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FOOD FOR PHYSICS > Waves > Flashcards

Waves Flashcards

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1
Q

What are progressive waves?

A

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

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2
Q

Properties of a progressive wave

A

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)

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3
Q

What is phase difference a measure of

A

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

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4
Q

What is polarisation

A

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

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5
Q

Why can polarisation only occur in transverse waves?

A

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

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6
Q

What does it mean when transverse waves are polarised

A

Vibrations are restricted to one direction

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

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7
Q

Applications of polarisers: Polaroid sunglasses

A

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

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8
Q

Applications of polarisations: Polaroid photography

A

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

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9
Q

Applications of polarisation : Polarisation of radio & microwave signals

A

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

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10
Q

How are standing waves produced

A

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

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11
Q

Do stationary waves store energy or not

A

Yes, unlike progressive waves which transfer energy

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12
Q

What are nodes

A

Regions where there is no vibration

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13
Q

What are anti nodes?

A

Regions where the vibrations are at their maximum amplitude

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14
Q

When are points in phase or out of phase

A

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

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15
Q

State the principle of superposition

A

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

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16
Q

Describe superposing IN PHASE

A

causes constructive interference.

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

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17
Q

Describe superposing in ANTI-PHASE

A

Causing destructive interference.

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

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18
Q

When is a stationary wave formed

A

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

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19
Q

Examples of stationary waves: stretched string

A

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

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20
Q

Examples of stationary waves: microwaves

A

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

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21
Q

Examples of stationary waves: sound waves

A

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

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22
Q

What are harmonics

A

Different wave patterns

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23
Q

What do harmonics depend on

A

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

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24
Q

Harmonics on a string
Frequency formula

A
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25
Q

Required Practical: Investigating Stationary Waves
Aims

A

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

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26
Q

Required Practical: Investigating Stationary Waves
Variables

A

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

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27
Q

Required Practical: Investigating Stationary Waves
Method

A
  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
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28
Q

Required Practical: Investigating Stationary Waves
Systematic errors

A

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

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29
Q

Required Practical: Investigating Stationary Waves
Random errors

A

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

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30
Q

When does interference occur

A

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

31
Q

What is coherence

A

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

32
Q

When are waves said to be coherent

A

If they have:

The same frequency
A constant phase difference

33
Q

What is the importance of coherence

A

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

34
Q

Define path difference

A

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

35
Q

What is constructive interference in the screen

A

Bright fringes
- the highest intensity is in the middle

36
Q

What is destructive interference in the screen

A

Dark fringes
- these have zero intensity

37
Q

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

A

Coherent (constant phase difference)

Monochromatic (single wavelength)

38
Q

What will the difference in wavelengths be for constructive interference

A

an integer number of whole wavelengths

39
Q

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

A

an integer number of whole wavelengths plus a half wavelength

40
Q

What does young’s double slit experiment demonstrate

A

how light waves can produce an interference pattern

41
Q

Describe young’s double slit experiment

A

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

42
Q

3.3.5 Required Practical: Young’s Slit Experiment & Diffraction Gratings

Variables

A

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

43
Q

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

A

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

44
Q

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

A
  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
45
Q

Interference by a Diffraction Grating practical aim

A

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

46
Q

Interference by a Diffraction Grating practical variables

A

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

47
Q

Interference by a Diffraction Grating practical
Method

A
  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
48
Q

Diffraction grating equation

A

nλ = d sin θ

49
Q

what is diffraction

A

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

50
Q

What are the features of the single slit diffraction pattern

A

A central maximum with a high intensity

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

51
Q

What is a diffraction grating

A

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

52
Q

What happens when monochromatic light is incident on a grating

A

a pattern of narrow bright fringes is produced on a screen

53
Q

Diffraction grating equation

A
54
Q

How to calculate angular separation (diffraction grating)

A

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

55
Q

What are diffraction gratings useful for

A

separating light of different wavelengths with high resolution

56
Q

What are diffraction gratings used in spectrometers to do

A

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

57
Q

What do diffraction gratings do in X-ray crystallography

A

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

58
Q

What are the only properties that change during refraction

A

speed and wavelength – the frequency of waves does not change

59
Q

How to calculate refracting index

A

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

60
Q

What is snells law

A

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

61
Q

Critical angle formula

A
62
Q

When does total internal reflection occur

A

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

63
Q

What are the two conditions for total internal reflection

A

The angle of incidence > the critical angle

The refractive index n1 is greater than the refractive index n2

64
Q

What are three main components that make up optical fibres

A

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

65
Q

What is importance of outer sheath in optic fibres

A

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

66
Q

What is importance of cladding in optic fibres

A

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

67
Q

Pulse broadening an absorption

When does absorption occur

A

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

68
Q

Pulse broadening an absorption

What is pulse broadening caused by

A

modal and material dispersion

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

69
Q

Reducing Pulse Broadening & Absorption
To reduce absorption:

A

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

70
Q

Reducing Pulse Broadening & Absorption

To reduce pulse broadening:

A

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

71
Q

Role of the single slit

A

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

72
Q

Role of the single slit

A

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

73
Q

Explain why coherent light is required

A

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

74
Q

Explain the formation of the fringes seen on the screen

A

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]

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