Wave Properties

What you'll learn

Why waves diffract at slits and obstacles, and when the spreading is large.
How superposition gives interference patterns, including Young’s double-slit experiment.
How to use $λ=aΔyD\lambda = \frac{a\Delta y}{D}$ and $dsin⁡θ=nλd\sin\theta = n\lambda$ to determine wavelength.
How stationary waves form, and how they differ from progressive waves.

Wave ideas you need first

A wave transfers energy by oscillations. For a progressive wave, the disturbance travels through space.

Definition

Wavelength, frequency and wave speed

The wavelength $λ\lambda$ is the distance between two neighbouring points in phase, such as crest to crest. The frequency $f$ is the number of oscillations per second, measured in hertz (Hz). For a wave travelling at speed $v$ ,

f\lambda

Definition

Wavefront

A wavefront is a line or surface joining points on a wave that are at the same stage of oscillation, or in phase.

Diffraction

Diffraction is the spreading of waves when they pass through a slit or around the edge of an obstacle. It happens for all waves: water waves, sound waves, microwaves and light.

The amount of diffraction depends on the wavelength compared with the size of the gap or obstacle.

$Diffraction at a slit for different wavelength-to-slit-width ratios$

If $λ\lambda$ is much smaller than the slit width, written $width\lambda \ll \text{slit width}$ , there is little diffraction and the wave continues mainly forwards.

If $λ\lambda$ is equal to or greater than the slit width, the wave spreads out strongly as roughly semicircular wavefronts. If $λ\lambda$ is less than the slit width, the main beam spreads through less than 180°.

Key Idea

When diffraction is greatest

Diffraction is most noticeable when the wavelength is similar to, or larger than, the size of the slit or obstacle.

Example

Comparing diffraction through a doorway

Sound of frequency 500 Hz passes through a doorway of width 0.80 m. Take the speed of sound as 340 m s $^{-1}$ .

Find the sound wavelength using $f\lambda$ :
$m\lambda = \frac{v}{f} = \frac{340\ \text{m s}^{-1}}{500\ \text{Hz}} = 0.68\ \text{m}$
Compare with the doorway width: 0.68 m is close to 0.80 m, so the sound diffracts strongly.
Visible light has $m\lambda \approx 5 \times 10^{-7}\ \text{m}$ , which is much smaller than 0.80 m, so visible light diffracts very little through the doorway.

Superposition and interference

When two waves meet, their displacements add.

Definition

Principle of superposition

When two or more waves overlap, the resultant displacement at a point is the vector sum of the individual displacements at that point.

This means two crests can combine to give a larger crest, or a crest and trough can cancel.

Sketch graphs showing constructive and destructive superposition

Interference is the pattern produced when waves superpose. For two waves from in phase sources:

Constructive interference occurs when the path difference is $nλn\lambda$ .
Destructive interference occurs when the path difference is $(n+12)λ\left(n + \frac{1}{2}\right)\lambda$ .

Here $n$ is an integer: 0, 1, 2, 3, …

Definition

Path difference

The path difference is the difference between the distances travelled by two waves from their sources to the same point.

Example

Classifying interference from path difference

Two in phase sources send waves of wavelength 0.50 m to point P. The distances travelled are 2.40 m and 2.15 m.

Calculate the path difference:
$m\Delta x = 2.40\ \text{m} - 2.15\ \text{m} = 0.25\ \text{m}$
Compare with the wavelength:
$m=0.50\frac{\Delta x}{\lambda} = \frac{0.25\ \text{m}}{0.50\ \text{m}} = 0.50$
The path difference is $12λ\frac{1}{2}\lambda$ , so the waves arrive in antiphase and give destructive interference.

Coherent sources

For a stable interference pattern, the sources must be coherent.

Definition

Coherent sources

Coherent sources are monochromatic, have wavefronts continuous across the beam, and, when comparing more than one source, have a constant phase relationship.

Monochromatic means single frequency, so for a given wave speed it also means single wavelength.

Examples of coherent sources include:

two slits illuminated by the same laser beam;
two loudspeakers driven by the same signal generator;
two water-wave dippers attached to the same oscillator.

Examples of incoherent sources include:

two separate filament lamps;
two ordinary independent light sources;
two lasers that are not phase-locked.

For two-source interference to be observed, the sources must have zero or constant phase difference and their oscillations must be in the same direction. For light, this means the same polarisation.

Common Mistake

Same colour is not enough

Two separate lamps emitting the same colour are not automatically coherent. Their phase relationship changes randomly, so any interference pattern averages out.

Young’s double-slit experiment

Young’s double-slit experiment showed that light can produce interference, giving strong historical evidence for the wave nature of light.

A single source illuminates two narrow slits. The two slits act as coherent sources because their light comes from the same original wavefront. Bright and dark fringes form on a screen.

Young double-slit interference with slit separation, screen distance and fringe spacing

Let:

$a$ be the slit separation;
$D$ be the distance from slits to screen;
$Δy\Delta y$ be the fringe spacing;
$λ\lambda$ be the wavelength.

For small angles, $sin⁡θ≈tan⁡θ≈θ\sin\theta \approx \tan\theta \approx \theta$ , so:

λ=aΔyD\lambda = \frac{a\Delta y}{D}

Common Mistake

Small-angle condition

The equation $λ=aΔyD\lambda = \frac{a\Delta y}{D}$ assumes the screen is far from the slits compared with the slit separation, so the fringe angles are small.

Example

Finding wavelength using Young’s double slits

A laser produces fringes using slits separated by 0.250 mm. The screen is 2.40 m away. The distance across 10 fringe spacings is 48.0 mm.

Convert the measured quantities:
$0.250\ \text{mm} = 2.50 \times 10^{-4}\ \text{m}$ $m\Delta y = \frac{48.0\ \text{mm}}{10} = 4.80\ \text{mm} = 4.80 \times 10^{-3}\ \text{m}$
Substitute into $λ=aΔyD\lambda = \frac{a\Delta y}{D}$ :
$m\lambda = \frac{\left(2.50 \times 10^{-4}\ \text{m}\right)\left(4.80 \times 10^{-3}\ \text{m}\right)}{2.40\ \text{m}}$
Calculate and give a sensible unit:
$nm\lambda = 5.00 \times 10^{-7}\ \text{m} = 500\ \text{nm}$

Practical: wavelength from Young’s double slits

Use a low-power laser, double slits and a screen. Never look into the laser beam or its reflections.

Good method:

Measure $D$ from the slits to the screen.
Use the manufacturer’s value for $a$ , or a measured value if available.
Measure across many fringes, then divide by the number of fringe spacings to find $Δy\Delta y$ .
Repeat and average.
A useful graph is $Δy\Delta y$ against $D$ , since $Δy=λDa\Delta y = \frac{\lambda D}{a}$ . The gradient is $λa\frac{\lambda}{a}$ .

Tip

Reducing percentage uncertainty

Do not measure one fringe spacing if you can measure 10 or 20. A larger measured distance gives a smaller percentage uncertainty.

Diffraction gratings

A diffraction grating has many equally spaced slits. The slit spacing is called the grating spacing $d$ .

If a grating has $N$ lines per metre, then:

\frac{1}{N}

Bright beams are called orders. The central maximum is the zero order, $n = 0$ . The first bright beam on either side is the first order, $\pm 1$ .

For adjacent slits, the path difference at angle $θ\theta$ is $dsin⁡θd\sin\theta$ . A bright order occurs when this path difference is a whole number of wavelengths:

dsin⁡θ=nλd\sin\theta = n\lambda

$Diffraction grating geometry and stationary wave node spacing$

A grating has very small $d$ , so the beams are much further apart than in Young’s experiment. The large number of slits also makes the bright beams much sharper, because destructive interference cancels light strongly between the maxima.

Example

Using a diffraction grating

Light of wavelength $nm589\ \text{nm}$ is incident normally on a grating with 600 lines per millimetre. Find the first-order angle and the highest possible order.

Convert the line density into lines per metre:
$\times 10^3\ \text{m}^{-1} = 6.00 \times 10^5\ \text{m}^{-1}$
So:
$\frac{1}{N} = \frac{1}{6.00 \times 10^5\ \text{m}^{-1}} = 1.67 \times 10^{-6}\ \text{m}$
Use $dsin⁡θ=nλd\sin\theta = n\lambda$ for the first order, where $n = 1$ :
$m=0.353\sin\theta = \frac{n\lambda}{d} = \frac{1 \times 589 \times 10^{-9}\ \text{m}}{1.67 \times 10^{-6}\ \text{m}} = 0.353$ $θ=20.7∘\theta = 20.7^\circ$
For the highest order, require $sin⁡θ≤1\sin\theta \le 1$ :
$\le \frac{d}{\lambda} = \frac{1.67 \times 10^{-6}\ \text{m}}{589 \times 10^{-9}\ \text{m}} = 2.83$
The highest possible order is therefore $n = 2$ .

Practical: wavelength from a diffraction grating

Use a laser or spectral lamp, a diffraction grating and a screen or spectrometer.

Good method:

Measure angles to matching orders on both sides and average them.
Use $\frac{1}{N}$ from the grating line density.
Calculate $λ=dsin⁡θn\lambda = \frac{d\sin\theta}{n}$ .
For better analysis, plot $sin⁡θ\sin\theta$ against $n$ . The gradient is $λd\frac{\lambda}{d}$ .

Common Mistake

Line density conversion

600 lines per millimetre is not 600 lines per metre. It is $600 \times 10^3\ \text{m}^{-1}$ .

Stationary waves

A progressive wave transfers energy from one place to another. A stationary wave has a fixed pattern of nodes and antinodes and transfers no net energy along the wave.

Definition

Nodes and antinodes

A node is a point of zero amplitude. An antinode is a point of maximum amplitude.

A stationary wave can be regarded as the superposition of two progressive waves of equal amplitude and frequency travelling in opposite directions. The distance between neighbouring nodes is called the internodal distance, and it is:

λ2\frac{\lambda}{2}

Key differences:

In a progressive wave, the wave profile moves; in a stationary wave, the pattern stays fixed.
In a progressive wave, energy is transferred; in a stationary wave, there is no net energy transfer.
In a progressive wave, all points usually have the same amplitude if there is no damping; in a stationary wave, amplitude depends on position.
In a stationary wave, adjacent loops are in antiphase.

Example

Finding the speed of sound from a stationary wave

A loudspeaker produces sound of frequency 1.00 kHz in a tube. A microphone detects adjacent antinodes separated by 0.170 m.

Adjacent antinodes are separated by $λ2\frac{\lambda}{2}$ , so:
$m\lambda = 2 \times 0.170\ \text{m} = 0.340\ \text{m}$
Use $f\lambda$ :
$\left(1.00 \times 10^3\ \text{Hz}\right)\left(0.340\ \text{m}\right)$
Calculate the speed:
$340\ \text{m s}^{-1}$

Practical: speed of sound using stationary waves

A loudspeaker connected to a signal generator sends sound along a tube. Reflections create a stationary wave. Move a microphone along the tube and use an oscilloscope or data logger to locate nodes and antinodes.

Measure the distance between several neighbouring nodes or antinodes, then divide by the number of gaps to reduce uncertainty. Since adjacent nodes or adjacent antinodes are separated by $λ2\frac{\lambda}{2}$ , find $λ\lambda$ and then use $f\lambda$ .

Exam technique

In the exam

Check which geometry you are using: Young’s fringes use $λ=aΔyD\lambda = \frac{a\Delta y}{D}$ , while gratings use $dsin⁡θ=nλd\sin\theta = n\lambda$ .
Convert all distances to metres before substituting, especially millimetres and nanometres.
For interference questions, decide whether the path difference is a whole number of wavelengths or a half-integer number of wavelengths.

Self review

Check yourself

Why does sound diffract through a doorway much more than visible light?
What two conditions must two sources satisfy for a stable interference pattern?
Why are diffraction grating maxima sharper than Young’s double-slit fringes?

What you'll learn

Why waves diffract at slits and obstacles, and when the spreading is large.
How superposition gives interference patterns, including Young’s double-slit experiment.
How to use $λ=aΔyD\lambda = \frac{a\Delta y}{D}$ and $dsin⁡θ=nλd\sin\theta = n\lambda$ to determine wavelength.
How stationary waves form, and how they differ from progressive waves.

Wave ideas you need first

A wave transfers energy by oscillations. For a progressive wave, the disturbance travels through space.

Definition

Wavelength, frequency and wave speed

f\lambda

Definition

Wavefront

A wavefront is a line or surface joining points on a wave that are at the same stage of oscillation, or in phase.

Diffraction

Diffraction is the spreading of waves when they pass through a slit or around the edge of an obstacle. It happens for all waves: water waves, sound waves, microwaves and light.

The amount of diffraction depends on the wavelength compared with the size of the gap or obstacle.

$Diffraction at a slit for different wavelength-to-slit-width ratios$

If $λ\lambda$ is much smaller than the slit width, written $width\lambda \ll \text{slit width}$ , there is little diffraction and the wave continues mainly forwards.

Key Idea

When diffraction is greatest

Diffraction is most noticeable when the wavelength is similar to, or larger than, the size of the slit or obstacle.

Example

Comparing diffraction through a doorway

Sound of frequency 500 Hz passes through a doorway of width 0.80 m. Take the speed of sound as 340 m s $^{-1}$ .

Find the sound wavelength using $f\lambda$ :
$m\lambda = \frac{v}{f} = \frac{340\ \text{m s}^{-1}}{500\ \text{Hz}} = 0.68\ \text{m}$
Compare with the doorway width: 0.68 m is close to 0.80 m, so the sound diffracts strongly.
Visible light has $m\lambda \approx 5 \times 10^{-7}\ \text{m}$ , which is much smaller than 0.80 m, so visible light diffracts very little through the doorway.

Superposition and interference

When two waves meet, their displacements add.

Definition

Principle of superposition

When two or more waves overlap, the resultant displacement at a point is the vector sum of the individual displacements at that point.

This means two crests can combine to give a larger crest, or a crest and trough can cancel.

Sketch graphs showing constructive and destructive superposition

Interference is the pattern produced when waves superpose. For two waves from in phase sources:

Constructive interference occurs when the path difference is $nλn\lambda$ .
Destructive interference occurs when the path difference is $(n+12)λ\left(n + \frac{1}{2}\right)\lambda$ .

Here $n$ is an integer: 0, 1, 2, 3, …

Definition

Path difference

The path difference is the difference between the distances travelled by two waves from their sources to the same point.

Example

Classifying interference from path difference

Two in phase sources send waves of wavelength 0.50 m to point P. The distances travelled are 2.40 m and 2.15 m.

Calculate the path difference:
$m\Delta x = 2.40\ \text{m} - 2.15\ \text{m} = 0.25\ \text{m}$
Compare with the wavelength:
$m=0.50\frac{\Delta x}{\lambda} = \frac{0.25\ \text{m}}{0.50\ \text{m}} = 0.50$
The path difference is $12λ\frac{1}{2}\lambda$ , so the waves arrive in antiphase and give destructive interference.

Coherent sources

For a stable interference pattern, the sources must be coherent.

Definition

Coherent sources

Coherent sources are monochromatic, have wavefronts continuous across the beam, and, when comparing more than one source, have a constant phase relationship.

Monochromatic means single frequency, so for a given wave speed it also means single wavelength.

Examples of coherent sources include:

two slits illuminated by the same laser beam;
two loudspeakers driven by the same signal generator;
two water-wave dippers attached to the same oscillator.

Examples of incoherent sources include:

two separate filament lamps;
two ordinary independent light sources;
two lasers that are not phase-locked.

For two-source interference to be observed, the sources must have zero or constant phase difference and their oscillations must be in the same direction. For light, this means the same polarisation.

Common Mistake

Same colour is not enough

Two separate lamps emitting the same colour are not automatically coherent. Their phase relationship changes randomly, so any interference pattern averages out.

Young’s double-slit experiment

Young’s double-slit experiment showed that light can produce interference, giving strong historical evidence for the wave nature of light.

A single source illuminates two narrow slits. The two slits act as coherent sources because their light comes from the same original wavefront. Bright and dark fringes form on a screen.

Young double-slit interference with slit separation, screen distance and fringe spacing

Let:

$a$ be the slit separation;
$D$ be the distance from slits to screen;
$Δy\Delta y$ be the fringe spacing;
$λ\lambda$ be the wavelength.

For small angles, $sin⁡θ≈tan⁡θ≈θ\sin\theta \approx \tan\theta \approx \theta$ , so:

λ=aΔyD\lambda = \frac{a\Delta y}{D}

Common Mistake

Small-angle condition

The equation $λ=aΔyD\lambda = \frac{a\Delta y}{D}$ assumes the screen is far from the slits compared with the slit separation, so the fringe angles are small.

Example

Finding wavelength using Young’s double slits

A laser produces fringes using slits separated by 0.250 mm. The screen is 2.40 m away. The distance across 10 fringe spacings is 48.0 mm.

Convert the measured quantities:
$0.250\ \text{mm} = 2.50 \times 10^{-4}\ \text{m}$ $m\Delta y = \frac{48.0\ \text{mm}}{10} = 4.80\ \text{mm} = 4.80 \times 10^{-3}\ \text{m}$
Substitute into $λ=aΔyD\lambda = \frac{a\Delta y}{D}$ :
$m\lambda = \frac{\left(2.50 \times 10^{-4}\ \text{m}\right)\left(4.80 \times 10^{-3}\ \text{m}\right)}{2.40\ \text{m}}$
Calculate and give a sensible unit:
$nm\lambda = 5.00 \times 10^{-7}\ \text{m} = 500\ \text{nm}$

Practical: wavelength from Young’s double slits

Use a low-power laser, double slits and a screen. Never look into the laser beam or its reflections.

Good method:

Measure $D$ from the slits to the screen.
Use the manufacturer’s value for $a$ , or a measured value if available.
Measure across many fringes, then divide by the number of fringe spacings to find $Δy\Delta y$ .
Repeat and average.
A useful graph is $Δy\Delta y$ against $D$ , since $Δy=λDa\Delta y = \frac{\lambda D}{a}$ . The gradient is $λa\frac{\lambda}{a}$ .

Tip

Reducing percentage uncertainty

Do not measure one fringe spacing if you can measure 10 or 20. A larger measured distance gives a smaller percentage uncertainty.

Diffraction gratings

A diffraction grating has many equally spaced slits. The slit spacing is called the grating spacing $d$ .

If a grating has $N$ lines per metre, then:

\frac{1}{N}

Bright beams are called orders. The central maximum is the zero order, $n = 0$ . The first bright beam on either side is the first order, $\pm 1$ .

For adjacent slits, the path difference at angle $θ\theta$ is $dsin⁡θd\sin\theta$ . A bright order occurs when this path difference is a whole number of wavelengths:

dsin⁡θ=nλd\sin\theta = n\lambda

$Diffraction grating geometry and stationary wave node spacing$

Example

Using a diffraction grating

Light of wavelength $nm589\ \text{nm}$ is incident normally on a grating with 600 lines per millimetre. Find the first-order angle and the highest possible order.

Convert the line density into lines per metre:
$\times 10^3\ \text{m}^{-1} = 6.00 \times 10^5\ \text{m}^{-1}$
So:
$\frac{1}{N} = \frac{1}{6.00 \times 10^5\ \text{m}^{-1}} = 1.67 \times 10^{-6}\ \text{m}$
Use $dsin⁡θ=nλd\sin\theta = n\lambda$ for the first order, where $n = 1$ :
$m=0.353\sin\theta = \frac{n\lambda}{d} = \frac{1 \times 589 \times 10^{-9}\ \text{m}}{1.67 \times 10^{-6}\ \text{m}} = 0.353$ $θ=20.7∘\theta = 20.7^\circ$
For the highest order, require $sin⁡θ≤1\sin\theta \le 1$ :
$\le \frac{d}{\lambda} = \frac{1.67 \times 10^{-6}\ \text{m}}{589 \times 10^{-9}\ \text{m}} = 2.83$
The highest possible order is therefore $n = 2$ .

Practical: wavelength from a diffraction grating

Use a laser or spectral lamp, a diffraction grating and a screen or spectrometer.

Good method:

Measure angles to matching orders on both sides and average them.
Use $\frac{1}{N}$ from the grating line density.
Calculate $λ=dsin⁡θn\lambda = \frac{d\sin\theta}{n}$ .
For better analysis, plot $sin⁡θ\sin\theta$ against $n$ . The gradient is $λd\frac{\lambda}{d}$ .

Common Mistake

Line density conversion

600 lines per millimetre is not 600 lines per metre. It is $600 \times 10^3\ \text{m}^{-1}$ .

Stationary waves

A progressive wave transfers energy from one place to another. A stationary wave has a fixed pattern of nodes and antinodes and transfers no net energy along the wave.

Definition

Nodes and antinodes

A node is a point of zero amplitude. An antinode is a point of maximum amplitude.

λ2\frac{\lambda}{2}

Key differences:

In a progressive wave, the wave profile moves; in a stationary wave, the pattern stays fixed.
In a progressive wave, energy is transferred; in a stationary wave, there is no net energy transfer.
In a progressive wave, all points usually have the same amplitude if there is no damping; in a stationary wave, amplitude depends on position.
In a stationary wave, adjacent loops are in antiphase.

Example

Finding the speed of sound from a stationary wave

A loudspeaker produces sound of frequency 1.00 kHz in a tube. A microphone detects adjacent antinodes separated by 0.170 m.

Adjacent antinodes are separated by $λ2\frac{\lambda}{2}$ , so:
$m\lambda = 2 \times 0.170\ \text{m} = 0.340\ \text{m}$
Use $f\lambda$ :
$\left(1.00 \times 10^3\ \text{Hz}\right)\left(0.340\ \text{m}\right)$
Calculate the speed:
$340\ \text{m s}^{-1}$

Practical: speed of sound using stationary waves

Exam technique

In the exam

Check which geometry you are using: Young’s fringes use $λ=aΔyD\lambda = \frac{a\Delta y}{D}$ , while gratings use $dsin⁡θ=nλd\sin\theta = n\lambda$ .
Convert all distances to metres before substituting, especially millimetres and nanometres.
For interference questions, decide whether the path difference is a whole number of wavelengths or a half-integer number of wavelengths.

Self review

Check yourself

Why does sound diffract through a doorway much more than visible light?
What two conditions must two sources satisfy for a stable interference pattern?
Why are diffraction grating maxima sharper than Young’s double-slit fringes?

What you'll learn

Wave ideas you need first

Diffraction

Superposition and interference

Coherent sources

Young’s double-slit experiment

Practical: wavelength from Young’s double slits

Diffraction gratings

Practical: wavelength from a diffraction grating

Stationary waves

Practical: speed of sound using stationary waves

Review Wave Properties by teaching Genie

Remember key concepts with flashcards

Wave Properties

What you'll learn

Wave ideas you need first

Diffraction

Superposition and interference

Coherent sources

Young’s double-slit experiment

Practical: wavelength from Young’s double slits

Diffraction gratings

Practical: wavelength from a diffraction grating

Stationary waves

Practical: speed of sound using stationary waves

Review Wave Properties by teaching Genie

Remember key concepts with flashcards

Wave Properties

What you'll learn

Wave ideas you need first

Diffraction

Superposition and interference

Coherent sources

Young’s double-slit experiment

Practical: wavelength from Young’s double slits

Diffraction gratings

Practical: wavelength from a diffraction grating

Stationary waves

Practical: speed of sound using stationary waves

Review Wave Properties by teaching Genie

Remember key concepts with flashcards

Wave Properties Revision Guide

Wave Properties

What you'll learn

Wave ideas you need first

Diffraction

Superposition and interference

Coherent sources

Young’s double-slit experiment

Practical: wavelength from Young’s double slits

Diffraction gratings

Practical: wavelength from a diffraction grating

Stationary waves

Practical: speed of sound using stationary waves

Review Wave Properties by teaching Genie

Remember key concepts with flashcards

Wave Properties Revision Guide