AS PHYSICS:WAVES

AS PHYSICS:WAVES

1. The NAture of Waves

WHAT IS A WAVE?

A wave is a disturbance that transfers energy from one place to another without transferring matter.

This energy transfer occurs through the oscillation of particles in the medium through which the wave travels. Energy is transferred in the same direction that the wave is travelling.

Energy transfer

Definitions

Features of a WAVE

Wave Diagram Features:

Other features:

• Period, t

• Frequency, f

• Phase

• Phase difference

Learn these definitions!

WAVE EQUATIONS

Frequency - Period Equation

The Wave Equation

This equation links wave speed, wave frequency, and wavelength:

• Exercise:

Use this equation to derive the units of frequency in SI base units.

• Question:

What does this equation tell us about the relationship between:- Speed of wave and frequency- Speed of wave and wavelength- Frequency and wavelength

Reveal

Reveal

WHAT IS AN OSCILLOSCOPE?

USING AN OSCILLOSCOPE to Find the speed of sound

PRACTICE EXAM question

ANSWER THE FOLLOWING QUESTION:

Answer

2. TYPES OF Waves

OVERVIEW OF TYPES OF WAVE

There are two types of wave:

1. Transverse waves

2. Longitudinal waves

These types of wave have both similarities and differences.

TRANSVERSE WAVES

DEFINING FEATURE

EXAMPLEs

GRAPHS

In transverse waves, oscillation is at right angles to the direction of travel

All electromagnetic waves are transverse.

There are two main ways of graphing transverse waves: 1. Displacement - distance 2. Displacement - time

LONGITUDINAL WAVES

DEFINING FEATURE

EXAMPLEs

How they work

In transverse waves, oscillation is in the same direction as the direction of travel

All sound waves are longitudinal.

Longitudinal waves create pressure variations in the medium that they are travelling through.

COMMON FEATURES OF ALL WAVES

REFLECTION

Waves are bounced back when they hit a boundary

REFRACTION

Waves change direction as they enter a different medium.

INTENSITY

The amount of energy carried by any wave is called its 'intensity'. It means how much energy is transferred per second.

WAVE INTENSITY

What is it?

Exercises:

A measure of the amount of energy that a wave is carrying.

1. Show how the intensity equation can be used to derive the units of intensity.

Definition

2. What does the intensity equation tell us about the relationship between:- Intensity and power- Intensity and area- Power and area

Answer

Intensity Equation

3. Light is hitting a 12 cm2 piece of paper at right angles to its surface. The powere received at the paper is 0.26W. Calculate the intensity of the light received.

Answer

Intensity and Amplitude

Intensity is proportional to amplitude squared

4. What happens to intensity when amplitude is a) halved b) trippled

Answer

COMMON FEATURES OF ALL ELECTROMagnetic Waves

SPEED IN A VACUUM

All EM waves travel at a speed of 3.00 x 108 ms-1 in a vacuum and slower speeds in all other media

THEY ARE TRANSVERSE WAVES

These transverse waves consist of vibrating electric and magnetic fields.The electric and magnetic fields are at right angles to each other and to the direction of travel.

REFRACTION, REFLECTION, DIFFRACTION, AND INTERFERENCE

Like all waves, EM waves can be refracted, reflected and diffracted, and can undergo interference.

WAVE EQUATION

Like all waves, EM waves obey the wave equation

POLARISATION

Like all transverse waves, EM waves can be polarised (as will be covered later on)

PRACTICE EXAM question

ANSWER THE FOLLOWING QUESTION:

Answer

3. Polarisation of waves

WHAT IS A POLARISED WAVE?

Polarized waves are waves in which oscillations occur in a single plane.

Plane polarized waves are waves in which the direction of oscillation is the same for all waves.

Polarisation happens to unpolarised transverse waves when they're reflected, refracted or scattered (but not when they're diffracted).

Definitions

Polarisation in action: ROPE EXAMPLE

• If you shake a rope to make a wave, you can move your hand up and down, side to side, or a mixture of directions. Any of these would be a transverse wave.

• BUT if you try to pass waves in the rope through a vertiical fence, the waves will only get through if the vibrations are vertical. The fence filters out vibrations in other directions. This 'filtering out' of certain directions is called polarising the wave.

• The plane in which a wave vibrates is called the plane of polarisation. In this case, the rope was polarised in the vertical plane by the fence.

• Polarising a wave so that it only oscillates in one direction is called plane polarisation.

• Polarisation can only happen for transverse waves.

WORKIng with Polarising Filters

• Ordinary light waves can be polarised using a polarising filter. What does this tell us about light waves?

Answer

• When light is sequentially passed through two filters, rotating the second filter alters the amount of light that passes through.

• When light is sequentially passed through two filters, rotating the second filter alters the amount of light that passes through.

• What happens when:

4. The axes are at right angles to each other?

1. The axes of the two filters are aligned?
2. You start rotating the second filter?
3. The axes of the filters are at 45o to each other?

5. You continue rotating the second filter beyond 90o?

6. The axes are at 180o to each other?

Rotation - Intensity Graph

Answers

POLARISATION examples

METAL GRILLES

polarising Light

TV And Radio signals

When light reflects, it is partially polarised. This is used in photography and Polaroid sunglasses

TV and radio signals are polarised by their transmitting aerials.

Metal grilles are used to polarise microwaves

PRACTICAL: Observing Polarisation of Microwaves

Step by Step Instructions

PRACTICE questions: multiple choice

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice

ANSWER THE FOLLOWING QUESTION:

PRACTICE question: short answer

ANSWER THE FOLLOWING QUESTIONs:

Answer

1. Describe what is meant by plane polarised light. Use a diagram to show the difference between unpolarised and plane polarised light.

[4 marks]

2. Unpolarised light passes through two polarising filters P₁ and P₂. The transmission axis of P₁ is vertical, and the transmission axis of P₂ is horizontal.

(a) If the intensity of the unpolarised light is 12 mW/m², calculate the intensity after passing through filter P₁

[1 mark]

(b) Describe what happens to the light intensity when it passes through the second filter P₂. Explain your answer.

[2 marks]

(c) If the second filter P₂ is slowly rotated from 0° to 360°, describe how the intensity of light reaching the detector would change. You do not need to calculate specific values.

[3 marks]

PRACTICE question: short answer

ANSWER THE FOLLOWING QUESTION:

Answer

3. A student performs an experiment to investigate plane polarisation using two polarising filters and a light sensor.

(a) Explain how the student could demonstrate that light from an ordinary light bulb is unpolarised

[2 marks]

(b) The student also wants to investigate whether light reflected from a glass surface is polarised. Describe an experimental setup the student could use.

[3 marks]

(c) Explain how the results would show whether the reflected light is polarised or not.

[2 marks]

PRACTICE question: LONG answer

ANSWER THE FOLLOWING QUESTION:

Answer

1. (a) When unpolarised light reflects from a glass surface, the reflected light becomes partially polarised. Explain why this happens.

[3 marks]

(b) Polaroid sunglasses have their transmission axes oriented vertically. Explain how this reduces glare from horizontal surfaces such as roads or water, referring to the polarisation of reflected light

[3 marks]

(c) The same sunglasses can make the sky appear darker in some directions than others. Explain this observation.

[2 marks]

PRACTICE question: long answer

ANSWER THE FOLLOWING QUESTION:

Answer

2. Television and radio transmissions use electromagnetic waves

(a) A TV aerial consists of a metal rod mounted horizontally. Explain why it needs to be oriented this way to effectively receive a horizontally polarised signal.

[3 marks]

(b) In cities, TV signals can reflect off buildings before reaching an aerial. Explain why this can sometimes cause reception problems, with reference to polarisation.

[3 marks]

PRACTICE question: long answer

ANSWER THE FOLLOWING QUESTION:

Answer

2. The door of a microwave oven has a metal mesh with small holes.

(a) Microwaves used in cooking have a wavelength of approximately 12 cm. The holes in the mesh of a microwave door are typically about 1 mm in diameter. Explain why visible light (wavelength 400-700 nm) can pass through the mesh but microwaves cannot.

[3 marks]

(b) A metal mesh consisting of parallel wires can act as a polarising filter for microwaves. Explain how this works.

[3 marks]

(c) A physics student constructs a simple polarisation demonstrator for microwaves using two metal grilles with parallel wires. When the grilles are aligned parallel to each other, microwaves pass through, but when one grille is rotated by 90°, no microwaves are detected. Explain these observations

[3 marks]

4. SUPERPOSITION & coherence

superposition AND COHERENCE

The principle of superposition describes how waves that meet at a point in space interact. It states that "when two or more waves with the same frequency arrive at a point, the resultant displacement is the sum of the displacements of each wave"

Coherence, meanwhile, refers to two or more waves having a fixed relationship. For waves to be coherent, they must have the same wavelength and phase difference.

Understanding wave phase

Question:

What is a wave 'phase'

In the graph below, which points are:

• The phase of a wave refers to its position within the wave cycle, relative to a fixed point in space, such as the origin.

Answer

1. In phase
2. Out of phase

'In-phase' means 'in step'

• Two points in a wave are 'in phase' if they're both at the same point in the wave cycle. This will happen if the points have the same displacement and velocity.

'Out of phase' means 'out of step'

• When two points in a wave are at opposite points in the wave cyle, they are 'out of phase'. This will happen when they have opposite displacement and velocity.

We can also talk about two different waves being 'in-phase' and 'out of phase'

Understanding phase difference

What is 'phase difference?

• Phase difference refers to the time gap or cycle difference between two points.
• It describes how much a single wave is in front or behind another wave.
• When two waves move and their cycles don't coincide, they have a phase difference.

How do we measure 'phase difference'?

• It's mathematically helpful to show one complete cycle of a wave as an angle of 360o (2 π radians), as a point on a wave moves 360o during one oscillation.
• Phase difference is measured in degrees (o) or radians.

Phase difference simulator

Why measure in degrees / radians?

APPLyying phase difference

How can we tell that two points / waves are 'in-phase'?

• If two points / waves have a phase difference of either zero or a multiple of 360o (2 π radians), we know that they're 'in-phase'.
• E.g. two waves with a phase difference of 720o (4π radians) will be 'in-phase'.

How can we tell that two points / waves are 'out of phase'?

• If two points / waves have a phase difference of odd-number multiples of 180o (π radians), we know that they're exactly 'out of phase'
• E.g. two waves with a phase difference of 540o (3π radians) will be exactly 'out of phase'

What if the phase difference isn't equal to a multiple of 180o (π radians)?

• If the phase difference isn't equal to a multiple of 180o (π radians), we know that the points / waves are partially out of phase.

• E.g. two waves with a phase difference of 270o (1.5π radians) will be partially 'out of phase'

How does superposition work?

What happens when waves meet?

• When waves meet, they don't bounce off each other like solid objects would. Instead, they pass right through one another. While they're overlapping, however, they combine their effects through a process called superposition.
• When the two or more waves cross each other, their displacements are added together. In scientific language, we say that the resultant displacement equals the vector sum of the individual displacements.

Waves moving 'in phase' amplify each other

Waves moving 'out of phase' cancel each other out

Answer

Comparison

What is Wave interference?

What is interference?

• Wave interference occurs when two or more waves overlap in the same space, i.e. when superposition takes place.
• There are two key types of interference: contructive and destructive.

What is constructive interference?

• When the peaks of two waves align, they add together to create a bigger wave.
• Think of two people pushing a swing at the same time - their combined effort makes the swing go higher than either person could achieve alone.

What is destructive interference?

• When the peak of one wave meets the trough (low point) of another, they cancel each other out, potentially creating a point of stillness.
• This is like two people pushing a swing from opposite sides with equal force - the swing stays still.

what is PATH DIfference?

What is 'path difference?'

• Path difference is the extra distance one wave travels compared to another wave before they meet at a point. This difference determines how the waves align when they overlap.

How does path difference relate to phase difference?

Path difference directly relates to phase difference:

• A path difference of one full wavelength (λ) equals a phase difference of 360° (or 2π radians)
• You can convert between them using the following formula:

Phase difference = (Path difference ÷ λ) × 360°

Path difference and interference

destructiveInterference

practical example:sound waves

Constructive Interference

Sound wave interference creates noticeable loud and quiet spots in everyday situations.

Contructive interference happens when two waves are in-phase

Destructive interference happens when two waves are out-of-phase.

How do we explain sound wave interference patterns?

What happens when sound waves intefere?

When two sound sources emit waves of the same frequency:

1. The waves spread out and overlap in the surrounding space (i.e. superposition takes place)
2. At certain points, the waves arrive in-phase (constructive interference) creating a louder sound
3. At other points, they arrive out of phase (destructive interference) creating quieter spots

What happens when sound waves intefere?

Click on the icon below to see how this interference pattern relates to path difference:

What is COherence?

What is wave coherence?

Wave coherence refers to how well multiple waves maintain a consistent relationship with each other over time and distance.

When are waves 'coherent'?

For waves to be coherent, they must have:

1. The same wavelength / frequency
2. A constant phase difference that doesn't change over time
3. The same amplitude (ideally, though not strictly required)
4. The same polarization (for transverse waves like light)

Think of coherence like synchronized swimmers: when waves are coherent, they maintain the same pattern and relationship to each other. The peaks and troughs line up in a predictable way, creating stable interference patterns.

Intereference patterns in practice

Not all overlapping waves will create noticeable interference patterns

For noticeable interference patterns to arise, several conditions must be met. Click on the icon to the righth to find out more.

In practice, waves rarely cancel out completely

You may still hear sound at points of destructive interference. Click on the icon to the right to find out more.

Sound wave interference has many practical applications

We use sound wave interference in a number of useful technologies. Click on the icon to the right to learn more about these.

PRACTICE questions - B GRade

ANSWER THE FOLLOWING QUESTION:

Question 1: Basic Wave Principles

(a) Define what is meant by the wavelength of a wave.

[2 marks]

(b) A sound wave has a frequency of 400 Hz and travels at 340 m/s.(i) Calculate the wavelength of the wave (ii) State what happens to the wavelength if the frequency is doubled

[4 marks]

(c) When two waves meet at a point:(i) State the principle that determines how they combine (ii) Draw diagrams to show how two pulses combine when they meet, one showing constructive and one showing destructive interference

[4 marks]

Answer

PRACTICE questions - B GRade

ANSWER THE FOLLOWING QUESTION:

Question 2: Interference Basics

(a) Two speakers are connected to the same signal generator.(i) Explain what is meant by coherent sources (ii) Explain why the speakers must be coherent to produce a clear interference pattern

[4 marks]

(b) A student walks between two speakers and notices alternating loud and quiet spots.(i) Calculate the wavelength of the sound if its frequency is 500 Hz and speed is 340 m/s (ii) Explain why the student hears these variations in volume

[4 marks]

Answer

PRACTICE questions - A GRade

ANSWER THE FOLLOWING QUESTION:

Question 3: Wave Interference

Two speakers produce sound waves of wavelength 0.8 m. The speakers are 2.4 m apart.

(a) Calculate how many positions of maximum volume (constructive interference) occur on a line between the speakers

[3 marks]

(b) Explain why these maxima occur at these positions in terms of path difference

[2 marks]

(c) The frequency of the sound is now increased(i) Explain what happens to the number of maxima between the speakers(ii) Explain why the pattern becomes harder to detect at higher frequencies

[4 marks]

Answer

PRACTICE questions - A GRade

ANSWER THE FOLLOWING QUESTION:

Question 4: Applications

A pair of noise-canceling headphones uses a microphone to detect incoming sound waves and a speaker to produce canceling waves.

(a) For a sound wave of frequency 200 Hz:

(i) Calculate its wavelength in air (speed = 340 m/s)(ii) Calculate the time delay needed between detection and cancellation if the microphone and speaker are separated by 3.4 cm

[4 marks]

(b) The noise-canceling system works well for low-frequency sounds but poorly for high-frequency sounds.

[3 marks]

(i) Suggest two reasons why this happens(ii) Calculate the wavelength of a 2000 Hz sound and use this to support one of your reasons

[5 marks]

Answer

PRACTICE questions - A* GRade

Answer

ANSWER THE FOLLOWING QUESTION:

Question 5: Wave Analysis and Problem Solving

Two loudspeakers are connected to the same signal generator producing sound waves of frequency 440 Hz at a speed of 340 m/s in air.

a) The speakers are positioned 2.0 m apart. A student walks along a line parallel to and 3.0 m away from the line joining the speakers.(i) Calculate the wavelength of the sound waves (ii) Calculate the distance between adjacent positions where the student hears maximum volume (iii) Explain why the student may still hear some sound at positions where destructive interference is expected

[6 marks]

b) The signal generator frequency is now changed to 880 Hz.(i) Without calculation, explain what happens to the distance between adjacent maximum volume positions (ii) Calculate the new distance between maxima to verify your answer (iii) Explain why this change makes it harder for someone to locate the exact positions of maximum and minimum volume

[6 marks]

PRACTICE questions - A* GRade

Answer

ANSWER THE FOLLOWING QUESTION:

Question 6: Applications and Analysis

(a) A noise-canceling system is being designed to reduce unwanted sound in a workshop. (i) Explain what conditions must be met for effective noise cancellation (ii) Calculate the time delay needed between detecting a sound wave and producing the canceling wave if the microphone and speaker are separated by 20 cm (speed of sound = 340 m/s)

[4 marks]

b) Two identical wave sources in a ripple tank produce waves of wavelength 2.0 cm. The sources are 8.0 cm apart.(i) Calculate the number of positions between the sources where complete destructive interference occurs (ii) Explain why the interference pattern becomes less distinct as the waves travel further from the sources (iii) Describe one way to verify experimentally that destructive interference is occurring at the predicted positions

[6 marks]

PRACTICE questions - A* GRade

Answer

ANSWER THE FOLLOWING QUESTION:

Question 7: Problem Solving and Analysis

a) In an experimental setup, two speakers produce waves with different amplitudes, where one amplitude is twice the other.(i) When the waves arrive in phase at a point P, calculate the ratio of the combined amplitude to the larger original amplitude (ii) When the waves arrive exactly out of phase at point Q, calculate whether complete silence occurs Show your reasoning clearly.

[4 marks]

(c) In terms of the principle of superposition:(i) Explain why the maximum amplitude when two identical waves meet is twice the original amplitude, but the maximum intensity is four times the original intensity (ii) Describe what this means in terms of the energy of the waves

[4 marks]

4. diffraction

WHAT IS diffraction?

Diffraction refers to the way waves spread out or bend around the edges of obstacles or apertures, especially when the size of the obstacle or aperture is comparable to the wavelength of the wave.

In this lesson we will learn about two types of diffraction:

1. At a slit: The wave spreads out after passing through the slit, with the spreading effect being more pronounced when the slit width is closer to the wavelength of the wave.

2. At an obstacle: The wave bends around the edges of the obstacle, allowing the wave to reach regions that would otherwise be in the "shadow" of the obstacle.

How can we use a ripple tank to investigate diffraction?

What is a ripple tank?

• A shallow tank of water that you can generate a wave in. The waves are generated by an oscillating paddle that continually dips into the water and creates regular waves with straight, parallel wavefronts.

How can we use a ripple tank to investigate diffraction at a slit?

• Objects are placed into the ripple tank to create a barrier with a gap in the middle of it.
• We can vary the gap to see the effects this has on how waves spread through the tank.

Key Point

How does huygens' construction explain diffraction through a slit?

What is Huygens' Construction?

Huygens' Construction is a general model of the propogation of waves that states that:

• Every point on a wavefront acts as a source of new circular "wavelets"
• These wavelets spread out in all directions
• The new wavefront is formed by the combined outer edges of all these wavelets

How does Huygens' Construction explain diffraction at a slit?

We can apply Huygens' Construction at each stage of the wave slit experiment:

1. Before the slit: The wave has many points acting as sources along its wavefront
2. At the slit: Only the points within the slit opening can continue creating wavelets
3. After the slit:

• If the slit is wide: Many points create wavelets that mostly reinforce in the forward direction, and spreading happens only at the edges.
• If the slit is narrow: Only a few points create wavelets, causing significant spreading in many directions. The wavelets from the few points in the slit spread out in a semicircular pattern

Diagrams

How does huygens' construction explain diffraction around an obstacle?

Is the basic principle the same as for diffraction through a slit?

Yes. The basic principle is the same:

• Each point on a wavefront creates circular wavelets that spread outward
• The new wavefront forms from the combined outer edges of these wavelets

What happens at an obstacle?

When a wave meets an obstacle, you get diffraction around the edges. Behind the obstacle is a 'shadow' where the wave is blocked. The wider the obstacle compared with the wavelength of the wave, the less diffraction you get, so the longer the shadow.

Explanation

what happens when light diffracts through a single slit?

What happens?

When light passes through a single narrow slit, it creates a pattern of bright and dark fringes on a screen:

How does Huygens' Construction explain this?

1. At the slit: According to Huygens' Construction, each point in the slit becomes a source of secondary wavelets
2. Path differences: Light from different parts of the slit travels slightly different distances to reach any point on the screen
3. Interference effects: At some points, waves arrive in phase (crest meets crest) creating bright fringesAt other points, waves arrive out of phase (crest meets trough) creating dark fringes

What pattern is created?

1. A bright central maximum appears directly opposite the slit
2. This is flanked by alternating dark and bright fringes on both sides
3. The central maximum is the brightest and widest
4. Secondary bright fringes become progressively dimmer and narrower

Key Factors

what happens when light diffracts through a diffraction grating?

What is a diffraction grating?

A diffraction grating is a surface with many parallel, closely-spaced slits. These slits are typically just a few wavelengths of light apart.

What happens when monochromatic light is shone through a diffraction grating?

An interference pattern appears on the screen behind the diffraction grating, comprised of dark and light bands.

How does increasing the number of slits in the grating impact the interference pattern?

• With just two slits, you get broad bright fringes with fairly gentle transitions
• With many slits (like in a diffraction grating), you get very sharp, narrow bright lines separated by wider dark regions

Explanation

Types of light

what terminology do we use to describe interference patterns?

When describing interference patterns, we use the following terminology

• The line of maximum brightness in the middle of the screen is called the zero order line
• The bright bands are referred to as maxima, and the dark bands, are referred to as minima

• The distance between consecutive maxima is called the fringe width
• The maxima on either side of the zero order line are called 'first order lines'
• The next pair of maxima out from the centre are called 'second order lines' and so on.

Whe use the following notation:

• We use n = 1 to refer to the first pair of maxima, n = 2 for the second pair, n = 3 for the third pair, etc.
• The distance between the slits in shown as 'd'
• The distance from the grating to the screen is shown as 'D'
• The angle from the horizontal zero order line is shown as 'θ'
• The number of slits per meter is shown as 'N', where

N = 1 / d

how can we use diffraction gratings to find the wavelength of light?

Experimental set-up

take measurements

calculate Wavelength

See how to set up the experiment

Discover how to calculate wavelength from experimental results

Learn what measurements to take from the experiment

let's put theory into practice

Watch the video to see how diffraction gratings are used to calculate wavelength in practice. Then click on the button below to learn more about the relationships in the formula nλ = d sin θ

what happens when we shine white light through gratings?

Let's find out!

Watch the video to see how diffraction gratings are used to calculate wavelength in practice. Then click on the button below to learn more!

PRACTICE questions: multiple choice 1

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice 2

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice 3

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice 4

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice 5

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice 6

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice 7

ANSWER THE FOLLOWING QUESTION:

PRACTICE questions: multiple choice 8

ANSWER THE FOLLOWING QUESTION:

PRACTICE question: short answer 1 - 4

ANSWER THE FOLLOWING QUESTIONs:

Answer

1. Explain in 2-3 sentences what is meant by diffraction and state one everyday example.
2. Using Huygens' construction, explain why a wave appears to spread out after passing through a narrow gap

[2 marks]

[3 marks]

3. Describe how the diffraction pattern changes when:

[2 marks]

(a) The width of a single slit is decreased

[2 marks]

(b) The wavelength of light is increased

4. Explain why radio waves of long wavelength can be received behind a mountain, while shorter wavelength signals cannot.

[3 marks]

PRACTICE question: short answer 5 - 6

ANSWER THE FOLLOWING QUESTIONs:

Answer

5. A ripple tank is set up with a barrier containing a gap. Describe what happens to the diffraction pattern when:

[2 marks]

(a) The gap width is reduced to approximately the same as the wavelength

[2 marks]

(b) The frequency of the wave generator is increased

6. Sketch the intensity pattern you would expect to see on a screen when light passes through a single narrow slit. Label the central maximum and indicate where the first minimum occurs on each side.

[3 marks]

PRACTICE question: LONG answer 1

Answer

ANSWER THE FOLLOWING QUESTION:

1. Describe an experiment using a ripple tank to investigate the diffraction of water waves through gaps of different widths. Include:

• The equipment you would use
• How you would set up the experiment
• The measurements or observations you would make
• How these observations relate to the principle of diffraction
• What conclusion you would expect to reach

[6 marks]

PRACTICE question: LONG answer 2

Answer

ANSWER THE FOLLOWING QUESTION:

2 (a) Explain using Huygens' construction how a wave diffracts when passing through a narrow slit. Include a diagram in your answer.

[4 marks]

(b) When light of wavelength 500 nm passes through a single slit, a diffraction pattern is formed on a screen 2.0 m away. The distance from the central maximum to the first minimum is 5.0 mm. Calculate the width of the slit.

[3 marks]

(c) Explain why we can hear around corners but not see around them. Use the concept of diffraction and appropriate wavelength values in your explanation.

[3 marks]

PRACTICE question: LONG answer 3

Answer

ANSWER THE FOLLOWING QUESTION:

[3 marks]

3 (a) Define diffraction and explain why it occurs according to wave theory.

(b) Describe the pattern formed when light passes through a single slit and explain how this pattern provides evidence for the wave nature of light. (c) Compare and contrast the diffraction effects for: (i) Sound waves passing through a doorway (ii) Light waves passing through a doorway (iii) Radio waves passing over a mountain

[4 marks]

Explain the differences in terms of the relative sizes of wavelengths and obstacles.

[6 marks]

OTHER EXAMPLES

1. Riplles on water 2. Waves on a string

Watch and learn

• Click on the link below to watch a video that shows interference between sound waves:

Set up equipment

1. Place a laser (common source of monochromatic light) securely on one end of an optical bench
2. Mount a diffraction grating on a holder in the path of the laser beam
3. Set up a white screen about 1-2 meters away from the grating
4. Ensure the laser beam passes perpendicularly through the grating
5. Check that the central maximum (brightest spot) appears in the center of the screen
6. Ensure the room is sufficiently dark to see the pattern clearly

ANSWER: SHORT Question 3

3. (a) The student could place a polarising filter between the light source and the detector, and measure the light intensity. Then rotate the filter by 90° and measure again. If the light is unpolarised, the intensity readings would be the same regardless of the filter's orientation.

(b) The student could set up a light source, a glass plate at an angle, and a rotatable polarising filter between the reflected light and the detector. The light source would shine on the glass plate, and the detector would be positioned to receive the reflected light through the polarising filter.

(c) If the reflected light is polarised, the intensity of light reaching the detector would vary significantly as the polarising filter is rotated. The maximum and minimum intensities would occur at filter orientations 90° apart. If the light is unpolarised, the intensity would remain constant as the filter is rotated.

Conditions Required for Noticeable Interference Patterns

• Coherence: The waves must maintain a constant phase relationship with each other.
• Similar amplitudes: The waves should have roughly similar strengths. If one wave is much stronger than the other, the interference effects will be less noticeable.
• Similar frequencies: The waves should have the same or very similar frequencies. Different frequencies create complex, changing patterns rather than stable interference.
• Suitable medium: The medium must allow waves to travel without too much distortion or absorption.
• Appropriate wavelength: The wavelength should be similar to the dimensions of the setup (like the distance between sources).

ANSWER: long answer Question 3

3 (a) Diffraction is the spreading of waves when they pass through a gap or around an obstacle. It occurs because, according to wave theory, each point on a wavefront acts as a source of new waves (Huygens' principle). When waves encounter a gap or obstacle, this limits which points can continue to produce new waves, causing the wave pattern to change and spread.

(b) When light passes through a single slit, it produces a pattern with:

• A bright central maximum
• Alternating dark and bright fringes on either side
• Decreasing intensity for fringes further from center

This pattern can only be explained if light behaves as a wave, with different parts of the wave from across the slit interfering constructively (bright fringes) or destructively (dark fringes).

(c) (i) Sound waves (λ ≈ 0.3-3 m) passing through a doorway (≈ 1 m wide) experience significant diffraction as the wavelength is comparable to the doorway width. Sound waves spread out into the room, allowing us to hear someone speaking in another room.

(ii) Light waves (λ ≈ 5×10^-7 m) passing through a doorway experience minimal diffraction as the wavelength is much smaller than the doorway width. Light travels mostly in straight lines, creating a clear shadow effect and preventing us from seeing around corners.(iii) Radio waves passing over a mountain: Long-wave radio (λ ≈ 1000 m) experiences significant diffraction around mountains as the wavelength is comparable to the obstacle size. Short-wave radio (λ ≈ 10 m) experiences less diffraction as the wavelength is much smaller than the mountain, creating more defined shadow zones.

ANSWER: long answer question 3

3. (a) or electromagnetic waves to pass through an aperture, the aperture size should be comparable to or larger than the wavelength. Visible light has wavelengths of 400-700 nm, much smaller than the 1 mm holes, so it passes through easily. Microwaves with a wavelength of 12 cm are much larger than the holes, so they cannot pass through and are reflected instead.

(b) When a mesh consists of parallel conducting wires, it acts as a polarising filter because it affects the different components of the electromagnetic wave differently. Microwaves with their electric field parallel to the wires induce currents in the wires, causing the energy to be reflected or absorbed. Microwaves with their electric field perpendicular to the wires pass through more easily. This creates a polarising effect similar to an optical polarising filter

(c) The first grille creates polarised microwaves by allowing only the component with electric field perpendicular to the metal wires to pass through. When the second grille has its wires parallel to the first, these polarised microwaves can pass through. When the second grille is rotated by 90°, its wires are now parallel to the electric field of the polarised microwaves, causing them to be blocked. This demonstrates that the metal wire arrangement acts as a polarising filter for microwaves.

If you struggled with this question, review 'Polarisation Examples' section 3. Click on the button to the left to return to this page.

When does this happen?

• When the path difference equals half a wavelength plus any whole number of wavelengths (0.5λ, 1.5λ, , etc.)

• Formula

Path difference = (n + ½)λ

• Results in minimum amplitude (complete cancellation)

Answers:

1. All of the light passes through - maximum intensity reaches the viewer
2. Amount of light passing through decreases - intensity declines
3. Amount of light passing through is halved - intensity is halves
4. No light passes through - intensity is nil
5. Amount of light passing through increases - intensity increases
6. Filters are re-alligned, so all light passes through - maximum intensity

ANSWER: Question 6

(a) (i) Required conditions:

• Waves must be of equal amplitude
• Must be exactly out of phase
• Must arrive at same time

(ii) Time delay = 0.20/340 = 0.59 ms

b) (i) Number of nodes = separation/λ = 8.0/2.0 = 4 positions (ii) Pattern less distinct because:

• Energy spreads out
• Multiple reflections occur
• Water resistance dampens waves

(iii) Could use:

• Depth sensor at predicted positions
• Video analysis of wave motion

[3 marks]

ANSWER: Short Questions 1 - 4

1. Diffraction is the spreading out of waves when they pass through a gap or around an obstacle. An everyday example is hearing sounds around corners.
2. According to Huygens' construction, each point on a wavefront acts as a source of new circular wavelets. When a wave meets a narrow gap, only the points within the gap can continue to produce wavelets. These wavelets spread out in all directions from each point, combining to form a curved wavefront that spreads beyond the gap's width.
3. (a) When the width of a single slit is decreased, the diffraction pattern becomes wider, with the central maximum and side fringes spreading out more.

(b) When the wavelength of light is increased, the diffraction pattern becomes wider, with increased spacing between the minima.

4. Radio waves diffract around obstacles when the wavelength is comparable to or larger than the obstacle size. Long wavelength radio (e.g., 1000m) can diffract significantly around a mountain as the wavelength is comparable to the mountain's size. Shorter wavelength radio (e.g., 10m) experiences much less diffraction because the wavelength is much smaller than the mountain, creating a "shadow zone" behind it.

Definitions

Oscillation: A repeated back-and-forth motion around an equilibrium position. Plane: a flat surface that extends into infinity in all directions. It has infinite width and length, zero thickness, and zero curvature.

ANSWER: long answer Question 1

1. a) When light reflects from a surface, the electric field components parallel and perpendicular to the plane of incidence behave differently. The component parallel to the surface (or perpendicular to the plane of incidence) reflects more strongly than the component perpendicular to the surface. This causes the reflected light to have more of its electric field oscillating parallel to the surface, making it partially polarised in this direction

(b) When light reflects from horizontal surfaces like roads or water, the reflected light becomes partially polarised in the horizontal direction (parallel to the surface). Polaroid sunglasses with vertical transmission axes block horizontally polarised light while allowing vertically polarised light to pass. This selectively reduces the reflected glare while allowing other light through.

(c) Sunlight becomes partially polarised as it scatters in the atmosphere. The degree of polarisation depends on the angle to the sun. When looking through polarising sunglasses, the sky appears darker in directions where the scattered light is more strongly polarised perpendicular to the transmission axis of the glasses.

If you struggled with this question, review 'Polarisation Examples' section 1. Click on the button to the left to return to this page.

Wave Phase Simulator

Use the link below to access a useful phase simulator. The dots on the circle represent the position of two points as a wave oscillates, while the main graph shows the displacement of each point as the wave travels.

• Click on the displacement - distance graph to select the phase of the blue and red dots
• Adjust the slider on the y-axis to change the amplitude
• Tick the 'y vs t' box to see the displacement - time graph for the two points

Experiment with moving the dots so that they are in-phase, completely out-of-phase, and partially in-phase. Note how this impacts the displacement - time graph:

• When the dots are in-phase (with phase difference of zero or multiples of 360 degrees), there will be a single line on the displacement - time graph, as the points have the same velocity at all times.
• When the dots are completely out of phase (phase difference in odd multiples of 180 degrees), the two lines on the displacement - time graph will be mirror images, as the points have opposite velocity at all times.
• When the dots are partially out of phase, there will be two lines on the displacement - time graph with a slight 'lag' between them. This is becuse the points always have slightly differing velocities.

How do we use this property?

• Refraction is used in lenses to correct eyesight.
• It is also used in magnifying glasses
• It can be used to create rainbows

OTHER EXAMPLES

1. Earthquake waves (P-waves) 2. Ultrasound waves

Metal Grilles

used for polarising microwaves

How does it work?

• The free electrons moving in the metal bar cancel out the electric field in the same direction as the grille by completely absorbing it.
• The horizontal electric field passes if the grilles are positioned vertically and vice versa
• This is the opposite of what happens in a polarising filter, where light only passes when the filter is aligned with (i.e. in the same direction as) the waves.

What is a metal grille?

• A metal grille is similar to a polarising filter, and commonly used for polarising microwaves
• Microwaves are electromagnetic waves with a longer wavelength than visible light

ANSWER: Question 4

(a) (i) λ = v/f = 340/200 = 1.7 m

(ii) Time delay = distance/speed = 0.034/340 = 0.0001 s = 0.1 ms

(b) (i) Low frequency sounds are easier to cancel because:Longer wavelengths mean position is less criticalProcessing time is small compared to wave period Wave patterns are less affected by small movements (ii) At 2000 Hz: λ = v/f = 340/2000 = 0.17 m This means half a wavelength (where cancellation changes to reinforcement) is just 8.5 cm, making precise positioning much more critical than at 200 Hz where λ/2 = 0.85 m

[3 marks]

ANSWER: Question 3

(a) (i) For constructive interference, path difference = nλ Maximum possible path difference = speaker separation = 2.4 m Number of complete wavelengths = 2.4/0.8 = 3 Therefore 4 maxima occur (including positions at both speakers)

(ii) Maxima occur at these positions because: The path difference is a whole number of wavelengths This means waves arrive in phase Their displacements add constructively to give maximum amplitude

[3 marks]

(b) (i) When frequency increases: Wavelength decreases (λ = v/f) More wavelengths fit between speakers Therefore more maxima occur

(ii) Pattern becomes harder to detect because: Maxima are closer together, so more difficult to see Small movements cause larger phase changes

ANSWER: Question 2

(a) (i) Coherent sources produce waves that maintain a constant phase difference with each other over time. (ii) Without coherence, the phase relationship between the waves would vary randomly, causing the positions of constructive and destructive interference to shift constantly. This would prevent a stable interference pattern from forming.

[3 marks]

(b) (i) λ = v/f = 340/500 = 0.68 m (ii) The volume variations occur because:

• Waves from both speakers travel different distances to reach each point
• Where waves meet in phase (path difference = nλ), constructive interference creates loud spots
• Where waves meet in antiphase (path difference = (n + ½)λ), destructive interference creates quiet spots

DID YOU KNOW?

• At points of compression, the molecules are closer together, increasing the pressure at that point.

• At points of rarefaction, the molecules are further apart, reducing the pressure at that point.

• It's hard to represent longitudinal waves graphically, They're most commonly represented as displacement - time graphs, but this can make them look like transverse waves.

WHY you still hear sounds at 'quiet points'

• Perfect cancellation rarely occurs: Completely destructive interference requires exactly equal amplitudes and precisely opposite phases, which is difficult to achieve in real environments.
• Room reflections: Sound bounces off walls, ceilings, and objects, creating additional sound paths that interfere with the primary pattern.
• Multiple frequencies: Most sounds contain many frequencies. Even if one frequency experiences destructive interference, others may not.
• Air movement: Small changes in temperature or air movement can slightly alter the path lengths and interference patterns.

ANSWER: Short Questions 1 and 2

1. Plane polarised light consists of electromagnetic waves oscillating in a single plane perpendicular to the direction of travel. In unpolarised light, the electric field vector vibrates in all possible planes perpendicular to the direction of propagation. In plane polarised light, the electric field vector vibrates in only one plane. [Diagram should show unpolarised light with electric field vectors in multiple planes vs. polarised light with electric field vectors in only one plane]
2. (a) Intensity after P₁ = 12 mW/m² × (1/2) = 6 mW/m²

(b) When the light passes through the second filter P₂, which has its transmission axis perpendicular to P₁, no light passes through. This is because the second filter only allows light oscillating parallel to its transmission axis to pass through, but all the light after P₁ is oscillating perpendicular to this axis

(c) As the second filter rotates:

• When its transmission axis is parallel to P₁ (at 0° or 180°), maximum light passes through
• When its transmission axis is perpendicular to P₁ (at 90° or 270°), no light passes through
• At intermediate angles, the light intensity varies between maximum and zero, with the intensity decreasing as the angle between the filters increases from 0° to 90°, then increasing as the angle increases from 90° to 180°, and so on.

TV and Radio Signals

polarised by transmission aerials

What happens?

• TV and analogue radio waves are sent by transmitting aerials. They are then picked up by receiving aerials like the ones you have a home.
• The intensity of the signal received by the receiving aerial depends on its orientation (vertical, horizonatal, etc).
• If the orientation of the two aerials is aligned, the intensity of the received signal is maximised.
• If the two aerials are at 90 degrees to each other, the intensity of the signal will be zero, i.e. no signal will be picked up!

Why does this happen?

• Transmitting aerials create polarised waves.
• Receiving aerials act like a polarising filter.
• This works like having two polarised filters.

What is monochromatic light?

• Monochromtic light is light of a single wavelength/color

What is 'white light'?

• White light is really a mixture of different colours
• Each colour has a different wavelength

Explaining diffraction grating interference patterns

When monochtomatic light hits the grating:

• The incoming light waves encounter the multiple parallel slits
• Each slit acts like a new source of light waves (following Huygens' principle)
• The waves from different grooves interfere with each other
• The interference creates a pattern where waves interfere constructively at certain angles and destructively at others, creating light and dark bands.
• The light bands are the same colour as the original light source (i.e. monochromatic)

Explaining the impact of number of slits

In a diffraction pattern, the number of slits has two important effects:

1. Brightness of the pattern: More slits means more light passes through, creating brighter interference patterns. Each slit contributes light waves that can constructively interfere.
2. Sharpness of the pattern: More slits create narrower, more precise bright spots (maxima) and darker regions between them. This is because the additional light waves (one from each slit) reinforce the pattern (like having more pixels on a screen)

PRACTICAL: Observing Polarisation of Microwaves

1. A microwave transmitter emits naturally polarised waves with a wavelength of around 3 cm. To check whether the microwaves are polarised, place a microwave receiver in front of the transmitter and rotate about the axis through them
2. The signal, which is detected using an ammeter or audio amplifier and loudspeaker will rise and fall with intensity as the receiver is rotated
3. When the transmitter and the polarisation axis of the receiver are perpendicular to each other, the signal will be 0
4. When the receiver has a maximum signal, a metal grille can be placed between the transmitter and receiver, acting as a second polariser
5. When the grille is rotated (for example, in 10° increments), the signal varies and is 0 when the metal rods are aligned with the electric field vector of the emitted microwaves
6. If the microwave transmitter is producing vertically plane-polarised radiation:
7. If the bars of the metal grille are horizontally orientated, very few of the microwaves will be absorbed and the ammeter reading will be high
8. If the bars of the metal grille are vertically orientated, all of the microwaves will be absorbed and the ammeter reading will be zero

ANSWER: long answer question 2

3. (a) A TV aerial receives signals by having electrons move up and down the aerial in response to the electric field of the incoming electromagnetic wave. For a horizontally polarised signal, the electric field oscillates horizontally. Therefore, the aerial must be horizontal to align with these oscillations and generate the maximum possible current.

(b) When TV signals reflect off buildings, the polarisation can change. A horizontally polarised signal may gain a vertical component after reflection. If the aerial is oriented to receive only horizontal polarisation, it will be less effective at detecting the vertical component of the reflected signal. This can lead to weaker reception or "ghost" images where the direct and reflected signals arrive with different timings and polarisations

If you struggled with this question, review 'Polarisation Examples' section 2. Click on the button to the left to return to this page.

Definitions

Oscillation: A repeated back-and-forth motion around an equilibrium position. Equilibrium position: The normal resting position of a particle in the medium Medium: The substance through which a wave travels (water, glass, etc.)

ANSWER: long answer Question 2

2. a) Explanation should mention:

Explanation should mention:

• Huygens' principle that all points on a wavefront act as sources of new wavelets
• Limited points within slit can produce wavelets
• These wavelets spread in all directions
• Combine to form new curved wavefront that spreads beyond the original width of the slit.

• Incident plane wavefront approaching slit
• Points within slit becoming sources of secondary wavelets
• Semicircular wavelets emanating from these points
• New wavefront formed by envelope of all these wavelets

b) For a single slit diffraction pattern, the first minimum occurs at an angle θ where: sin θ = λ/a

Wavelength (λ) = 500 nm = 500×10^-9 mDistance to screen = 2.0 m Distance from central maximum to first minimum = 5.0 mm = 5.0×10^-3 m

θ = tan^-1(5.0×10^-3 m / 2.0 m) = 0.00250 radFor small angles, tan θ ≈ sin θ, therefore sin θ = 0.00250 a = λ/sin θ a = 500×10^-9 m / 0.00250 = 2.0×10^-4 m = 0.20 mm

c) Explaination should include:

• Sound waves have wavelengths of approximately 0.3-3 m while light has wavelengths around 400-700 nm (4×10^-7 to 7×10^-7 m).
• Since diffraction is significant when wavelength is comparable to obstacle size, sound waves readily diffract around corners of doors and walls (which are comparable to their wavelength), allowing us to hear around corners.
• Light waves have wavelengths thousands of times smaller than typical obstacles, so they experience negligible diffraction and travel essentially in straight lines, preventing us from seeing around corners.

ANSWER: long answer Question 1

1. Key points:

• Equipment: Ripple tank with water, wave generator, barriers with different gap widths, strobe light
• Setup: Fill tank with water, position barrier with gap across tank, generate regular waves
• Measurements: Observe pattern of waves after passing through gaps, possibly photograph or trace patterns
• Observations should include: With wide gaps (several wavelengths wide), minimal diffraction with waves continuing straight; with gaps comparable to wavelength, significant spreading out of waves in semicircular pattern
• Conclusion: Diffraction is most pronounced when gap width is comparable to wavelength

Key Relationships for diffraction gratings

The equation nλ = d sin θ reveals several important relationships in diffraction:

1. Wavelength affects diffraction angle: Longer wavelengths (like red light) diffract at larger angles than shorter wavelengths (like violet light). This is why prisms and gratings spread white light into a spectrum, with red bending more than blue.
2. Spacing matters: The closer together the slits or grooves (smaller d), the wider the spacing between bright spots in your pattern. This is why gratings with more lines per mm create more spread-out spectra.
3. Higher orders spread further: As n increases (1st order, 2nd order, etc.), the angle θ increases too. This means higher-order maxima appear further from the center of your pattern.
4. Maximum angle limit: Since sin θ can't exceed 1, there's a maximum possible value for n. This explains why you only see a limited number of diffraction orders in an experiment.

Key Factors

• The narrower the slit, the wider the diffraction pattern becomes
• This occurs because the wavelets spread out more when constrained to a smaller opening
• The pattern depends on the relationship between the slit width and the wavelength of light

For Large Obstacles (much larger than wavelength)

• Wavelets can only form around the edges of the obstacle
• The obstacle blocks most of the wave, creating a distinct shadow region
• Only slight bending occurs at the edges
• The wave largely travels in straight lines around the obstacle

For Small Obstacles (comparable to wavelength)

• Wavelets form around all sides of the obstacle
• These wavelets spread and recombine behind the obstacle
• This causes waves to bend significantly around the object
• The shadow region becomes less distinct

For Very Small Obstacles (smaller than wavelength)

• Wavelets easily spread around all sides
• They recombine almost completely behind the obstacle
• The obstacle creates minimal disturbance to the overall wave pattern
• Almost no shadow region is formed

practical applications of sound wave interference

• Noise-canceling headphones: These generate sound waves that are out-of-phase to ambient noise, creating destructive interference. The headphones identify the frequency and amplitude of the external sounds, then creates a coherent soundwave that is exactly opposite (180 degrees path difference vs the noise).
• Concert hall design: Architects use interference principles to enhance sound quality and eliminate "dead spots."
• Directional speakers: Arrays of speakers use interference to focus sound in particular directions.
• Sound level meters: Some devices use interference to measure sound intensity accurately.
• Medical ultrasound: Interference patterns help create clearer images in ultrasound scanning.

How White Light Creates Spectra Through a Diffraction Grating

When white light passes through a diffraction grating, it creates colorful spectra because:

1. White light contains all colors: White light is a mixture of all visible wavelengths (red, orange, yellow, green, blue, indigo, violet).
2. Different wavelengths bend differently: The diffraction equation (nλ = d sin θ) shows that each wavelength diffracts at a specific angle. Longer wavelengths (red) diffract at larger angles than shorter wavelengths (violet).
3. Sorting by wavelength: As white light passes through the grating, it gets sorted by wavelength. This creates a rainbow-like spectrum on each side of the central white maximum.
4. Multiple orders: You'll see several spectra (1st order, 2nd order, etc.) on each side of the central maximum, with each higher order spreading the colors even more.
5. Complete separation: Unlike a prism, where colors can overlap slightly, a good diffraction grating can completely separate close wavelengths, making it excellent for detailed spectroscopy.

Sound Wave Interference and Path Difference:

SPRING EXAMPLE:

The wave consist of alternate compressions and rarefactions of the medium it is travelling through. This is why longitudinal waves can't travel through a vacuum.

OTHER WORDS FOR intensity

• Brightness for light
• Loudness for sound

When does this happen?

• When the path difference equals a whole number of wavelengths (0λ, 1λ, 2λ, etc.)

• Formula:

Path difference = nλ

• Results in maximum amplitude where the waves combine.

ANSWER: Short Questions 5 - 6

5. (a) When the gap width is reduced to approximately the same as the wavelength, the diffraction becomes much more pronounced, with waves spreading out in an almost semicircular pattern after passing through the gap.

(b) When the frequency of the wave generator is increased, the wavelength decreases, resulting in less pronounced diffraction (the pattern becomes narrower with the waves spreading out less).

6. Graph should show:

• intensity vs position
• a tall central maximum flanked by smaller secondary maxima,
• zero intensity at the minima positions

ANSWER: Question 1

(a) The wavelength of a wave is the distance between two consecutive points on a wave that are in phase (or the distance between two consecutive peaks or troughs).

(b) (i) Using v = fλλ = v/f = 340/400 = 0.85 m (ii) If frequency is doubled, wavelength is halved (since v = fλ and v remains constant)

[3 marks]

(ii) [Two clearly labeled diagrams required showing]:Constructive interference: Two pulses with same displacement adding to give larger resultant Destructive interference: Two pulses with opposite displacements canceling out

calculate Wavelength

1. Use the formula nλ = d sin θ (For small angles, you can approximate sin θ ≈ tan θ)
2. Rearrange to find wavelength: λ = d sin θ/n
3. Take multiple readings for different orders
4. Calculate the average wavelength value
5. Calculate percentage uncertainty using the measurement uncertainties

Displacement - Distance

Displacement - Time

Measuring Phase Difference

Why do we use degrees / radians?

Think of a wave as something that repeats over and over - like a roller coaster that goes up and down. One complete cycle (up and down and back to the starting point) represents 360 degrees or 2π radians. Here's why it makes sense:

• It's a universal measurement: Whether we're talking about sound waves, light waves, or ocean waves, they all have cycles. Using degrees or radians gives us a consistent way to compare waves of different types.
• It's proportional: If one wave is a quarter cycle ahead of another, we can say it's 90 degrees (or π/2 radians) out of phase. If it's halfway through its cycle compared to another, it's 180 degrees (or π radians) out of phase.
• It's circular in nature: Wave motion repeats, just like going around a circle. When you complete 360 degrees (or 2π radians), you're back where you started - this matches perfectly with the repeating nature of waves.

Example:

take measurements

1. Mark the position of the central maximum (0th order) on the screen
2. Mark the positions of the 1st order maxima on both sides of the central spot
3. If visible, mark the 2nd and 3rd order maxima as well
4. Measure the distance (L) from the grating to the screen with a meter rule
5. Measure the distance (x) from the central maximum to each of the other maxima
6. For each maximum, calculate the angle θ using: tan θ = x/L

Polarising Light

for photography & sunglasses

How is this used?

• In photography, attaching a polarising filters to the camera lense removes unwanted reflections.
• In Polaroid sunglasses, polarising filters are incorporated into lenses to remove glare

What happens?

• When you direct a beam of unpolarised light at a reflective surface, then view the reflected ray though a polarising filter, the intensity of light leaving the filter changes with the orientation of the filter.

Why does this happen?

• This happens because, at certain angles, light is partially polarised when it is reflected.
• This is shown in the image to the right.

This video explains some of the key concepts relating to polarisation. Note that you are not required to do any of the calculations in the video at AS level.

Watch

Did you know?

• Light from an ordinary bulb is non-coherent (random), while laser light is highly coherent (synchronized).
• Coherent waves are essential for applications like holograms, interferometry, and many optical technologies where predictable wave interactions matter.

ANSWER: Question 7

(c) (i) When amplitudes are A and 2A: Maximum amplitude = 3A (in phase) (ii) When out of phase: Minimum amplitude = A (larger wave minus smaller) Therefore complete silence does not occur

(c) (i) When waves superpose: Amplitudes add linearly (2×) Energy/intensity proportional to amplitude squared (4×) (ii) Energy is conserved because:

• High energy at maxima balanced by low energy at minima
• Average energy same as individual waves

[3 marks]

EXAMPLEs

1. You can see the reflection of ligt in a mirror. 2. The reflection of water can be demonstrated in a ripple tank.

How do we know that waves transfer energy?

ANSWER: Question 5

a) (i) λ = v/f = 340/440 = 0.77 m(ii) For maxima along line parallel to speakers:Distance between maxima = λD/d= (0.77 × 3.0)/2.0 = 1.16 m

(iii) Some sound still heard because:

• Perfect cancellation requires exact amplitude matching
• Room reflections add extra wave paths
• Head size means ears receive different phases

[3 marks]

(b) (i) When frequency doubles:

• Wavelength halves
• Distance between maxima halves

(ii) New distance = 1.16/2 = 0.58 m (iii) Harder to locate because:

• Maxima closer together
• Head movement causes larger phase changes
• Multiple frequencies present in real sounds

Answer