Module 4: C11 - Waves 1

Question 1

What is a Progressive Wave

Answer 1

A progressive wave is a means of transferring energy from one place to another without a transfer of matter between the two points.

Question 2

What are the 2 types of Progressive Waves

Answer 2

There are two types of progressive wave; a transverse wave and a longitudinal wave.

Question 3

What are Transverse Waves

Answer 3

Transverse waves are waves where the direction of oscillation of a wave is perpendicular to the direction of motion of the wave.

(Transverse waves have peak and troughs where the oscillating particle are at a maximum displacement from their equilibrium position.)

Question 4

What are Longitudinal Waves

Answer 4

Longitudinal Waves are waves where the direction of oscillation of a wave is parallel to the direction of motion of the wave.

Question 5

Explain what is meant by a progressive wave

Answer 5

A progressive wave is where energy is transferred from one place to another without a transfer of matter between the two points. (They can transfer energy and information)

Question 6

Describe, in terms of vibrations, the difference between a longitudinal and transverse wave. Give one example of each wave

Answer 6

Transverse waves are waves where the direction of motion is perpendicular to the oscillation of the wave (such as light waves), meanwhile longitudinal waves are waves where the direction of motion is parallel to the oscillation of the wave (such as sound waves and P-waves)

Question 7

What are examples of Transverse Waves

Answer 7

• Waves on the surface of the water
• Any electromagnetic wave - radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays
• Waves on stretched strings
• S-waves produced in earthquakes

Question 8

What are examples of Longitudinal Waves

Answer 8

• Sound waves
• P-waves produced in earthquakes

Question 9

What happens with Longitudinal waves when they travel through a medium

Answer 9

Longitudinal waves are often called compression waves. When they travel through a medium they create a series of compressions and rarefactions.

Question 10

Displacement:

• Symbol
• Unit
• Definition

Answer 10

Symbol:
s

Unit:
m

Definition:
Distance from the equilibrium position in a particular direction; a vector, so it can have either positive or negative value.

Question 11

Amplitude:

• Symbol
• Unit
• Definition

Answer 11

Symbol:
A

Unit:
m

Definition:
Maximum displacement from the equilibrium position (can be positive or negative)

Question 12

Wavelength:

• Symbol
• Unit
• Definition

Answer 12

Symbol:
λ

Unit:
m

Definition:
Minimum distance between two point in phase on adjacent waves, for example, the distance from one peak to the next or from one compression to the next.

Question 13

Period of Oscillation:

• Symbol
• Unit
• Definition

Answer 13

Symbol:
T

Unit:
s

Definition:
The time taken for one oscillation or time taken for wave to move one whole wavelength past a given.

Question 14

Frequency:

• Symbol
• Unit
• Definition

Answer 14

Symbol:
f

Unit:
Hz

Definition:
The number of wavelengths passing a given point per unit of time

Question 15

Wave Speed:

• Symbol
• Unit
• Definition

Answer 15

Symbol:
v (or c)

Unit:
ms^-1

Definition:
The distance travelled by the wave per unit of time

Question 16

Wave Equation (involving frequency, wave length, and wave speed)

Answer 16

Wave Speed = Frequency x Wavelength

v = f λ

Question 17

What is a Wave Profile

Answer 17

A graph showing the displacement of particles in the wave against the distance along the wave is called wave profile.

Question 18

What can the Wave Profile be used to determine in a wave

Answer 18

The wave profile can be used to determine the wavelength and amplitude of both types of waves. As the displacement of the particle in the wave is continuously changing, the wave profile changes shape over time.

Question 19

Worked Example: Finding the wavelength of a musical note

A flute produces a high-pitched note that has a time period of 0.45ms. The speed of sound through air is 330ms^-1. Calculate the wavelength of the note produced.

Answer 19

Step 1: Identify the correct equation to calculate the speed of the wave.

v = f λ

As f = 1/T substituting into the first equation gives v = λ/T => λ = vT.

Step 2: Substitute in known values in SI units and calculate the wavelength of the note.

λ = 330 x 0.45x10^-3 = 0.15m

Question 20

What is Phase Difference

Answer 20

Phase difference describes the difference between the displacements of the particles along a wave, or the difference between the displacements of the particles on different waves.

Note: It is most often measured in degrees or radians.

Question 21

What is the equation for Phase Difference

Answer 21

Phase Difference ϕ = 2π x/ϕ

Question 22

What is Phase Difference

Answer 22

Phase difference describes the difference between the displacement of particles along a waves or the difference between the displacements of particles on different waves. It is most often measured in degrees or radians, with complete cycle or wave representing 360° or 2 π radians

Question 23

What does it mean when Particles are described as being ‘in Phase’

Answer 23

If particles are oscillating perfectly in step with each other (they both reach their maximum positive displacement at the same time) then they are described as in phase. They have a phase difference of zero.

If two particles are separated by a distance of one whole wavelength, we say their phase difference is 360°, or 2π radians. If they are two complete cycles out of step their phase difference is 720° or 4 π radians.

Question 24

What does it mean when Particles are described as being ‘in Antiphase’

Answer 24

If particles are oscillating completely out of step with each other (one reaches its maximum positive displacement at the same time as the other reaches its maximum negative displacement) then they are described as being antiphase. They have a phase difference due of 180°, or π radians.

Two particles can have any phase difference as a phase difference depends on the separation of particles in terms of the wavelength.

Question 25

What is the Law of Reflection?

Answer 25

The law of reflection applies whenever waves are reflected. It states that the angle of incidence is equal to the angle of reflection.

Question 26

What is Reflection

Answer 26

Reflection occurs when a wave changes direction at a boundary between two different media, remaining in the original medium. Example: mirrored surface. The law of reflection: The angle of incidence is equal to the angle of reflection when measured to the normal.

Question 27

What is Refraction?

Answer 27

Refraction occurs when a wave changes direction as it changes speed when it passes from one medium to another. The amount of refraction depends on the refractive index of the material.

Question 28

How does the Refraction of Water Waves affect Wave Speed and Wavelength?

Answer 28

The speed of water waves is affected by changes in the depth of the water, which gives us an easy way to investigate the refraction of water waves. When a water wave enters shallower water, it slows down and the wavelength gets shorter

Question 29

Why do wave refract towards and away from the normal?

Answer 29

If a wave slows down it will refract towards the normal, if it speeds up it refracts away from the normal. Sound waves normally speed up when they enter a denser medium, whereas electromagnetic waves, like light, normally slow down. This results in the waves refracting in different directions.

Question 30

Equation for Refractive Index

Answer 30

n = c/v (Refractive Index = Speed of Light in Vacuum / Speed of Light in Material) Where n is the refractive index of the material (it has no units), c is the speed of light through a vacuum (3.00x10^8) and v is the speed of light through the material in ms^-1 If n=1 then the speed of light through the material is the same as the speed of light through a vacuum.

Question 31

Equation to show what happens when light travels from one medium to another

Answer 31

n1sinΘ1 = n2sinΘ2

Question 32

Worked Example: The speed of light through Olive Oil Determine the speed of light through Olive Oil (For Olive Oil, n=1.47)

Answer 32

Step 1: Identify the correct equation. n = c/v v = c/n Step 2: Substitute in known values in SI units and calculate the speed of light through the Olive oil. v = 3.0x10^8 / 1.47 = 2.04x10^8ms^-1 The speed of light through the material will always be less than the speed of light through a vacuum.

Question 33

Worked Example: Calculating the angle of Refraction A ray of light travels from water to crown glass. The light strikes the boundary between the two at an angle of 40.0° to the normal. Calculate the angle of refraction. (For down glass, n=1.52)

Answer 33

Step 1: Identify the correct equation n sin Θ = k Apply this equation to a ray of light travelling from water into glass. n.water sinΘ.water = n.glass sinΘ.glass Rearrange for sinΘ.glass sinΘ.glass = (n.water sinΘ.water) / (n.glass) Step 2: Substitute in known values and calculate sinΘ.glass sinΘ.glass = [1.33 x sin(40)] / 1.52 = 0.562… Θ.glass = 34.2° This angle is less than 40.0°, as the light has slowed down when it entered the glass from the water, bending towards the normal.

Question 34

What is the Special case if the Refraction Law

Answer 34

n = sinΘ1 / sinΘ2

Question 35

Equation made by Snellius (Snell’s Law)

Answer 35

n sinΘ = k

Question 36

Example Question: The speed of light in ethanol is measured as 220x10^6ms^-1. Calculate the refractive index of ethanol.

Answer 36

n = 3x10^8 / 220x10^6 n = 1.36

Question 37

Example Question: A ray of light travels from diamond into water. The light strikes the boundary between the diamond and the water at 20°C. Calculate the angle of refraction. (Diamond: n = 2.42 Water: n = 1.33)

Answer 37

n1sin(Ø1) = n2sin(Ø2) 2.42 sin(20) = 1.33 sin(Ø) 0.622 = sinØ Ø = 38.5°

Question 38

What is Diffraction

Answer 38

The spreading out of a wave as it passes through an gap/aperture or when they encounter an obstacle is called diffraction. The speed, wavelength, and a frequency of a wave do not change when diffraction occurs.

Question 39

When will Diffraction occur

Answer 39

Diffraction will only occur if the wavelength of the wave is of a similar size to the size of the gap/aperture.

Question 40

What is the difference between refraction and diffraction?

Answer 40

Refraction occurs when a wave changes direction as it changes speed when it passes from one medium to another, whereas diffraction is when a wave spreads out after passing through a gap when there is an obstruction in the way (but doesn’t change direction).

Question 41

What condition ensures that diffraction is increased?

Answer 41

Diffraction is increased when the size wavelength is closer to the size of the gap/aperture in the obstruction.

Question 42

What is Polarisation

Answer 42

Polarisation means that the particles oscillate along one direction only, which means that the wave is confined to a single plane.

Question 43

What is Unpolarised Light?

Answer 43

Light which oscillates in many different planes.

Question 44

What is Polarised Light?

Answer 44

Light which oscillates in one plane only

Question 45

Why can’t Longitudinal Waves be Polarised?

Answer 45

In longitudinal waves, the oscillations are always parallel to the direction of energy transfer, so longitudinal waves cannot be plane polarised. Their oscillations are already limited to only one plane (the direction of energy transfer)

Question 46

What type of wave can be Polarised

Answer 46

Transverse Waves

Question 47

Explain why electromagnetic waves can be polarised but sound waves cannot be polarised

Answer 47

Electromagnetic waves can be polarised, but sound waves cannot, as they are longitudinal waves, where the oscillations are always parallel to the direction of energy transfer, meaning they can’t be polarised. (Only transverse waves can be polarised)

Question 48

How does Partial Polarisation take place

Answer 48

Unpolarised light can also undergo polarisation by reflection off of non- metallic surfaces. The extent to which polarisation occurs is dependent upon the angle at which the light approaches the surface and upon the material that the surface is made of. Non-metallic surfaces such as asphalt roadways, snowfields and water reflect light such that there is a large concentration of vibrations in a plane parallel to the reflecting surface. A person viewing objects by means of light reflected off of non-metallic surfaces will often perceive a glare if the extent of polarisation is large.

Question 49

Why is it not possible to polarise a sound wave?

Answer 49

You can’t polarise a sound wave, as it is a longitudinal wave, where the oscillations are already always parallel to the direction of energy transfer. (Sound Waves & P-Waves)

Question 50

Explain why the diffraction of sound is regularly observed, but the diffraction of light is observed less frequently

Answer 50

This will be because sound waves have a larger wavelength compared to visible light, meaning sound waves are more likely to meet gaps similar in size to it’s wavelength, therefore it’s will diffract more often.

Question 51

Explain why it is possible to revive long-wavelength radio signals at the bottom of some valleys in which the higher-frequency TV signal cannot be received?

Answer 51

Long wavelength radio signals can be diffracted around valleys, allowing them to reach most areas, as the gap of the valley is a similar size to length of the wave. Therefore longer wavelength signals can be received better than shorter wavelength signals that would not diffract as easily.

Question 52

What is Intensity

Answer 52

Intensity is the radiant power passing through a surface per unit area measured in Wm-2. I = P/A

Question 53

Equation for Intensity

Answer 53

Intensity = Power / Area I = P/A

Question 54

Calculate the intensity when a power of 400 W is received over a cross-sectional area of 20m^2

Answer 54

I = P/A P=400W A=20m^2 400/20 = 20 Intensity = 20Wm^-2

Question 55

What is the relationship between Intensity and Distance (using the Intensity Equation)

Answer 55

The total radiant power P at a distance r from the source is spread out over an area equal to the surface area of the sphere (A = 4πr^2). I = P/A I = P/4πr^2 Therefore the relationship between Intensity and Distance follows the Inverse Square Law

Question 56

Worked Example: Finding the Power of the Sun What is the total power output of the Sun if the intensity of radiation received by the upper atmosphere is 1400 Wm-2? The average distance between the Earth and the Sun is 150 million km.

Answer 56

Step 1: Identify the correct equation to calculate the power from the Sun. I = P/A = P/4πr^2 Step 2: Rearrange to make the power the subject P = I x 4πr^2 Substitute in known values: P = 4000 x 4π x (150x10^9)^2 = 4.0x10^26 W.

Question 57

How can you tell there is an Inverse Square Relationship between Intensity and Distance from the Equation I = P/4πr^2

Answer 57

We can see from the equation that the intensity has an inverse square relationship with the distance from the source (I∝1/r^2). If the distance doubles, the intensity decrease by a factor of 4 (2^2), and if the distance increases by a factor of 100 the intensity will be 100^2 times smaller.

Question 58

Relationship between Intensity and Amplitude

Answer 58

Intensity ∝ (Amplitude)^2

Question 59

How are Intensity and Amplitude related (+what is their relationship)

Answer 59

When ripples travel out across the surface of a pond the intensity drops as the energy becomes more spread out. This causes a drop in amplitude. That is, the ripple height decreases the further the wave is from the source. Decrease amplitude means a reduced average speed of the oscillating with half the speed, and a quarter of the kinetic energy (Ek = 1/2mv^2). So for any wave the intensity is directly proportional to the square of the amplitude. Double the amplitude of a wave and the intensity will quadruple. Intensity ∝ (Amplitude)^2

Question 60

What are the 4 Properties of Electromagnetic Waves

Answer 60

● All EM waves travel at the speed of light. (In a vacuum c = 3.0 x 108 ms-1) Therefore the wave equation for EM waves is: c=f𝝀 ● EM waves can travel through a vacuum, no medium is required to travel through. ● EM waves with high frequency have high energy ● They are all transverse waves

Question 61

State two properties shared by all electromagnetic waves which distinguish them from other waves.

Answer 61

All EM waves travel at the speed of light (3x10^8ms^-1), and they are able to travel in a vacuum, meaning no medium is required for them to travel. Also EM waves with high frequency also have high energy too

Question 62

Worked Example: A radio station transmits at a frequency of 107.3MHz. Calculate the wavelength of the radio wave to an appropriate number of significant figures.

Answer 62

Step 1: Identify the correct equation to calculate the speed of the wave. Rearranging this equation for λ gives λ = c / f Step 2: Substitute in the known values in SI units and calculate the wavelength. λ = 3x10^8 / 107.3x10^6 = 2.80m (3sf)

Question 63

Order of the EM spectrum from highest wavelength to lowest

Answer 63

In order of reducing wavelengths λ Radio Waves >10^6 - 10^-1 Microwaves 10^-1 - 10^-3 Infrared 10^-3 - 7x10^-7 Visible Light 7x10^-7 - 4x10^-7 Ultraviolet 4x10^-7 - 10^-8 X-Rays 10^-8 - 10^-13 Gamma Rays 10^-10 - <10^-16

Question 64

The wavelength ranges of which two rays overlap?

Answer 64

X-rays and gamma rays overlap at a certain point. At this point they are classified on where they come from

Question 65

Example Question: The human eye can detect EM waves with wavelengths from around 400 to 700nm. Calculate the minimum and maximum frequencies of light that the eye can detect.

Answer 65

λ = 700 x 10^-9 c = fλ f = c/f 3x10^8 / 7x10^-7 = 4.3x10^14 Hz λ = 400 x 10^-9 c = fλ f = c/λ 3x10^8 / 4x10^-7 = 7.5x10^14Hz Therefore, the human eye can detect frequencies between 4.3x10^14 - 7.5x10^14

Question 66

Example Question: The Earth is on average 150 million km from the Sun. Calculate the time taken for light to travel from the Sun to the Earth.

Answer 66

c = 3x10^8ms^-1 s = d/t => t = d/s 1.5x10^11 / 3x10^8 = 500 seconds

Question 67

What is a use for Polarisation of electromagnetic waves?

Answer 67

One use of the polarisation of electromagnetic waves is in communications transmitters. In order to reduce interference between different transmitters, some transmit vertically plane polarised waves and other nearby transmit horizontally plane polarised waves. An aerial aligned to detect vertically polarised radio waves will suffer less interference from horizontally polarised waves and vice versa.

Question 68

How do Polarising Filters work?

Answer 68

Polarised electromagnetic waves can be polarised using filters called polarisers. The nature of the polariser depends on the part of the electromagnetic spectrum to be polarised, but each filter only allows waves with a particular orientation through.

Question 69

* What is Malus’ Law

Answer 69

The intensity of a beam of plane polarised light after passing through a rotatable polarised varies as the square of the cosine of the angle through which the polariser is rotated from the position that gives maximum intensity.

Question 70

Equation for Intensity (involving cos)

Answer 70

I = Io cos^2 Θ

Question 71

When using 3 polarising filters, observe what happens when: a) Filter 1 and 3 in the same plane, rotate filter 2 b) Filter 1 and 3 at 90° to each other. Rotate filter 2

Answer 71

a) When this happens you can see more b) The light intensity decreases and you can’t see anything

Question 72

Example Question: A plane polarised radio wave is transmitted from a vertical aerial to a nearby receiving aerial. The entire receiving aerial is rotated slowly through 180° in the direction shown by the arrow. Explain clearly what will be observed on the ammeter and how the detected signal varies.

Answer 72

From 0-90° the current will decrease before fall to 0 at 90°, as Malus’ Law states that I = Iocos^2Ø, where cos^2(90)=0, before the current starts to increase from 90-180°, where it reaches its maximum at 180°, where cos^2(180) = 1.

Question 73

What happens when you place 2 Polaroid filters over each other and rotate them?

Answer 73

Unpolarised light passing through the first filter becomes plane polarised. If the second filter (sometimes called the analyser) is in the same plane as the first, then the light passes through it unaffected. However, if the second Polaroid is slowly rotated the intensity of the light transmitted through it drops. When the second filter has turned through 90°, no light is transmitted and the intensity falls to zero.

Question 74

A metal grille is placed between a source of plane polarised microwaves and a receiver. Describe and explain the effect of rotating the grille through 180° around the axis of the beam on the intensity recorded by the receiver

Answer 74

As the grille is rotated the intensity falls. It falls to zero when the gaps in the grille are in the opposite plane to the microwaves (after 90°). As the grille is rotated further the intensity creases again. It reaches the maximum value again when the gaps in the grille are in the same plane as the microwaves (after 180°).

Question 75

Suggest why the metal sheet in the door of microwave ovens contains little holes, rather than a series of slits.

Answer 75

Holes will not allow any orientation of plane polarised microwaves through.

Question 76

A student holds a polarising filter in front of a laptop screen and then rotates it. At a particular angle, the laptop screen appears to go dark. a) Suggest what you can deduce about the nature of light emitted from the laptop screen from the student’s observation b) Explain how the laptop screen can be viewed once again through the filter

Answer 76

a) It must must be plane polarised b) Rotate the Polaroid further until it is at 90° from the minimum intensity. In this orientation the Polaroid is aligned in the same plane as the light emitted from the screen.

Question 77

Example Question: A beam of polarised light is directed normally at a polarising filter of cross-sectional area 9.0x10^-4m^2. The polarising filter is slowly rotated in a plane at right angles to the beam. The transmitted intensity, I, plotted against the angle Ø resembles figure 3 with a maximum intensity of 20Wm^-2 a) Calculate the power of light transmitted through the filter at Ø = 0°.

Answer 77

a) 20 x 9.0x10^-4 = 1.8x10^-3 W

Question 78

What is Total Internal Reflection (TIR)

Answer 78

The total internal reflection (TIR) of light occurs at the boundary between two different media. When the light strikes the boundary at a large angle to the normal, it is totally internally reflected. All the light is reflected back into the original medium. There is no light energy refracted out of the original medium.

Question 79

What are the 2 Conditions required for Total Internal Reflection (TIR)

Answer 79

● The original material must have a higher refractive index than the surrounding material. (e.g. glass to air will cause TIR) ● The angle of incidence must be greater than the critical angle.

Question 80

What is the Critical Angle

Answer 80

Critical angle: The angle of incidence at the boundary between two media that will produce an angle of reflection of 90°.

Question 81

By using n1Sinθ1 = n2Sinθ2 , determine the relationship between the refractive index of the medium and the critical angle when light travel into air

Answer 81

At the critical angle C, θair is 90° n sinC = nair sin90° Both the refractive index of air and sin90 are equal to 1, so this becomes n sinC = 1x1 sinC = 1/n From this we can see that the greater the refractive index the lower the critical angle.

Question 82

Equation to find the Critical Angle

Answer 82

SinC = 1/n

Question 83

Worked Example: Crown Glass Crown glass has a refractive index of 1.52. Determine the critical angle between crown glass and air.

Answer 83

Step 1: Identify the correct equation to use. Sin C = 1/n => C = sin^-1(1/n) Step 2: Substitute in known values and calculate the critical angle. C = sin^-1(1/1.52) = 41.1° If light strikes the internal surface of crown glass at above 41.1° then it will be totally internally reflected.

Question 84

What are Optical Fibres and their uses

Answer 84

Optical fibres are designed to totally internally reflect pulses of visible light (or occasionally infrared) travelling through them. They have many uses, including transmitting data for fast broadband connections and images from inside patients during keyhole surgery.

Question 85

How does a simple Optical Fibre work?

Answer 85

A simple optical fibre has a fine glass core surrounded by a glass cladding with a lower reflective index. Light travelling through the fibre is contained within the core because of total internal reflection at the core/boundary.

Question 86

How can you determine the refractive index from the critical angle?

Answer 86

A simple experiment to determine the refractive index of a material can be carried out by carefully measuring the critical angle of a semi-circular block. Directing the ray of light towards the centre of the semi-circular block ensures that light enters the block at 90° to the boundary and does not change direction, so the critical angle can be measured accurately.