Superposition

The Principle of Superposition

• The principle of superposition states that:
When two or more waves meet, the resultant displacement is the vector sum of the
displacements of the individual waves
• This principle describes how waves which meet at a point in space interact
• When two waves with the same frequency and amplitude arrive at a point, they superpose
either:
o in phase, causing constructive interference. The peaks and troughs line up on
both waves. The resultant wave has double the amplitude
o or, in anti-phase, causing destructive interference. The peaks on one wave line
up with the troughs of the other. The resultant wave has no amplitude
Constructive and destructive interference

Waves in superposition can undergo constructive or destructive interference

 • The principle of superposition applies to all types of waves i.e. transverse and
longitudinal
Worked example
Two overlapping waves of the same types travel in the same direction. The variation with x and y
displacement of the wave is shown in the figure below.

Use the principle of superposition to sketch the resultant wave.

Answer:
• The graph of the superposition of both waves is in black:

• To plot the correct amplitude at each point, sum the amplitude of both graphs at that
point
o E.g. A point A, each graph has a value of 0.7. Therefore, the same point with the
resultant superposition is 2 × 0.7 = 1.4
 Stationary Waves
• Stationary waves, or standing waves, are produced by the superposition of two waves of
the same frequency and amplitude travelling in opposite directions
• This is usually achieved by a travelling wave and its reflection. The superposition
produces a wave pattern where the peaks and troughs do not move
Formation of a stationary wave

Formation of a stationary wave on a stretched spring fixed at one end

Stretched strings
• Vibrations caused by stationary waves on a stretched string produce sound
o This is how stringed instruments, such as guitars or violins, work
• This can be demonstrated by a length of string under tension fixed at one end and
vibrations made by an oscillator:
Standing wave experiment
 Stationary wave on a stretched string kept taut by a mass and pulley system

• As the frequency of the oscillator changes, standing waves with different numbers of
minima (nodes) and maxima (antinodes) form
Microwaves
• A microwave source is placed in line with a reflecting plate and a small detector between
the two
• The reflector can be moved to and from the source to vary the stationary wave pattern
formed
• By moving the detector, it can pick up the minima (nodes) and maxima (antinodes) of the
stationary wave pattern

Using microwaves to demonstrate stationary waves

Air Columns
• The formation of stationary waves inside an air column can be produced by sound waves
o This is how musical instruments, such as clarinets and organs, work
• This can be demonstrated by placing a fine powder inside the air column and a
loudspeaker at the open end
• At certain frequencies, the powder forms evenly spaced heaps along the tube, showing
where there is zero disturbance as a result of the nodes of the stationary wave
Stationary waves in an air column

Stationary waves can be seen in air columns using dry power

• In order to produce a stationary wave, there must be a minima (node) at one end and a
maxima (antinode) at the end with the loudspeaker
Exam Tip
Always refer back to the experiment or scenario in an exam question e.g. the wave produced by
a loudspeaker reflects at the end of a tube. This reflected wave, with the same frequency,
overlaps the initial wave to create a stationary wave.
Formation of Stationary Waves
• A stationary wave is made up of nodes and antinodes
o Nodes are where there is no vibration
o Antinodes are where the vibrations are at their maximum amplitude
• The nodes and antinodes do not move along the string. Nodes are fixed and antinodes
only move in the vertical direction
 • Between nodes, all points on the stationary are in phase
• The image below shows the nodes and antinodes on a snapshot of a stationary wave at a
point in time
Nodes and antinodes on a stationary wave

Nodes are points of zero amplitude, anti-nodes are points of maximum amplitude
• L is the length of the string
• 1 wavelength λ is only a portion of the length of the string
Worked example
A stretched string is used to demonstrate a stationary wave, as shown in the diagram.

Which row in the table correctly describes the length of L and the name of X and Y?

Length L Point X Point Y

A 5 wavelengths Node Antinode

B Antinode Node
2 wavelengths

C Node Antinode
2 wavelengths

D 5 wavelengths Antinode Node

Answer: C
Step 1: Determine the number of wavelengths in the length of the string

• The string has 2 wavelengths

o This rules out A and D
Step 2: Determine points X and Y
• X is a point of 0 displacement - a node
• Y is a point of maximum displacement - an antinode
o Therefore, the correct row is C

Measuring Wavelength
• Stationary waves have different wave patterns depending on the frequency of the
vibration and the situation in which they are created

Two fixed ends

• When a stationary wave, such as a vibrating string, is fixed at both ends, the simplest
wave pattern is a single loop made up of two nodes and an antinode
• This is called the fundamental mode of vibration or the first harmonic
• The particular frequencies (i.e. resonant frequencies) of standing waves possible in the
string depend on its length L and its speed v
• As you increase the frequency, the higher harmonics begin to appear
• The frequencies can be calculated from the string length and wave equation
Wavelength and frequencies of different harmonics
 Diagram showing the first three modes of vibration of a stretched string with corresponding
frequencies

• The nth harmonic has n antinodes and n + 1 nodes

One or two open ends in an air column
• When a stationary wave is formed in an air column with one or two open ends, we see
slightly different wave patterns in each
Harmonics in an air column
Diagram showing modes of vibration in pipes with one end closed and the other open or both
ends open
• Image 1 shows stationary waves in a column which is closed at one end
o At the closed end, a node forms
o At the open end, an antinode forms
• Therefore, the fundamental mode is made up of a quarter wavelength with one node and
one antinode
o Every harmonic after that adds on an extra node or antinode
o Hence, only odd harmonics form
• Image 2 shows stationary waves in a column which is open at both ends
o An antinode forms at each open end
• Therefore, the fundamental mode is made up of a half wavelength with one node and two
antinodes
o Every harmonic after that adds on an extra node and an antinode
o Hence, odd and even harmonics can form
• In summary, a column length L for a wave with wavelength λ and resonant
frequency f for stationary waves to appear is as follows:
Air Column Length & Frequencies Summary Table
Worked example
A standing wave is set up in a column of length L when a loudspeaker placed at one end emits a
sound wave of frequency f. The column is closed at the other end. The speed of sound is 340 m
s−1.

For a column of length 7.5 m, what is the frequency of the second lowest note produced?
Answer:
Step 1: Determine the positions of the nodes and antinodes
• One end of the column is closed, and the loudspeaker represents an open end
• Hence, an antinode forms at the loudspeaker (open end) and a node forms at the closed
end
• The fundamental frequency represents the lowest note - this would be 1 node and 1
antinode
• So, the second-lowest note must have 2 nodes and 2 antinodes
Step 2: Write an expression for the length of the sound wave in the column
• In the column, there is a quarter wavelength and a half wavelength, or

• Therefore the length of the column is:

• Note: for a column with an open and closed end, , this would represent the
third harmonic (n = 3)
Step 3: Determine the wavelength of the second lowest note

Step 4: Calculate the frequency using the wave equation

• Note: you could combine steps 3 and 4 by using the expression

Exam Tip
The fundamental counts as the first harmonic or n = 1 and is the lowest frequency with half or
quarter of a wavelength. A full wavelength with both ends open or both ends closed is
the second harmonic. Make sure to match the correct wavelength with the harmonic asked for in
the question!
What is Diffraction?
• Diffraction is the spreading out of waves when they pass an obstruction
o This obstruction is typically a narrow slit (an aperture)
• The extent of diffraction depends on the width of the gap compared with the wavelength
of the waves
o Diffraction is the most prominent when the width of the slit is approximately
equal to the wavelength
Diffraction through a gap

Diffraction: when a wave passes through a narrow gap, it spreads out

• Diffraction is usually represented by a wavefront as shown by the vertical lines in the

diagram above
• The only property of a wave that changes when its diffracted is its amplitude
o This is because some energy is dissipated when a wave is diffracted through a gap
• Diffraction can also occur when waves curve around an edge:
Diffraction by a barrier
 When a wave goes past the edge of a barrier, the waves can curve around it
• Any type of wave can be diffracted i.e. sound, light, water
Worked example
When a wave is travelling through air, which scenario best demonstrates diffraction?
A. UV radiation through a gate post
B. Sound waves passing a steel rod
C. Radio waves passing between human hair
D. X-rays passing through atoms in a crystalline solid
Answer: D
• Diffraction is most prominent when the wavelength is close to the aperture size
• UV waves have a wavelength between 4 × 10-7 – 1 × 10-8 m so won’t be diffracted by a
gate post
• Sound waves have a wavelength of 1.72 × 10-2 – 17 m so would not be diffracted by the
diffraction grating
• Radio waves have a wavelength of 0.1 – 106 m so would not be diffracted by human hair
• X-rays have a wavelength of 1 × 10-8 – 4 × 10-13 m which is roughly the gap between
atoms in a crystalline solid
o Therefore, the correct answer is D
Exam Tip
When drawing diffracted waves, take care to keep the wavelength constant. It is only the
amplitude of the wave that changes when diffracted.
Diffraction Experiments
 • As discussed above, the effects of diffraction are most prominent when the gap size is
approximately the same or smaller than the wavelength of the wave
• As the gap size increases, the effect gradually gets less pronounced until, in the case that
the gap is much larger than the wavelength, the waves are no longer spread out
Size of the gap and diffraction

The size of the gap (compared to the wavelength) affects how much the waves spread out
• Ripple tanks are used a common experiment to demonstrate diffraction of water waves
Diffraction in a ripple tank

Wave effects may all be demonstrated using a ripple tank

 • The diagram below shows how the wavelengths differ with frequency in a ripple tank
o The higher the frequency, the shorter the wavelength
o The lower the frequency, the longer the wavelength
High and low frequency in the ripple tank

Ripple tank patterns for low and high frequency vibration

Exam Tip
Familiarising yourself with the wavelength of electromagnetic waves is essential for identifying
which wave will cause the greatest diffraction effect for a giving gap width.
 Interference & Coherence
• Interference occurs when waves overlap and their resultant displacement is the sum of
the displacement of each wave
• This result is based on the principle of superposition and the resultant waves may be
smaller or larger than either of the two individual waves
• Interference of two waves can either be:
o In phase, causing constructive interference. The peaks and troughs line up on
both waves. The resultant wave has double the amplitude
o In anti-phase, causing destructive interference. The peaks on one wave line up
with the troughs of the other. The resultant wave has no amplitude
Constructive and destructive interference

Waves in superposition can undergo constructive or destructive interference

 • At points where the two waves are neither in phase nor in antiphase, the resultant
amplitude is somewhere in between the two extremes
• Waves are coherent if they have the same frequency and constant phase difference
Coherent vs. incoherent waves

Coherent v non-coherent wave. The abrupt change in phase creates an inconsistent phase
difference

• Coherence is vital in order to produce an observable interference pattern

• Laser light is an example of a coherent light source, whereas filament lamps produce
incoherent light waves
Worked example
The diagram shows the interferences of coherent waves from two point sources.

Which row in the table correctly identifies the type of interference at points X, Y and Z.

X Y Z

A Constructive Destructive Constructive

B Constructive Constructive Destructive

C Destructive Constructive Destructive

D Destructive Constructive Constructive

Answer: B
• At point X:
o Both peaks of the waves are overlapping. This is constructive interference and
rules out options C and D
• At point Y:
o Both troughs are overlapping so constructive interference occurs there
• At point Z:
o A peak of one of the waves meets the trough of the other. This
is destructive interference (Row B)
Exam Tip
Think of ‘constructive’ interference as ‘building’ the wave and ‘destructive’ interference as
‘destroying’ the wave.
 Demonstrating Two Source Interference
• Interference of sound, light and microwaves can be demonstrated with slits or diffraction
gratings

Using Water Waves

• Two-source interference in can be demonstrated in water using ripple tanks
• The diagram below shows diffracted circle shaped water waves from two point sources
eg. dropping two pebbles near to each other in a pond
Wavefront interference

Water waves interference pattern from a ripple tank

• The two waves interfere causing areas of constructive and destructive interference
• The lines of maximum displacement occur when all the peaks and troughs line up with
those on another wave

Using Sound Waves

• Two source interference for sound waves looks very similar to water waves
Sound wave interference
 Sound wave interference from two speakers

• Sound waves are longitudinal waves so are made up of compressions and rarefactions
• Constructive interference occurs when two compressions or two rarefactions line up and
the sound appears louder
• Destructive interference occurs when a compression lines up with a rarefaction and vice
versa. The sound is quieter
o This is the technology used in noise-cancelling headphones

Using Microwaves
• Two source interference for microwaves can be detected with a moveable microwave
detector
Microwave interference experiment
 Microwave interference experiment

• Constructive interference: regions where the detector picks up a maximum amplitude

• Destructive interference: regions where the detector picks up no signal

Using Light Waves

• For light rays, such as a laser light through two slits, an interference pattern forms on the
screen
Light interference experiment
 Laser light interference experiment

• Constructive interference is shown as bright fringes on the screen

o The highest intensity is in the middle
• Destructive interference is shown as the dark fringes on the screen
o These have zero intensity

Two Source Interference Fringes

• For two-source interference fringes to be observed, the sources of the wave must be:
o Coherent (constant phase difference)
o Monochromatic (single wavelength)
• When two waves interfere, the resultant wave depends on the phase difference between
the two waves
• This is proportional to the path difference between the waves which can be written in
terms of the wavelength λ of the wave
• As seen from the diagram, the wave from slit S2 has to travel slightly further than that
from S1 to reach the same point on the screen. The difference in distance is the path
difference
Path difference for constructive and destructive interference
 Path difference for constructive and destructive interference is determined by wavelength

• For constructive interference (or maxima), the difference in wavelengths will be

an integer number of whole wavelengths
• For destructive interference (or minima) it will be an integer number of whole
wavelengths plus a half wavelength
o n is the order of the maxima/minima since there is usually more than one of these
produced by the interference pattern
• An example of the orders of maxima is shown below:
Double slit interference pattern
Interference pattern of light waves shown with orders of maxima
• n = 0 is taken from the middle, n = 1 is one either side and so on
Worked example
Two coherent sources of sound waves S1 and S2 are situated 65 cm apart in air as shown below.

The two sources vibrate in phase but have different amplitudes of vibration. A microphone M is
situated 150 cm from S1 along the line normal to S1 and S2.
The microphone detects maxima and minima of the intensity of the sound. The wavelength of the
sound from S1 to S2 is decreased by increasing the frequency.
Determine which orders of maxima are detected at M as the wavelength is increased from 3.5 cm
to 12.5 cm.
Answer:
Step 1: Calculate the path difference
• Using pythagoras' theorem
Path difference =
Step 2: State the path difference for constructive interference
• Path difference = nλ
o n = 0, 1, 2, 3...
Step 3: Determine the wavelength
• 13 = nλ

• Only orders 2 and 3 are within the wavelength range of 3.5 cm to 12.5 cm
o Therefore 2 and 3 are the orders where the maxima are detected
Exam Tip
The path difference is more specifically how much longer, or shorter, one path is than the other.
In other words, the difference in the distances. Make sure not to confuse this with the distance
between the two paths.
Double Slit Interference
• Young’s double slit experiment demonstrates how light waves produced an interference
pattern
• The experiment is shown below
Young's double slit experiment
Young’s double-slit experiment arrangement creates an interference pattern when light is
passed through slits A and B
• When a monochromatic light source is placed behind a single slit, the light is diffracted
producing two light sources at the double slits A and B
• Since both light sources originate from the same primary source, they are coherent and
will therefore create an observable interference pattern
• Both diffracted light from the double slits create an interference pattern made up of bright
and dark fringes
• The wavelength of the light can be calculated from the interference pattern and
experiment set up
• These are related using the double-slit equation:

• Where:
o λ = wavelength of source (m)
 o a = distance between the centres of the slit (m)
o x = fringe width (distance between successive bright fringes) (m)
o D = distance between the slits and the screen (m)

Double slit interference equation with a, x and D represented on a diagram

• The interference pattern on a screen will show as ‘fringes’ which are dark or bright bands
• Constructive interference is shown through bright fringes with varying intensity (most
intense in the middle)
• Destructive interference is shown from dark fringes where no light is seen
• A monochromatic light source makes these fringes clearer and the distance between
fringes is very small due to the short wavelength of visible light
Worked example
A laser is placed in front of a double-slit as shown in the diagram below.

The laser emits light of frequency 750 THz. The separation of the maxima P and Q observed on
the screen is 15 mm. The distance between the double slit and the screen is 4.5 m.
Calculate the separation of the two slits.
Answer:
Step 1: Calculate the wavelength of the light

Step 2: Rearrange for λ and substitute in values

Step 3: State the double slit equation

Step 4: Rearrange for a and substitute in the values

• This is the separate between 9 bright fringes, so x needs to be divided by 9

Exam Tip
Since a, x and D are all distances, it's easy to mix up which they refer to. Labelling the double
slit diagram in the way given in the notes above will help to remember the order i.e. a and x in
the numerator and D underneath in the denominator.
 The Diffraction Grating Equation
• A diffraction grating is a plate on which there is a very large number of parallel, identical,
close-spaced slits
• When monochromatic light is incident on a grating, a pattern of narrow bright fringes is
produced on a screen
Diffraction grating

Diagram of diffraction grating used to obtain a fringe pattern

• The angles at which the maxima of intensity (constructive interference) are produced can
be deduced by the diffraction grating equation:

• Where:
o d = spacing between adjacent slits (m)
o θ = angular separation between the order of maxima (degrees)
o n = order of maxima (n = 0, 1, 2, 3...)
o λ = wavelength of light source (m)
• Exam questions sometimes state the lines per m (or per mm, per nm etc.) on the grating
which is represented by the symbol N
• d can be calculated from N using the equation

Angular Separation
 • The angular separation of each maxima is calculated by rearranging the grating equation
to make θ the subject
• The angle θ is taken from the centre meaning the higher orders are at greater angles
Angular separation

Angular separation depends on the order of maxima

• The angular separation between two angles is found by subtracting the smaller angle
from the larger one
• The angular separation between the first and second maxima n1 and n2 is θ2 – θ1
Orders of Maxima
• The maximum angle to see orders of maxima is when the beam is at right angles to the
diffraction grating
o This means θ = 90o and sin θ = 1
• The highest order of maxima visible is therefore calculated by the equation:

• Note that since n must be an integer, if the value is a decimal it must be rounded down
o E.g If n is calculated as 2.7 then n = 2 is the highest-order visible
Worked example
An experiment was set up to investigate light passing through a diffraction grating with a slit
spacing of 1.7 µm. The fringe pattern was observed on a screen. The wavelength of the light is
550 nm.
Calculate the angle α between the two second-order lines.
Answer:
Step 1: List the known quantities
• Order of maxima, n = 2
• Diffraction slit spacing, d = 1.7 µm = 1.7 × 10–6 m
• Wavelength, λ = 550 nm = 550 × 10–9 m
Step 3: Rearrange for θ and substitute in the values

Step 4: Calculate α
• θ is the angle from the centre to the second-order line (β on the diagram)

Exam Tip
Take care that the angle θ is the correct angle taken from the centre and not the angle taken
between two orders of maxima.
Determining the Wavelength of Light
Method
 • The wavelength of light can be determined by rearranging the grating equation to make
the wavelength λ the subject
• The value of θ, the angle to the specific order of maximum measured from the centre, can
be calculated through trigonometry
• The distance from the grating to the screen is marked as D
• The distance between the centre and the order of maxima (e.g. n = 2 in the diagram) on
the screen is labelled as h - the fringe spacing
• Measure both these values with a ruler
• This makes a right-angled triangle where angle θ can be described as the ratio

Order of maxima and wavelength

The wavelength of light is calculated by the angle to the order of maximum

• Remember to find the inverse of tan to find

• This value of θ can then be substituted back into the diffraction grating equation to find
the value of the wavelength (with the corresponding order n)
Improving the experiment and reducing uncertainties
• The fringe spacing can be subjective depending on its intensity on the screen. Take
multiple measurements of h (between 3-8) and find the average
• Use a Vernier scale to record h, in order to reduce percentage uncertainty
• Reduce the uncertainty in h by measuring across all fringes and dividing by the number
of fringes
• Increase the grating to screen distance D to increase the fringe separation (although this
may decrease the intensity of light reaching the screen)
• Conduct the experiment in a darkened room, so the fringes are clearer
• Use grating with more lines per mm, so values of h are greater to lower percentage
uncertainty