligned}

The path difference is one wavelength, so the two waves arrive in phase. The sound at P is loud. A second point with a path difference of 0.34 m would be quiet, because 0.34 m is half a wavelength.

Watch out. A path difference is measured in metres, so compare it with the wavelength, which is also in metres. A common trap is to compare a path difference of 0.68 m with the frequency of 500 Hz. Those numbers cannot be compared. Divide the path difference by the wavelength to get a number of wavelengths: $\dfrac{0.68}{0.68} = 1$ , so the waves are in phase.

5. Coherence

Two sources give a steady interference pattern only if they are coherent. Coherent sources have the same frequency and a constant phase difference between them. Two loudspeakers driven by the same signal generator are coherent. Two separate lamps are not, because each emits light in short, unrelated bursts, so the pattern shifts many times a second and the eye sees only an average.

A laser is different. All the light from one laser is coherent, which is why a laser is used for a clear interference pattern.

6. Two slits and the fringe pattern

Light from two narrow slits is coherent, because the two slits are illuminated by the same source. The waves travel different distances to the screen, so the path difference varies across the screen. Bright fringes are the places where the path difference is a whole number of wavelengths.

For a slit separation $s$ and a screen distance $D$ much larger than $s$ , the bright fringes are equally spaced. The distance from one bright fringe to the next is the fringe spacing $w$ :

w = \frac{\lambda D}{s}

The same geometry applies to any two coherent sources, such as two loudspeakers. The bright fringes lie at angles given by

s\sin\theta = n\lambda

For a small angle, $\sin\theta \approx \theta$ in radians, and the two equations agree.

Worked example 5. A laser of wavelength 633 nm passes through two slits 0.25 mm apart. The screen is 2.50 m away. Find the fringe spacing.

\begin{aligned} w &= \frac{\lambda D}{s} \\ &= \frac{633 \times 10^{-9} \times 2.50}{0.25 \times 10^{-3}} \\ &= 6.33 \times 10^{-3}\,\text{m} \end{aligned}

The fringes are 6.33 mm apart. That is about the width of a fingernail.

Worked example 6. In a two-slit experiment the slits are 0.50 mm apart and the screen is 3.00 m away. The fringe spacing is 3.60 mm. Find the wavelength of the light.

\begin{aligned} w &= \frac{\lambda D}{s} \\ \lambda &= \frac{ws}{D} = \frac{3.60 \times 10^{-3} \times 0.50 \times 10^{-3}}{3.00} \\ &= 6.0 \times 10^{-7}\,\text{m} \end{aligned}

The wavelength is $6.0 \times 10^{-7}\,\text{m}$ , or 600 nm. This is how a two-slit experiment measures the wavelength of light.

7. Try it: two sources

The figure draws the interference pattern from two coherent sources. Three sliders set the wavelength, the distance between the sources, and the distance to the screen.

Predict first. Do not move the sliders yet. Write down your answers.

If you halve the wavelength, what happens to the spacing of the bright fringes?
If you move the sources closer together, what happens to the fringe spacing?
If you move the screen further away, what happens to the path difference at a fixed point on the screen?

Test. Set the wavelength to 600 nm, the source separation to 0.25 mm and the screen distance to 2.50 m. Watch the pattern and read the fringe spacing. Then change one slider at a time and check each prediction.

Explain. Why does making the sources closer together spread the fringes further apart? Write two or three sentences. Use the idea that the path difference must still be a whole number of wavelengths, but a smaller separation makes the waves travel more nearly equal distances.

Progress check

Return to the concert hall. The two loudspeakers are 1.20 m apart and are driven in phase at 1.00 kHz. The speed of sound is $340\,\text{m}\,\text{s}^{-1}$ .

Find the wavelength of the sound.
Find the angle at which the listener hears the first loud band to the side of the centre line. Use $s\sin\theta = n\lambda$ with $n = 1$ .
Explain why a listener on the centre line hears a loud sound.
Explain why the pattern would vanish if the two speakers were driven by separate signal generators set to nearly the same frequency.

If question 1 goes wrong, use $\lambda = \dfrac{v}{f}$ . If question 2 goes wrong, divide the wavelength by the separation before taking the inverse sine. If question 3 goes wrong, find the path difference on the centre line and compare it with the wavelength.

You should now be able to

describe a progressive wave and say what it carries;
tell a transverse wave from a longitudinal wave;
use $v = f\lambda$ to find a speed, a frequency or a wavelength;
find the period from a frequency, and a frequency from a period;
find the phase difference between two points on a wave;
state the principle of superposition;
use a path difference to decide whether two waves reinforce or cancel;
explain what makes two sources coherent;
use $w = \dfrac{\lambda D}{s}$ for the fringes from two slits;
find the angle of a bright fringe from $s\sin\theta = n\lambda$ .