Waves

Question 1

What are progressive waves?

Answer 1

A wave that transfers energy from one point to another without transferring the medium itself

Question 2

Properties of a progressive wave

Answer 2

Displacement (x) of a wave is the distance of a point on the wave from its equilibrium position
It is a vector quantity; it can be positive or negative

Amplitude (A) is the maximum displacement of a particle in the wave from its equilibrium position

Wavelength (λ) is the distance between points on successive oscillations of the wave that are in phase

These are all measured in metres (m)

Question 3

What is phase difference a measure of

Answer 3

how much a point or a wave is in front or behind another

Question 4

What is polarisation

Answer 4

Particle oscillations occur in only one of the directions perpendicular to the direction of wave propagation

Question 5

Why can polarisation only occur in transverse waves?

Answer 5

because transverse waves oscillate in any plane perpendicular to the propagation direction

Question 6

What does it mean when transverse waves are polarised

Answer 6

Vibrations are restricted to one direction

These vibrations are still perpendicular to the direction of propagation / energy transfer

Question 7

Applications of polarisers: Polaroid sunglasses

Answer 7

Polaroid sunglasses are glasses containing lens with polarising filters with transmission axes that are vertically oriented
- the glasses do not allow any horizontally polarised light to pass through

When light is reflected from a reflective surface it undergoes partial plane polarisation
- This means if the surface is horizontal, a proportion of the reflected light will oscillate more in the horizontal plane than the vertical plane

polaroid sunglasses are useful in reducing the glare on the surface of the water (or any reflective surface) as the partially-polarised light will be eliminated by the polarising filter

So objects under water surface are viewed more clearly

Question 8

Applications of polarisations: Polaroid photography

Answer 8

Polarising filters also enable photographers to take photos of objects underwater

This is because the light reflected on the surface of the water is partially polarised in the horizontal plane

This glare is eliminated by the polarising lens

However, the light from the underwater object is refracted by the surface of the water, not reflected, so it is not plane-polarised

Therefore, the light from the underwater object is more intense than the glare and shows up much more brightly in the photo

Question 9

Applications of polarisation : Polarisation of radio & microwave signals

Answer 9

Radio and television services are broadcast either horizontally-polarised or vertically-polarised

Therefore, the reception aerial needs to be mounted flat (horizontal), or on its side (vertical),

The particular orientation of an aerial will depend on the transmitter it is pointing towards and the polarity of the services being broadcast

Question 10

How are standing waves produced

Answer 10

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

Question 11

Do stationary waves store energy or not

Answer 11

Yes, unlike progressive waves which transfer energy

Question 12

What are nodes

Answer 12

Regions where there is no vibration

Question 13

What are anti nodes?

Answer 13

Regions where the vibrations are at their maximum amplitude

Question 14

When are points in phase or out of phase

Answer 14

Points between nodes are in phase with each other

Points that have an odd number of nodes between them are out of phase

Points that have an even number of nodes between them are in phase

Question 15

State the principle of superposition

Answer 15

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

Question 16

Describe superposing IN PHASE

Answer 16

causes constructive interference.

The peaks and troughs line up on both waves and the resultant wave has double the amplitude

Question 17

Describe superposing in ANTI-PHASE

Answer 17

Causing destructive interference.

The peaks on one wave line up with the troughs of the other. The resultant wave has no amplitude

Question 18

When is a stationary wave formed

Answer 18

Two waves travelling in opposite directions along the same line with the same frequency superpose

Question 19

Examples of stationary waves: stretched string

Answer 19

Vibrations caused by stationary waves on a stretched string produce sound
This is how stringed instruments, such as guitars or violins, work

At specific frequencies, known as resonant frequencies, a whole number of half wavelengths will fit on the length of the string

As the resonant frequencies of the oscillator are achieved, standing waves with different numbers of minima (nodes) and maxima (antinodes) form

Question 20

Examples of stationary waves: microwaves

Answer 20

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

Question 21

Examples of stationary waves: sound waves

Answer 21

Sound waves can be produced as a result of the formation of stationary waves inside an air column

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

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

Question 22

What are harmonics

Answer 22

Different wave patterns

Question 23

What do harmonics depend on

Answer 23

on the frequency of the vibration and the situation in which they are created

Question 24

Harmonics on a string
Frequency formula

Answer 24

Question 25

Required Practical: Investigating Stationary Waves Aims

Answer 25

The overall aim of the experiment is to measure how the frequency of the first harmonic is affected by changing one of the following variables: The length of the string The tension in the string Strings with different values of mass per unit length

Question 26

Required Practical: Investigating Stationary Waves Variables

Answer 26

Independent variable = either length, tension, or mass per unit length Dependent variable = frequency of the first harmonic Control variables If length is varied = same masses attached (tension), same string (mass per unit length) If tension is varied = same length of the string, same string (mass per unit length) If mass per unit length is varied = same masses attached (tension), same length of the string

Question 27

Required Practical: Investigating Stationary Waves Method

Answer 27

1. Set up the apparatus by attaching one end of the string to the vibration generator and pass the other end over the bench pulley and attach the mass hanger 2. Adjust the position of the bridge so that the length L is measured from the vibration generator to the bridge using a metre ruler 3. Turn on the signal generator to set the string oscillating 4. Increase the frequency of the vibration generator until the first harmonic is observed and read the frequency that this occurs at 5. Repeat the procedure with different lengths Repeat the frequency readings at least two more times and take the average of these measurements 6. Measure the tension in the string using T = mg Where m is the amount of mass attached to the string and g is the gravitational field strength on Earth (9.81 N kg–1) 7. Measure the mass per unit length of the string μ = mass of string ÷ length of string - Simply take a known length of the string (1 m is ideal) and measure its mass on a balance

Question 28

Required Practical: Investigating Stationary Waves Systematic errors

Answer 28

An oscilloscope can be used to verify the signal generator’s readings The signal generator should be left for about 20 minutes to stabilise The measurements would have a greater resolution if the length used is as large as possible, or as many half-wavelengths as possible This means measurements should span a suitable range, for example, 20 cm intervals over at least 1.0 m

Question 29

Required Practical: Investigating Stationary Waves Random errors

Answer 29

The sharpness of resonance leads to the biggest problem in deciding when the first harmonic is achieved This can be resolved by adjusting the frequency while looking closely at a node. This is a technique to gain the largest response Looking at the amplitude is likely to be less reliable since the wave will be moving very fast When taking repeat measurements of the frequency, the best procedure is as follows: Determine the frequency of the first harmonic when the largest vibration is observed and note down the frequency at this point Increase the frequency and then gradually reduce it until the first harmonic is observed again and note down the frequency of this If taking three repeat readings, repeat this procedure again Average the three readings and move onto the next measurement

Question 30

When does interference occur

Answer 30

when waves overlap and their resultant displacement is the sum of the displacement of each wave

Question 31

What is coherence

Answer 31

At points where the two waves are neither in phase nor in antiphase, the resultant amplitude is somewhere in between the two extremes

Question 32

When are waves said to be coherent

Answer 32

If they have: The same frequency A constant phase difference

Question 33

What is the importance of coherence

Answer 33

Coherence is vital in order to produce an observable, or hearable, interference pattern Laser light is an example of a coherent light source, whereas filament lamps produce incoherent light waves When coherent sound waves are in phase, the sound is louder because of constructive interference

Question 34

Define path difference

Answer 34

The difference in distance travelled by two waves from their sources to the point where they meet

Question 35

What is constructive interference in the screen

Answer 35

Bright fringes - the highest intensity is in the middle

Question 36

What is destructive interference in the screen

Answer 36

Dark fringes - these have zero intensity

Question 37

For two-source interference fringes to be observed, the sources of the wave must be…

Answer 37

Coherent (constant phase difference) Monochromatic (single wavelength)

Question 38

What will the difference in wavelengths be for constructive interference

Answer 38

an integer number of whole wavelengths

Question 39

What will be the number of wavelengths for destructive interference (or minima)

Answer 39

an integer number of whole wavelengths plus a half wavelength

Question 40

What does young’s double slit experiment demonstrate

Answer 40

how light waves can produce an interference pattern

Question 41

Describe young’s double slit experiment

Answer 41

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

Question 42

3.3.5 Required Practical: Young's Slit Experiment & Diffraction Gratings Variables

Answer 42

Independent variable = Distance between the slits and the screen, D Dependent variable = Fringe width, w Control variables Laser wavelength, λ Slit separation, s

Question 43

3.3.5 Required Practical: Young's Slit Experiment & Diffraction Gratings Aim

Answer 43

to investigate the relationship between the distance between the slits and the screen, D, and the fringe width, w

Question 44

3.3.5 Required Practical: Young's Slit Experiment & Diffraction Gratings Method

Answer 44

1. Set up the apparatus by fixing the laser and the slits to a retort stand and place the screen so that D is 0.5 m, measured using the metre ruler 2. Darken the room and turn on the laser 3. Measure from the central fringe across many fringes using the vernier callipers and divide by the number of fringe widths to find the fringe width, w 4. Increase the distance D by 0.1 m and repeat the procedure, increasing it by 0.1 m each time up to around 1.5 m 5. Repeat the experiment twice more and calculate and record the mean fringe width w for each distance D

Question 45

Interference by a Diffraction Grating practical aim

Answer 45

s to calculate the wavelength of the laser light using a diffraction grating

Question 46

Interference by a Diffraction Grating practical variables

Answer 46

Independent variable = Distance between maxima, h Dependent variable = The angle between the normal and each order, θn (where n = 1, 2, 3 etc) Control variables Distance between the slits and the screen, D Laser wavelength λ Slit separation, d

Question 47

Interference by a Diffraction Grating practical Method

Answer 47

1. Place the laser on a retort stand and the diffraction grating in front of it 2. Use a set square to ensure the beam passes through the grating at normal incidence and meets the screen perpendicularly 3. Set the distance D between the grating and the screen to be 1.0 m using a metre ruler 4. Darken the room and turn on the laser Identify the zero-order maximum (the central beam) 5. Measure the distance h to the nearest two first-order maxima (i.e. n = 1, n = 2) using a vernier calliper 6. Calculate the mean of these two values Measure distance h for increasing orders Repeat with a diffraction grating with a different number of slits per mm

Question 48

Diffraction grating equation

Answer 48

nλ = d sin θ

Question 49

what is diffraction

Answer 49

the spreading out of waves when they pass an obstruction This obstruction is typically a narrow slit known as an aperture

Question 50

What are the features of the single slit diffraction pattern

Answer 50

A central maximum with a high intensity Subsidiary maxima equally spaced, successively smaller in intensity and half the width of the central maximum

Question 51

What is a diffraction grating

Answer 51

a plate on which there is a very large number of parallel, identical, close-spaced slits

Question 52

What happens when monochromatic light is incident on a grating

Answer 52

a pattern of narrow bright fringes is produced on a screen

Question 53

Diffraction grating equation

Answer 53

Question 54

How to calculate angular separation (diffraction grating)

Answer 54

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 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

Question 55

What are diffraction gratings useful for

Answer 55

separating light of different wavelengths with high resolution

Question 56

What are diffraction gratings used in spectrometers to do

Answer 56

Analyse light from stars Analyse the composition of a star Chemical analysis Measure red shift / rotation of stars Measure the wavelength / frequency of light from a star Observe the spectra of materials Analyse the absorption / emission spectra in stars

Question 57

What do diffraction gratings do in X-ray crystallography

Answer 57

X-rays are directed at a thin crystal sheet which acts as a diffraction grating to form a diffraction pattern This is because the wavelength of x-rays is similar in size to the gaps between the atoms This diffraction pattern can be used to measure the atomic spacing in certain materials

Question 58

What are the only properties that change during refraction

Answer 58

speed and wavelength – the frequency of waves does not change

Question 59

How to calculate refracting index

Answer 59

C = the speed of light in a vacuum m/s Ces - the speed of light in a substance m/s

Question 60

What is snells law

Answer 60

Snell’s law relates the angle of incidence to the angle of refraction, it is given by: n1 sin θ1 = n2 sin θ2 Where: n1 = the refractive index of material 1 n2 = the refractive index of material 2 θ1 = the angle of incidence of the ray in material 1 θ2 = the angle of refraction of the ray in material 2

Question 61

Critical angle formula

Answer 61

Question 62

When does total internal reflection occur

Answer 62

The angle of incidence is greater than the critical angle and the incident refractive index n1 is greater than the refractive index of the material at the boundary n2

Question 63

What are the two conditions for total internal reflection

Answer 63

The angle of incidence > the critical angle The refractive index n1 is greater than the refractive index n2

Question 64

What are three main components that make up optical fibres

Answer 64

An optically dense core, such as plastic or glass A lower optical density cladding surrounding the core An outer sheath Since the refractive index of the core is more than the refractive index of the cladding, this allows TIR to occur ncladding < ncore

Question 65

What is importance of outer sheath in optic fibres

Answer 65

Prevents physical damage to the fibre Strengthens the fibre Protects the fibre from the outside from scratches

Question 66

What is importance of cladding in optic fibres

Answer 66

It protects the core from damage It prevents signal degradation through light escaping the core, which can cause information from the signal to be lost It keeps signals secure and maintains the quality of the original signal It prevents scratching of the core It keeps the core away from adjacent fibre cores hence preventing crossover of information to other fibres It provides the fibre with strength and prevents breakage given that the core needs to be very thin

Question 67

Pulse broadening an absorption When does absorption occur

Answer 67

Part of the signal’s energy is absorbed by the fibre The signal is attenuated by the core This reduces the amplitude of the signal, which can lead to a loss of information

Question 68

Pulse broadening an absorption What is pulse broadening caused by

Answer 68

modal and material dispersion The consequence of pulse broadening is that different pulses could merge, resulting in a completely distorted final pulse

Question 69

Reducing Pulse Broadening & Absorption To reduce absorption:

Answer 69

Use a core which is extremely transparent Use of optical fibre repeaters so that the pulse is regenerated before significant absorption has taken place

Question 70

Reducing Pulse Broadening & Absorption To reduce pulse broadening:

Answer 70

Make the core as narrow as possible to reduce the possible differences in path length of the signal Use of a monochromatic source so the speed of the pulse is constant Use of optical fibre repeaters so that the pulse is regenerated before significant pulse broadening has taken place Use of single-mode fibre to reduce multipath modal dispersion

Question 71

Role of the single slit

Answer 71

(The single slit makes) the waves from the light source coherent / identical / in phase (before they hit the double slit)

Question 72

Role of the single slit

Answer 72

(The single slit makes) the waves from the light source coherent / identical / in phase (before they hit the double slit)

Question 73

Explain why coherent light is required

Answer 73

Coherent light is needed to observe clear interference fringes OR The waves (from the double slit) must have a constant phase difference to observe clear interference fringes

Question 74

Explain the formation of the fringes seen on the screen

Answer 74

Interference fringes are formed [1 mark] When light from the two slits overlap/superpose [1 mark] The light from the two slits is coherent [1 mark] The bright/blue fringes are formed where light from the two slits interfere constructively/crest meets crest/reinforces [1 mark] The dark fringes are formed where light from the two slits interfere destructively/crest meets trough/cancels [1 mark] Constructive interference/reinforcement occurs when light waves are in phase/phase difference is a whole number of wavelengths [1 mark] OR Destructive interference/cancellation occurs when light waves are out of phase of 180°/in anti-phase/phase difference is an odd number of half wavelengths [1 mark]