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Physics > Waves > Flashcards

Waves Flashcards

1
Q

What a progressive wave

A

Carries energy from one place to another without transferring any material

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2
Q

What are some ways to tell a wave carries energy

A

-Electromagnetic waves cause things to heat up
-X rays and gamma rays knock electrons out of their orbits, causing ionisation
- Loud sounds cause large oscillations of air particles which can make things vibrate
- wave power can make things vibrate

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3
Q

Reflection

A

The wave is bounced back when it hits a boundary, e.g. you can see the reflection of light in mirrors

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4
Q

Refraction

A

The wave changes direction as it enters a different medium

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5
Q

Diffraction

A

The wave spreads out as it passes though a gap or round an obstacle, e.g. you can hear sound round a corner

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6
Q

Displacement

A

Measured in meters(m)
How far a point on the wave has moved from its undisturbed position

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7
Q

Amplitude

A

Measured in meters(m)
The maximum magnitude of displacement, I.e. the distance from the undisturbed position to the crest or the trough

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8
Q

Wavelength

A

Measured in meters(m)
The length of one whole wave oscillation or wave cycle, e.g. the distance between two crests, (or troughs)

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9
Q

Period

A

T, measured in seconds(s)
Time taken for one whole wave cycle

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10
Q

Frequency

A

F, measured in hertz(Hz)
The whole number of wave cycles(oscillations) per second passing a given point. Or the number of oscillations given out from a source per second

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11
Q

Phase

A

A measurement of the position of a certain point along a wave cycle
Measured in angles( degrees or radians)
Points in phase have the same displacement and velocity

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12
Q

Phase difference

A

The amount by which one wave lags behind another wave
Measured in angles( degrees or radians)

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13
Q

Frequency and period

A

F = 1/T
F = frequency (hertz)
T = period (seconds)

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14
Q

Calculating wave speed

A

c = d/t
c = wave speed in ms-1
d = distance in meters(m)
t = time in seconds (s)

C = f λ
λ = wavelength in meters(m)
f = frequency in hertz(Hz)
c = wave speed in ms-1

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15
Q

Electromagnetic wave speed in a vacuum

A

All electromagnetic waves, including light, travel at the speed of light in a. Vacuum
C = 3.00 x 10^8 ms-1 in a vacuum

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16
Q

Transeverse waves

A

In transverse waves the displacement of particles or field is at right angles to the direction of energy propagation ( transfer).
All electromagnetic waves are transverse
Other examples are ripples on water, waves in strings, and some types of earthquake shock wave (S-waves)

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17
Q

Longitudinal waves

A

The diplacement of the particles or fields is along the direction of energy propagation
E.g. sound waves or P-waves

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18
Q

Polarised waves

A

A wave that oscillates in one direction
For example: ordinary light waves are a mixture of different of vibration. A polarising filter can be used to polarise light and other waves. It only transmits vibrations in one direction

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19
Q

Measuring the speed of sound with microphones

A

Use two microphones in a straight line a distance apart.
The microphones must have separate inputs so the signal from each can be recorded separately
Use the signal generator to produce a sound from the loudspeaker and use a computer to record the time between the first and second microphone
Then use speed = distance / time

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20
Q

Measuring wave speed in water

A

Use a ripple tank
Record the depth of water using a ruler
Use a ripple tank dipper to create vibrations with a regular frequency in the tank
Dim main lights in room and turn on the strobe light(a light that flashes periodically)
Increase the frequency of the strobe light from zero until the waves appear to be standing still
When this happens the frequency of the strobe light is equal to the frequency of the water waves
Use a ruler on white paper below tank to measure the

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21
Q

Polarising filter

A

Ordinary light waves are a mixture of different directions of vibration
A polarising filter can be used to polarise light and other waves. It only transmit vibrations in one direction.
Polarisation can only happen for transverse waves. It provides evidence for the nature of transverse waves

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22
Q

Two polarising filters

A

If you have two polarising filters at right angles to each other, then no light will get through
The second filter blocks out all of the light when the transmissions axis is at right angles to the plane of polarisation.
Otherwise, it just reduces the intensity of the light passing through it(but still allows some light through it)

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23
Q

Nature of electromagnetic waves

A

In 1808, Etienne-Louis Malus discovered that light was polarised by reflection
At that time light was thought as longitudinal wave, so polarisation was hard to explain. In 1817, Young suggested that light was a transverse wave consisting of vibrating electric and magnetic fields perpendicular to the direction of energy transfer. This explained why light could be polarised

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24
Q

Polarisation in the real world

A

Most light you see is unpolarised - the vibrations are in all possible directions. But light reflected off some surfaces is partially polarised - some of it is made to vibrate in the same direction
The amount of polarisation depends on the angle of the incident light

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25
Q

Glare reduction

A

When light reflected by surfaces such as water, glass or Tarmac enters the eye, it can cause glare. The fact that reflected light is partially polarised allows us to filter some of it out with polarising filters.
If you view partially-polarised reflected light through a polarising filter at a right angle, you can block out some of the reflected light, while letting through light which vibrates at the angle of the filter.
This reduces the intensity of light enetering your eye
This effect is used to reduce unwanted reflections in photography, and in Polaroid sunglasses to reduce glare

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26
Q

Improving TV signals

A

TV signals are polarised by the orientation of the rods on the transmitting aerial.
To receive a strong signal, you have to line up the rods on the receiving aerial with the rods on the transmitting aerial- if they aren’t aligned, the signal strength will be lower, so the rods on the TV aerial are all horizontal

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27
Q

Improving radio signals

A

Radio signals are the same as TV signals-
If you try turning a radio and then moving the aerial around, your signal will come and go as the transmitting and receiving aerials go in and out of alignment

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28
Q

Superposition of waves

A

Superposition happens when two or more waves pass through each other.
At the instant that waves cross, the displacements due to each wave combine the the wave continues on its way

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29
Q

Principle of superposition

A

Says when two or more waves cross, the resultant displacement equal the vector sum of the individual displacements

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30
Q

Constructive interference

A

When two waves meet, if their displacements are in the same direction, the displacements combine to give a bigger displacement.
A crest plus a crest gives a bigger crest
A trough plus a trough gives a bigger trough

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31
Q

Destructive interference

A

If a wave with a positive displacement (crest) meets a wave with a negative displacement (trough), they’ll undergo destructive interference and cancel eachother out

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32
Q

Total destructive interference

A

If two waves with equal and opposite displacements meet (e.g. a crest and trough with equal magnitudes), they cancel each other out completely

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33
Q

Phase difference maths

A

One complete cycle of a wave = 360 degrees (2pie radians)
Two points with a phase difference of zero or a multiple of 360 are in phase
Points with a phase difference of odd-number multiples of 180 (pie radians, or half a cycle) are exactly out of phase

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34
Q

Stationary (stationary) wave

A

The superposition of two progressive waves with the same frequency (or wavelength) and amplitude, moving in opposite directions. Unlike progressive waves, no energy is transmitting by a stationary wave

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35
Q

Progressive wave

A

A moving wave that carries energy from one place to another without transferring any material

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36
Q

Resonant frequency and

A

A frequency at which a stationary wave is formed because an exact number of waves are produced in the time it takes for a wave to get to the end of the vibrating medium and back again

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37
Q

Demonstrating stationary waves

A

Set up a driving oscillator at one end of a stretched string with the other end fixed
The wave generated by the oscillator is reflected back and forth
For most frequencies the resultant pattern is a jumble. However, if the oscillator happens to produce an exact number of waves in the time it takes for a wave to get to the end and back again, then the original and reflected waves reinforce each other
The frequencies at which this happens date called resonant frequencies and it causes a stationary wave where the overall pattern doesn’t move along - it just vibrates up and down, so the string form oscillating ‘loops’
These stationary waves are transverse waves, so each particle vibrates at right angles to the string.

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38
Q

Nodes and antinodes

A

Nodes - amplitude = zero, stay perfectly still
Antinodes - points of maximum amplitude
At resonant frequencies, an exact number of half wavelength fits into the string
At a node there is total detsructive interference - the displacement of two wave cancel each other out
At an antibody there is constructive interference - the displacements of the two waves combine to make a bigger displacements

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39
Q

First harmonic

A

This stationary wave is vibrating at its lowest possible resonant frequency, called the first harmonic
It has one loop with a node at each end. One half wavelength fits onto the string, and so the wavelength is double the length of the string

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40
Q

Second harmonic

A

Has twice the frequency of the first harmonic.
There are two loops with a node in the middle and one at each end
Two half wavelengths fit onto the string, so the wavelength is the length of the string

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41
Q

Third harmonic

A

The third harmonic is three times the frequency of the first harmonic
1 and a half wavelengths fit on the string

42
Q

Other harmonics

A

You can have as many harmonics as you like
An extra loop and extra node are just added with each one, the number of (lambda) that fit goes up by a half, and the frequency increase by the value of the frequency of the first harmonic

43
Q

Stationary microwaves

A

You can set up a stationary wave by reflecting a microwave beam at a metal plate
The superposition of the wave and its reflection produces a stationary wave
You can find the nodes and antinodes by moving the probe between the transmitter and the reflecting plate
The meteor loudspeaker receives no signal at the nodes and maximum signal at the antinodes

44
Q

Stationary sound waves

A

Powder in a tube of air can show stationary sound waves
A loudspeaker produces stationary sound waves in the glass tube
The powder laid along the bottom of the tube is shaken away from the antinodes but left undisturbed at the nodes
The distance between each pile of powder (node) is (lambda/2), so the speed of sound c = f x lambda is equal to c = f x 2d = 2df
So the speed of sound can be calculated by measuring d and knowing the frequency of the signal generator

45
Q

Mass per unit length

A

To find the mass per unit of length, you must measure the mass and length of the strings to different types using a mass balance, and a ruler.
then find the mass per unit length of each string using
ų = M/L

46
Q

Tension

A

Tension on a string (N) equals acceleration due to gravity (ms-2) times the total mass of the masses (kg)

47
Q



A

Turn on the signal generator and vary the frequency at which the vibration transducer vibrates
find the first harmonic
the frequency of the signal generator tells you the frequency of the first harmonic
You can then begin to investigate how each factors affects the frequency of the first harmonic on a string, keep all other factors constant, and make one of the following changes :
-Move the vibrations, transducer, towards, or away from the police to change the length of the vibrating string
-Add masses to change the tension on the string
-use a range of string samples of varying masses change the mass per unit length

48
Q

Factors effecting resonant frequency

A

-The longer, the string, the lower, the resonant frequency - as the half wavelength is longer
-The heavier, the string, the lower, the resonant frequency - as waves travel more slowly down the string.
For a given length a lower velocity, c, makes a lower frequency, f
- The lower detention on the string, the lower, the resident frequency - as waves travel more slowly down a loose string

49
Q

What affects resonant frquency

A

Length, mass per unit length and tension all affect the resonant frequency

50
Q

Calculating resonant frequency

A

F = 1/2L root T/ų

51
Q

Diffraction

A

The way that waves spread out as they come through a narrow gap or go round obstacles is called diffraction.
All waves diffract, but it’s not always easy to observe 

52
Q

How wavelength and the size of the gap affect diffraction

A

The amount of defraction depends on the wavelength of the wave compared with the size of the gap-
-When the gap is a lot bigger than the wavelength, diffraction is unnoticeable
-You get noticeable diffraction through a gap several wavelengths wide
- The most diffraction is when the gap is the same size as the wavelength
- If the gap is smaller than the wavelength, the waves are mostly just reflected back

53
Q

Diffraction through doorway example

A

When sound passes through a doorway, the size of the gap and the wavelength are usually roughly equal, so a lot of diffraction occurs. That’s why you have no trouble hearing someone through an open door to the next room, even if the other person is out your line of sight.
The reason you can’t see them is that when light passes through the doorway, it is passing through a gap around a few million times bigger than its wavelength - the amount of diffraction is tiny.
To get a diffraction with lights, you must shine it through a very narrow slit

54
Q

Diffraction around obstacles

A

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 and so the longer the shadow

55
Q

Monochromatic light

A

light of a single wavelength (and frequency), and so a single colour

56
Q

What happens if you use light that isn’t monochromatic

A

If you use like that isn’t monochromatic, different wavelength will diffract by different amounts and the pattern produced won’t be very clear

57
Q

Diffraction of monochromatic light

A

Light shone, through a narrow slit with effect, and sometimes produce a diffraction pattern.
To observe a clear fraction pattern, you should use a monochromatic coherence, light source

58
Q

Laser light

A

You can put a colour filter in front of white light to make it a single wavelength, but you get clearer a fraction patterns if you use a laser
Laser light is monochromatic and coherent - it has a single wavelength (and frequency) and so single colour, so it’s really useful for looking at the diffraction of light

59
Q

What if the wavelength of the lights wave is roughly similar to the size of the aperture(slit)

A

If the wavelength of the lights wave is roughly similar to the size of the aperture(slit), you get a defraction pattern of light and dark hinges

60
Q

Bright fringes

A

Bright fringes are due to constructive interference, where waves from across the width of the slits arrive at the screen in phase

61
Q

Dark fringes

A

The dark fringes are due to total destructive interference, where waves from across the width of the slits arrive at the screen, completely out of phase

62
Q

Diffraction of white light

A

White light is actually a mixture of different colours, each with different wavelengths.
When white light is shown through a single, narrow slit, all of the different wavelengths are diffracted by different amount.
This means that instead of getting clear fringes(as you would get with a monochromatic light source) you get a spectre of colours

63
Q

Intensity of light

A

The central maximum in a single slit light diffraction pattern is the brightest part of the pattern.
This is because the intensity of light is highest in the centre
Intensity is the power per unit area
For monochromatic light, all photons have the same energy, so an increase in the intensity means an increase in the number of photons per second.
So there are more photons per unit area, hitting the central maximum per second than the other bright fringes

64
Q

Width of the central maximum

A

The width of the central maximum varies with the width of the slits, and the wavelength of the light being diffracted
-Increasing the slit width decreases the amount of diffraction.
This means the central maximum is narrower, and the intensity of the central maximum is higher
- Increasing wavelength increases the amount of diffraction.
This means the central maximum is wider, and the intensity of the central maximum is lower

65
Q

What is two source interference

A

To source interferences when waves from two sources interfered to produce a pattern in order to get a clear interference patterns. The waves from the two sources must be monochromatic and coherent.

66
Q

Coherence

A

Two waves are coherent if they are the same wavelength and frequency and a fixed phase difference between them.
A light sources, coherent and phase, the troughs and the crests line up - this causes constructive interference and and a very intense beam

67
Q

Interference with waves of a different wavelength and frequency

A

Interference still happens when you’re observing waves of a different wavelength and frequency, but it happens in a jumble.
If the sources are coherence, clear, patterns of constructive instructive interference are seen
Whether you get constructive or destructive interference at a point of depends on how much further wave has travelled, and the other wave to get to that point.

68
Q

Path difference

A

The amount by which the path travelled by one wave is longer than the path travelled by the other wave is called path difference

69
Q

Constructive interference with two coherent wave sources in phase

A

At any point and equal distance from the two sources in phase, you’ll get constructive interference. These points are known as maxima.
You could also get constructive interference at any point where the path difference is a whole number of wavelengths.
At these points, the two waves are in phase and reinforce each other, which is why you get constructive interference 

70
Q

Total destructive interference with two coherent waste, sources in phase

A

At points where path difference is half a wavelength, 1 1/2 wavelengths, 2 1/2 wavelengths, etc., the waves arrive out of phase and you’ll get total destructive interference. These points are known as minima

71
Q

Why is it easier to demonstrate to source and fear for either sound or water?

A

It’s easier to demonstrate to source interference for either sound or water because it got relatively large wavelength. This makes it easier to detect interference patterns.

72
Q

Demonstrating two-source interference

A

The trick is to use the same oscillator to drive both sources
For water, one vibrator drives two dippers
For sound, one oscillator is connected to two loudspeakers

73
Q

Laser safety

A

Working with lasers are very dangerous because laser light is focused into a very direct powerful being on monochromatic light
If you looked at a laserbeam directly, your islands would focus it onto your retina which would damage it permanently

74
Q

Laser safety precautions

A

-Never shine the laser towards a person
-wear laser safety goggles
-Avoid shining the laserbeam at reflective surfaces
-Have a warning sign on display
-turn the laser off when it’s not needed

75
Q

The double slit formula

A

w - fringe spacing (meters)
Lambda - wavelength (meters)
s - spacing between slits (meters)
D - distance from slits to the screen
(Meters)
w = (lambda)D/s
w = French spacing, the distance between two adjacent maxima or two adjacent minima in metres

76
Q

Single-source double-slit set-up

A

To see two source interference with lights. You can either use to coherent, light sources, or you can shine a laser through two slits.
Remember a laser is a source of monochromatic and coherent light. This means you can effectively create two coherent sources by shining as single laser through mounted card, containing two slits
This is known as a single source double slit set up

77
Q

Double slit interference of light

A

Slits have to be the same size as the wavelength of the laser lights, so that it is diffracted
This makes the lights from the slits act like to coherent point sources
You get a pattern of lights and dark fringes, depending on whether constructive or destructive interference, is taking place

78
Q

Young’s double slit experiment

A

You might see the double slip, interference of light experiment, referred as to young double split experiment
Thomas Young was the first person to carry it out, Although he used the source of white light. Instead of a laser he then came up with the equation to work out, wavelength of the light from the experiment.

79
Q

Using a white light source, instead of a coherent, monochromatic laser in the double slits, interference of light experiment

A

If you were to use a white light sauce, instead of a coherent, monochromatic laser, the diffraction pattern would be less intense with wider maxima
The pattern would also contain different colours with a central white fringe, because the white light is made up of a mixture of frequencies

80
Q

Evidence for the wave nature of light

A

Towards the end of the 17th century, too important theories of lights were published one by Isaac Newton, and the other by Huygens
newton’s theory suggested that light was made up of tiny particles which he called corpuscles
Huygens put forward a theory using waves
The corpuscles theory could explain reflection and refraction, but diffraction and interference are both uniquely wave properties
It could be shown that light shown interference patterns that would help settle the argument once for all
Young’s double slicks experiments over 100 years later, provided the necessary evidence
It showed that light could both distract and interfere

81
Q

Opticle density of materials

A

Light goes fastest in a vacuum it travels slower in other materials as it interacts with the particles in them
The more optically dense material is the more light slows down when it enters it.
The optically density of a material is measured by its refractive index - the higher materials optical density, the higher its refractive index

82
Q

Refractive index of a material equation

A

The absolute refractive index of a material, n, is the ratio between the speed of lighting, vacuum C and the speed of light in that material Cs
n = C/Cs

83
Q

The refractive index of a boundary

A

The relative refractive index between two materials, 1n2, is the ratio of the speed of lights in a material, one to the speed of light in material two
1n2 = C1/C2
Combining this with n = C/Cs
You get 1n2 = n2/n1

84
Q

Snells law of refraction

A

If light is passing through a boundary between two materials, you can use the lower refraction to calculate unknown angles or refractive indices
The angle that incoming light makes the normal , θ1, it’s called the angle of incidence
The angle of the refracted ray makes with the normal, θ2, is the angle of refraction
Snow law refraction for a boundary between two material is given by :
n1sin θ1 = n2sin θ2

85
Q

What happens when light enters a more/less density

A

When light enters, a more dense, medium it bends towards the normal
E.g. if n1<n2>θ2
When light enters, the less dense medium, it bends away from the normal
E.g. if n1>n2 then θ1<θ2</n2>

86
Q

Critical angle

A

The angle of incidence at which the angle of refraction is 90°

87
Q

Critical angle formula

A

Sin θc = n2/n1 where n1 > n2

θc = critical angle
n2 = refractive index of less optically dense material
n1 = refractive index of more optically, dense material

88
Q

Optical fibres

A

An optical fibre is a very thin flexible tube of glass or plastic fibre that can carry light signals over long distances and round corners using TIR

89
Q

Total internal reflection

A

At angles of incidence greater than the critical angle refraction cant happen
that means all the lights is reflected back into the material
This affect is called total internal reflection - TIR

90
Q

Step index optical fibre configuration

A

Step index optical fibres have a high refractive index (optically dense) course, surrounded by cladding with lower reflective index (less optically dense) to allow TIR
The cladding also protects the fibre from scratches, which could allow light to escape

91
Q

What happens when light is shown at one end of an optical fibre

A

Light is on in at one end of the fibre.
The fibre, so narrow, the light, always hits the boundary between the fibre and cladding and angle greater than the critical angle.
So all the light is totally internally reflected from boundary to boundary until it reaches the other end
It doesn’t matter what shape the fibre is in either - TIR always occurs until light comes out of the other end

92
Q

Advantages of using optical fibre cables, instead of the old system of using copper cables

A

-The signal can carry more information because light has a high frequency
-The light doesn’t heat up the fibre - almost no energy is lost as heat
-There is no electrical interference
-Fibre optic cables are much cheaper to produce
-signal can travel along way very quickly and with minimal signal loss, although some does occur

93
Q

Signal degradation in optical fibres

A

Information is sent down optical fires as pulses of lights that can make up a signal
The signal can be degraded by absorption or by dispersion
Signal degradation can cause cause information to be lost

94
Q

Absorption

A

Absorption is where some of the signals energy is absorbed by the material the fibre is made from.
This energy loss results in the amplitude of the signal being reduced
Therefore a worse signal

95
Q

Dispersion

A

There are two types of dispersion, which can degrade an optical signal
-Model and material dispersion
Both types of dispersion cause pulse broadening, the signal is broader than the initial signal
Broaden pulses can overlap each other leading to information loss

96
Q

Modal dispersion

A

Modal dispersion is caused by light rays, entering the optical fibre at different angles
This causes them to take different paths down the fibre with rays, taking a path straight down the middle of the fibre arriving quicker than rays taking a longer reflected path
Model dispersion can be reduced by using a single mode fibre - in which light is only allowed to follow a very narrow path

97
Q

Material dispersion

A

Material dispersion is caused by different amounts of refraction experienced by different wavelengths of light
Different wavelength, slow down by different amounts in a material
As light consists of many different wavelengths, this causes some parts of the signal to take a longer time to travel down the fibre than others
Using monochromatically can stop material dispersion
Optical fibre repeaters can be used to regenerate the signal every so often to help reduce signal degradation from both absorption and dispersion

98
Q

Diffraction grating

A

A slide or other thin object that contains lots of equally spaced slits very close together used to show diffraction patterns of waves

99
Q

Derivation of diffraction grading equation

A

You can calculate the wavelength of the lights being used in a fraction, greeting experiment, using the diffraction grating equation
dsin θ = n(lambda)
d = distance between slits(metres)
n = order of maximum
θ = I’m going to the normal made by the maximum in degrees or radians

100
Q

Conclusions you can gather from the diffraction grating equation

A

-if Lambda is bigger, sin θ is bigger
This means that the larger, the wavelength, the more the pattern will spread out
-If d it’s bigger, sin θ smaller
This means that the course of the grating, the less the pattern will spread out
-Value of sin θ greater than one are impossible
So, if for a certain n you get a result of more than one for sign sin θ, you know that the order doesn’t exist

101
Q

Interference with diffraction grating

A

A diffraction grating contains lots of equally space slits very close together and so can be used to do this
When monochromatic light,(all the same wavelength) is passed through a diffraction grating with hundreds of slits per millimetre at normal incidence(right angles to the grating), the interference pattern is really sharp, because there are so many beams reinforcing the pattern
Sharper fringes make for more accurate measurements

102
Q

Zero order line

A

The line of maximum brightness at the centre of a diffraction grating interference pattern
It’s in the same direction as the incident beam

Physics (6 decks)

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