Waves and Optics

Question 1

Define the term “mechanical waves”

Answer 1

Mechanical waves are vibrations which pass through a substance.

Question 2

Define the term “EM waves”

Answer 2

EM waves are oscillating electric and magnetic fields that progress through space without the need for a substance.

Question 3

Define the term “longitudinal waves”

Answer 3

Longitudinal waves are waves in which the direction of vibrations of the particles is parallel to the direction in which the wave travels (e.g. sound waves and primary (P) seismic waves).

Question 4

Define the term “transverse waves”

Answer 4

Transverse waves are waves in which the direction of the vibration is perpendicular to the direction in which the wave travels (e.g. EM waves and secondary (S) seismic waves).

Question 5

Define the term “plane-polarised wave”

Answer 5

A plane polarised wave is a transverse wave where the vibrations stay in one plane only.

The plane of polarisation of an EM waves is defined as the plane in which the electric field oscillates.

Question 6

State how unpolarised light can be polarised.

Answer 6

If unpolarised light is passed through a polaroid filter, the transmitted light is polarised as the filter only allows through light which vibrates in a certain direction, according to the alignment of its molecules.

Question 7

Describe what happens when unpolarised light is passed through 2 polaroid filters

Answer 7

If unpolarised light is passed through 2 polaroid filters, the transmitted light intensity changes if 1 polaroid is turned relative to the other.
Light intensity is at minimum when the second filter is rotated 90 degrees to the first filter - because the polarised light from the first filter can’t pass through the second.
As you rotate the second filter further, light intensity increases until it reaches a maximum, where the 2 filters are parallel to each other.

Question 8

How do polaroid sunglasses work?

Answer 8

Polaroid sunglasses reduce the glare of light reflected by water or glass. The reflected light is polarised and the intensity is reduced when it passes through the polaroid sunglasses.

Question 9

Define the term “displacement”

Answer 9

The displacement of a vibrating particle is its distance and direction from its equilibrium position.

Question 10

Define the term “amplitude”

Answer 10

The amplitude of a wave is the maximum displacement of a vibrating particle.
For a transverse wave, this is the height of a wave crest.

Question 11

Define the term “wavelength”

Answer 11

The wavelength of a wave is the least distance between 2 adjacent vibrating particles with the same displacement and velocity at the same time.

Question 12

Define the term “period”

Answer 12

The period of a wave is the time for one complete wave to pass though a fixed point.

Question 13

Define the term “frequency”

Answer 13

The frequency of a wave is the number of cycles of vibrations of a particle per second, or the number of complete waves passing a point per second.

Question 14

Define the term “phase”

Answer 14

The phase of a vibrating particle at a certain time is the fraction of a cycle it has completed since the start of the cycle.

Question 15

Define the term “phase difference”

Answer 15

The phase difference between 2 particles vibrating at the same frequency is the fraction of a cycle between the vibrations of the 2 particles. One cycle = 360 degrees.
For 2 points at distance d apart along a wave:
Phase difference in radians = 2dpi / wavelength

Question 16

Define the term “wave fronts”

Answer 16

Wave fronts are lines of constant phase.

Question 17

Describe what happens during reflection

Answer 17

Straight waves directed at a certain angle to a hard flat surface reflect off at the same angle. The angle between the reflected wave front and the surface is the same as the angle between the incident wave front and the surface.

Question 18

Explain what causes waves to refract when they pass across a boundary

Answer 18

When waves pass across a boundary at which both the wave speed and wavelength changes. If the wave fronts approach at an angle to the boundary, they change direction as well - this is refraction.

Question 19

Describe the direction light waves bend when they travel into the glass from the air, and out of glass into the air

Answer 19

When a light ray is directed into a glass block at an angle, it changes direction because light waves travel more slowly in glass than in air. The light bends towards the normal.
When the light ray is directed out of a glass block at an angle, it changes direction away from the normal.

Question 20

What is diffraction?

Answer 20

Diffraction occurs when waves spread out after passing through a gap, or round an obstacle.

Question 21

How can you maximise diffraction?

Answer 21

• The narrower the gap, the more the waves spread out

- The longer the wavelength, the more the waves spread out

Question 22

Why do satellite TV dishes need to be designed carefully?

Answer 22

The bigger the dish, the stronger the signal it can receive, because more radio waves are reflected by the dish onto the aerial.
A bigger dish reflects the radio waves to a smaller focus, because it diffracts the waves less. The dish needs to be aligned more carefully than a smaller dish, otherwise the radio waves will not focus onto the aerial.

Question 23

Define the term “superposition”

Answer 23

Superposition is the effect of two waves adding together, when the waves combine for an instant.

Question 24

What does the principle of superposition state?

Answer 24

The principle of superposition states that when two waves meet, the total displacement at a point is equal to the sum of the individual displacements at that point.

Question 25

What happens during reinforcement?

Answer 25

When a crest meets a crest - a supercrest is formed.
When a trough meets a trough - a super trough is formed.
Maximum amplitudes are achieved when waves reinforce each other.

Question 26

What happens during cancellation?

Answer 26

When a crest meets a trough of the same amplitude, the resultant displacement is 0 - since the 2 waves cancel each other out.
If the waves are not of the same amplitude, the resultant is called a minimum.

Question 27

Define the term “progressive waves”

Answer 27

Progressive waves are waves which travel through a substance or through space, if electromagnetic.

Question 28

Define the term “stationary waves”

Answer 28

Stationary waves are a wave pattern with nodes and antinodes formed, when 2 or more progressive waves of the same frequency and amplitude pass through each other.

Question 29

Define the term “interference”

Answer 29

Interference is the formation of points of cancellation and reinforcement when 2 coherent waves pass through each other.
Since there is constant frequency and constant phase difference, these points happen at fixed positions.

Question 30

Why are coherent sources of waves needed for interference?

Answer 30

Coherent wave sources produce an interference pattern where they overlap, because they vibrate at the same frequency with a constant phase difference.
If the phase difference randomly changed, the points of cancellation and reinforcement would move at random - no interference pattern seen.

Question 31

Describe an experiment to test diffraction and interference using microwaves

Answer 31

• The transmitter produces microwaves of wavelength 3cm. The receiver is connected to a suitable meter.
• Place the receiver in the path of the microwaves beam from transmitter. Move the receiver gradually away from the transmitter.
• Place a metal plate between the transmitter and the receiver to show microwaves can’t pass through metal
• Use metal plates to make two slits to allow diffraction. Direct the transmitter at the slits and use the receiver to find points of cancellation and reinforcement.

Question 32

Define the term “first harmonic”

Answer 32

The first harmonic is a pattern of stationary waves on a string when it vibrates at its lowest possible frequency.

Question 33

Why do stationary waves that vibrate freely not transfer energy to their surroundings?

Answer 33

The amplitude of vibration is 0 at the nodes so there is no energy at the nodes. The amplitude of vibration is a maximum at the antinodes, so there is maximum energy at the antinodes. Since nodes and antinodes are at fixed positions, no energy is transferred in a freely vibrating stationary wave pattern.

Question 34

Explain why nodes are formed in fixed positions

Answer 34

• When 2 progressive waves are in phase, they reinforce each other to produce a large wave.
• A quarter of a cycle later, the 2 waves have each moved one quarter of a wavelength in opposite directions. They are in antiphase so they cancel each other.
• After a further quarter cycle, the 2 waves are back in phase, and the resultant is a large wave.
• The points where there is no displacement are in fixed positions .

Question 35

Describe a general stationary wave pattern

Answer 35

• The amplitude of a vibrating particle varies with position from 0 at a node to maximum amplitude at an antinode.
• Phase difference between 2 vibrating particles is 0 if they are between adjacent nodes or separated by an even number of nodes.
• Phase difference is 180 degrees if the 2 particles are separated by an odd number of nodes.

Question 36

State how the frequency differs between stationary and progressive waves

Answer 36

In stationary waves, all particles except those at the nodes vibrate at the same frequency.

In progressive waves, all particles vibrate at the same frequency.

Question 37

State how the amplitude differs between stationary and progressive waves

Answer 37

In stationary waves, the amplitude varies from 0 at the nodes to a maximum at the antinodes.

In progressive waves, the amplitude is the same for all particles.

Question 38

State how the phase difference between 2 particles differs between stationary and progressive waves

Answer 38

In stationary waves, the phase difference is equal to mpi, where m is the number of nodes between the 2 particles.

In progressive waves, the phase difference is equal to 2dpi/wavelength, where d is the distance apart.

Question 39

Explain what condition must be satisfied at both ends of the vibrating string

Answer 39

A string must be tied to a mechanical vibrator which is connected to a frequency generator. The other end of the string passes over a pulley and supports a weight, which keeps the tension in the string constant.

Question 40

Describe the simplest possible stationary wave pattern that can be formed

Answer 40

The first harmonic pattern of vibration is seen at the lowest possible frequency that gives a pattern.This has an antinode in the middle and 2 nodes at either end. Since the length L of the vibrating section is between adjacent nodes, and the distance between them is half a wavelength.
Wavelength = 2L
Frequency = c/2L

Question 41

Explain how a stationary wave pattern forms on a vibrating string

Answer 41

A progressive wave is sent out by a vibrator. The crest reverses its phase when it reflects at the fixed end, and travels back along the string as a trough. When it reaches the vibrator, it reverses and changes phase again. If 2 crests reinforce each other, the amplitude of the wave increases. The time taken for a wave to travel along and back = time taken for a whole number of cycles.

Question 42

How can you increase the pitch?

Answer 42

• Raising the tension

- Shortening the length of the string

Question 43

What happens to the light ray at a boundary between 2 transparent substances?

Answer 43

The light ray bends towards the normal if it passes into a more dense substance.
The light ray bends away from the normal if it passes into a less dense substance.

Question 44

What is Snell’s Law?

Answer 44

Snell’s Law is a law stating that the ratio of the sines of the angles of incidence to refraction of a wave is constant when it passes between 2 given media.

Question 45

Compare glass to air refraction with air to glass refraction

Answer 45

The angle of refraction of the light ray emerging from the block is the same as the angle of incidence of the light ray entering the block. This is because the 2 sides of the block where refraction occurs are parallel.
Refractive index of glass = n
Refractive index when leaving block = 1/n

Question 46

Relate the refractive index to the speed of light waves

Answer 46

n = c/cs where cs is speed of light in a substance
n = w/ws where ws is speed of light in a substance

Question 47

Explain why a glass prism splits white light into the colours of a spectrum

Answer 47

• White light is composed light with a continuous range of wavelengths, from red with the largest wavelength to violet with the shortest.
• The glass prism refracts light by different amounts dependent on the wavelength. Shorter wavelength = greater refraction.
• Each colour in the white light is refracted by different amounts.
• This dispersive effect occurs because the speed of light in glass depends on wavelength.
• Violet light travels more slowly that red light, so n is greater for violet light that red light.

Question 48

Define the term “critical angle”

Answer 48

The angle of incidence of a light ray must exceed the critical angle for total internal reflection to occur.
At this angle, the light ray refracts along the boundary.

Question 49

State the conditions for total internal reflection

Answer 49

1. The incident substance has a larger refractive index than the other substance.
2. The angle of incidence must exceed the critical angle.

Question 50

Relate the critical angle to the refractive index

Answer 50

At the critical angle, the angle of refraction is 90 degrees since the light ray emerges along the boundary.
Sine of critical angle = n2/n1
where n1 is the incident substance and n2 is the refractive substance.

Question 51

Why do diamonds sparkle when white light is directed at them?

Answer 51

• When white light enters a diamond, it’s split into the colours of the spectrum.
• Diamond has a very high refractive index so it separates all the colours more than any other substance.
• The high refractive index gives a small critical angle, so the light ray may experience TIR many times before it emerges.
• The colours spread out more and more, causing the diamond to sparkle.