Waves

Question 1

W1-5
1. What is a progressive wave?
2. Which type of wave can be polarised?
3. What does it mean if a wave is polarised?

Answer 1

1. A progressive wave is a wave that transfers energy from one point to another, without transferring matter.
2. The transverse waves can be polarised because its displacement vector is perpendicular to the energy transfer direction (1), while the longitudinal waves travel parallel to the direction of energy transfer (1), so cannot restrict vibrations of the displacement vector to a single plane. (1)
3. Polarisation limits a transverse wave to a single perpendicular plane of oscillation.

Question 2

1. What’s the difference between a transverse and a longitudinal wave?
2. List some examples of transverse waves
3. List some examples of longitudinal waves

Answer 2

1. A transverse wave oscillates perpendicular to the direction of energy transfer, while a longitudinal wave oscillates parallel to the direction of energy transfer.
2. Examples of transverse waves include: light waves (or any electromagnetic waves), water waves.
3. Examples of longitudinal waves include: sound waves, p-waves (earthquakes)

Question 3

W1-5
1. How can light be polarised?
2. How can microwaves be polarised?

Answer 3

1. When the second polariser is at 0° or 180° it aligns with the axis of the first polariser and also the plane of polarisation for the light. Therefore the light that has already been polarised can pass through.
   When the second polariser is at 90° (perpendicular) to the first it blocks all of the light as it has already been plane polarised perpendicular to the axis of the polariser. The intensity of the unpolarised electromagnetic wave reduces after it passes through a polarising filter, with maximum intensity 0 or 180 degrees, and minimum (zero) intensity 90 and 270 degrees.
2. Microwaves are polarised by passing through a metal grille, where wires are aligned parallel to each other.

Question 4

1. What does ‘coherent’ mean?
2. What does a graph of displacement against time of a wave look like, and what would another wave with a phase difference of 𝜋/2 radians look like?
3. What does the term antiphase mean?

Answer 4

1. Waves are coherent if they have the same wavelength and frequency, as well as there being a fixed phase difference between them.
2. It will look like a sine wave graph, with displacement on the y-axis and time on the x-axis. For a wave with a wavelength of 2𝜋 radians, the second wave with a phase difference of 𝜋/2 radians will start at the place where the amplitude of the first wave is.
3. Antiphase is when two waves have a phase difference of 180 degrees or 𝜋 radians.

Question 5

Define the following:
1. Amplitude
2. Wavelength
3. Frequency
4. Time period
5. Displacement
6. Speed

Answer 5

1. Amplitude: the maximum displacement reached from equilibrium.
2. Wavelength: the shortest distance between two points in a phase.
3. Frequency: the number of complete oscillations per second.
4. Time period: the time taken for one complete oscillation.
5. Displacement (x) of a wave is the distance of a point on the wave from its equilibrium position. It is a vector quantity; it can be positive or negative.
6. Speed (v) is the distance travelled by the wave per unit time.

Question 6

1. Define node
2. Define antinode
3. What do the diagrams for the first 3 harmonics look like?

Answer 6

1. A node is a point of minimum amplitude.
2. An antinode is a point of maximum amplitude.
3. The first harmonic (also called the fundamental frequency) has one antinode and two nodes. The second harmonic has two antinodes and three nodes. The third harmonic has three antinodes and four nodes.

Question 7

Define all of the terms for the formula of the first harmonic
f= (1/2l)(Sqrt(T/μ))

Answer 7

f= (1/2l)(Sqrt(T/μ))
f = frequency in Hz
l = length in m
T = tension in N
μ = mass per unit length in kgm^-1

Question 8

1. Describe the motion of a point on a stationary wave
2. What is one wavelength in degrees and radians?
3. What is the formula for the time period of a wave?

Answer 8

1. The point oscillates repeatedly about an equilibrium position.
2. One wavelength = 2𝜋
3. Time period of a wave = 1 / frequency

Question 9

1. Explain how a stationary wave is formed
2. Explain how a sequence of equally spaced maxima and minima are formed from stationary waves

Answer 9

1. Two progressive waves travel in opposite directions along the string. The waves have the same frequency and the same amplitude, and the two waves superpose with each other. (3 marks) The node is the point of minimum or no disturbance, which is where destructive interference occurs, and the waves cancel each other out. The antinode is the point of maximum amplitude, and is where constructive interference occurs. (6 marks)
2. There is superposition when two waves of the same frequency travel in opposite directions. There are maxima where waves are in phase or interfere constructively, minima where are in out of phase or interfere destructively.

Question 10

Explain the following differences between an undamped progressive wave and a stationary transverse wave: amplitude, phase, energy transfer, maxima and wave speed

Answer 10

Difference in amplitude:
Each point along the wave has the same amplitude for a progressive wave, but varies for stationary waves.
Difference in phase:
In a progressive wave, adjacent points vibrate with a different phase. In a stationary wave, between nodes all particles vibrate in phase.
Difference is energy transfer:
In progressive waves, energy is transferred from one point to another. In stationary waves, energy is stored, not transferred.
Difference in maxima and minima:
Stationary waves have nodes and antinodes, while progressive waves do not.
Difference in wave speed:
A progressive wave’s speed is the speed at which the wave moves through a medium, while in a stationary wave each point on the wave oscillates at a different speed. The overall wave does not move.

Question 11

1. State the physical conditions that are necessary for a stationary wave to form on the string
2. State how you know a wave on a string is transverse
3. How do you describe a phase difference between two waves?

Answer 11

1. The waves are travelling in opposite directions, the waves have similar amplitudes, the same frequency, and reflected at the end of the string.
2. The displacement is perpendicular to the rest position of the string.
3. You write the degrees that one point is out of phase with another point.

Question 12

1. Explain how a stationary wave is produced when a stretched string is plucked.
2. Explain what the observer would see as a second polarising filter placed in front of the first, is rotated around 360 degrees.
3. What must the orientation of two polarising filters be in order to see light on the other side?

Answer 12

1. The waves are travelling in opposite directions because of reflection at the end of a string, the two waves superpose and interfere with each other.
2. If you rotate a polarising filter that is currently rotated at letting the maximum light intensity through, you will have the minimum light intensity at 90 and 270 degrees; you will have the maximum light intensity at 180 and 360 degrees.
3. If you place two polarising filters in front of each other, in order to see the light on the other side, the direction of the transmission axis must be the same on both filters.

Question 13

1. What are the uses of polaroid filters?
2. How is energy transmitted through a longitudinal wave?
3. What is particle displacement?

Answer 13

1. It is used in polaroid (sun)glasses and cameras to reduce glare and enhance the image. It is used in a microscope to identify minerals and rocks. It is used in a polarimeter to analyse chemicals and different types of sugar.
2. There is a region in the transmitting medium where compression causes particles to vibrate more. Energy is transferred by collisions between molecules. The compression causes a compression further along the medium.
3. Particle displacement is the distance of a particle from equilibrium position in the direction of wave energy transfer.

Question 14

1. Why are turntables used in microwaves, and what will happen if the turntable is removed?
2. How can you calculate the frequency of an electromagnetic wave (such as a microwave) when given the distance between the antinodes?
3. How does the distance between maxima and minima of waves relate to the wavelength?

Answer 14

1. This is so the position of the antinode (where the maximum energy is) continually changes location. If the turntable is removed, the food will only melt where it receives microwaves with the maximum amplitude. The food will heat up at the antinodes of the microwaves.
2. All electromagnetic waves travel at the speed of light, which is given in the formula sheet. The wavelength is the distance between the first and the third antinode. Rearrange the formula speed = frequency x wavelength to find the frequency.
3. The distance between two maxima or two minima is one wavelength. The distance between a maxima and a minima is half a wavelength.

Question 15

What is wave interference?

Answer 15

Interference happens when two coherent waves are superimposed on one another. Coherent means that the two waves must have the same frequency and wavelength, and a fixed phase difference. Interference occurs when waves overlap and their resultant displacement is the sum of the displacement of each wave.

Question 16

W1
Explain the use of stationary waves in microwave ovens

Answer 16

A microwave source is placed in line with a reflecting plate and a small detector between the two.
The reflector can be moved to and from the source to vary the stationary wave pattern formed.
By moving the detector, it can pick up the minima (nodes) and maxima (antinodes) of the stationary wave pattern.

Question 17

Explain how stationary waves work with sound

Answer 17

Sound waves can be produced as a result of the formation of stationary waves inside an air column.
This is how musical instruments, such as clarinets and organs, work.
This can be demonstrated by placing a fine powder inside the air column and a loudspeaker at the open end.
At certain frequencies, the powder forms evenly spaced heaps along the tube, showing where there is zero disturbance as a result of the nodes of the stationary wave. In order to produce a stationary wave, there must be a minima (node) at one end and a maxima (antinode) at the end with the loudspeaker.

Question 18

W1
Explain how stationary waves work with stretched string.

Answer 18

Vibrations caused by stationary waves on a stretched string produce sound.
This is how stringed instruments, such as guitars or violins, work. This can be demonstrated by a length of string under tension fixed at one end and vibrations made by an oscillator.
At specific frequencies, known as resonant frequencies, a whole number of half wavelengths will fit on the length of the string.
As the resonant frequencies of the oscillator are achieved, standing waves with different numbers of minima (nodes) and maxima (antinodes) form.

Question 19

W1
Longitudinal waves

Question 20

Define and explain refraction

Answer 20

Refraction: The change in direction of a wave when it passes through a boundary between mediums of different density.
Refraction is caused by a change in speed of different parts of the wavefront as they hit the boundary of the medium (a transparent material). When a wave refracts, its speed and wavelength change, but its frequency remains the same. When the light ray is incident on the boundary at 90°: The wave passes straight through without direction, because the whole wavefront enters the boundary at the same time.

Question 21

What are the conditions for refraction?

Answer 21

The angles of incidence and refraction are measured from the normal line. This is drawn at 90° to the boundary between the two media.
When light passes from a less optically dense medium to a more optically dense medium, (e.g. air → glass): The refracted light has a lower speed and a shorter wavelength than the incident light, therefore resulting in a smaller angle of refraction bending towards the normal and vice versa into a more dense medium.

Question 22

1. Explain the refractive index and what is the refractive index formula?
2. What is the Snell’s law formula?

Answer 22

1. The refractive index, n of a material tells us how optically dense it is. The refractive index of air is n = 1. Media that are more optically dense than air will have a refractive index of n > 1.
   The refractive index of a material is calculated using the equation: n = c/c(small s)
   Where: n = refractive index of material; c = the speed of light in a vacuum (m s–1); cs = the speed of light in a substance (m s–1)
2. The refractive index of a material is calculated using the equation: n1 sin theta1 = n2 sin theta2
   Where:
   n1 = the refractive index of material 1 (the material that the ray goes through first)
   n2 = the refractive index of material 2 (the material that the ray goes through second)
   θ1 = the angle of incidence of the ray in material 1
   θ2 = the angle of refraction of the ray in material 2

Question 23

1. Explain the Critical Angle
2. What is the critical angle formula?

Answer 23

1. The larger the refractive index of a material, the smaller the critical angle. As the angle of incidence is increased, the angle of refraction also increases, until the angle of incidence reaches the critical angle.
   When the angle of incidence = critical angle then: Angle of refraction = 90°; The refracted ray is refracted along the boundary between the two materials.
   When the angle of incidence < critical angle then: the ray is refracted and exits the material.
   When the angle of incidence > critical angle then: the ray undergoes total internal reflection.
2. Sin theta c = n2 / n1
   Where: n1 = refractive index of material 1; n2 = refractive index of material 2; θc = critical angle of material 1
   The formula finds the critical angle of the denominator of the fraction
   The critical angle can also be calculated using the angles of incidence and refraction:
   Sin theta c = n2 / n1 = sin theta 1 / sin theta 2

Question 24

Explain TIR

Answer 24

Total internal reflection is a special case of refraction that occurs when: The angle of incidence within the denser medium is greater than the critical angle (I > θc), and the incident refractive index n1 is greater than the refractive index of the material at the boundary n2 (n1 > n2). Total internal reflection follows the law of reflection; angle of incidence = angle of reflection. TIR is more likely in optically dense materials.

Question 25

1. Define diffraction 2. Describe the conditions for diffraction 3. List some examples of diffraction

Answer 25

1. Diffraction: the spreading out of waves after they pass through a narrow gap or around an obstruction. After passing through a gap: the waves spread out so they have curvature, and the wave has a smaller amplitude because the barrier on either side of the gap absorbs wave energy. The only property of a wave that changes when it diffracts is its amplitude, while the wavelength of the wave remains the same. 2. No diffraction occurs when gaps are very small or very big in comparison to the wavelength of the waves. When the wavelength of the wave and the width of the gap are similar in size, then diffraction occurs. When the wavelength is bigger than the gap, more diffraction occurs, and the wave spreads out more after passing through. 3. Radio waves moving in between or around buildings Water waves moving through a gap into a harbour

Question 26

Describe the single slit monochromatic diffraction pattern

Answer 26

The diffraction pattern of light passing through a single slit is a series of light and dark fringes on a screen. The bright fringes are areas of maximum intensity, produced by the constructive interference of each part of the wavefront as it passes through the slit The dark fringes are areas of zero or minimum intensity, produced by the destructive interference of each part of the wavefront as it passes through the slit The central maximum is: Much wider and brighter than the other bright fringes Much wider than that of the double-slit diffraction pattern On either side of the wide central maxima are the much narrower and less bright maxima. These get dimmer as the order increases.

Question 27

1. Describe the single slit intensity pattern for monochromatic light 2. Describe the intensity pattern for white light

Answer 27

1. The features of the single slit intensity pattern are: The central maximum is the central bright fringe with the greatest intensity The dark fringes are regions with zero intensity As you move away from the central maxima either side, the intensity of each bright fringe reduces. 2. A source of white light diffracted through a single slit will produce the following intensity pattern: The central maxima is equal in intensity to that of monochromatic light The non-central maxima are wider and less intense The fringe spacing between the maxima get smaller The amount of red wavelengths in the pattern increases with increasing maxima, n increases from n = 1, 2, 3... The amount of blue wavelengths decrease with increasing maxima

Question 28

Describe the single slit diffraction pattern for white light

Answer 28

The central maximum is bright white because constructive interference from all the colours happens here. It is wider and brighter than the other bright fringes, and wider than the central maximum of the double-slit diffraction pattern. All the other maxima are composed of a spectrum: where the shortest wavelength (violet) will appear closest to the central maximum as it is diffracted less, while the longest wavelength (red) will appear furthest from the central maximum as it is diffracted more.

Question 29

1. Explain the effect of wavelength on fringe width 2. Explain the effect of slit width on fringe width

Answer 29

1. An increased wavelength will increase in the angle of diffraction, which increases the width of the bright minima. Red light (with the longest wavelength) will produce a diffraction pattern with wide fringes, while blue light will produce narrow fringes. 2. If the slit was made narrower: The angle of diffraction is greater, so the waves spread out more beyond the slit The intensity graph will show that: the intensity of the maxima decreases, so the width of the central maxima increases, and the spacing between fringes is wider.

Question 30

1. What is diffraction grating? 2. What is the diffraction grating equation?

Answer 30

1. A diffraction grating is a piece of optical equipment that also creates a diffraction pattern when it diffracts monochromatic light into bright and dark fringes, and white light into its different wavelength components. A diffraction grating consists of a large number of very thin, equally spaced parallel slits carved into a glass plate. Diffraction gratings are useful because they create a sharper pattern than a double slit. This means their bright fringes are narrower and brighter while their dark regions are wider and darker. 2. The regions where constructive interference occurs are also the regions of maximum intensity. Their location can be calculated using the diffraction grating equation: d sin θ = n λ Where: n is the order of the maxima, the number of the maxima away from the central (n = 0); d is the distance between the slits on the grating (m); θ is the angle of diffraction of the light of order n from the normal as it leaves the diffraction grating (°); λ is the wavelength of the light from the source (m).

Question 31

1. How do you calculate slit spacing? 2. What is angular separation?

Answer 31

1. Diffraction grating sizes are determined by the number of lines per millimetre (lines / mm) or lines per m. This is represented by the symbol N. d is the distance between the slits on the grating. d can be calculated from N using the equation: d = 1 / N 2. The angular separation of each maxima is calculated by rearranging the grating equation to make θ the subject. The angle θ is taken from the centre meaning the higher orders of n are at greater angles. The angular separation between two angles is found by subtracting the smaller angle from the larger one. The angular separation between the first and second maxima at n1 and n2 is θ2 – θ1.

Question 32

1. What is the order of maxima? 2. How do you calculate the angle between two order lines?

Answer 32

1. The maximum angle of diffraction with which maxima can be seen is when the beam is at right angles to the diffraction grating. This means θ = 90 degrees and sin θ = 1. The highest order of maxima visible is therefore calculated by the equation: n = d / λ. Since n is an integer number of maxima, if the value obtained is a decimal it must be rounded down to determine the highest-order visible, e.g If n is calculated as 2.7 then n = 2 is the highest-order visible. 2. Calculate the two angles between the order lines and the normal, then add these together (do not directly measure the angle between the lines).

Question 33

What are the applications of diffraction gratings?

Answer 33

They are used in spectrometers to analyse light from stars, to determine their composition, and carry out chemical analysis to observe the spectra of materials. They are also used in X-ray crystallography: X-rays are directed at a thin crystal sheet which acts as a diffraction grating to form a diffraction pattern. This is because the wavelength of X-rays is similar in size to the gaps between the atoms. This diffraction pattern can be used to measure the atomic spacing in certain materials.

Question 34

1. Define and explain the principle of superposition 2. Define interference and explain the types of interference 3. Define coherence and give some examples

Answer 34

1. Superposition: two or more waves arrive at the same point and overlap, and their amplitudes combine. The principle of superposition states that: When two or more waves overlap at a point, the displacement at that point is equal to the sum of the displacements of the individual waves. 2. Interference is the effect of superposition, which can be seen clearly when waves overlap in phase or antiphase. When two waves are in antiphase, their combined effect means they cancel each other out; this is destructive interference When two waves are in phase, their combined effect makes the resultant wave amplitude larger; this is constructive interference. 3. Coherence occurs when waves are the same frequency and have a constant phase difference. Examples of interference from coherent light sources are: Monochromatic laser light; Sound waves from two nearby speakers emitting sound of the same frequency.

Question 35

1. Define path difference 2. What is the difference between path and phase difference?

Answer 35

1. Path difference is defined as: The difference in distance travelled by two waves from their sources to the point where they meet. (TIP: The path difference is how much longer, or shorter, one path to a point is than the other. In other words, the difference in the distances. Make sure not to confuse this with the distance between the two paths where the slits start.) 2. Path difference compares the amount of progress made by waves along a path, while phase difference compares the distance between the phases (peaks and troughs) of coherent waves that are normally travelling parallel to each other at a point.

Question 36

1. What are the properties of lasers, and what is the diffraction pattern produced by a laser? 2. What are the safety issues with lasers?

Answer 36

1. Lasers form light that is coherent and monochromatic. It consists of areas of constructive interference – the bright fringes, and areas of destructive interference – the dark fringes. 2. Lasers produce a very high-energy beam of light, which can cause permanent eye damage or even blindness. Therefore, you must take the following precautions: never look directly at a laser or its reflection; don’t shine the laser towards a person; wear laser safety goggles.

Question 37

1. What are the conditions for constructive and destructive interference? 2. Explain wavefront diagrams

Answer 37

1. The condition for constructive interference is: path difference = n lambda The condition for destructive interference is: path difference = (n + ½) lambda Where: lambda is the wavelength of the wave in metres, and n is any integer. 2. A curved line represents each wavefront (peak or trough), where you can count the number of wavelengths to determine the type of interference that occurs. This is done by counting the number of wavelengths from the point to the centre of each individual wavefront.

Question 38

1. Explain interference using sound waves 2. Explain interference using microwaves

Answer 38

1. Sound waves are longitudinal waves made up of compressions and rarefactions. Constructive interference occurs when the compressions and rarefactions from each wave line up and the sound appears louder. Destructive interference occurs when a compression from one wave lines up with a rarefaction from the other and vice versa. The two waves cancel each other out, so zero sound is heard. This is the technology used in noise-cancelling headphones. 2. Two-source interference for microwaves (and other electromagnetic waves) can be detected with a moveable microwave detector. The detector picks up a maximum amplitude or intensity in regions of constructive interference. The detector picks up a minimum or zero amplitude, so no signal in regions of destructive interference.

Question 39

1. What is the formula of intensity variation with amplitude? 2. How do you calculate the number of quiet spots heard between two speakers in phase in a specified distance?

Answer 39

1. The intensity (power per unit area) of a wave is proportional to the energy transferred by the wave. The intensity of a wave is proportional to the square of the amplitude I ∝ A^2 Where: I = intensity of the wave in W m–2; A = amplitude of the wave in metres (m) 2. Calculate the wavelength (with the speed of sound in air and frequency given). Use the destructive interference formula to calculate the smallest path difference (n) – the location of the first quiet spot. Then add the path difference multiple times up to and including the maximum distance.

Question 40

W1-5 How to calculate the superposition and find the resultant wave of waves that are: 1. In phase? 2. Out of phase?

Answer 40

1. For waves that are in phase, you add together the amplitude to get the total amplitude. If one amplitude is above the rest point and the other is below the rest point, then subtract the one below the rest point from the one above. If the value is positive, then the amplitude is above the rest point. If negative, then the amplitude is below the rest point. 2. For waves that are out of phase, where they intersect will be their maximum and minimum amplitudes. Multiply the amplitude of where they intersect by 2 to find the maximum amplitude.

Question 41

W1-5 1. What is the name of the effect that causes a wave to return, and what do you need to remember when calculating the speed for these waves? 2. Describe what the observer would see if one polarising filter is placed in front of another and rotated around 360 degrees.

Answer 41

1. A reflection is when a sound wave changes direction when it hits the boundary between two different materials. When calculating the speed of these waves, you need to remember to double the distance, or half the time, as the wave is travelling away from the observer and back again. 2. There is a variation in intensity between maximum and minimum amount of light allowed through. There are two maxima and two minima in a 360 degrees rotation.

Question 42

W1-5 1. How you would describe the motion of a particle during a complete transverse wave cycle just before it moves vertically upwards 2. What do you need to remember when calculating the speed of sound based on a reflection?

Answer 42

1. The oscillations are perpendicular to the direction of wave travel. (1) It oscillates from equilibrium to maximum positive displacement, and back to equilibrium. (1) (Do not state that the particle moves up and down). 2. As the wave is travelling to the wall and back, don’t forget to half the time or double the distance.

Question 43

Define all of the terms in the formula for the double slit experiment.

Answer 43

The fringe separation formula is: w = λD / s λ is the wavelength of light. D is the distance from the slits to the screen. s is the slit spacing.