Waves

Question 1

What is a progressive wave?

Answer 1

A wave that transfers energy from one place to another without transferring any matter. The transfer of energy is in the same direction as the wave is travelling.

Question 2

State four ways you can demonstrate a progressive wave transfers energy.

Answer 2

EM waves cause things to heat up which transfers energy.

X-rays and gamma rays knock electrons out of their orbits, causing ionisation.

Loud sounds cause large oscillations in air particles which can make them vibrate.

Wave power can be used to generate electricity as waves can carry energy.

Question 3

Define wavelength

Answer 3

The distance between two identical points on a wave.

Question 4

Define amplitude.

Answer 4

The maximum displacement of a wave from its equilibrium position.

Question 5

Define period.

Answer 5

The time taken for a wave to complete one full cycle.

Question 6

Define frequency

Answer 6

The number of complete oscillations of a wave per second.

Question 7

What is the wave speed equation in symbols?

Answer 7

v = f x λ

Question 8

How is time period linked to frequency?

Answer 8

f = 1/T

Question 9

What is a transverse wave?

Answer 9

A wave whose oscillations are perpendicular to the direction of energy transfer.

Question 10

What is a longitudinal wave?

Answer 10

A wave whose oscillations are parallel to the direction of energy transfer.

Question 11

Describe a longitudinal wave in terms of pressure variation and the displacement of molecules.

Answer 11

Longitudinal waves in a fluid consists of areas of compression and rarefaction. These compressions and rarefactions creature variations in pressure in the medium.

Areas of high pressure (compression) are where molecules in the medium are close together.

Areas of low pressure (rarefaction) are where molecules in the medium are far apart.

Areas of rarefaction move into vacant spaces creating vacant spaces behind them. Areas of compression push neighbouring particles into vacant spaces. This allows the movement of the wave.

Question 12

What is a wavefront?

Answer 12

A line joining all points in a wave that are in phase and are drawn perpendicular to the direction of wave travel.

Question 13

What is a standing wave?

Answer 13

A wave that stores energy at the nodes and antinodes instead of transferring it.

Question 14

What conditions are needed to produce a stationary wave?

Answer 14

Two progressive waves must be travelling in opposite directions in the same medium, in the same plane.

The waves must be coherent; same frequency, same wavelength.

The two waves superpose together.

Question 15

What is a node?

Answer 15

It is a point at which the displacement in a standing wave is at its minimum due to the wave being in antiphase, causing total destructive interference.

Question 16

What is an antinode?

Answer 16

It is a point at which the displacement in a standing wave is at its maximum due to the wave being in phase, causing constructive interference.

Question 17

Define superposition.

Answer 17

When the total displacement of a wave at any point is equal to the vector sum of the individual displacements.

Question 18

Define coherence

Answer 18

Same type of waves that have same wavelength, same frequency and fixed phase difference.

Question 19

Define interference

Answer 19

The pattern observed when two or more waves superpose with each other (could be the pattern of light, sound, microwaves etc).

Question 20

What is the fundamental frequency equation in symbols?

Answer 20

v/2L or (√ (T/μ))/2L

Question 21

What factors affect the speed of a wave on a string?

Answer 21

Tension.

Mass per unit length.

Question 22

What is meant by the intensity of wave?

Answer 22

The power transferred by a wave per unit area.

Question 23

What is the relationship between intensity and amplitude of a wave?

Answer 23

I α A^2

Question 24

What is the equation for the intensity of a wave?

Answer 24

I = P/A

Question 25

What are the properties of all electromagnetic waves?

Answer 25

All of them are transverse.

All of them travel at the same speed in a vacuum.

All of them can be refracted, reflected, diffraction, and undergo interference.

All of them obey the relationship v = fλ.

All of them carry energy.

All of them consist of vibrating magnetic and electric fields at right angles to each, and to the direction of travel.

Question 26

What can happen when waves meet an interface between two materials?

Answer 26

They reflect, refract, transmit and absorb.

Question 27

Define interface.

Answer 27

The boundary between two different media.

Question 28

Define refractive index.

Answer 28

A measure of how much the speed and direction of light changes between two different
media.

Question 29

How can you calculate refractive index?

Answer 29

c / v

“c” being the speed of light

“v” being the speed of light in the material.

Question 30

Describe how to find the refractive index of a glass.

Answer 30

Use a “ray box”/”LVU” with a slit screen and lens. Shine the ray through a glass block at an angle. Draw crosses on the rays entering and leaving the block. Join up the crosses to show the three rays. Measure the angle of incidence and angle of refraction when entering/leaving the block. Use the equation for Snell’s law (where the refractive index of air = 1) to find the refractive index of glass. Or, take readings for different angles and plot a graph of sin i / sin r. The gradient will be the refractive index.

Question 31

How do you reduce uncertainty in finding the refractive index of a solid material

Answer 31

The lens/slit screen makes the beam narrower which reduces the percentage uncertainty in the position of the beam.

Using larger angles rather than smaller ones reduces the percentage uncertainty in the angle measurements.

Question 32

What is the critical angle?

Answer 32

The greatest angle of incidence in an optically dense medium where the angle of refraction is 90°

Question 33

What is Snell’s Law?

Answer 33

n1 sinθ1 = n2 sinθ2

Question 34

What is total internal reflection?

Answer 34

When the incident ray is greater than the critical angle all light is reflected back into the material.

Question 35

Under what conditions does total internal reflection occur.

Answer 35

Light must be travelling from a more dense medium into a less dense medium (i.e. into a medium of lower refractive index).

Question 36

How could you predict if total internal reflection will occur at an interface?

Answer 36

sin C = 1 / n

Question 37

What is the principle focus?

Answer 37

In a converging lens, all rays converge onto the point called the principle focus.

In a diverging lens all (virtual) rays appear to have diverged/come from the principle focus.

Question 38

What is meant by the focal length of a lens?

Answer 38

The distance between the lens and the principle focus.

Question 39

What does it mean if the focal length of a lens is negative?

Answer 39

The lens is a diverging lens.

Question 40

Why is a converging lens’ focal length always positive?

Answer 40

Because the principal focus is always in front of the lens.

Question 41

What are the properties of a converging lens?

Answer 41

It is also called a “convex lens” and the image is always real and inverted except when the object is in front of the principal focus.

Question 42

What are the properties of a diverging lens?

Answer 42

It is also called a “concave lens” and the image is always smaller, virtual and upright wherever the object is positioned.

Question 43

State the lens equation.

Answer 43

1 / f = (1 / u) + (1 / v)

“u” being the distance between the object and the lens.

“v” being the distance between the image and the lens. It is also positive if the image is real and negative if the image is virtual.

“f” is only negative for a diverging lens.

Question 44

Under what conditions is the lens equation true?

Answer 44

The lens must be thin.

Question 45

What does the power of a lens tell you?

Answer 45

The higher the power of the lens, the more it will refract light.

Question 46

What is power measured in?

Answer 46

Dioptres (D)

For diverging lens it is -D

Question 47

How can you calculate the power of lens?

Answer 47

P = 1 / f

Question 48

How can the power of a compound lens be calculated? (lots of thin lenses in combination).

Answer 48

P = P1 + P2 + P3 + P4 + …

The lenses must be very close together/touching.

Question 49

How can you calculate magnification?

Answer 49

m = v / u

hi / ho (image height / object height)

Question 50

What is a polarised wave?

Answer 50

A wave that only oscillates in one direction/one plane of polarisation which includes the direction of energy transfer.

Question 51

What is perpendicular to an unpolarised light wave?

Answer 51

An electric and magnetic field.

Question 52

What waves can be polarised and why can light waves be polarised?

Answer 52

Only transverse.

Light is a mixture of many different oscillations of vibrations and can be polarised, which is evidence that light is a transverse wave.

Question 53

Describe what happens when two polarising filters are used and one is rotated.

Answer 53

The first filter polarises the light and all components of oscillations perpendicular to the plane of polarisation are absorbed.

Initially, the second filter is aligned so the planes of polarisation of both filters are parallel. This is when the light is most intense.

As the second filter is rotated, the amount of light that passes through it reduces.

If the second filter is 90° to the first, no light will pass through as it is all absorbed and the light is least intense.

As you continue to rotate the filter the amount of light passing through the filter combination will then increase.

When the total rotation of the second filter is 180°, the planes of polarisation of both filters are parallel again.

Question 54

Why is reflected or refracted light partially polarised?

Answer 54

Because most of the vertical oscillations are absorbed which leaves mostly the horizontal oscillations/planes.

Question 55

Why do TV and radio aerials have to be carefully aligned?

Answer 55

TV and radio aerials are polarised by the orientation of the rods on the aerial. The rods are either aligned in the same plane either horizontally or vertically depending on the transmitting aerial they are pointing at.

To receive a strong signal, you must line up the rods on the receiving aerial with the rods on the transmitting aerial. If they are not aligned the signal strength/intensity will be lower.

If the TV or radio aerial is moved around after tuning it, the signal will come and go, as the transmitting and receiving aerials go in and out of alignment.

Question 56

How are microwaves polarised?

Answer 56

The wavelength of microwaves is significantly longer than the wavelength of visible light, so the polarising filters they are used with will not work for light. Instead, microwaves can be polarised using metal grilles.

When the plane of polarisation of the metal grilles is at right angles (90°) to the plane of polarisation of the microwaves, the intensity drops to zero, because the grille is absorbing all of the energy.

Question 57

What is diffraction?

Answer 57

When a wave is spread out after passing through a narrow gap or around an obstacle.

Question 58

What waves can diffract?

Answer 58

All of them.

Question 59

What is Huygen’s construction?

Answer 59

A model for predicting the future movement of waves/a wavefront if we know the current position of the wavefront.

Question 60

How is Huygen’s construction used to explain diffraction.

Answer 60

Every point on a wave front may be considered to be a point source of secondary spherical wavelets that spread out in the forward direction with the wave speed. The new wavefront is the tangential line that joins all of these secondary wavelengths.

Using Huygen’s construction shows how, if the gap is the same size as the wavelength there is more diffraction. If the gap is bigger than the wavelength there is less diffraction/spreading out.

Question 61

Why is it easier to diffract sound than light?

Answer 61

The wavelength of visible light is small, so the gap needs to be similarly small to diffract it (i.e. a diffraction grating). The wavelength of sound is longer so the gap can be larger and it is easier to diffract. This is why sound can diffract through a doorway, but light cannot.

Question 62

What would be seen on a screen if laser light is passed through a diffraction grating?

Answer 62

A series of monochromatic (since laser light is monochromatic) bright, single maxima spread out horizontally where it is brightest at the centre.

Question 63

What would be seen on a screen if white light is passed through a diffraction grating?

Answer 63

A bright/high intensity central maxima (white light as it is not diffracted) followed by maxima on either side. Each maxima is a spectrum. The coloured light is diffracted by different amounts due to the different wavelengths of each colour.

Question 64

What is the diffraction grating equation?

Answer 64

dsinθ = nλ

“d” is the distance between the slits on the grating.

“n” is the order of the spectrum spot.

“θ” is the angle between the zero order and the n order.

“λ” is the wavelength of the wave.

Question 65

Why when a laser beam is shone through a diffraction grating is a pattern shown?

Answer 65

As light passes through two slits it diffracts, producing a coherent source from each slit. The two coherent waves meet together on the screen, where there is interference, either constructive (forming a maximum) or destructive (forming a minimum).

To form a bright spot, the waves superpose and the displacements of the two waves sum to produce a maximum.

The path difference between the two waves will be a whole number of wavelengths, nλ, where n is an integer.

This produces a phase difference of 2π showing that they are in phase.

Question 66

What would happen to the pattern displayed by a diffraction grating if the split spacing or wavelengths increased or decreased?

Answer 66

If d increases, sin θ becomes smaller, θ becomes smaller so the pattern is less spread out.

If d decreases, sin θ becomes larger, θ becomes larger so the pattern is more spread out.

Sin θ increases therefore θ increases so the pattern will spread out/maximas will be further apart.

Sin θ decreases therefore θ decreases so the pattern will be less spread out/maximas will be closer together.

Question 67

Which experiments provide evidence that waves behave as waves.

Answer 67

Reflection, refraction, interference, diffraction, I α A^2.

Question 68

Which experiments provide evidence that waves behave as particles?

Answer 68

Photoelectric effect, line spectra.

Question 69

Which experiments provide evidence that particles behave as particles?

Answer 69

Conservation of energy, conservation of momentum.

Question 70

Which experiments provide evidence that particles behave as waves?

Answer 70

Electron diffraction.

The pattern of light and dark bands/rings is consistent with what would be observed in light passing through a diffraction grating.

Question 71

Describe the electron diffraction experiments.

Answer 71

A filament (the cathode) is heated and emits electrons by thermionic emission.

These electrons are accelerated towards an anode by a large accelerating voltage.

There is a hole in the anode to produce a fine beam of electrons that hit the thin polycrystalline graphite target.

The experiment is in a vacuum, to prevent collisions between the electrons and particles in the air.

Electrons that hit the fluorescent screen cause visible light to be emitted.

A diffraction pattern is observed on the fluorescent screen of a series of light and dark rings.

Question 72

What is the de Broglie wave-particle duality theory?

Answer 72

Louis de Broglie discovered that electrons could show wave-like properties. The de Broglie equation links a wave property the (de Broglie wavelength) with a moving particle property (momentum).

λ = h / p

The faster the electrons, the greater their momentum and the shorter their de Broglie wavelength.

Question 73

What happens to the diffraction pattern seen if the accelerating voltage in the electron diffraction tube is increased?

Answer 73

As the accelerating voltage increases, the velocity of the electrons increases, the diameter of the rings decreases, therefore their de Broglie wavelength must decrease.

Question 74

Why is electron diffraction useful to investigate crystalline structure, but cannot be used for something like a car?

Answer 74

Diffraction only occurs if the (de Broglie) wavelength is the same size as the gap/space it interacts with.

The de Broglie wavelength of electrons is similar to the spacing between the layers of atoms in graphite. So, diffraction is observed.

For particles with more mass (e.g. a tennis ball), the gap needed is too small to be possible, or it would need to be moving too slow to be possible for the gap to be similar to the de Broglie wavelength.

Question 75

What is an ultrasound transducer?

Answer 75

It can both transmit and receive ultrasound pulses at the same time.

Question 76

What happens when ultrasound meets a boundary between two media?

Answer 76

The boundary between two media is called an interface. When waves (e.g. ultrasound) meet an interface, some energy is transmitted, some is reflected, and some may be absorbed.

The amount of energy transmitted/reflected/absorbed depends on the difference in densities between the two materials.

Question 77

What is the pulse-echo technique?

Answer 77

A short pulse of ultrasound is produced by a transducer. It passes through the body (for example). When it meets an interface, some is reflected back to the transducer. The transducer records the time it takes for the pulse to return. This is used to calculate the distance (using v=s/t). Note, the distance is there and back, so the time should be halved.

A computer uses this information to build an image of different tissues in the body.

Used for ultrasound foetal scans, sonar in boats.

Question 78

What does resolution mean?

Answer 78

The ability to distinguish one object from another.

Question 79

Describe and explain how to ensure a clear ultrasound image.

Answer 79

Gel is applied to the skin to remove air between the transducer and the skin because air has a substantially lower density than skin, so almost all the ultrasound would be reflected at the skin surface. This increases the proportion of ultrasound waves entering the body.

Question 80

How does pulse duration and wavelength affect the image produced in the pulse-echo technique?

Answer 80

The ultrasound transducer cannot transmit and receive pulses at the same time – there must be a gap in time between them (called the listening time). Shorter pulses mean it is less likely that the transmitted and reflected pulses will overlap. The listening time is longer. This increases the amount of information that can be gathered, making the image clearer.

For an object to be resolved, the wavelength of the ultrasound must be similar size to the object. Shorter wavelengths diffract less, so produce clearer, higher resolution images.

Question 81

Define a photon

Answer 81

A discrete packet/particle of electromagnetic radiation.

Question 82

What are the two equations to calculate the energy of a photon?

Answer 82

E = h f

E = (h x c) / 𝜆

Question 83

What is the photoelectric effect?

Answer 83

When electrons are released from the surface of metal due to incident electromagnetic radiation. There is a 1 to 1 interaction between a single photon and an electron. The energy of the photon depends on its frequency (E = h f). If the frequency of the photon is above the threshold frequency of the metal, electrons will gain enough energy to be released OR If the photon energy is greater than the work function, electrons will be released.

Question 84

What is a quantum?

Answer 84

A “packet”.

Question 85

What is a photoelectron?

Answer 85

An electron released from the surface of a metal due to incident electromagnetic radiation.

Question 86

Define work function.

Answer 86

The minimum energy required to release an electron from the surface of a metal.

Question 87

Define threshold frequency.

Answer 87

The minimum frequency required to release an electron from the surface of a metal.

Question 88

Define threshold wavelength

Answer 88

The maximum wavelength required to release an electron from the surface of a metal,

Question 89

How can you calculate work function?

Answer 89

𝜙 = ℎ 𝑓𝑜

𝜙= ℎ𝑐 / 𝜆𝑜

2nd equation is found by rearranging 𝑐=𝑓𝜆 for f, and substituting this in.

Question 90

What is Einstein’s photoelectric effect equation?

Answer 90

ℎ 𝑓 = 𝜙 + 0.5mv^2

(Photon energy = work function + Ek)

Question 91

What is the electronvolt?

Answer 91

An electronvolt is a measure of energy. It is much smaller than a joule.

Question 92

How can you convert between electronvolts and joules?

Answer 92

1eV = 1.6 x 10^-19 J

Question 93

What is the effect of increasing the intensity of radiation below the threshold frequency?

Answer 93

Even though more photons hit the surface per second, per m^2, no photoelectrons are released/it has no effect, because the photons do not have enough energy to overcome the work function.

Question 94

What is the effect of increasing the intensity of radiation above the threshold frequency?

Answer 94

More photons hit the surface per second, per m^2, therefore more photoelectrons are released per second, per m^2. The maximum kinetic energy of the photoelectrons stays the same as the photons do not have more energy.

Question 95

Why is it important that air is not allowed to enter the photocell?

Answer 95

Air oxidises the metal plate. This increases the work function of the metal. So, less photoelectrons will be released, and the current will be less

Photons are absorbed by the air. Less photons hit the metal plate. Therefore there are fewer one to one interactions between photons and photoelectrons. Therefore less photoelectrons are released, and the current is less

Photoelectrons collide with the air so less photoelectrons reach the anode therefore the current is less

Question 96

What does the graph from the photoelectron experiment show?

Answer 96

The threshold frequency is the y intercept.

The gradient is “h” (Planck’s constant).

Question 97

Why does the photoelectric effect provide evidence for the particle nature of electromagnetic radiation?

Answer 97

Light travels as particles, called photons. The energy of the photons is linked to the frequency of EM radiation. There is a 1 to 1 interaction between a single photon and an electron where the photon is absorbed, and the electron gains the energy. If it gains energy lower than the work function, the metal will heat up. If it gains energy more than the work function, it will leave the surface of the metal. This supports the experimental evidence from the photoelectric effect.

Question 98

Why did the wave theory of light not support the photoelectric effect?

Answer 98

According to wave theory, energy α intensity α A^2. Wave theory states that brighter, more intense light (with a bigger amplitude) has more energy. If wave theory was correct, the energy would be spread out over the surface, and each electron could gradually accumulate energy until there is enough to emit an electron. If wave theory was correct, the greater the intensity of the light, the more energy it has so the kinetic energy of the electrons should increase. This is not supported by the photoelectric effect, where intensity does not affect the kinetic energy of the photoelectrons. Wave theory also does not explain the threshold frequency where electrons should be emitted at all frequencies.

Question 99

What happens to the gold leaf when the zinc plate gets negatively charged?

Answer 99

The gold leaf falls because the photos of UV have enough energy to overcome the work function. Photoelectrons are lost from the zinc and the overall excess negative charge on the electroscope decreases, therefor the repulsive force between the two parts decreases. Hence the leaf falls.

Question 100

What happens to the gold leaf when the zinc plate gets negatively charged?

Answer 100

The gold leaf stays in its original position. No photoelectrons are lost from the zinc. As the plate is positively charged, there is a strong attractive force keeping the negatively charged electrons on the plate. The photons do not have enough energy to overcome this.

Question 101

Why is it the maximum kinetic energy of photoelectrons in Einstein’s equation?

Answer 101

Work function is the minimum energy needed to release an electron from the surface of a metal. This would correspond to the maximum kinetic energy of photoelectron near the surface. Electrons deeper down in the metal will need more energy than the work function, so will be emitted with less kinetic energy.

Question 102

What does ground state mean?

Answer 102

The lowest energy level of an atom.

Question 103

How can an electron move up and down energy levels?

Answer 103

Electrons can move up and down energy levels (transitions) by either losing exactly the right amount of energy, or gaining exactly the right amount of energy. This energy is in the form of a photon.

ΔE = E2 – E1 = hf = hc/λ

Question 104

Describe and explain how a continuous spectrum could be produced.

Answer 104

White light is passed through a prism or a diffraction grating. The different colours of light refract (prism) or diffract (grating) by different amounts.

The spectrum is continuous – there are no gaps.

Question 105

Describe and explain how line emission spectra are formed.

Answer 105

A glass tube/source contains atoms of only one type of gas. These atoms are ….

excited due to high temperatures

this causes their electrons to move to higher energy levels

the electrons at these higher energy levels are unstable

they de-excite/relax to a lower energy level. To do this they emit a photon of energy equal to the difference in energy between the two energy levels

A line emission spectra consists of a series of bright lines against a dark background, some of which may not be visible light

The line emission spectrum only has certain photon energies present, therefore only certain wavelengths are being emitted

Each line corresponds to a photon emitted when the electron de-excites between one energy level and another

There are discrete energy levels in an atom, so only certain transitions are possible, so only certain photon energies are possible corresponding to specific wavelengths

Question 106

Describe and explain the line absorption spectra.

Answer 106

White light is passed through a cool gas

At low temperatures, most of the electrons in the gas are in the ground state

Photons of the correct energy and wavelength (that correspond to the difference in electron energy levels for the atom) are absorbed by the electrons in the cool gas

The electrons are excited to a higher energy

Those energies/wavelengths are missing from the continuous spectra as they have been absorbed

Question 107

How do line emission and line absorption spectra compare?

Answer 107

The bright lines present in the emission spectra match the lines absent in the absorption spectra.