3. Waves Flashcards

Question

Can mechanical waves travel in a vacuum?

Answer 1

Waves that can travel through a vacuum

Answer 2

3 x 10⁸ m/s

Answer 3

The wave can be reflected, transmitted or absorbed

Answer 4

Its temperature increases

Answer 5

In all planes

Answer 6

Have the oscillations in one plane only

Answer 7

* by absorbing all planes of oscillation except one | * by reflection

Answer 8

Many tiny crystals, all lined up

Answer 9

All of them except the vertical one

Answer 10

Horizontal

Answer 11

Reflection off the water surface is absorbed, because its plane of polarisation is perpendicular to that of the Polaroid

Answer 12

The plane of polarisation of the water is perpendicular to that of the Polaroid

Answer 13

No light will get through

Answer 14

Transverse waves can be polarised; longitudinal cannot

Answer 15

Electric and magnetic field waves produced by moving charges

Answer 16

* sunglasses | * alignment of aerials for transmission and reception

Answer 17

When two waves of the same type are in the same place at the same time

Answer 18

By adding displacements of each separate wave

Answer 19

The adding together of waves

Answer 20

At a point where two or more waves meet, the instantaneous displacement (amplitude) is the vector sum of the individual displacements due to each wave at that point

Answer 21

When waves are in phase and the same frequency

Answer 22

nλ (where n is a whole number of wavelengths)

Answer 23

When waves are in antiphase and have the same frequency

Answer 24

(1+n/2)λ i.e. an odd number of wavelengths

Answer 25

The waves have a constant phase difference

Answer 26

* the waves are of similar amplitude (↑ contrast between maxima and minima) * the waves have similar frequencies - otherwise the interference patterns create change so fast that they are difficult to detect * the waves have a constant phase difference i.e. they are phase linked

Answer 27

* light produced by a laser * sound from two loudspeakers connected in parallel * light emerging from two apertures illuminated by the same source

Answer 28

Sources that have synchronised phase changes, as well as same frequency and λ

Answer 29

On stationary waves, points that are always at equilibrium and 0 oscillation

Answer 30

On stationary waves, points of maximum oscillation

Answer 31

* vibrator moves up and down - sends travelling wave down cord * wave reflected at end, so 2 travelling waves overlap and interfere * has antinodes and nodes; distance between nodes = 1/2λ

Answer 32

Where the band vibrates with large amplitude

Answer 33

* stationary - all particles (except nodes) have the same frequency * travelling - all particles have the same frequency

Answer 34

* stationary - varies from 0 (nodes) to maximum (antinodes) | * travelling - same for all particles

Answer 35

When the frequency driving the system matches the natural frequency of the system

Answer 36

* stationary - mπ (m = no. of nodes between the two particles) * travelling - 2πx/λ (x = distance apart)

Answer 37

* stationary - energy stored and not transferred | * progressive - energy transferred

Answer 38

Where energy is stored rather than transmitted - formed when 2 coherent waves travelling in opposite directions interfere to produce nodes and antinodes

Answer 39

* ↑ tightness/tension * ↓ length of string * ↓ thickness of string

Answer 40

1st harmonic frequency f depends on tension T in wire, its length l and its mass per unit length

Answer 41

``` f = 1/2l x √T/μ f= frequency l= length (m) T= tension (m) μ= mass per unit length( kg per meter) ```

Answer 42

A longitudinal wave travels down the tube, and is reflected at the opposite end - forming a stationary wave

Answer 43

At both ends

Answer 44

Air acts as a barrier outside

Answer 45

2nd harmonic (2f₁)

Answer 46

All of them

Answer 47

Node at closed end, anti-node at open end

Answer 48

Amplitude ↓ gradually from the maximum at the open to zero at the closed end

Answer 49

3rd harmonic (3f₁)

Answer 50

5th harmonic (5f₁)

Answer 51

Odd harmonics

Answer 52

There needs to be a whole number of stationary wave loops fitting into the length of the string

Answer 53

Equidistant parallel fringes alternating between: * maxima (constructive interference) * minima (destructive interference)

Answer 54

They interfere

Answer 55

They emit identical waves which start in phase

Answer 56

* use 2 coherent sources | * use single source with double slits

Answer 57

They have the same frequency, wavelength and synchronised phase changes

Answer 58

w = λD / s w= fringe spacing (m) λ= wavelength (m) D= slit to screen distance (m) (capital d is always the bigger distance) s=spacing between slits (m)

Answer 59

To obtain a clear interference pattern it requires two coherent waves of monochromatic light- light is usually emitted in bursts of waves, after which is a random phase change

Answer 60

In bursts of waves, each burst lasting 10⁻⁹s, after which there's a random phase change

Answer 61

A source of a single wavelength

Answer 62

A sodium lamp

Answer 63

The distance between neighbouring bright fringes

Answer 64

Wavelength is decreased so distance between adjacent fringes decreases

Answer 65

The central fringe is white, with red edges. Other fringes will be spectra with the blue end towards the middle of the overlap area

Answer 66

D↑ so distance between adjacent fringes increases

Answer 67

The maxima will become minima and vice versa

Answer 68

Wider interference so there are more dots, but fainter as there is less light through (x ↑)

Answer 69

The waves don't fully cancel out

Answer 70

* ↑ D - measurement easier and more accurate but fringe intensity decreased * ↓ a - practical limit to this * ↑ wavelength

Answer 71

When a wave passes through a gap and spreads out

Answer 72

Diffraction ↑

Answer 73

When the gap with is similar to the wavelength of the wave

Answer 74

* only 1 slit but more than 1 wave * single slit can be thought of as large no. of sources next to each other * each 'source' produces coherent wave - overlap and interfere

Answer 75

* short λ (e.g. blue light) - narrow diffraction pattern | * long λ (e.g. red light) - broad diffraction pattern

Answer 76

* white light yields less clear patterns (as position of dark bands depends on λ) * colours appear; only central band is white

Answer 77

* single slit - central max. fringe that is twice the width of the other fringes * double slit pattern has equally spaced fringes

Answer 78

Diffraction grating - as not much light gets through the double slits so are dim and unclear

Answer 79

A set of slits for light waves to pass through

Answer 80

``` dsinθ = nλ d= slit seperation (m) theta= angle n= order of maxima λ= wavelength (m) ```

Answer 81

There is a limit to the number of spectra that be obtained

Answer 82

The waves that produce the bright spot straight on - paths are all the same length, so phase difference is zero

Answer 83

It will give a maximum when sinθ=1, so cancels out

Answer 84

Diffraction grating

Answer 85

* double slits - fringes formed are slightly blurred → large errors * diffraction grating - images are clear and measurements accurate, also final result is an average of several calculations

Answer 86

Analysis of spectra

Answer 87

A long, thin, cylindrical core of glass, encased in a cladding of glass of lower refractive index

Answer 88

A change in the direction of light as it passes across a boundary between two transparent substances

Answer 89

It doesn't refract

Answer 90

A measure of the optical density of a material relative to air

Answer 91

n=c/v n = refractive index c = velocity of light in vacuum v = velocity of light in the medium

Answer 92

The ratio of the sines of the angles of incidence and refraction are constant when it passes between two given media

Answer 93

n₁sinθ₁ = n₂sinθ₂ n= refractive index theta= angle

Answer 94

* light passes from more to less dense medium | * angle of incidence > critical angle

Answer 95

When light passes from a more to less dense medium, and the angle of incidence > critical angle, all light is reflected back to the less dense medium

Answer 96

The limiting angle of incidence, as the angle of refraction cannot exceed 90°

Answer 97

``` sinx= n2/n1 x = critical angle n2= the material it diffracts into n1= the first material ```

Answer 98

By total internal reflection, only escaping when it reaches the other end

Answer 99

A medical instrument that uses optical fibres to look inside the body

Answer 100

* a coherent bundle of fibres (lens system) | * an incoherent bundle of fibres (light delivery system)

Answer 101

Angle of incidence will be less than critical angle and light will escape

Answer 102

Digestive, respiratory and female reproductive systems

Answer 103

Can diagnose patients without an incision, often without anesthetic

Answer 104

As long as θ is larger than the critical angle i.e. sin θ > n of cladding ÷ n of core

Answer 105

* endoscopes | * lasers - burn tissue to heal wound

Answer 106

Where the fibres stay in the same relative position along their length

Answer 107

* scratches can cause light to leak * two fibres touching can cause light to pass from one to the other - 'cross talk' * dispersion

Answer 108

Using cladding

Answer 109

In 'pulses' or 'bursts'

Answer 110

• absorption - some energy absorbed so pulse has lower amplitude. • dispersion - causes pulse broadening. 2 types: modal- light enters at different angles and hence takes a different path. material- light travels at different speeds.solved by using monochromatic light.

Answer 111

* material (chromatic) dispersion - light kaing different paths * modal (multipath) dispersion- light being different speeds

Answer 112

A pulse can take a variety of different paths through a fibre, meaning a single pulse can spread out over time

Answer 113

* use monomode fibres with a core diameter of only a few wavelengths, so light travels via one path * cladding

Answer 114

Refractive index of cladding is only slightly lower than the refractive index of the core, so the critical angle is larger than is would be at a glass-air boundary → only small range of angles that can be transmitted

Answer 115

10 micrometers (1-10 x 10⁻⁶ m)

Answer 116

Using monochromatic light (red will travel faster than blue)

Answer 117

Red - it travels faster

Answer 118

Dispersion

Answer 119

Blue light travels more slowly in glass than red light

Answer 120

When the pulses are very short and close together

Answer 121

In communications

Answer 122

In endoscopes

Answer 123

Pitch is a term used to describe how high or low a note seems to be. The pitch of a note depends on the frequency. A high frequency produces a high pitched note and a low frequency produces a low pitched note.

Answer 124

A beat pattern is a wave whose amplitude is changing at a regular rate. Observe that the beat pattern (drawn in green) repeatedly oscillates from zero amplitude to a large amplitude, back to zero amplitude throughout the pattern.

Answer 125

'dsinΘ =nλ' equation?

Answer 126

The oscillation of particles or fields.

Answer 127

A wave that carries energy from place to place without transferring any material.

Answer 128

One complete vibration of a wave.

Answer 129

How far a point on the wave has moved from its undisturbed position. Unit: metres

Answer 130

The maximum magnitude of displacement.

/ distance from the undisturbed position to the crest or trough

Unit: metres

Answer 131

The time taken for a whole cycle (vibration) to pass a given point. Unit: seconds

Answer 132

The length of one whole wave cycle, from crest to crest or trough to trough. Unit: metres

Answer 133

The number of cycles (vibrations) per second passing a given point. Unit: hertz

Answer 134

A measurement of the position a certain point along the wave cycle.

Answer 135

The amount one wave lags behind another.

Answer 136

Angles (degrees or radians) or as fractions of a cycle.

Answer 137

• Displacement - x
• Amplitude - A
• Wavelength - Lambda
• Period - T
• Frequency - f

Answer 138

When a wave is bounced back when it hits a boundary.

Answer 139

When a wave changes direction as it enters a different medium.

Answer 140

Frequency = 1 / Time period

f = 1 / T

Answer 141

Wave speed = Frequency x Wavelength

c = f x lambda

Answer 142

The speed of light in a vacuum - 3.0 x 10^8 m/s

Answer 143

Wave speed = Distance travelled / Time taken

c = d / t

Answer 144

Transverse

Answer 145

• EM Waves
• Water waves

Answer 146

Microphones = separate inputs so signals can be recorded separately.

Answer 147

1) Displacement against distance along the path of a wave
2) Displacement against time for a POINT as the wave passes

(Note: 1 is just a standard graph of what a wave looks like. 2 is what happens to a specific point as a wave passes through it.)

Answer 148

Displacement - distance: Wavelength

Answer 149

Displacement - time: Time period

Answer 150

... magnetic and electric fields.

Answer 151

The label on the x axis. This may be distance or time, depending on what the graph is showing.

Answer 152

At right angles to the direction of energy transfer.

Answer 153

• Sound
• Pressure

Answer 154

• Compressions
• Rarefactions

Answer 155

Rarefactions

Answer 156

• Transverse - Usually no
• Longitudinal - Usually yes

Answer 157

• Displacement against time.
• This can it look like a transverse wave!

Answer 158

Parallel to the direction of energy transfer.

Answer 159

A wave that only oscillates in one direction (e.g. only up and down).

Answer 160

• Transverse - Yes
• Longitudinal - No

Answer 161

• Transverse - Perpendicular to the direction of energy transfer.
• Longitudinal - Parallel to the direction of energy transfer.

Answer 162

Causing a transverse to only vibrate in one direction (e.g. up and down) usually by passing it through a polarisation filter.

Answer 163

It can be polarised by reflection. A longitudinal wave could not do this, so light must be a transverse wave.

Answer 164

Which waves are transverse. For example, light can be polarised, so it must be transverse.

Answer 165

They are a mixture of different directions of vibration. This means that they can be polarised by allowing only some of these directions to pass through a filter.

Answer 166

A panel that polarised waves by only allowing a specific direction of vibration to pass through.

Answer 167

It becomes partially polarised. This means some of it vibrates in the same direction.

Answer 168

No light will get through.

Answer 169

It instead reduced the intensity of the light passing through it (but still allows some light through).
By rotating the filter we can see the change in intensity:

Answer 170

No - most light we see is unpolarised

Answer 171

Light reflected off some surfaces is partially polarised - some of it is made to vibrate in the same direction (Figure 9).

When light reflected off surfaces like water, glass or tarmac enters the eye, it can cause glare.

Answer 172

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 the right angle, you can block out some of the reflected light, while still letting through light which VIBRATES at the angle of the filter.

This reduces the intensity of light entering your eye.

Answer 173

Reducing unwanted reflections in photography, and in polaroid sunglasses to reduce glare.

Answer 174

The angle of the incident light

Answer 175

• Partially polarised light is reflected into a polarising filter at the correct angle.
• This blocks out unwanted glare.

Answer 176

• Broadcasting aerial has rods, which emit polarised waves
• TV aerials on homes have horizontal rods
• These rods must be lined up in order to get maximum signal strength
• The same thing happens with radio aerials

Answer 177

• Polaroid sunglasses
• TV and radio signals

Answer 178

When two or more waves pass through each other and their displacements combine.

Answer 179

When two or more waves cross, the resultant displacement equals the vector sum of the individual displacements.

Answer 180

Add the individual displacements at each point along the x-axis and then plot these.

Answer 181

• Constructive interference
• The amplitude of the wave is increased (i.e. the crest or trough gets bigger).

Answer 182

• Destructive interference
• The displacements cancel themselves out.

Answer 183

Add the displacements of the two waves (=resultant)

Answer 184

They are both at the same point in the wave cycle. They are likely to be 360*, 720*, etc. out of phase. They are the same wavelength and velocity

Answer 185

• Same velocity
• Same displacement

Answer 186

2π radians

Answer 187

Radian -> 1 radian is equal to 180/π.

Answer 188

Multiply by π/180.

Answer 189

Multiply by 180/π.

Answer 190

• 180*
• π radians.

Answer 191

• 90*
• 1/2 π radians

Answer 192

• 270*
• 3/2 π radians

Answer 193

• 360*
• 2π radians

Answer 194

The fraction of a cycle it has completed since the start of a cycle.

Answer 195

Thee fraction of a cycle between the vibrations of the particles, measured in either degrees or radians.

/ Difference in their positions in a wave's cycle

Answer 196

Distance between two maximums = phases difference of 1 wavelength

Answer 197

Degrees or radians.

Answer 198

... in phase.

Answer 199

... exactly out of phase.

Answer 200

When they have the same:
• Wavelength
• Frequency
And have a fixed phase difference between them.

Answer 201

When the two sources are coherent (have the same wavelength and frequency and have a fixed phase difference between them).

Answer 202

• How much further a wave has travelled compared to another
• This is used when looking at the type of interference between two waves that will occur at a certain point (see diagram pg 27 of revision guide).

Answer 203

At a whole number of wavelengths.

Path difference = nλ

Answer 204

At a whole number of wavelengths and a half.

Path difference = nλ + 0.5λ

Answer 205

• the waves are of similar amplitude (↑ contrast between maxima and minima)

• the waves have similar frequencies - otherwise the interference patterns create change so fast that they are difficult to detect

• the waves have a constant phase difference i.e. they are phase linked

Answer 206

• light produced by a laser

• sound from two loudspeakers connected in parallel

• light emerging from two apertures illuminated by the same source

Answer 207

The superposition of two progressive waves with the same frequency (wavelength) moving in opposite directions.

Answer 208

A progressive wave.

Answer 209

• Vibration generator is attached to a piece of string at one end, while the string is fixed at the other end.
• The frequency of the generator is varied until a resonant frequency is found.

Answer 210

• At most frequencies, the pattern on the string is a jumble
• If the vibration generator produces an exact number of waves in the time it takes a wave to get to the end and back, the original and reflected waves reinforce each other. This produces a stationary wave. - The overall pattern doesn't move along, it just vibrates up and down.

Answer 211

Waves intefere with it's reflection.

Only happens at specific frequencies.

Nodes must occur at the point of return.

Answer 212

Where the amplitude of the vibration is zero.

Total destructive inteference

Answer 213

Where the maximum amplitude of the wave is.

constructive interence

Answer 214

Oscillating loops

Answer 215

When an exact number of half wavelengths fit onto the string.

Answer 216

1 Loop = 1/2 wavelength = First harmonic
2 Loops = 1 wavelength = Second harmonic
3 Loops = 1.5 wavelengths = Third harmonic

Answer 217

• When the stationary wave is vibrating at the lowest possible resonant frequency.
• One loop is on the string, with a node at each end.

Answer 218

1/2 a wavelength of the wave

Answer 219

1 wavelength
(when 2 half wavelengths (loops) fit on a string, the wavelength is the length of the string)

Answer 220

1.5 wavelengths

Answer 221

an extra loop and an extra node, the number of wavelengths goes up by 1/2.

At the a^th harmonic the number of antinodes = a. number of nodes is a + 1.

At the a^th harmonic, a/2 wavelengths fit on the string

Answer 222

a x first harmonic frequnecy.

a = number of antinodes or the amount of harmonics or just the amount of bumps

Answer 223

Closed tube: the first harmonic has 1/4 wavelength.

(still increases by 1/2 a wavelength but the first harmonic starts at 1/4 wavlenghs)

Answer 224

3/4 wavelengths (0.75)

(still increases by 1/2 a wavelngth but the first harmonic starts at 1/4 wavlenghs)

Answer 225

5/4 wavelngths (1.25)

(still increases by 1/2 a wavelngth but the first harmonic starts at 1/4 wavlenghs)

Answer 226

deep sound

Answer 227

Antinode at entrance and exit, normal number of wavelngths in a harmonic (1st = 1/2, 2nd = 1, 3rd = 1.5)

Answer 228

Open tubes have more wavelength in the tube (in order to have an antinode at both ends).

This means they have a smaller wave in the same size tube.

When you hit a tube on a table with its 1 lid off, you get a deeper pitch than if it had both sides open and being hollow.

Watch TLPhysics video about closed and open tube harmonics.

Answer 229

Pg 28 of revision guide.

Answer 230

• Microwaves beam reflected of a metal plate - the superposition of the wave and it's reflection produces a stationary wave.
• The nodes and antinodes are found by moving the probe between the transmitter and reflecting plate.
• The meter or loudspeaker receievs no signal at the nodes and maximum signal at the antinodes.

Answer 231

A loudspeaker produces sound waves in a glass tube
Glass tube with a speaker at the end is set up
Lycopodium powder is laid along the bottom of the tube
The powder is shaken away from the antinodes and left undisturbed at the nodes
Distance, d, between each pile of power (node) is λ /2. ( λ = 2d)
The speed of sound = c = fλ .
c=fλ is the same as c = f x 2d which is c = 2df
You can find the speed of sound by measring d and knowing the frequnency of the signal generator

Answer 232

• First = f
• Second = 2f
• Third = 3f

Answer 233

f = c / λ

Where:
f = Harmonic frequency
c = Speed of wave on string
λ = The wavelength of the wave given in terms of the length of the string (e.g. first harmonic: λ = 2L)

Answer 234

f = c / 2L.

(½ wavlength = 1 length of string.

Therefore, 1 wavelngth = 2 lengths (2L))

When substituting L for wavelength, always start with “1 length of string = x wavelengths” then rearange to find 1 wavelength in terms of “lengths of string (L)"

Answer 235

f = c / L

(1 wavlength = 1 length of string)

Answer 236

f = 3c / 2L
Because

1.5 wavelngths = 1 length of string

1 wavelength = 1/1.5 = ⅔ lengths of string

f = c / (2/3 L)

rearange to get f = 3c / 2L

Answer 237

Phase difference (radians) = 2πd / λ

Where d = the distance apart of the particles in wavelengths (λ).

(e.g. d might equal 1/4 λ if there is a quarter of a cycle difference)

Remember the period of a wave is 2π

Answer 238

1) Measure the mass and length of the string using a mass balance and ruler. Work out the mass per unit length (μ = M/L) in kg/m.
2) Set up the equipment as shown. This involves connecting a vibration generator (connected to a signal generator) to a piece of string attached to a pulley and some masses. Clamp the entire setup to the bench.
3) Measure the length (l) of the string between the vibration generator and the pulley. Work out the tension in the string using (T = mg) where m is the mass of the masses on the end of the string.
4) Turn on the signal generator and adjust the frequency until the first harmonic is found.

Depending on the experiment, you can chose to either change the mass (per unit length), the length or the tension; of the string

Chose one to change and keep the rest the same.

Answer 239

The resonant frequencies.

Answer 240

• Length of the vibrating string - longer the string the lower the resonant frequency - because the half wavelength is longer (c = fλ, f increases for a fixed c)
• Tension in the string - waves travel more slowly if the string is loose and there is less tension (lower c = lower f)
• Type of string (different μ) - heavier string (more mass per unit length) - waves more slowly down the string (lower c = lower f)

Answer 241

• μ = Mass per unit length of string
• Μ = Mass of the string
• L = Length of vibrating string
• T = Tension in the string
• m = Mass of the masses in the end of the string
• g = Gravitational field strength

Answer 242

Newtons (N)

Answer 243

Pg 29 of revision guide.

Answer 244

• Keep the type of string and tension the same
• Move the vibration transducer towards or away from the pulley

Answer 245

• Keep the string type and length the same
• Add or remove masses to vary tension

Answer 246

• Keep the vibrating string length and tension the same
• Use different string samples to vary μ (different masses of string with the same length)

Answer 247

• The longer the string, the lower the resonant frequency.
• Because the half wavelength at the resonant frequency is longer.

Answer 248

• The heavier (greater μ) the string, the lower the resonant frequency.
• Because waves travel more slowly down the string. A lower wave speed, c, makes a lower frequency, f.

Answer 249

• The higher the tension, the higher the resonant frequency.
• Because waves travel more quickly on a taut string. A higher wave speed, c, makes a higher frequency.

Answer 250

f = (1 / 2l) x root(T / μ)

Where:
l = String length (m)
T = Tension in string
μ = Mass per unit length of string (kg/m)

See page 29 of revision guide.

Answer 251

Pg 29 of revision guide

Answer 252

do it - some of the next questions may be a repeat

Answer 253

The spreading out of waves when passing through a gap (or going around an object).

Answer 254

The wavelength of the wave compared to the size of the gap.

Answer 255

When the gap is the same size as the wavelength.

Answer 256

It is increased.

Answer 257

It is decreased.

Answer 258

Diffraction is unnoticeable.

Answer 259

The waves are mostly just reflected back.

Answer 260

The doorway is a gap of a similar size to the wavelength of sound, so it diffracts to the listener. However, the gap is much larger than the wavelength of light, so the diffraction is not noticeable.

Answer 261

• Blue is on the inner side, while red is on the outer side
• This is because red light has a longer wavelength, so it diffracts more

Answer 262

The fringes become less bright.

Answer 263

The power per unit area.

Answer 264

• It is twice as wide
• The outer fringes are all of the same width

Answer 265

• Laser
• Vapour lamps and discharge tubes

Answer 266

No, as long as they have a constant phase difference.

Answer 267

W = 2Dλ/a

Where:
• W - Width of the central maximum
• D - Distance between the slit and screen
• λ - Wavelength
• a - Slit width
(All units in m)

Answer 268

Single-slit interference
(W = 2Dλ/a)

Answer 269

Two coherent sources.

Answer 270

They have the same frequency/wavelength AND constant phase difference.
This is achieved by both speakers being connected to same signal (generator)

Answer 271

The sound waves from the two speakers superpose (at a point) (not interfere)
At A (and B) the two waves are in phase (they have zero phase difference) and a maximum is produced.
Moving away from A introduces a path difference/phase difference = waves are out of phase (and amplitude decreases)
Minima's are formed when there is destructive interference (odd number of half wavelength = path difference or π/ 180 degrees = phase difference or antiphase)
(Moving on towards B the waves move back in phase)

Answer 272

Path difference between two maxima's = 1 wavelength.

1 path difference = 0.12 m

1 wavelength = 0.12 m

0.12 x 2960 = 360m/s

Answer 273

As frequency increases, so does wavelength.
Therefore, the separation of maxima's (along the line AB) increases
Maximum moves (from B) towards C so amplitude of sound gets larger/louder (then quieter). OR Maximum moves further along path/beyond C so amplitude of sound gets quieter.

Answer 274

Connect both dippers/loudspeakers to the same vibrator/oscillator.

Answer 275

Use two microwave transmitter cones attached to the same signal generator.

Use a microwave receiver probe - you get an alternating pattern of strong and weak signals.

Answer 276

Young's double-slit experiment

Answer 277

Two coherent sources of light

Answer 278

No, due to the various frequencies of light in it.

Answer 279

Pg 32 of revision guide.

Answer 280

Because they are all of the same width, unlike in the single slit pattern.

Answer 281

P.d. = nλ
Where n is an integer.

Answer 282

P.d. = nλ + 0.5λ
Where n is an integer.

Answer 283

The distance from the centre of a bright fringe to the centre of the next one.

Answer 284

If you looked at the beam directly, your eye would focus it onto your retina, which would be permanently damaged.

Answer 285

1) Never shine the laser towards a person.
2) Wear laser safety goggles.
3) Avoid shining the beam at a reflective surface.
4) Have a warning sign on display.
5) Turn the laser off when it's not needed.

Answer 286

• Replace the laser and slits with 2 microwave transmitter cones attached to the same signal generator
• Replace the screen with a receiver probe
• Move the probe along where the screen was and you'll get an alternating pattern of strong and weak signals

Answer 287

w = λD/s

Where:
w = Fringe spacing
λ = Wavelength
D = Distance from slits to screen
s = Slit separation

Answer 288

Measure several fringes and divide by the number of fringe widths between them.

Answer 289

• When dividing to find 'w', remember to divide by the number of fringe WIDTHS between them, not the number of fringes.
• e.g. 10 bright lines only have 9 fringe widths between them.

Answer 290

Single slit:
• Widest central maximum + equal outer fringes
• Brightest central maximum + decreasing intensity of outer fringes
Double slit:
• All fringes of equal width
• Decreasing intensity of outer fringes

Answer 291

Because it's multiplied by the single slit diffraction pattern for either of the slits separately.

Answer 292

The blue light creates a smaller fringe separation. This makes the pattern appear more compact.

Answer 293

• White central fringe - Every colour contributes at the centre
• Inner fringes tinged with blue on the inside and red on the outer side - Red fringes are more spaced out than blue fringes.
• After a few fringes, no clear fringe pattern - The different colour's fringe patterns have all blended

Answer 294

• It was evidence for light interference and diffraction.
• This was important in the debate between Newton's particle (corpuscle) theory of light and Huygen's wave theory of light.
• It supported Huygen's theory (even though the debate was raging again 100 years later).

Answer 295

• The same shaped pattern is observed, except the bright bands are brighter and the dark bands are darker
• This gives a sharper pattern

Answer 296

There are many beams reinforcing the pattern.

Answer 297

It allows for more accurate measurements.

Answer 298

• Each slit must be sufficiently narrow to diffract the light enough
• The two slits must be close enough for the diffracted waves to overlap

Answer 299

The waves from different points across the slit interfere to reinforce or cancel each other.
(See pg 84 of textbook for a very good explanation!)

Answer 300

• Diffracted light waves from adjacent slits reinforce each other in certain directions only and cancel out in all other directions.
• It works just like with double slit interference, except with many more slits.
• More slits result in more sharp lines, so the are several distinct, sharp lines produced by a diffraction grating.

Answer 301

d sin θ = nλ

Answer 302

d is the slit spacing, not the number of slits per metre, which is how the data may be given.

Answer 303

• 300 slits/mm = 300,000 slits/m
• Therefore, d = 1/300,000

Answer 304

See pg 34 of revision guide + have a go.

Answer 305

It is more spread out.

Answer 306

It is more compact.

Answer 307

Round it down to the next integer.

Answer 308

• θ can never be greater than 90. Therefore, the greatest value that sinθ can have is sin(90), which is 1.
• So, replace sinθ with 1 in the equation.
• This leaves d = nλ. Solve for n.
• Round n DOWN to the nearest integer.

Answer 309

• The wavelength of x-rays is similar to the spacing between atoms in crystalline solids.
• So x-rays directed at a thin crystal form a diffraction pattern -> The crystal acts like a diffraction grating.
• Looking at the diffraction pattern, the spacing of the atoms can be calculated.

Answer 310

To discover the structure of DNA.

Answer 311

• A measure of optical density
• A ratio of the speed of light in a vacuum compared to the speed of light in the material.

Answer 312

In a vacuum.

Answer 313

It interacts with the particles in the material.

Answer 314

slows down when entering it

Answer 315

its refractive index

Answer 316

Higher optical density = higher refractive index

Answer 317

the absolute refractive index of a material, n, is the ratio between the speed of light in a vaccuum, c, and the speed of light in that material cs (subscript s)

Answer 318

n = c/cs
(NOTE: This is just a specific case of the equation for relative refractive index)

Answer 319

cs (subscript s)

Answer 320

The ratio of the speed of light in material 1 to the speed of light in material 2.

Answer 321

₁n_2.

(it means the relative refractive index of a boundary, going from material 1 to material 2)

Answer 322

3.00 x 10^8 m/s

Answer 323

• Absolute refractive index is the ratio of the speed of light in a vacuum compared to the speed of light in the material.
• Relative refractive index is the ratio of the speed of light in material 1 to the speed of light in material 2.
(NOTE: Absolute refractive index is just a case of the relative refractive index)

Answer 324

1n2 = c1/c2
or
1n2 = n2/n1

Answer 325

• n = c/c_s
• ₁n₂ = c₁/c₂ (It's logical from the definition!)
•₁n₂ = n₂/n₁

Answer 326

• Absolute - Property of one material only.
• Relative - Property of the interface between two materials. It is different for every possible pair.

Answer 327

They are essentially the same since they both have a refractive index of about 1.

Answer 328

The angle that incoming light makes to the normal.
Symbol: θ₁

Answer 329

The angle that the refracted ray makes to the normal.
Symbol: θ₂

Answer 330

They are measured from the NORMAL, not the boundary.

Answer 331

Towards the normal.

Answer 332

Away from the normal.

Answer 333

n1 x sinθ1 = n2 x sinθ2

Answer 334

The angle of incidence at which the angle of refraction is 90* so that the light is refracted along the boundary.

Answer 335

sinθ_c = n₂/n₁ = ₁n₂

Answer 336

In this case the angle of incidence is the critical angle: sinθ1 = sinθc
At the critical angle, the angle of refraction is 90*
So sinθ2 = sin90 = 1
Use snells law : n1 x sinθ1 = n2 xsinθ2
n1 x sinθc = n2 x 1
sinθc = n2/n1

Answer 337

There is partial reflection observed always, even when there is no total internal reflection.
(See pg 73 of textbook)

Answer 338

When light going from a more optically dense to a less optically dense material hits the boundary at an angle greater than the critical angle and is completely reflected.

Answer 339

1) Incident substance has a larger refractive index than the other substance
2) Angle of incidence exceeds the critical angle

Answer 340

A very thin flexible tube of glass or plastic fibre that can carry light signals over long distances using TIR

Answer 341

Step index optical fibres used
The fibre has a very high refractive index (optically dense), but is surrounded by a cladding with a lower refractive index (less optically dense) -> This enables TIR + Protects the fibre.
The fibre is narrow -> Light always hits the boundary at an angle greater than the critical angle.
Light that enters at one end is totally internally reflected to the other end

Answer 342

• Thin -> Ensures light hits at angle above the critical angle + Prevents modal dispersion
• Cladding of lower refractive index -> Protects the fibre from scratches + Ensures TIR happens

Answer 343

• Protects the fibre from scratches
• Ensures TIR happens (by having a lower refractive index)

Answer 344

• Ensures light hits at angle above the critical angle |
• Prevents modal dispersion

Answer 345

• Endoscopes
• Communications

Answer 346

Light has high frequency = signal can carry lots of information
Light does'nt heat up the fibre = no energy is lost
No electrical interference
Fibre-optics are cheap to produce
Signal can travel very far, very quickly, with minimal signal loss.

Answer 347

The disruption and changing of a signal as it passes through an optical fibre.

Answer 348

• Absorption
• Dispersion

Answer 349

• Some of the signal's energy is lost through absorption by the material of the fibre.
• This reduces the amplitude.

Answer 350

Pulse broadening

Answer 351

The recieved signal is boader than the initial signal. Broadened pulses can overlap each other, leading to information loss

Answer 352

• Modal dispersion
• Material dispersion

Answer 353

• Light rays enter the fibre at different angles and so take different paths -> Some arrive later than others. Rays taking a path straight down the middle of the fibre arrive faster than rays taking a longer path.
• This causes pulse broadening.

Answer 354

Caused by different amounts of refraction experienced by different wavelengths of light
Different wavelengths slow down by different amounts in the material so they travel at different speeds in the fibre -> Some arrive later than others.
This causes some parts of the signal to take a longer time to travel down the fibre than others.
This causes pulse broadening.

Answer 355

• Use a highly transparent fibre to stop absorption.
• Use an optical fibre repeater

Answer 356

• Use a very narrow fibre -> Path difference is small.
• Use a single-mode fibre -> Only allows light to take one path.
• Use an optical fibre repeater

Answer 357

• Use monochromatic light
• Use an optical fibre repeater

Answer 358

• A device that boosts and regenerates the signal every so often
• This reduces signal degradation by absorption and dispersion

Answer 359

Broadened pulses can overlap, causing confusion and information loss

Answer 360

• Has 2 bundles of fibres.
• One is used to illuminate the area of the body.
• A lens forms an image on the end of the other bundle
• The fibres then take this image back to the other end, where it can be viewed.

Answer 361

The bundle must be coherent, so that the image on the other end is not muddled up.

Answer 362

In all planes

Answer 363

When the fibre ends are in the same relative positions (i.e. the fibres arrange themselves in the same order to recreate the original image).

Answer 364