Name…………………
Class…………..
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G482
Electrons, Photons and Waves
Module 2.4:
Waves
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G482 Module 2.4 Waves
1.
Wave equation
2.4.1 Wave motion
Candidates should be able to:
(a) describe and distinguish between progressive longitudinal and transverse
waves;
(b) define and use the terms displacement, amplitude, wavelength, period,
phase difference, frequency and speed of a wave;
(c) derive from the definitions of speed, frequency and wavelength, the wave
equation v = fλ;
(d) select and use the wave equation v = fλ;
Progressive,
longitudinal,
transverse,
displacement,
amplitude,
wavelength, period,
phase difference,
frequency, speed,
velocity
2.
Reflection and
Refraction
(e) explain what is meant by reflection, refraction and diffraction of waves such
as sound and light.
Reflection, refraction,
speed, echo, sonar
3.
Diffraction
(e) explain what is meant by reflection, refraction and diffraction of waves such
as sound and light.
Diffraction,
wavelength,
interference,
constructive,
destructive,
transverse,
longitudinal.
4.
Electromagnetic
Waves
2.4.2 Electromagnetic waves
Candidates should be able to:
(a) state typical values for the wavelengths of the different regions of the
electromagnetic spectrum from radio waves to γ-rays;
(b) state that all electromagnetic waves travel at the same speed in a vacuum;
(c) describe differences and similarities between different regions of the
electromagnetic spectrum;
Wavelength,
frequency, spectrum,
speed, vacuum.
5.
Uses of
Electromagnetic
Waves
2.4.2 Electromagnetic waves
Candidates should be able to:
(d) describe some of the practical uses of electromagnetic waves;
(e) describe the characteristics and dangers of UV-A, UV-B and UV-C radiations
and explain the role of sunscreen (HSW 6a);
UV-A, UV-B, UV-C,
radiation, sunscreen
6.
Polarization and
Malus’ Law
(f) explain what is meant by plane polarised waves and understand the
polarisation of electromagnetic waves;
(g) explain that polarisation is a phenomenon associated with transverse waves
only;
(h) state that light is partially polarised on reflection;
(i) recall and apply Malus’s law for transmitted intensity of light from a polarising
filter.
Plane, polarise,
transverse,
longitudinal,
reflection, transmit,
intensity, filter.
7.
Interference
2.4.3 Interference
Candidates should be able to:
(a) state and use the principle of superposition of waves;
(b) apply graphical methods to illustrate the principle of superposition;
(c) explain the terms interference, coherence, path difference and phase
difference;
(d) state what is meant by constructive interference and destructive
interference;
Superposition,
interference,
coherence, path
difference, phase
difference,
constructive,
destructive
8.
2 Source
Interference
(e) describe experiments that demonstrate two source interference using sound,
light and microwaves;
(f) describe constructive interference and destructive interference in terms of
path difference and phase difference;
Interference,
constructive,
destructive, path
difference, phase
difference, intensity,
power, amplitude
9.
Intensity
(g) use the relationships intensity = power/cross-sectional area
intensity = k x amplitude2 ;
intensity, power,
amplitude
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10.
Young’s Slits
(h) describe the Young double-slit experiment and explain how it is a classical
confirmation of the wave-nature of light (HSW 1);
(i) Select and use the equation
λ = ax
D for electromagnetic waves;
(j) describe an experiment to determine the wavelength of monochromatic light
using a laser and a double slit (HSW 1);
Young double slit,
wavelength, slit width,
slit separation,
distance to screen,
monochromatic (laser)
11.
Diffraction gratings
(k) describe the use of a diffraction grating to determine the wavelength of light
(the structure and use of a spectrometer are not required);
(l) select and use the equation dsinθ = nλ;
(m) explain the advantages of using multiple slits in an experiment to find the
wavelength of light.
Diffraction, diffraction
grating, wavelength,
frequency, coherence,
path difference, phase
difference.
12.
Stationary Waves
2.4.4 Stationary waves
Candidates should be able to:
(a) explain the formation of stationary (standing) waves using graphical
methods;
(b) describe the similarities and differences between progressive and stationary
waves;
(c) define the terms nodes and antinodes;
Progressive,
stationary, graphical,
node, antinode
13.
Stationary Waves
and resonance
(d) describe experiments to demonstrate stationary waves using microwaves,
stretched strings and air columns;
(e) determine the standing wave patterns for stretched string and air columns in
closed and open pipes;
(f) use the equation: separation between adjacent nodes (or antinodes) = λ/2;
(g) define and use the terms fundamental mode of vibration and harmonics;
(h) determine the speed of sound in air from measurements on stationary waves
in a pipe closed at one end.
Standing, progressive,
node, antinode
14.
G482 Module 4:
2.4 Waves Test
Review topic
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Lesson 21 notes – The wave equation
Objectives
(a) describe and distinguish between progressive longitudinal and transverse
waves;
(b) define and use the terms displacement, amplitude, wavelength, period,
phase difference, frequency and speed of a wave;
(c) derive from the definitions of speed, frequency and wavelength, the wave
equation v = fλ;
(d) select and use the wave equation v = fλ;
The Anatomy of a Wave
A transverse wave is a wave in which the particles of the medium are
displaced in a direction perpendicular to the direction of energy transport. A
transverse wave can be created in a rope if the rope is stretched out
horizontally and the end is vibrated back-and-forth in a vertical direction. If a
snapshot of such a transverse wave could be taken so as to freeze the shape
of the rope in time, then it would look like the following diagram.
The dashed line drawn through the center of the diagram represents the
equilibrium or rest position of the string. This is the position that the string
would assume if there were no disturbance moving through it. Once a
disturbance is introduced into the string, the particles of the string begin to
vibrate upwards and downwards. At any given moment in time, a particle on
the medium could be above or below the rest position. Points A and F on the
diagram represent the crests of this wave. The crest of a wave is the point on
the medium which exhibits the maximum amount of positive or upwards
displacement from the rest position. Points D and I on the diagram represent
the troughs of this wave. The trough of a wave is the point on the medium
which exhibits the maximum amount of negative or downwards displacement
from the rest position.
The wave shown above can be described by a variety of properties. One such
property is amplitude. The amplitude of a wave refers to the maximum
amount of displacement of a a particle on the medium from its rest position. In
a sense, the amplitude is the distance from rest to crest. Similarly, the
amplitude can be measured from the rest position to the trough position. In
the diagram above, the amplitude could be measured as the distance of a line
segment which is perpendicular to the rest position and extends vertically
upward from the rest position to point A.
The wavelength is another property of a wave which is portrayed in the
diagram above. The wavelength of a wave is simply the length of one
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complete wave cycle. If you were to trace your finger across the wave in the
diagram above, you would notice that your finger repeats its path. A wave has
a repeating pattern. And the length of one such repetition (known as a wave
cycle) is the wavelength. The wavelength can be measured as the distance
from crest to crest or from trough to trough. In fact, the wavelength of a wave
can be measured as the distance from a point on a wave to the corresponding
point on the next cycle of the wave. In the diagram above, the wavelength is
the horizontal distance from A to F, or the horizontal distance from B to G, or
the horizontal distance from E to J, or the horizontal distance from D to I, or
the horizontal distance from C to H. Any one of these distance measurements
would suffice in determining the wavelength of this wave.
A longitudinal wave is a wave in which the particles of the medium are
displaced in a direction parallel to the direction of energy transport. A
longitudinal wave can be created in a slinky if the slinky is stretched out
horizontally and the end coil is vibrated back-and-forth in a horizontal
direction. If a snapshot of such a longitudinal wave could be taken so as to
freeze the shape of the slinky in time, then it would look like the following
diagram.
Because the coils of the slinky are vibrating longitudinally, there are regions
where they become pressed together and other regions where they are
spread apart. A region where the coils are pressed together in a small amount
of space is known as a compression. A compression is a point on a medium
through which a longitudinal wave is travelling which has the maximum
density. A region where the coils are spread apart, thus maximizing the
distance between coils, is known as a rarefaction. A rarefaction is a point on
a medium through which a longitudinal wave is travelling which has the
minimum density. Points A, C and E on the diagram above represent
compressions and points B, D, and F represent rarefactions. While a
transverse wave has an alternating pattern of crests and troughs, a
longitudinal wave has an alternating pattern of compressions and rarefactions.
As discussed above, the wavelength of a wave is the length of one complete
cycle of a wave. For a transverse wave, the wavelength is determined by
measuring from crest to crest. A longitudinal wave does not have crest; so
how can its wavelength be determined? The wavelength can always be
determined by measuring the distance between any two corresponding points
on adjacent waves. In the case of a longitudinal wave, a wavelength
measurement is made by measuring the distance from a compression to the
next compression or from a rarefaction to the next rarefaction. On the diagram
above, the distance from point A to point C or from point B to point D would be
representative of the wavelength.
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Deducing and using the wave equation
Justification/deduction of the wave equation v = f λ.
Lets use an example of the coaches of a train are going past;
You count how many coaches go by in a second and you know the length of
one – so you multiply the two together to get the train’s speed.
Apply this to waves: count the number of waves passing each second (=
frequency), and multiply by the length of each (= wavelength) to find the
speed.
speed = distance/time = l/T= l / (1/f) = f λ.
So v = f λ
Which becomes:
as c is the speed of light.
c=fλ
You must practice and be able to rearrange this to find f or λ.
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Lesson 21 – The wave equation
1
(a) Define the following terms associated with waves.
(i) frequency f
.......................................................................................................................................
...................................................................................................................................[1]
(ii) wavelength λ
...................................................................................................................................
...................................................................................................................................[1]
(b) Use the definitions in (a) to deduce an equation for the speed v of a wave
in terms of λ and f.
(c)
[3]
(i) The speed of sound in air is about 340 m s–1 while light travels at a
speed of 3.0 × 108 m s–1. Calculate the time interval between seeing a
flash of lightning, 1.0 km away, and hearing the sound of thunder
caused by the lightning.
time interval = ....................................................... s [3]
(ii) Describe how observers may estimate their distance away from
the point of a flash of lightning.
.......................................................................................................................................
.......................................................................................................................................
...................................................................................................................................[1]
(d) State two differences, other than their speeds, between sound and light
waves.
.......................................................................................................................................
.......................................................................................................................................
...................................................................................................................................[2]
[Total: 11]
Turn over for question 2
2
Label both of the diagrams appropriately with the key words below. You can
use some of the keywords on both diagrams. You will gain one mark for each correct
use of the word (maximum of once per diagram)
Transverse, Longitudinal, Amplitude, Wavelength, compression, rarefaction, crest,
trough
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Total [9]
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Lesson 22 questions – reflection and refraction
1 The diagram shows a ray of light striking a flat mirror.
a Add a normal line which strikes the mirror at point X.
(1)
b What is the value of the angle between the reflected ray and the normal?
…………………………………………………………………………………………
……
…………………………………………………………………………………………
(1)
2 The diagram shows a series of water waves reflecting off a barrier in a ripple tank.
Complete the diagram to show how the reflected waves move.
(3)
3 The distance from the Earth to the Moon can be found by reflecting a ray of laser
light off the surface of the Moon, and measuring the time it takes to travel to the
Moon and back.
If the return journey takes 2.7 s, how far is the Moon from the Earth?
(Speed of light in space = 300 000 km/s)
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(2)
4
The diagram shows what happens when light waves travel from air into glass.
Their wavelength and speed change; their frequency is unaffected.
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a
Does their wavelength increase or decrease?
…………………………………………………………………………………………
(1)
b
Does their speed increase or decrease?
…………………………………………………………………………………………
(1)
5
Explain why non-swimmers should be wary of diving into a pool of water
even if they think they can see the bottom?
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (2)
6
Sketch on the path of the light as it is refracted through the transparent block.
Label the diagram fully.
(4)
7
Explain what refraction is and how it occurs.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
Material
Speed of light
air
300 000 km/s
water
225 000 km/s
diamond
124 000 km/s
perspex
201 000 km/s
glass
197 000 km/s
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8
i) a) Plot a bar chart
to show the figures in the table
above on the graph paper
below.
(3)
b) Which ma
……………………………….
(1)
RAB Plymstock School
ii)
When a ray of light goes from air into one of the other materials in the table:
a
which material refracts light rays the most?
…………………………………………………………………………………………
(1)
b
which material refracts light rays the least?
…………………………………………………………………………………………
(1)
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Lesson 23 notes – Diffraction
Objectives
(e) explain what is meant by reflection, refraction and diffraction of waves
such as sound and light.
The wavelength of sound waves may be several metres.
If the wavelength is of a similar size to a gap in a door, then the wave will
diffract as shown below.
If the wavelength does not match the size of the gap,
then only a little diffraction will occur at the edge of the wave.
The part of the wave which hits the wall in the above two pictures is reflected
straight back on itself.
The amount of spread of the wave will depend on the path difference and the
constructive or destructive interference that occurs because of this.
The Huygens' Principle says that every point on a wavefront acts like a new
source so each transparent slit becomes a new source.
So the diffraction of light would look like this on a screen:
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So because of interference there will be maxima where there is constructive
interference and minima where there is destructive interference.
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Lesson 23 questions – diffraction
(a) State two differences between sound and light waves.
1.
.......................................................................................................................................
.......................................................................................................................................
2.
.......................................................................................................................................
...................................................................................................................................[2]
(b) The diffraction of waves can be demonstrated using a ripple tank.
(i) Describe how plane, transverse water waves can be produced in the ripple tank.
.......................................................................................................................................
.......................................................................................................................................
...................................................................................................................................[2]
(ii) Explain how the wavelength of the water waves can be increased.
.......................................................................................................................................
...................................................................................................................................[1]
(iii) State how the speed of the water waves can be reduced.
.......................................................................................................................................
...................................................................................................................................[1]
(c) Fig. 5.1 shows plane water waves in a ripple tank approaching a narrow gap.
On the diagram, draw the pattern of the wavefronts emerging from the gap. [3]
Fig. 5.1
(d) Explain why the diffraction of sound waves is much more likely to be noticeable
than the diffraction of light.
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
...................................................................................................................................[3]
[Total: 12]
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Lesson 24 notes – The EM spectrum
Objectives
(a) state typical values for the wavelengths of the different regions of the electromagnetic
spectrum from radio waves to γ-rays;
(b) state that all electromagnetic waves travel at the same speed in a vacuum;
(c) describe differences and similarities between different regions of the
electromagnetic spectrum;
Deducing and using the wave equation
Justification/deduction of the wave equation v = f λ.
Lets use an example of the coaches of a train are going past;
You count how many coaches go by in a second and you know the length of
one – so you multiply the two together to get the train’s speed.
Apply this to waves: count the number of waves passing each second (=
frequency), and multiply by the length of each (= wavelength) to find the
speed.
speed = distance/time = l/T= l / (1/f) = f λ.
So v = f λ
Which becomes:
as c is the speed of light.
c=fλ
You must practice and be able to rearrange this to find f or λ.
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The EM Spectrum
Here is a review table of the information you should be able to recall.
Type of ray:
Gamma rays:
X-rays:
Ultraviolet:
Visible
light:
Production:
Emitted during
radioactive
decay
Produced by
firing electrons
at a metal
target
Emitted by
the Sun
Emitted by
the Sun
Uses:
Medicine in
chemotherapy
Medicine for
looking at
bones
Tanning
Seeing
Hazards:
Causes cancer
by damaging
cells
Causes cancer
by damaging
cells
Can cause
skin cancer
Intense light
can
damage
your sight
Wavelength
(m):
x10-12
x10-10
x10-8
7 x10-7 to
4x10-7
Frequency
(Hz):
x10 20
x10 18
x10 15
x10 14
Photon
Energies
(eV):
400 k
4k
4
0.4
Type of ray:
Infra red:
Production:
Emitted by hot
objects
Uses:
Conventional
cooking
Micro-waves:
Radio-waves:
Produced by
Produced by
changing currents changing currents in a
in a conductor
conductor
Microwave
cooking and
communications
Communication and
media
Hazards:
Can burn
Can burn
Currently not
considered to be
hazardous
Wavelength
(m):
x10-5
x10-3 to x10-2
x1
Frequency
(Hz):
x10 12
x10 10
x10 8 to x10 10
Photon
Energies
(eV):
4m
40 m
4
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Lesson 25 notes – EM Spectrum
Objectives
(a) state typical values for the wavelengths of the different regions of the electromagnetic
spectrum from radio waves to γ-rays;
(b) state that all electromagnetic waves travel at the same speed in a vacuum;
(c) describe differences and similarities between different regions of the electromagnetic
spectrum;
(d) describe some of the practical uses of electromagnetic waves;
(e) describe the characteristics and dangers of UV-A, UV-B and UV-C radiations and
explain the role of sunscreen (HSW 6a);
Infrared cooking
Thermal radiation is infrared waves. These waves don’t need particles to
transfer the energy; when they hit a surface they make the particles in that
surface vibrate and so heat it up. If you heat something up enough you can
cook it. This is how grills and toasters work.
If you look inside a toaster when it is on you will see that the wire inside is
glowing red-hot and it is emitting infrared waves.
A grill or toaster works by emitting these infrared waves, which are then
absorbed by the surface layer of the thing you are cooking. The surface
particles vibrate increasing their kinetic energy and so heat is transferred to
the centre by conduction, cooking the rest of the food.
Infrared waves
Grill
Infrared waves hit surface
particles
Heat conducted
through food
Microwave cooking
Microwaves produce microwaves
in the Magnetron; this gives out a
stream of the electromagnetic
waves. These electromagnetic
waves will interact with particles
that
have
charges and make them vibrate, increasing their
kinetic energy. Typically, microwaves interact
with water molecules or fat molecules in the
first few centimetres of a food. They vibrate
and rub against neighbouring molecules and the
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friction between them heats them up. The heat is then transferred to the
centre of the food by conduction.
Ultraviolet and The Ozone Layer
Objectives
Be able to explain how darker skins reduce cancer risk:
Be able to calculate how long a person can spend in the sun without burning
from knowledge of the sun protection factor.
Describe how the ozone layer protects the Earth from ultraviolet radiation and
that environmental pollution from CFCs is depleting the layer.
The Ozone Layer
About 20 kilometres thick, this giant
umbrella is made up of a layer of ozone
(O3) gas.
This gas is found some 15 to 35
kilometres above the earth's surface in
the upper atmosphere or "stratosphere".
Ozone gas absorbs UV rays.
UV can be harmful because it can cause
damage to skin cells and harm our eyes.
Gases that are very useful to us in things
like refrigerators called CFCs
(Chlorofluorocarbons) react with Ozone
gas (O3).
•This causes the Ozone layer to thin (we
call these thinned areas “holes”).
•These gases have now been banned
and the Ozone layer is reforming slowly.
Sunburn
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UVB rays are very short
wavelengths of UV that all
get absorbed by the ozone
layer.
UVB rays are short
wavelengths of UV light
that reach the epidermis
(the outer layer of skin).
UVA rays are longer
wavelengths of UV light
that penetrate the dermis
(the second layer of skin).
Exension
Skin Types
Darker skins reduce cancer risk by:
absorbing more ultraviolet
radiation in the epidermis
letting less ultraviolet radiation
reach underlying body tissues where it is
more damaging.
Example Calculation
Albert will get sunburn if he stays in the sun for 30 minutes. How many
hours can he stay in the sun if he uses sunscreen with an SPF of 15?
Time (in minutes)
= 30 x 15
= 450 minutes / 60 (to get hours)
= 7.5 hours
Radio communications
Objectives
Recognise common uses of wireless technology.
• Radio;
• mobile phones.
• laptop computers.
Explain how long-distance communication depends on the reflection of waves
from the Ionosphere or by being received and re-transmitted from satellites.
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Recognise that radio stations with similar transmission frequencies often
interfere.
History
http://pass.maths.org.uk/issue8/features/phones/index.html
Uses
Wireless is a term used to describe telecommunications in which
electromagnetic waves (rather than some form of wire) carry the signal over
part or the entire communication path. Common examples of wireless
equipment in use today include:





Cellular phones and pagers: provide connectivity for portable and
mobile applications, both personal and business.
Global Positioning System (GPS): allows drivers of cars and trucks,
captains of boats and ships, and pilots of aircraft to ascertain their
location anywhere on earth.
Cordless computer peripherals: the cordless mouse is a common
example; keyboards and printers can also be linked to a computer via
wireless.
Cordless telephone sets: these are limited-range devices, not to be
confused with cell phones.
Satellite television: allows viewers in almost any location to select from
hundreds of channels.
Long Distance Communication
Radio waves travel in straight lines; they can also be reflected and refracted
since they are a wave and part of the electromagnetic spectrum. They have
long wavelengths and do not lose energy easily.
These properties allow it to carry information for hundreds of miles.
The ionosphere
The ionosphere is a set of ionized layers in the
upper portions of the atmosphere that span the
altitude range between about 75 and several
hundred kilometers above Earth's surface.
Depending on the
angle of incident
radio waves, they
can be reflected
back towards the
Earth over large
open expanses
such as the
Atlantic Ocean to
places where the
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radio waves would normally be stopped from getting because of the curvature
of the Earth (the Earth is round).
Satellites
Although slightly refracted through the ionosphere, radio
waves can be transmitted to geostationary satellites and then
retransmitted around the world as shown in the diagram.
Mobile phones
Objectives
Recognise that microwaves are used to transmit information over large
distances that are in line of sight.
Describe how diffraction and interference of microwaves can cause signal
loss.
Mobile phone range
Mobile signals have a range of about 35km but can be absorbed and reflected
by different materials. A concrete wall or metal lift can stop the signal
completely. Other materials will also absorb them and signal strengths on
your mobile can vary due to its position because of this.
Hillsides will stop transmission so mobile masts must be positioned on the
tops of hills in line of sight of any users.
Diffraction
Diffraction occurs around objects like buildings and hills and the diagram
shows how, as the signal is spread out its strength reduces. The lighter the
line, the weaker the signal.
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Interference
Interference can happen when 2 signals with the same frequency are used
close to each other. It can also happen when waves are received and then
reflect off a nearby surface so you pick them up more than once.
Dangers of Mobile Phones?
There may or may not be dangers to residents near to the site of a mast or to users of
mobile phones.
The energy associated with microwaves and infrared depend on their frequency. The
higher the frequency, the higher energy the wave has.
The following are links to websites about the dangers where you can make up your
own mind:
http://www.telegraph.co.uk/news/main.jhtml?xml=/news/2007/07/25/nmasts125.xml
http://www.thehealthierlife.co.uk/article/3096/mobile-phone-health-risks.html
http://web.ukonline.co.uk/faderuk/Health/Concerns/Why_Worry/why_worry.html
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http://www.thelancet.com/journals/lanonc/article/PIIS1470204500002382/fulltext
(free registration needed)
TIR and Optical Fibres
Objectives
You will be able to:
Describe how light and infrared radiation can both travel along an optical fibre
from one end to another by Total Internal Reflection (TIR).
Describe the transmission of light in optical fibres
• optical fibres allow the rapid transmission of data;
• optical fibres allow the transmission of data pulses using light.
Describe the application of total internal reflection in fibre optics.
Describe advantages of using optical fibres to allow more information to be
transmitted:
• multiplexing;
• lack of interference.
• no corrosion (unlike copper coaxial cable)
Optical Fibres
Optical fibres use Total internal
reflection to send pulses of light at
specific frequencies from one end
to the other. Bundles of very thin
pieces of glass can send large
amounts of information from one
place to the other.
Every time the light strikes the side
of the glass fibre, the angle that it
makes with the side must be
greater than the critical angle for
that material in order for total
internal reflection to occur.
An electrical signal is digitally
coded at one end of the tube and that energy transformed into light energy
(visible or infra-red light). It is then sent down the fibre and the decoded at the
other end, where it is transformed into an electrical signal at the other end to
be used.
It has the advantage over copper wires because there is no magnetic
interference and the information sent down it can be multiplexed. Metal wires
will have to boost their signals more often and can also corrode.
There are two types of multiplexing, time division and frequency division. They
both involve splitting the signal up and interleaving it with other signals. Using
specific codes they can then be interpreted at the other end of the cable and
stuck back together to make the full communication.
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Optical fibres are used for telecommunications, phone links and the internet.
They are also used in medicine for laproscopic (keyhole) surgery.
Extension
Other Uses
Optical fibres can be used for the purposes of illumination, often carrying light
from outside to rooms in the interiors of large buildings.
Another important application of optical fibres is in sensors. If a fibre is
stretched or squeezed, heated or cooled or subjected to some other change
of environment, there is usually a small but measurable change in light
transmission. Hence, a rather cheap sensor can be made which can be put in
a tank of acid, or near an explosion or in a mine and connected back, perhaps
through kilometres of fibre, to a central point where the effects can be
measured.
Optical fibres can also be used as simple light guides. At least one fancy
modern car has a single high intensity lamp under its bonnet, with optical
fibres taking the light to a series of mini-headlamps on the front. Less high
tech versions carry light from bulbs to the glove compartment etc.
As light is not affected noticeably by electromagnetic fields. It also does not
interfere with other instruments that do use electricity. For this reason, fibreoptics are also becoming very important for short-range communication and
information transfer in applications situations like aircraft. This application is
now being extended into motor cars, and plastic optical fibres will soon (say in
5-8 years time) be very common for transmitting information around the car.
So we can see that optical fibres are not just passive light pipes. Researchers
are finding ways in which they can make the fibres become the active
elements of the circuit, e.g. amplifiers or filters. This means that the
information could remain in light form from one end of a link to the other,
removing the limitations of the electronics in circuits and enabling more of the
theoretical information carrying capacity to be used.
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Lesson 25 questions – The EM spectrum
1)a) State 2 features of the EM spectrum that are the same to all of its sections.
…………………………………………………………………………………………..
…………………………………………………………………………………………
(2)
b)
The table below is part completed and summarises features of different
sections of the electromagnetic spectrum. Complete the table with brief
statements for the six sections of the spectrum shown. A mark is available for
the quality of written communication.
radiation
Typical
wavelength
Method of
production
A use
Gamma
(3)
10-10 m
Stopping high
speed electrons
at a target
(2)
10-8 m
(3)
From very hot
objects
Light
Infra-red
10m
High frequency
oscillation of
electrons
Sight,
photography
(1)
Heat from the
Sun
(2)
(2)
c)
A mobile telephone company transmits microwave signals to an orbiting
satellite at a frequency of 1.6 x 109Hz. Calculate the wavelength λ of the microwaves.
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
(3)
Total [19]
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Lesson 26 notes - Malus's Law
Objectives
(f) explain what is meant by plane polarised waves and understand the polarisation of
electromagnetic waves;
(g) explain that polarisation is a phenomenon associated with transverse waves only;
(h) state that light is partially polarised on reflection;
(i) recall and apply Malus’s law for transmitted intensity of light from a polarising filter.
At the beginning of the nineteenth century the only known way to generate
polarized light was with a calcite crystal. In 1808, using a calcite crystal, Malus
discovered that natural incident light became polarized when it wasreflected
by a glass surface, and that the light reflected close to an angle of incidence
of 57° could be extinguished when viewed through the crystal. He then
proposed that natural light consisted of the s- and p-polarizations, which were
perpendicular to each other.
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Since the intensity of the reflected light varied from a maximum to a minimum
as the crystal was rotated, Malus proposed that the amplitude of the reflected
beammust be A = A0 cosθ. However, in order to obtain the intensity, Malus
squared the amplitude relation so that the intensity equation I(θ) of the
reflected polarized light was
where I0 = A02. this equation is known as Malus’s Law. A normalized plot of
Malus’s Law is shown below.
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Lesson 26 questions – Malus’s Law
1.
(i)
Define the term plane-polarisation of visible light waves.
............................................................................................................
............................................................................................................
[1]
(ii)
Explain why sound waves cannot be plane-polarised.
............................................................................................................
............................................................................................................
............................................................................................................
............................................................................................................
[2]
[Total 3 marks]
2.
Fig. 1 shows a student observing a parallel beam of plane-polarised light
that has passed through a polarising filter.
eye
plane-polarised light
polaroid
Fig. 1
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(i)
Fig. 2 shows how the intensity of the light reaching the student
varies as the polarising filter is rotated through 360o in its own
plane.
intensity
0
0
90º
180º
270º
360º
angle of rotation
Fig. 2
Suggest why there is a series of maxima and minima in the
intensity.
...........................................................................................................
............................................................................................................
............................................................................................................
............................................................................................................
[2]
(ii)
Hence explain how sunglasses using polarising filters reduce glare.
............................................................................................................
............................................................................................................
............................................................................................................
............................................................................................................
[2]
[Total 4 marks]
3.
State an example of plane-polarisation that does not involve visible light
and state how the polarised wave may be detected.
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
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[Total 2 marks]
Lesson 27 notes – Interference
Objectives
(a) state and use the principle of superposition of waves;
(b) apply graphical methods to illustrate the principle of superposition;
(c) explain the terms interference, coherence, path difference and phase
difference;
(d) state what is meant by constructive interference and destructive
interference;
Interference
•
When 2 or more waves interact with each other to distort the
waveform of the other(s).
Path difference
•
Path difference is the difference in path length for two waves to reach
the same point.
Superposition
The principle of superposition may be applied to waves whenever two (or more)
waves travelling through the same medium at the same time. The waves pass through
each other without being disturbed. The net displacement of the medium at any point
in space or time, is simply the sum of the individual wave displacements. This is true
of waves which are finite in length (wave pulses) or which are continuous sine waves.
Constructive and destructive interference
Where a crest meets a crest we get constructive interference (reinforcement)
Where a crest meets a trough we get destructive interference
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Conditions for interference
•
•
The waves must be of similar types! (can’t be light and sound)
The wave sources must maintain a constant phase relation with one
another – they must be coherent. Superposition will occur whether
waves are coherent or not, they will just form a fixed pattern if
coherent.
Using Pythagoras, the path length of A = 5m therefore the path of the
reflected ray = 10m.
So the path difference = 2m.
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Constructive interference
Constructive interference occurs when the path difference equals n λ
Destructive interference
Destructive interference occurs when the path difference equals (n + ½) λ
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Lesson 28 notes - 2 source interference
Objectives
(e) describe experiments that demonstrate two source interference using sound, light and
microwaves;
(f) describe constructive interference and destructive interference in terms of path
difference and phase difference;
(g) use the relationships intensity = power/cross-sectional area and intensity = k x
amplitude2 ;
Waves on a pond
Imagine dropping a pebble into a pond. It would make these kind of concentric
circles:
If you dropped 2 pebbles in a pond the waves would interfere with each other
like this:
The lines are where there would be destructive interference. In between this
there is constructive interference because of the path difference between the
waves.
You can imagine that if these two sources were speakers and you walked
along the edge you would hear loud and soft sounds. Likewise if they were
light waves there would be light and dark areas, these areas are called fringes
and will be described in more detail in lesson 29.
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Two Source Sound Interference
Two speakers are set approximately 1 meter apart and produced identical
tones.
In the diagram, the compressions of a wavefront are represented by a thick
line and the rarefactions are represented by thin lines.
These two waves interfere in such a manner as to produce locations of some
loud sounds and other locations of no sound. The loud sounds are heard at
locations where compressions meet compressions or rarefactions meet
rarefactions and the "no sound" locations appear wherever the compressions
of one of the waves meet the rarefactions of the other wave.
So if you slowly walk across the room parallel to the plane of the speakers,
you hear loud sounds as you approached anti-nodal locations and virtually no
sound as you approached nodal locations. (there would be some because of
reflections from the walls)
Young’s Slits Experiment
http://www.colorado.edu/physics/2000/schroedinger/two-slit2.html
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Lesson 29 notes - Intensity
Objectives
(g) use the relationships intensity = power/cross-sectional area and intensity = k x
amplitude2 ;
Definition of Intensity
The intensity of any wave is the rate at which it transmits power per unit area
through some region of space.
I=P
A
The unit of intensity is the watt per square meter — a unit that has no special
name.
For waves that radiate in all directions, the intensity goes down as the area
covered by the wave goes up. This is called the Inverse Square Law:
Since the intensity is the power / area. As the sound is spread over larger
areas, the intensity goes down. You can find the area of a sphere by using the
equation 4r2.
So the equation for intensity becomes:
I=P
4r2
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Energy and Amplitude
The diagram shows how the total energy of the system stays constant whilst
KE and PE change.
For the mass and spring system, the work done stretching a spring by an
amount x is the area under the force extension graph = 1/2 kx2. The PEextension graph is a parabola.
The kinetic energy will be zero at +A and a maximum when x = 0, so its graph
is an inverted version of the strain energy graph. At any position kinetic +
elastic strain energy is a constant E, where E = KEmax = PEmax.
PEmax  A2, so the total energy E of a wave is proportional to (amplitude) 2.
Energy stored in a stretched spring
area below graph
= sum of force 
change in displacement
extra area
F1 x
F1
total area
1 Fx
2
0
x
0
unstretched
extension x
force F1
work F1 x
no force
larger force
Energy supplied
small change x
energy supplied = F x
F=0
x=0
F = kx
x
stretched to extension x by force F:
energy supplied = 12 Fx
spring obeys
Hooke’s law: F = kx
energy stored in stretched
spring = 12 kx2
Energy stored in stretched spring is 12 kx2
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Since power is the energy transferred in a given time, therefore intensity is
also proportional to (amplitude) 2.
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Lesson 29 questions - Intensity
1
a)
Define the Intensity of a wave
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
b)
Calculate the light intensity 1.45m from a 100 Watt light bulb. Assume
that the light radiates equally in all directions.
Intensity = ………………….. unit…………… (4)
Total [6]
2
Comparing brightness
A low-power lamp at close range can often appear as bright as a high-power lamp
much further away. Similar reasoning is used to calculate the brightness of stars.
Two filament lamps
Here you compare two incandescent filament lamps; one marked 2.5 V, 0.3 A and a
second
100 W, 240 V. Assume that approximately 20% of the power dissipated by the lamps
results in visible photons being emitted uniformly, and that both are being run at
normal brightness.
a)
Estimate the brightness of the 100 W lamp from 3 metres away, which is the
intensity in W m–2
Intensity = …………………… W m–2 (2)
b)
Calculate how close you must be to the 2.5 V lamp to see it as being the same
intensity?
Distance = ……………………… m (2)
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3
Two floodlights
This question is about a 3 W bicycle headlight and a 300 W exterior security
floodlight. Both the floodlight and the headlight have approximately the same shaped
beam, a cone that covers approximately 1 / 10 of a sphere.
a)
Estimate the intensity of the bicycle lamp from 20 metres away.
Intensity = …………………. W m–2 (3)
b)
Your pupil has a diameter of approximately 0.01 m.
What is the power of the light that enters your eyeball per second?
Power =…………………….. Watts (2)
c)
The floodlight looks just as bright as the bicycle lamp.
How far away is it?
Distance = …………………….metres (4)
Total [10]
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Lesson 30 notes – Young’s Slits
Objectives
(h) describe the Young double-slit experiment and explain how it is a classical confirmation of
the wave-nature of light (HSW 1);
(i) Select and use the equation
λ = ax
D for electromagnetic waves;
(j) describe an experiment to determine the wavelength of monochromatic light using a laser
and a double slit (HSW 1);
The Wave Nature of light
Look at the diagram below. This is Thomas Young’s experiment that he
carried out in 1801.
He fired coherent light of one colour towards a screen with two small slits in it.
He did this to discover if light was a wave or if it was a particle. If it was a
particle it would just form two fringes of light as in the diagram below.
If light was a wave though and the slits were of the right size (about the same
size as the wavelength of the light) it would diffract and behave as in the
diagram below:
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At some points the two sets of waves will meet crest to crest, at other spots
crest meets trough.
Where crest meets crest, there will be constructive interference and the
waves will make it to the viewing screen as a bright spot.
Where crest meets trough there will be destructive interference that cancel
each other out.
When this experiment is performed what we actually see is shown in the
diagram below:
So therefore light must be a wave because particles cannot diffract.
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Analysing the experiment
First bright
fringe
Slit P
x
a
O
Slit Q
λ
Screen
D
To create the first bright fringe, the path difference of light from slit Q must
one wavelength further than light from slit P so that constructive interference
occurs.
For small values of the fringe width, x, the two shaded triangles are similar
and so a/ λ = D/x.
So rearranging:
λ = ax/D
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Lesson 30 questions – Young’s Slits
1
Explain what is meant by the principle of superposition of two waves.
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
[Total 2 marks]
2
(a) A detector is moved in front of two identical coherent wave sources and
detects regions of constructive and destructive interference. Explain the terms
(i) coherence
.......................................................................................................................................
...................................................................................................................................[1]
(ii) path difference.
.......................................................................................................................................
...................................................................................................................................[1]
(b) Fig. 4.1 shows two identical monochromatic light sources S1 and S2
placed in front of a screen. The sources emit light in phase with each other.
Fig. 4.1
(i) State, in terms of the path difference of the waves, the conditions
necessary to produce:
1 constructive interference at point P on the screen
.......................................................................................................................................
...................................................................................................................................[1]
2 destructive interference at point Q on the screen.
.......................................................................................................................................
...................................................................................................................................[1]
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(ii) The light sources S1 and S2 are 0.50 mm apart. They each emit light of
wavelength 4.86 × 10–7 m. An interference pattern is produced on the screen placed
2.00 m from the sources. Calculate the distance between two neighbouring bright
fringes on the screen.
distance = ......................................................m [3]
(iii) Suggest how the appearance of the interference pattern would change if
coherent white light sources were used instead of the monochromatic sources.
.......................................................................................................................................
.......................................................................................................................................
...................................................................................................................................[2]
[Total: 9]
3
(a)
In an experiment to try to produce an observable interference
pattern, two monochromatic light sources, S1 and S2, are placed in
front of a screen, as shown in the diagram below.
screen
P
y
S1
O
a
S2
D
(i)
In order to produce a clear interference pattern on the screen,
the light sources must be coherent. State what is meant by
coherent.
..................................................................................................
..................................................................................................
..................................................................................................
[2]
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(ii)
In the diagram, the central point O is a point of maximum
intensity. Point P is the position of minimum intensity nearest
to O. State, in terms of the wavelength λ, the magnitude of
the path difference S1P and S2P.
..................................................................................................
[1]
(b)
In an experiment to try to produce an observable interference
pattern, two monochromatic light sources, S1 and S2, are placed in
front of a screen, as shown in Fig. 1.
screen
P
y
S1
O
a
S2
D
Fig. 1
In another experiment, a beam of laser light of wavelength 6.4 ×
10–7 m is incident on a double slit which acts as the two sources in
the diagram above.
(i)
Calculate the slit separation a, given that the distance D to
the screen is
1.5 m and the distance between P and O is 4.0 mm.
a = ............................................................ m
[3]
(ii)
Sketch on the axes of Fig. 2 the variation of the intensity of
the light on the screen with distance y from O.
intensity
–16 –14 –12 –10 –8 –6 –4 –2
0
0
2
4
6
8 10 12 14 16 y/mm
Fig. 2
[2]
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[Total 8 marks]
Lesson 31 notes - Diffraction Gratings
Objectives
(k) describe the use of a diffraction grating to determine the wavelength of light (the
structure and use of a spectrometer are not required);
(l) select and use the equation dsinθ = nλ;
(m) explain the advantages of using multiple slits in an experiment to find the wavelength
of light.
Introduction
A diffraction grating is an object that has lots of slits that light can pass
through. It can be used to split white light up into a spectrum.
Multiple slits



If the path difference is exactly 1 wavelength like in the diagram they will
constructively interfere to produce a bright fringe. From the diagram you can
see that this happens when: (/d)=sin. If the path difference is just slightly
out then destructive interference will occur and no light will be seen. So if
white light is shone through dispersion occurs. A spectrum of fringes will be
seen as below with white in the centre where the path difference in zero, violet
nearest the centre and red nearest the edge:
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Measuring the wavelength of monochromatic light
n=3 3rd order
maximum

Monochromatic source
Diffraction grating
n=2 2nd order
maximum
n=1 1st order
maximum
undeviated light.
central maximum
zeroth order
n=1 1st order
maximum
n=2 2nd order
maximum
n=3 3rd order
maximum
So this is what you would see on a screen. An undeviated central maximum
and then dark and light fringes because of constructive and destructive
interference.
The maxima occur because the path difference is equal to a number of
wavelengths, ie, 2, 3, etc, rather than just . The number of the maximum is
called the order of maximum, n.
So from above and rearranging, we get:
n=dsinn.
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Lesson 31 questions – Diffraction Gratings
These questions give you practice in using the grating formula n  = d sin n.
A grating is labelled '500 lines per mm'.
1.
Calculate the spacing of the slits in the grating.
2.
Monochromatic light is aimed straight at the grating and is found to give a first-order
maximum at 15º. Calculate the wavelength of the light source.
3.
Calculate the position of the first-order maximum when red light of wavelength
730 nm is shone directly at the grating.
4.
The longest visible wavelength is that of red light with  = 750 nm. The shortest
visible wavelength is violet where  = 400nm. Use this information to calculate the
width of the angle into which the first-order spectrum is spread out when white light is
shone onto the grating.
A grating is illuminated with a parallel beam of light of wavelength 550 nm. The first-order
maximum is in a direction making an angle of 20º with the straight-through direction.
5.
Calculate the spacing of the grating slits.
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6.
What would be the angle of the first-order maximum if a grating of slit spacing of
2.5 10
7.
–6
m were used with the same light source?
Calculate the wavelength of light that would give a second-order maximum at  = 32º
–6
with a grating of slit spacing 2.5  10 m.
Hints
1.
What must the gap be between the centre of each line in order to fit 500 lines into
1 mm? Remember to express your answer in metres.
2.
This is about the first-order minimum so use the formula n = d sin 
3.
the
Rearrange the formula n = d sin 
arcsin (or sin–1) to give an answer in degrees.
4.
Use the same method as question 3 to obtain the position of first-order maxima for
red and violet light. The dispersion is simply the angle of maximum of red light minus
the angle of maximum of violet light.
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 the subject. Remember to take
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Lesson 31 notes – Stationary Waves
Objectives
2.4.4 Stationary waves
Candidates should be able to:
(a) explain the formation of stationary (standing) waves using graphical methods;
(b) describe the similarities and differences between progressive and stationary waves;
(c) define the terms nodes and antinodes;
Stationary (Standing) Waves
Stationary waves occur when two progressive waves of the same wavelength
and frequency interfere constructively to produce a wave pattern that seems
to oscillate about fixed points.
This can be two waves from different sources or as the diagram below shows,
can be from one source and its reflected wave.
Nodes and Antinodes
The diagram shows that nodes are positions where there is complete destructive
interference while at antinodes there is complete constructive interference.
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Lesson 32 notes – stationary waves and resonance
Objectives
(d) describe experiments to demonstrate stationary waves using microwaves, stretched
strings and air columns;
(e) determine the standing wave patterns for stretched string and air columns in closed
and open pipes;
(f) use the equation: separation between adjacent nodes (or antinodes) = λ/2;
(g) define and use the terms fundamental mode of vibration and harmonics;
(h) determine the speed of sound in air from measurements on stationary waves in a
pipe closed at one end.
Guitar Strings
A guitar string has a number of frequencies at which it will naturally vibrate.
These natural frequencies are known as the harmonics of the guitar string and
are associated with a standing wave pattern.
The diagram below depicts the standing wave patterns for the lowest three
harmonics or frequencies of a guitar string.
The wavelength of the standing wave for any given harmonic is related to the
length of the string.
The first harmonic is sometimes called the fundamental mode because it is
the simplest wave that can be formed.
Length-Wavelength
Harmonic
Pattern
# of Loops
Relationship
1st
1
L=1/2•
2nd
2
L=2/2•
3rd
3
L=3/2•
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4th
4
L=4/2•
5th
5
L=5/2•
6th
6
L=6/2•
n
L=n/2•
-nth
Standing waves in a pipe closed at one end
We can do the same as above for a resonating tube and be able to find
wavelength or length of tube given the other and the harmonic.
Harmonic
Pattern
Length of resonating
air (L) -Wavelength
Relationship
1st
L = (1 / 4) •
2nd
L = (3 / 4) •
3rd
L = (5 / 4) •
nth
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L = (2n-1) / 4 •
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Substituting in v=f into allow us to work out the speed of sound if we know
the resonating frequency.
Resonating column with one end closed
Apparatus
Resonance Tube (eg perspex tube, about 75 cm long and about 5 cm internal
diameter), loudspeaker with similar diameter (0.1 W, higher impedance better
for typical oscillator), oscillator (preferably matching loudspeaker power
rating/impedance or low impedance audio amplifier may be necessary), tie
clip microphone/preamp, oscilloscope, thermometer, two test leads (BNC - 4
mm). The schematic experimental set up is as shown in figure 2.
Figure 2
Setting up the apparatus








Set up the resonance tube, microphone, oscillator and oscilloscope as
shown in figure 1.
Remove the piston from the tube and ensure the microphone is
mounted in the Resonance Tube.
Connect the low impedance (50 W or less) output of the oscillator to
the loudspeaker (32 W) and to channel one on the oscilloscope.
Connect the microphone output to channel two on the oscilloscope.
Switch on the microphone amplifier (remember to switch off the
microphone amplifier at the end of the laboratory period).
Trigger the oscilloscope on the signal going to the loudspeaker
(channel one).
Set the oscillator to approximately 100 Hz and adjust the amplitude
until the sound from the speaker can be heard.
Adjust the oscilloscope to show two steady signals, if you have any
difficulties consult one of the staff running the laboratory.
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Experiment
Closed Tube Method
Insert the piston into the tube so that the tube length is now between 0.50 and
0.70 m.
Set the oscillator frequency to about 1 kHz and then adjust the oscillator to
obtain a resonance.
Move the microphone along the tube, make a note of the positions where the
signal reaches a maximum and a minimum.
Sketch the wave amplitude along the tube. Mark on your sketch nodes,
antinodes and their positions. Make sure you mark the open and closed ends
of the tube on your sketch. Does your sketch agree with the displacement
amplitude patterns for the closed tube shown in figure 1? Which
overtone/harmonic does you sketch show?
The separation between adjacent nodes (weakest signal) and between
adjacent antinodes (strongest signal) is equal to l/2 where is the wavelength.
Calculate several times from your data.
Measure the frequency of the resonance using the oscilloscope.
Calculate the speed of sound using v = fl , how does this compare with your
previous value?
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Lesson 32 questions – Stationary waves and
resonance
1.
State and explain one difference between a progressive and a standing
wave.
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
[Total 2 marks]
2.
(a)
In an investigation of standing waves, a loudspeaker is positioned
above a long pipe containing water, causing sound waves to be
sent down the pipe. The waves are reflected by the water surface.
The water level is lowered until a standing wave is set up in the air
in the pipe as shown in Fig. 1. A loud note is heard. The water level
is then lowered further until a loud sound is again obtained from the
air in the pipe. See Fig. 2.
c
c
l1
l2
pipe
pipe
water
Fig. 1
water
Fig. 2
The air at the open end of of the pipe is free to move and this
means that the antinode of the standing wave is actually a small
distance c beyond the open end. This distance is called the end
correction.
A student writes down the following equations relating the two
situations shown.
l1 + c = λ/4
©2011 science-spark.co.uk
l2 + c = 3λ/4
RAB Plymstock School
(i)
Draw the standing wave in the pipe shown in Fig. 2 which
corresponds to the equation l2 + c = 3λ/4.
[1]
(ii)
On your diagram, label the positions of any displacement
nodes and antinodes with the letters N and A respectively.
[1]
(iii)
Use the two equations to show that l1 – l2 = λ/2.
[1]
(iv)
The following results were obtained in the experiment.
frequency of sound = 500Hz l1 = 0.170 m l2 = 0.506 m
Calculate the speed of sound in the pipe.
speed = ………………..m s–1
[3]
(b)
The student repeats the experiment, but sets the frequency of the
sound from the speaker at 5000 Hz.
Suggest and explain why these results are likely to give a far less
accurate value for the speed of sound than those obtained in the
first experiment.
In your answer, you should make clear the sequence of steps in
your argument.
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Total [10 marks]
©2011 science-spark.co.uk
RAB Plymstock School