3.3.1 Waves

Question 1

Define a progressive wave

Answer 1

A moving wave that carries energy from one point to another without transferring any matter

Question 2

What type of waves can be progressive?

Answer 2

Transverse and longitudinal

Question 3

Define displacement and amplitude of a wave.

Answer 3

Displacement: the distance a point on the wave has moved from its equilibrium position

Amplitude: the maximum magnitude of displacement (relative to the position of equilibrium)

Question 4

Define wavelength.

Answer 4

The length between two adjacent points moving in phase
(crest to crest or trough to trough)

Question 5

Define the period and frequency of a wave.

Answer 5

Period: the time taken for a whole cycle to complete / pass a given point

Frequency: the number of cycles produced / passing a given point
per second

F = 1/T

Question 6

Define phase and phase difference.

Answer 6

Phase: a measurement of the position of a certain point along the wave cycle

Phase difference: the amount one wave/point on a wave lags behind another

Both are measured in degrees or radians or fractions of a cycle

Question 7

What is the difference between transverse and longitudinal waves and give examples of each.

Answer 7

Transverse waves oscillate particles/fields perpendicular to the direction of energy transfer.
e.g. EM waves, S waves

Longitudinal waves oscillate particles/fields parallel to the direction of energy transfer.
e.g. sound waves, P waves, ultra sound

Question 8

What are unpolarised and polarised waves?

Answer 8

Unpolarised waves have oscillations in all planes that are perpendicular to the direction of wave travel.

Polarised waves have oscillations in only one plane that is perpendicular to the direction of wave.

Question 9

How can waves be polarised?

Answer 9

Waves are passed through a polariser (which only transmits oscillations in a certain plane) reducing the intensity of the wave by 50%.

Alternatively when waves are reflected off a reflective surface they undergo partial plane polarisation. For example light hitting a horizontal reflective surface such as water will oscillate more in the horizontal plane

(Only transverse waves can be polarised)

Question 10

Describe how polarisers are used in sunglasses.

Answer 10

The lenses contain vertically oriented polarisers that absorb all horizontally polarised light and reduce the intensity of light from the sun by 50%. This makes them more effective at reducing glare from reflective surfaces but also reducing light intensity in general.

Question 11

Describe how polarisers are used in Polaroid photography.

Answer 11

Using vertically oriented polarisers in the lenses glare can be effectively reduced and Color is intensified. It also makes photographing objects underwater easier as glare from the water surface is greatly reduced whereas the light from the subject is still relatively intense as it’s not partially polarised.

Question 12

Describe the application of polarising radio and microwave signals

Answer 12

Radio and television services are broadcast polarised. Therefore the reception aerial needs to be mounted in the same plane (depending on the transmitter it’s pointing towards)

Question 13

What are the rules for superposition?

Answer 13

The two waves must be of the same type
The waves must be coherent meaning they have the same frequency and wavelength and a fixed phase difference

Question 14

Describe a stationary wave

Answer 14

Waves that store energy instead of transferring it and don’t move

Question 15

How are stationary waves formed

Answer 15

The superposition of two progressive waves with the same speed, frequency (and wavelength) and similar amplitude moving In opposite directions

(Usually when a progressive wave is reflected back by a boundary)

Question 16

Give an example of transverse stationary waves

Answer 16

Microwaves (the appliance)

Microwaves are produced by a source and reflect off a far wall causing superposition.
At nodes there is no energy transfer and at antinodes there is a maximum energy transfer.
Food is spun through the nodes and antinodes to transfer heat evenly.

Question 17

Give an example of longitudinal stationary waves

Answer 17

Sound waves in a tube

Speakers produce sound waves which reflect off the end of the tube forming a stationary wave.
Powder will collect in heaps at nodes where there is no energy transfer.

Question 18

What are the key features of a stationary wave?

Answer 18

All oscillating points have the same frequency and time period
Points of positive amplitude are in antiphase with points of negative amplitude

Question 19

For a wave with N antinodes on a string of L length what is the;
Number of wavelengths, Wavelength and frequency?

Answer 19

No of wavelengths: N/2
Wavelength: 2L/N
Frequency: NV/2L = Nf

Question 20

What factors affect the resonant frequency of a string?

Answer 20

Length. f = v/2L so f is inversely proportional to L
Mass per unit length(mew). f is inversely proportional to root mew
Tension. f is proportional to root T

Question 21

Define path difference

Answer 21

The difference in the distance travelled by two waves from their source to the point where they meet

Question 22

With reference to path difference when do destructive and constructive interference occur?

Answer 22

Destructive occurs when there is n.5 path difference

Constructive interference occurs when there is n.0 path difference

Question 23

What are the requirements for youngs double slit experiment and how are they met?

Answer 23

The light must be coherent and monochromatic.

Laser light can be used as it already meets these conditions.Alternatively non coherent light can be used if it’s first passed through a single slit with equal distance to each of the double slits.

Question 24

Define fringe spacing

Answer 24

The distance between the center of two adjacent bright(or dark) fringes.

Question 25

What is the youngs double slit equation and when is it applicable?

Answer 25

Wavelength(gamma) = (w * D) / s W = fringe spacing D = distance between slits and screen S = slit separation Only applicable when D is a magnitude of ten greater than slit separation.

Question 26

Describe the effect of using different wavelengths of light in youngs double slit experiment

Answer 26

Fringe separation is directly proportional to wavelength therefore red light will have a greater fringe separation compared to blue light.

Question 27

Describe the effect of using white light in Youngs double slit experiment

Answer 27

Each wavelength of light produces its own interference pattern causing the central bright fringe to be white and the outer bright fringes to show a spectrum of colours with violet closest to the centre and red furthest. The outer bright fringes are wider so may merge and make the dark fringes no longer visible.

Question 28

What are the safety issues with lasers and how are they combated?

Answer 28

They can cause permanent eye damage. To combat this never look directly at a laser or its reflection, place a laser on warning outside the room, stand behind the laser.

Question 29

How do sound waves interfere?

Answer 29

Constructive interference occurs when the compressions and rarefactions line up with themselves Destructive interference occurs when the compressions line up with rarefactions

Question 30

How do microwaves interfere?

Answer 30

Two coherent sources can be produced using two transmitters connected to the same signal generator or a single transmitter and a double slit. Constructive interference occurs when the waves are in phase and destructive interference occurs when the waves are in antiphase.

Question 31

When does the most diffraction occur?

Answer 31

When the gap is the same size as the wavelength.

Question 32

Describe the single slit diffraction pattern for monochromatic light

Answer 32

If the wavelength of the light is about the same size as the slit there will be a central bright fringe with alternating dark and bright fringes on either side.the central bright fringe is double the width of the outer fringes.

Question 33

For a given slit width how does changing the wavelength of light affect the diffraction?

Answer 33

A longer wavelength results in a greater diffraction effect, so more spread out fringes and a wider central maxima

Question 34

Describe the diffraction pattern for single slit diffraction of white light

Answer 34

There is a central bright white fringe with outer bright fringes with violet nearest the centre and red furthest. The central bright fringe is still double the width. The outer bright fringes are wider than those of monochromatic light

Question 35

What affects the width of the central bright fringe?

Answer 35

Increasing the slit width will make the central maxima narrower but have a greater intensity. Increasing the wavelength will make the central maxima wider but have a lesser intensity.

Question 36

What is a diffraction grating?

Answer 36

A plate with hundreds of parallel slits per millimetre. When monochromatic light is passed through this a pattern similar to Youngs double slit experiment is produced with brighter and narrower maxima and darker and wider dark areas. (This is because there are so many beams from each of the slits reinforcing the pattern)

Question 37

Describe the lines of maximum brightness of the monochromatic light diffraction grating pattern.

Answer 37

The line at the centre is called the zero order line with a path difference of 0 lambda The lines on either side of this are called the first order lines with a path difference of 1 lambda The next pair out are called second order lines with a path difference of 2 lambda And so on

Question 38

How is the distance between sits calculated?

Answer 38

d = 1/N d = distance between slits N = slits per meter

Question 39

Derive the diffraction grating equation for monochromatic light

Answer 39

For each order Sin(theta) = opposite / hypotenuse Derived using a triangle For the 1st order Sin(theta) = lambda / distance between slits For the 2nd order Sin(theta) = 2lambda / distance between slits For nth order Sin(theta) = (n * lmabda) / distance between slits d * sin(theta) = n * lambda

Question 40

What is the effect of changing wavelength and distance between slits on the angle and its resultant pattern in the diffraction grating equation for monochromatic light?

Answer 40

If wavelength increases the angle increases causing a more spread out pattern If distance between the slits is increased then the angle decreases and the pattern becomes less spread out

Question 41

What are the key features of the diffraction grating pattern for white light?

Answer 41

The zero order maxima is white All other orders are spectrums with violet closest to centre and red furthest( because sin(theta) is directly proportional to lambda)

Question 42

Give examples of the applications of diffraction gratings.

Answer 42

Identifying elements: Astronomers can analyse light from stars to determine their composition and chemists can pass light through an unknown sample to identify it by its absorption spectra. Diffraction gratings are used rather than prisms because they are more accurate. X ray crystallography: The wavelength of x rays is of a similar scale to the spacing between atoms in crystalline solids. This means a crystal can act like a diffraction grating and the spacing between the atoms can be found from the diffraction pattern. This technique was used to discover the structure of DNA

Question 44

Describe refraction

Answer 44

When a wave crosses a boundary between two mediums at an angle. The wave will speed up if it’s travelling to a less optically dense medium and vice versa. If it’s going faster then the wave will bend away from the normal and again vice versa. (FAST)

Question 45

What is the approximate refractive index of air?

Answer 45

1 - same as in a vacuum ( rounded down to make calculations easier)

Question 46

What are the conditions for total internal reflection?

Answer 46

n1 > n2 (it must be travelling to a less optically dense medium) The angle of incidence must be greater that the critical angle Some internal refraction may occur below the critical angle but it won’t be TIR

Question 47

What are optical fibres used for?

Answer 47

Communications e.g. high speed j tenet transmission Medical imaging - endoscopy for diagnosing digestive system problems

Question 48

How do step index optical fibres work?

Answer 48

They have a core surrounded by cladding. The core is made from plastic or glass which has a greater refractive index than the cladding allowing for TIR. The cladding protects the core from scratches and provides strength whilst also preventing signals being transferred between cores.

Question 49

Describe how signal degradation is caused by absorption in optical fibres.

Answer 49

As the signal travels some of its energy is lost through absorption by the material the core is made of. This results in the amplitude of the signal being reduced. To reduce this problem the core should be made from a low absorption material

Question 50

Describe modal dispersion in optical fibres.

Answer 50

When light rays enter the core at different angles they take different paths. The rays that take longer paths take more time to reach the other end. To prevent this a single mode fibre can be used which only lets light take one(or few) path as the core is very narrow and there is a small difference between the refractive index of the core and cladding.

Question 51

Describe material dispersion in optical fibres.

Answer 51

When light consisting of different wavelengths is used the waves will travel at different speeds in the fibre due to their differing refractive indexes. For example red light travels faster than violet light in the core. To stop this problem monochromatic light can be used.

Question 52

How is signal degradation caused by both types of dispersion in optical fibres?

Answer 52

Both cause pulse broadening and broadened pulses can overlap with each other causing the signals to be mixed up.

Question 53

How can signal degradation be reduced?

Answer 53

An optical fibre repeater can be used to boost and regenerate the signal every so often reducing all types of signal degradation.