1.world communicates

copyright 2005-2006 keep it simple sciencePreliminary Physics Topic 1

Parramatta High School SL#606247

1

but first, lets revise...

Preliminary Physics Topic 1

THE WORLD COMMUNICATESWhat is this topic about?To keep it as simple as possible, (K.I.S.S.) this topic involves the study of:1. THE NATURE OF WAVES

2. THE PROPERTIES OF SOUND WAVES3. ELECTROMAGNETIC WAVES

4. REFLECTION & REFRACTION5. DIGITAL COMMUNICATION & DATA STORAGE

...in the context of communications

ENERGYEnergy is what causes changes and does work.The familiar forms of energy include

HEAT

ELECTRICITY

KINETIC (energy in a moving object) POTENTIAL (energy stored such as the

chemical energy in petrol).

Some forms of energy move around as WAVES. Awave is a carrier of energy. In a wave, energy moves,but matter does not. For example, compare these 2situations:

When the wind blows, both matter(the air) and energy move....

THIS IS NOT A WAVE!

In this case, only the energy moves. The air only vibrates as the SOUND WAVES move.

Waves Carry EnergyWithout the Transfer of Matter

TYPES of WAVESExamples of energy which moves around as waves include

SOUND

LIGHT

RADIO SIGNALS

WATER WAVES

X-RAYS

MICROWAVES

... and many more

ENERGY CONVERSIONS

Energy can be converted from one form to another. Inyour mobile phone the SOUND of your voice is convertedto ELECTRICAL signals then transmitted as RADIOwaves to your friend, whose phone converts it back again.

SOUND ELECTRICAL RADIO

In this topic you will learn about waves and theirproperties and features, and how they are used for

communication.

PhotoHelen Lee



2

WaveMeasurements

mechanicalEM wavestransverselongitudinal wavelength

amplitudefrequencyperiod

Light & MirrorsReflection inCommunication

Total InternalReflection

& Critical Angle

GeneralTypes

of Waves Graphing Waves

(displacement-ttime)

Velocity

Frequency & PitchAmplitude & Loudness

WaveEquationV = f

Natureof Sound

Waves Principleof Superposition

Dangers ofEMR

Production&

Detectionof EMR

TheEMR

Spectrum

Inverse SquareLaw

I 1 d2

RefractionSnells Law

Sin i = v1 = nSin r v2

Lawof

Reflection

EMRin

Communication

TTHHEE WWOORRLLDDCCOOMMMMUUNNIICCAATTEESS

The Natureof

Waves

Propertiesof

SoundWaves

ElectromagneticWaves(EMR)

Reflection&

RefractionDigital

Communication&

Data Storage

CONCEPT DIAGRAM (Mind Map) OF TOPICSome students find that memorizing the OUTLINE of a topic helps them learn and remember the concepts andimportant facts. As you proceed through the topic, come back to this page regularly to see how each bit fits thewhole. At the end of the notes you will find a blank version of this Mind Map to practise on.



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Waves Carry Energy

Waves carry energy, without the transfer of matter.

This can occur in 1 dimension:

... or in 2 dimensions:

Ripples spreading on thesurface of a pond.

...or in 3 dimensions,

such as when light radiates in all directions from a glowingobject.

Waves & Mediums

Mechanical waves are those which need a medium totravel through. For example, a water wave must have waterto travel in. Sound waves need air, or water, or somesubstance to move in. They CANNOT travel in a vacuum.

Electromagnetic (EM) waves do NOT need a medium...they can travel through a vacuum, and in fact travel fastestin a vacuum. EM waves include light, radio waves, ultra-violet and other types, and are studied in detail in a latersection.

Describing Waves

A wave is a vibration. In a mechanical wave, the particles(atoms & molecules) in the medium vibrate to transmit thewave energy. In EM waves the vibration occurs in electricand magnetic fields.

Consider a wave in a rope which has been given a single up-and-down twitch:

Energy moves along the rope, but the rope itself doesnt goanywhere. Particles of the medium (the rope fibres)vibrate up-and-down as the energy moves across.

This form of a wave, where the medium vibrates at rightangles to the direction that the energy moves, is called aTransverse wave.

If the rope is wiggled constantly up-and-down, you get notjust one pulse, but a periodic wave with one pulse followinganother.

TRANSVERSE Energy flow

WAVESVibration in medium

1. THE NATURE OF WAVES

Pulses moving along a slinky spring compression

CRESTA PULSE WAVEvibration of medium

TROUGHEnergy moves

rope

AA PPEERRIIOODDIICC,, TTRRAANNSSVVEERRSSEE WWAAVVEE

WAVELENGTH

AMPLITUDE

AMPLITUDETROUGH

CREST

Energy moves

TRANSVERSE WAVES

THE VIBRATION OF THE WAVEIS AT RIGHT ANGLES

TO THE DIRECTION OF ENERGY FLOW

PhotoPhilippPilz



4

Longitudinal waves are when the particles of the mediumvibrate back-and-forth in the same line as the energymoves. For example, when a series of compressions andrarefactions are sent along a slinky spring.

Energy moves

LLOONNGGIITTUUDDIINNAALL WWAAVVEE IINN AA SSPPRRIINNGG

compression rarefaction

Spring vibrates

LONGITUDINAL WAVES

THE VIBRATION OF THE WAVEIS IN THE SAME LINE

AS THE DIRECTION OF ENERGY FLOW

Wavelength = the distance from one crest to the next. (orfrom one trough to the next, or from one compression tothe next) The S.I. unit is the metre (m).

The Greek letter lambda is used as a symbol for wavelength.

Amplitude (a or A) = the distance that a particle in themedium is displaced from its rest position at a crest ortrough. i.e. the maximum displacement distance.

Frequency (f) = the rate at which the wave is vibrating.The number of waves that pass a given point in 1 second,or the number of complete vibrations per second.

S.I. unit is the hertz (Hz) 1 Hz = 1 wave/sec.

Wave Measurements

All periodic waves,

whether Longitudinal or Transverse, Mechanical or Electromagnetic,

can be described and measured by their:-

Period (T) = the time (in seconds) for one completevibration to occur.

Note that there is a simple relationship between Frequencyand Period... they are reciprocals.

Velocity (v) = the speed of the wave, in metres/sec.(ms-1)

There is a simple relationship between Velocity, Wavelengthand Frequency:

Velocity = Frequency x Wavelength

TRY THE WORKSHEET, at the end of this section.

THE WAVE EQUATION

V = fWAVELENGTH

AMPLITUDE

Energy moves

T = 1 and f = 1 f T



5

Graphing WavesA good way to represent a wave is by using a graph.

Imagine a floating cork bobbing up and down as a series ofripples move across the water surface (i.e. a periodic wave).

If you graph the (up-down) displacement of the corkagainst time, the graph will look something like this:

Be careful! The graph looks like a wave, so its tempting totry to read the wavelength from the horizontal scale... butit wont work! The horizontal scale is TIME, not length.

What you CAN read from a Displacement-Time graph:

Amplitude. The vertical scale measures the displacement ofthe cork from the equilibrium position (i.e. the flat watersurface).So, at 0 sec, the cork was in the equilibrium position.

at 0.1 sec, it was approx. 2cm upwards.at 0.2 sec, it was 3cm upwards... and so on.

Its maximum displacement was 3cm either above or below(d= -3cm) equilibrium, so the Amplitude = 3cm (0.03m)

Period. Since the horizontal scale is time, you can easilyread from the graph how long it takes for one complete up-and-down cycle. On this graph T = 0.8s

From Period, calculate Frequency: f = 1 / T= 1 / 0.8= 1.25Hz

If the speed of the wave was known, then you couldcalculate the wavelength, or vice versa.

e.g. if the ripples are 0.45m apart: (i.e. = 0.45m)V = freq. x wavelength

= 1.25 x 0.45So, velocity = 0.56 ms-1

TRY THE WORKSHEET (next page)

Graphing a Longitudinal Wave

You might think these Displacement-Time graphs wouldntwork for a Longitudinal wave where the particles vibrateback-and-forth rather than up-and-down.However, the graph of a longitudinal wave can be exactlythe same... you just have to realise that the displacementis sideways displacement from the equilibrium position,instead of up-down.Amplitude, Period and Frequency can all be determined inexactly the same way.

Relationship BetweenWavelength & Frequency

You may have carried out a First Hand Investigation inclass to see how a change in Frequency (at constantvelocity) affects the wavelength. Maybe you used a slinkyspring, or watched the water waves in a ripple tank.

You would have found...

INCREASING DECREASE inthe FREQUENCY WAVELENGTH

and

DECREASING INCREASE inthe FREQUENCY WAVELENGTH

(If VELOCITY is the same)

Cork bobs up and down

Ripples

Time (s)1.00.5

0-3

+3

Disp

lace

men

t (c

m)

One period= 0.8 s

LongerWavelength

To have the same speed, the shorter waves must vibrate faster

LowerFrequency

ShorterWavelength

HigherFrequency

Worksheet 1

Part A Fill in the blank spaces

Waves carry a)...................................... without thetransfer of b)................................ Mechanicalwaves require a c)............................. to travel in.Examples are d)...................... and ............................Electromagnetic waves do not need a mediumand can travel in a e).........................................Examples include f)............................... and...................................

A g)................................... wave is when thevibration and the movement of energy areh)........................................................In a Longitudinal Wave, the vibration and theenergy movement are i)..............................................

j)........................................ is the distance fromcrest to crest.

Amplitude is the k)...................................................

Frequency is the number of l)................................per second. The SI unit is the m)................. (........)

n)............................................ is the time for onecomplete vibration. This is theo).................................... of frequency.

Velocity is the speed of the wave and is equal top).......................... multiplied by q).............................

On the graph of a wave, showing Displacement vTime, the vertical scale shows ther)..................................... of the wave, while thehorizontal allows you to read the value of thes)........................................ and then easily calculatethe t)......................................................

For waves travelling at the same velocity,increasing the frequency wouldu).................................... (increase/decrease) thev)................................................, and vice-versa.

Part B Practice ProblemsWave Measurements and the Wave Equation

Example Problem 1A water wave in the ocean has a wavelength of 85m, anda velocity of 4.5ms-1.a) Find the frequency. b) What is the period?Solutiona) V = f

4.5 = f x 85f = 4.5 / 85

= 0.053 Hz (5.3 x 10-2 Hz)(i.e. only a fraction of a wave passes by each second.)

b) T = 1 / f= 1 / 0.053= 19 s

(i.e. it takes 19 seconds for 1 complete wave, crest to crest,to pass by)

Example Problem 2A sound wave has a period of 2.00x10-3s. (= 0.002s)Sound travels in air at a velocity of 330ms-1.a) What is the frequency of the wave?b) Find the wavelength.Solutiona) f = 1 / T

= 1 / 0.002= 500Hz (i.e. 500 vibrations per sec.)

b) V = f 330 = 500 x

= 330 / 500= 0.66m (i.e. 66cm from crest to crest)

TRY THESE

1. a) Find the velocity of a sound wave in water if itvibrates 280 times per second and has a wavelength of5.20m.b)What is the period of this wave?

2. An earthquake shockwave travels through rock at avelocity of 2,500 ms-1. Its frequency is 0.400 Hz.What is the wavelength?

3. What is the wavelength of a sound wave withfrequency 1200Hz? Sound travels in air at 330ms-1.

4. An ocean water wave in deep water travels at a velocityof 6.50ms-1. Its period is 16.0s.a) What is the frequency?b) Wavelength?c) As the wave enters shallower water it keeps the samefrequency but slows down to only 2.20 ms-1. Whathappens to the wavelength?

continued...

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Copyright 2005-2006 keep it simple sciencePreliminary Physics Topic 1

COMPLETED WORKSHEETSBECOME SECTION SUMMARIES

Worksheet 1 continued

5.a) Red light has a wavelength of 7.00x10-7m, and travels at3.00x108ms-1. What is the frequency?b) Blue light has a wavelength of 3.00x10-7m and travelsat the same speed. What is the frequency?

6.Radio signals travel at the speed of light.(3.00x108ms-1) A radio station has a frequency of530 kHz (=530,000Hz).a) What is the period of the waves?b) What is the wavelength?

7.In World War II, one major technological advance was thedevelopment of centimetric RADAR... Radar thatworked on radio waves of only a few centimetres inwavelength. (Previous radars worked at wavelengths of ametre or longer)Compare the frequency of a radio wave 2.50m long, withone 2.50cm long.(Assume they both travel at the speed of light)

Part C Wave Graph Exercise

The following sketch graph shows 3 different waves P,Q and R. For each wave;

i) What is the Amplitude?ii) State the (approx) displacement at time t=0.3siii) What is the Period of each wave?iv) What is the Frequency of each wave?v) Given that wave P has a wavelength of0.50m, calculate its velocity.

vi) Waves Q & R both travel with a velocityof 9.5ms-1. Find their wavelengths.

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Remember that for full marksin calculations, you need to show

FORMULA, NUMERICAL SUBSTITUTION,APPROPRIATE PRECISION and UNITS

8.When a guitar string is plucked, a wave vibration runsback and forth through the string. The string is 0.96mlong and it is found that exactly 8 complete wavelengthsfit along the string at a time. The vibration frequency is384Hz.How fast do the waves travel through the string?

9.X-rays are very short wavelength EM waves which travelat the speed of light. If the wavelength is 1.50x10-11 metre,a) find the frequency.b What is the period of the X-rays?

PQ

R0.1

0.1

-00.1

0

0.05

Time (s)

Disp

lace

men

t (m

)

Sound WavesSound waves are Mechanical (they need a medium)

and Longitudinal (vibrate back-and-forthin the line of the energy flow)

SOUND Energy moves

WAVES

Particles vibrate

Instead of crests and troughs, a series of compressionsand rarefactions pass through the medium as a soundtravels. The atoms and molecules are alternately squashedtogether and then stretched apart as the energy flowsthrough.

In a compression the air pressure is higher, and lower in ararefaction.

Velocity of SoundSound travels at different speeds in different mediums.In air, sound travels at about 330-350 ms-1, (about 1,200km/hr) depending on temperature and density.

The denser the air, the slower the speed of sound.

In liquids and solids, sound travels much faster......about 1,500ms-1 in water...about 5,000ms-1 in most metals.

FREQUENCY = PITCH

When you hear sounds of different pitch that is the wayyour brain interprets sound waves of different frequency.

Low Frequency = Low PitchHigh Frequency = High Pitch

AMPLITUDE = LOUDNESS or VOLUME

Sound waves with different amplitudes are interpreted byyour brain as sounds of different loudness or volume.

Larger Amplitude = Louder SoundSmaller Amplitude = Quieter Sound

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2. THE PROPERTIES OF SOUND WAVES

Sound Travels

Compression CompressionRarefaction Rarefaction

Time

Compression. Higher air pressure

Rarefaction. Lower pressureDisp

lace

men

t fro

mth

e eq

uilib

rium

ECHOES ...ECHOES ...ECHOES

Like all waves, sound can travel through a medium like air,strike another medium (say, a brick wall) and bounce back.The REFLECTED wave will be heard as an echo.

Some animals can send out sound waves and pick up the echoes to help locate their prey, or to navigate, in environments where they cant see very well.(Such as murky water, or in darkness.)

Squeaks of sound

Echoes from insect

SONAR SOund Navigation And Ranging

BAT

Humans have invented SONAR technologies for thingssuch as depth sounding and detecting underwaterobjects... fish or submarines, it all works the same way.

The time delay betweensending a sound ping andreceiving the echo, givesdepth and distance

FishFinders

DepthSounding

Anti-submarineWarfare

USES OFSONAR



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The Principle of SuperpositionAll waves have the ability to pass through other waveswithout being affected. For example, you could shine a redspotlight across a beam of blue light, and each colour andbeam will emerge on the other side exactly the same.

However, for the instant that the 2 waves are superimposedupon each other, they do interact and interfer with eachother.

Very simply, the displacement of the two waves addtogether at every point where the waves coincide.

In this case, the waves A&B were in phase (crest co-incided with crest, trough with trough) so the result wasconstructive interference... the resultant has an amplitudewhich is the sum of A+B.

However, if the waves are out of phase (for example, ifcompression coincided with rarefaction) then there isdestructive interference... the opposite amplitudes maycancel each other out.

Theoretically, if 2 sound waves had the same amplitude andwere perfectly out of phase they could cancel out totally...imagine having 2 sounds that add up to SILENCE! (or 2 lights that combine to form DARKNESS!)

In practice, this only happens over short distances or timeperiods to give interference patterns and beat sounds.

resultant A+B

To find aresultant, addthe displace-ments of A&B atconvenientpoints (circled)

wave Awave B

Disp

lace

men

t

Resultant

Add positive &negative displacementsat the circled points wave A

wave BDisp

lace

men

t

When an in-coming wave meets areflected wave going back out, theiramplitudes add together.

The Principle of Superposition

Technically, a breaking wave is not a wave at all.Once it breaks, the water begins moving forward (which allows you to catchit) and so both energy and matter are flowing forward... NOT a wave!True water waves are the swells which you cannot catch.

Photo: Chris Potter

PPhhoottoo:: IIggoorr KKaassaalloovviicc

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Worksheet 2

Part A Fill in the blanks. Check your answers at the back

Sound waves are a)....................................... and b).................................... A sound wave consists of a series of high pressurec)................................................... and lower pressure d)........................................................... travelling through the medium.In air, the speed of sound is about e)............. ms-1, but it is much f).................................. (higher/lower) in water or in solidssuch as metals.

The pitch of a sound is related to the g).................................... of the wave. The amplitude of the wave determines theh)............................. of the sound we hear. Echoes occur when sounds i)............................................. Some animals use echoesfor j)............................................ Humans use the technology of k)............................................ for depth sounding andl)..............................................When 2 or more waves coincide, they will interfere with each other. The m).................................................. wave can be foundby adding together the separate wave n)....................................................

Part B Principle of Superposition ExerciseFind the resultant of these 2 waves by adding the displacements at the circled points, then join the sum points withan even curve. IF THESE ARE SOUND WAVES, WHAT WOULD YOU HEAR? (check your answer )


-ve

Disp

lace

men

t

+

ve

EM WavesElectromagnetic waves are Transverse waves which doNOT require a medium to travel through...they travel through a vacuum at 3.00x108ms-1, the speedof light. They can travel through many other substancesat slightly slower speed. For example, light can travelthrough glass or water at speeds of around 2.5x108ms-1. Inair, the speed is so close to the speed in a vacuum that, forsimplicity, (K.I.S.S. Principle) we take it to be the same.

EM radiation (EMR) does not require a medium becausethe waves propagate as vibrations of electric and magneticfields, not as vibrating particles.

MEMBERS OF THE EMR SPECTRUM

Radio (and TV) waves

microwaves

infra-red

(heat radiation)

visible LIGHT

ultra-violet

X-rays

Gamma rays

Although we tend to think of these as 7 different types ofradiation, you must realise that they are really all the samething, just at different wavelengths and frequencies.

Production of EM WavesAll EM waves are produced in basically the same way:vibration or oscillation of electrically charged particles.For example....Radio waves are produced by electric currents runningback-and-forth in a conducting wire.

Infra-red waves are made by molecules vibrating rapidlybecause of the heat energy they contain.

Light is emitted when electrons rapidly jump down froma higher to a lower orbit around an atom.

Gamma waves come from the vibrations of chargedparticles within an atomic nucleus, during a nuclearreaction in the atom.

Detection & Reception of EM WavesJust as all EM waves are produced in the same basic way,they are all received or detected in the same basic way too...by a phenomenon called Resonance. When waves strikesomething and are absorbed, they may cause sympatheticvibrations within it.

In cartoons and the movies (not in real life) the operasinger hits a high note and all the wine glasses begin tovibrate and then shatter... a fictional example of resonance.

Some real examples...

When radio waves hit a suitable aerial wire or antenna, theycause some electrons in the metal to oscillate back-and-forth in sympathy with the wave. These oscillations areamplified electronically and the signal converted to soundin the speaker, allowing you to listen to the radio.

When infra-red waves hit your skin they cause certainmolecules to begin to resonate and vibrate. This sets offnerve messages to the brain and you feel warmth or heaton your skin.

In a camera the light causes resonance in chemicals in thefilm. Chemical reactions occurwhich permanently alter the filmso that an image appears whendeveloped later. Different filmcan be sensitive to infra-red,(photos in the dark) or X-raysfor medical uses.

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3. ELECTROMAGNETIC WAVESW

avel

engt

h de

crea

sing

very

sho

rtv.

long

Freq

uenc

y in

crea

sing

very

hig

hlo

wWhen the fat lady sings...



12

Danger of High Frequency EMRHigh frequency EMR (ultra-violet, X-ray & gamma) can bevery dangerous to living things.

A little UV gives you a suntan, but long-term exposureleads to skin damage, premature skin ageing, and isimplicated in causing skin cancers, including melanoma.

The Sun produces dangerous quantities of UV radiation,but luckily most of it is absorbed by the ozone layer inthe upper atmosphere of the Earth.

Ozone is a form of oxygen which has 3 atoms permolecule (O3) instead of the normal 2 (O2). The ozonemolecules resonate well at the frequency of UV and soabsorb it strongly.

The Sun only produces small amounts of the even moredangerous X-rays and gamma radiation. Once again, mostis absorbed in the upper atmosphere, this time by ordinaryoxygen and nitrogen gases.

Infra-red and light radiation penetrate well, (although about30% is reflected) and while some radio frequencies getthrough, many get absorbed or reflected.

The Inverse Square LawAs any form of radiation spreads out from its source itsintensity gets less. For example, a sound becomes quieter ifyoure further from the source, or a light is not so bright asyou move further from it.

Mathematically, the relationship is that the intensity (I)(such as brightness of light) is inversely proportional to theSQUARE of the distance (d) from which it is viewed.

This diagram explains why:

At distance d from the light source, some light energyfalls on an area of x2 units. At twice that distance (2d) thesame amount of light would fall on an area of 4x2. Thebrightness of the light must be only 1/4 as much (since thesame amount of light is falling on 4 times the area.)

So, twice the distance 1/4 as bright3 times the distance 1/9 as bright

10 times the distance 1/100 as bright

...or if you move closer it will getter brighter:at half the distance, 4 times brighterat 1/3 the distance, 9 times brighter

...and so on.

x

lightsource

distance d

distance

2d

2x

Intensity 1 (distance)2

I 1 d2

meansproportionalto

UV Rays

Ozone O3Absorbs UV

Oxygen O2does notAbsorb UV

TRY THE WORKSHEET PROBLEMSAT THE END OF THIS SECTION

Square with sidestwice as long.

Area = 4x2

Same amount oflight falls on 4times the area

Square

Area =x2

Earths surface

Sunozone layer

upper atmosphere

X-ray & gamma

UV

somereflected

radioinfrared & light



13

EMR & CommunicationHumans rely on sound waves for communicating by directspeech, but all our modern communication technologiesrely on EMR.

Radio & microwaves carry radio and TV broadcasts,telephone long-distance links, mobile phone networks, andsatellite links for telephone (including internet) and TV.If you have Satellite TV, the dish on your roof is anantenna to receive radio waves directly from an orbitingsatellite.

Add to that, 2-way radio for military uses, CB amateurs andboating, shipping and aircraft communications, and youbegin to realise how many radio waves are zapping around.

Light is being increasingly used in the form of LASERbeams carried in optical fibres for telephone and internetcommunication.

How a Wave Carries InformationHow can a voice or piece of music be carried by a wave?The key feature is Modulation of the wave. There are 3common ways to modulate the wave to carry information...

Whats special about LASER LIGHT?

It is one, pure frequency of light.The waves are all in phase and sothey interfere constructively toform a very intense, tight beam.A laser beam will stay inside anoptical fibre and not leak out ordissipate for long distances.A laser can be turned on & offvery rapidly, so its perfect forhigh speed digital communication.

Frequency Modulation (FM)

The amplitude stays constantwhile the frequency (andwavelength) vary within a fixedrange. The information (voice,music etc) is coded in thevariations of frequency.

FM radio gives much betterfidelity and is superior, compared

to AM, for the quality ofsound (eg for music)received.

Pulse Modulation(Digital)

To carry informationin digital form thewave must switchrapidly between 2

different states, representing the1 and 0 of digital codes. Thewave can be switched rapidly onand off (as in the diagram) orswitched back-and-forth betweendifferent phase states... phasemodulation.

Amplitude Modulation (AM)The frequency (and wavelength)of the wave stays constant whilethe amplitude varies.

The changing amplitudecorresponds to the informationbeing carried... whether voice ormusic, or whatever.

WAVEMODULATION

Carrierwave

AMsignal

FMsignal

Digitalsignal

Digital 1 0 1 1 0 0 1 0 1 1 message (a byte of data)

Noinformationcarried

Amplitudechanges.Frequencyconstant

Freq. changes.Amp.constant

Wave pulseson and off

This diagram compares the effect ofAM, FM & Digital Modulation

on the same carrier wave

Laser Light Show

Photo: Keith Syvinski



14

Case Study: MOBILE (CELL) PHONESWhen you use a mobile phone, the sound of your voicegoes into a microphone and almost instantly pops out theother end into your friends ear. What happens inbetween?

In the future we will need to switch morecommunications to use the laser light / optical fibremethod wherever possible, and to make better use ofthe RF bands. For example, it is possible to use thesame frequency channel for several differentpurposes as long as the different signals are modulateddifferently and as long as the radio receivers aresophisticated enough to pick out only the desiredsignal and ignore the others.

One thing is for sure... humans will keepcommunicating and the need for new services willkeep expanding. So far, our technology has alwaysmanaged to keep up, and it will probably continue todo so.

Modern communication systems have developedrapidly and new features and capabilities seem to comeout every day. It seems that the entire system isunlimited and that it can continue to expand andimprove forever.

Well perhaps it can, but NOT while continuing to usethe radio end of the EMR spectrum. Each stationor channel must operate on a different frequency orelse signals can jam or interfere with each other.

The simple fact is that there are now so many radio &TV stations, mobile phone networks, aircraft andshipping channels, military, police and emergencyservice channels, etc. etc. all using the RF (RadioFrequency) part of the EMR spectrum, that it isbecoming difficult to keep expanding services withoutinterfering with existing channels.

1. The SOUND energy ofyour voice is converted toELECTRICAL signals by themicrophone. The electricalsignal is used to digitallymodulate a RADIO wave.

2. The digital RADIO signal istransmitted by your phone and receivedby the local cell antenna.

3. If your call is going to aperson in another location(a different cell) thesignal is converted into amodulatedMICROWAVE andbeamed, via hilltop relaytowers, to the correct area.(Alternatively, it might besent as a modulated LaserLIGHT beam throughoptical fibres).

4. In the other cell area, the signalis converted back to a modulatedRADIO signal and transmitted.5. Your friends phone receives the

RADIO signal, amplifies it as anELECTRICAL signal and this isconverted to SOUND waves in theirearphone.

SOUND ELECTRICITY RADIO MICROWAVE RADIO SOUND(or LASER LIGHT)

ENERGYY CHANGES

Discussion:LIMITATIONS OF COMMUNICATION CHANNELS

Worksheet 3

Part A Fill in the blank spaces

Electromagnetic waves are a)....................................... waveswhich b).............................. (do/do not) require ac).................................... to travel in. They all move at thespeed of light, which is d)..................................... ms-1 in avacuum. The members of the EMR spectrum (in order ofincreasing frequency) are:e)............................................, f)..................................................,g)........................................, h)..............................................,i)............................................, j)..................................................,and k)...............................................

All EM waves are produced when electrical chargesl)........................................... They are all detected/received bythe process of m)................................................. This is whenthe wave is absorbed by a substance and causes electrons ormolecules to n).................................................... insympathy with the wave.

High frequency EMR, such as o).............................................is dangerous to life. Luckily, the p)................................... layerof the atmosphere absorbs most of the dangerousq)........................... rays from the Sun.

All forms of radiation decrease in intensity in proportion tothe r).......................................... from the source, so ifdistance is doubled, the intensity will drop tos)................................ (fraction)

EM waves are very important in modern communications.The wave types used are mainly t).......................................and ..................................................., but light is being usedmore and more in the form of u).........................................carried inside v)....................................... fibres.

Waves carry information by the process ofw).................................... This can be done in 3 ways: AM stands for x).................................................................,in which the information is carried as fluctuations in they)................................................ of the wave. FM stands for z)................................................................,in which the signal is carried by variations inaa).............................................. of the wave. Digital signals are carried by ab).....................................modulation in which the carrier wave rapidly switchesbetween 2 states (e.g. on and off).

The energy changes occurring in a mobile phone call are asfollows: Sound waves of your voice are converted to anac)......................................... signal. This is used to modulatea ad)........................................... wave transmitted to the localcell antenna. Next, the signal is sent viaae)......................................... link, or LASER beam to anothercell station. Here it is transmitted again as aaf)...................................... signal. The receiving phoneconverts this to an ag)................................... signal and finallyit is converted back to sound waves.

Part B Practice ProblemsThe Inverse Square Law

Example Problem:At a distance of 5m the brightness of a light is measuredto be 36 units. How bright would it be if viewed from15m?

Answer:Since the distance is 3x further, then intensity will be1/9. So new brightness = 36/9 = 4 units.

TRY THESE:1.At a distance of 10m from a light, the brightness(intensity) is 48 units. What intensity would it have atdistance:a) 20m?b) 40m?c) 100m?d) 5m?

2.How much stronger would a radio signal be if you movedfrom 100km, to 25km distance from the transmitter?

3.At 2m from a flame the brightness is 32 units. At whatdistance would the brightness be 2 units?

4.One light bulb (at a certain distance) gives I units oflight intensity. To get the same light intensity at doublethe distance, how many identical bulbs need to beswitched on?

15




When a Wave Hits a BoundaryWhen a wave is travelling through one medium and thenstrikes a different medium, one of 3 things can happen atthe boundary:

It is quite possible that all 3 things can happen at once. Forexample, if a beam of light is travelling through air, andthen strikes a glass window:

the glass ABSORBS some of the light. some REFLECTS off the glass some is TRANSMITTED through the glass.

ReflectionThe Law of Reflection is very simple:Whatever angle a ray of light hits the surface, it willbounce off again at the same angle.

OR, more technically:

Angle of = Angle ofIncidence Reflection

io = ro

The trickiest bit is how the angles are measured. They mustbe measured between the rays and the NORMAL... animaginary line at right angles to the surface.

What if the Surface Isnt Flat?The Law of Reflection isstill obeyed, as shown:

Reflection of Light from Curved Mirrors

Concave mirrors (go in like a CAVE) reflect light to aFocus, or focal point.

Concave mirrors can give ENLARGED images if viewedfrom the right distance, such as a household shaving mirroror make-up mirror, which gives a magnified reflection ofyour face. This is also the basis of a reflecting telescope.

Convex mirrors reflect light so the rays divergeoutwards, as if coming from a focus behind the mirror.

Convex mirrors produce smaller (diminished) images, butgive a wider-angle view. An example of use is the sidemirrors on a car which give you a wide-angle view into thedrivers blind-spot.

16



4. REFLECTION & REFRACTION

The Incident rays P,Q & Rare parallel.

Each obeys the Law ofReflection, but thereflected rays go indifferent directions.

The Normal for each rayis shown as a dotted line

P QR

Example: Light waves travelling in air, then hitting glass.

AAbbssoorrbbeedd eenneerrggyybbeeccoommeess hheeaatt

AABBSSOORRPPTTIIOONNooff tthhee eenneerrggyy

RREEFFLLEECCTTIIOONN((bboouunncceess ooffff))

TTRRAANNSSMMIISSSSIIOONN iinnttoo tthheenneeww mmeeddiiuumm,, wwiitthh ppoossssiibbllee RREEFFRRAACCTTIIOONN

((cchhaannggee ooff ddiirreeccttiioonn))

Normalline

Incident ray

Reflect

ed ray

iooroo

Reflectivesurfacesuch as amirror

FFooccuuss

VViirrttuuaallFFooccuuss



17

Reflections in CommunicationsWave reflection from the ionosphere can help with longdistance radio communications. It works best with thelonger wavelength AM signals.

The Ionosphere is a zone in the upper atmosphere wherethe air molecules are partly ionized (electrically charged) by radiations from the Sun. The ionized gases act as areflective surface to radio waves of certain wavelengths.

TV signals and FM (shorter wavelengths) radio do notreflect so well and generally you need to be in line ofsight from the transmitter to get good reception.

Another example involves how Microwaves are transmittedand received. Microwaves are used to relay TV programs toregional transmitters and to relay long distance phone calls(including internet) from city to city.

At the transmission end, a curved reflector keeps the wavesin a tight beam aimed at the next relay station. The receiverhas a similar dish to focus the waves into the receivingantenna.

RefractionWhen waves enters a new medium several things happen:

the waves change their speed (get faster or slower) their wavelength changes (gets longer or shorter)

Note that frequency does NOT change! they change direction (unless they hit at 90o to surface)

When a light wave enters a more dense medium:(Example: going from air into glass)

velocity slows wavelength shorter (but frequency remains unchanged) refracts towards the normal.

io > ro

When going from a more dense, to a less dense medium theopposite changes occur.

velocity speeds up wavelength gets longer refracts away from the normal.

io < ro

When a light ray refracts, its wavelength changes, butfrequency stays the same. Since COLOUR is

determined by frequency, there is no colour changeduring refraction.

Transmitter

EARTH

Ionosphere

Receiver

Transmitterdish

Microwavebeam travelsbetween relaystations

Receiverdish

Your satellite TV dish is a reflector too

Microwave Reflector Dishes

Angle ofIncidence

Angle ofRefraction

ReceivingAntenna

ReflectorDish

ioro

Incident Ray

Refracted Ray

Air Glass

ioro

Incident Ray

Glass Air

Refracted Ray

normal

normal

PhotoHelen Lee



18

Snells LawYou may have carried out an investigation in class using aRay Box Kit to measure angles of incidence and anglesof refraction.

When you graph the angles the result is a curve.

This is not much use for defining any relationship that mayexist.

In 1621, Snell discovered that if you graph the Sine ratiosof the angles, the points lie in a straight line. You may havedone the same with your experimental data.

The fact that its a straight line means there is a directrelationship between Sin i and Sin r.

The gradient of the line is not only the ratio between theSine of the angles, but is also equal to the ratio of velocitiesof the wave in the 2 mediums involved. This special ratiois known as the REFRACTIVE INDEX (n)

This is now called Snells Law:

Refractive IndexWhen waves enter a new medium, and then exit it again, therefractions that occur on the way in, are the opposite ofwhat happens on the way out.

For example, this light ray goes from air, into glass and outinto air again.

Refractive index (air -> glass) ang = sin45 / sin28 = 1.5andRefractive index(glass -> air) gna = sin28 / sin45 = 0.66

These 2 values are RECIPROCALS !! ...and this willalways be the case... the index of refraction going in is thereciprocal of the index coming out.

Angle of refraction, ro

Angl

e of

inc

iden

ce, i

o

Sin ro

Sin

io

Gradie

nt = ris

e = Sin

i

run

Sin r

TRY THE WORKSHEET at the end of this section

Sine (angle incidence) = velocity (medium 1) = nSine (angle refraction) velocity(medium 2)

Sin i = V1 = nSin r V2

1n2 = 1 2n1

3 beams of light beingrefracted through aperspex block.

Pencil appears brokenat the water surface, dueto refraction of the lightby which we see it.

4455o

4455o

2288o

2288o

Refractionair -> glass

Refractionglass -> air

normal



19

Total Internal Reflection & the Critical Angle

Consider the situation when waves are going from a moredense medium into a less dense medium, such as lightgoing from glass into air.

The waves refract away from the normal.

Now think about increasing the incident angle as shown inthis series of diagrams.

There comes an angle (called the Critical Angle, (c))where the angle of refraction = 90o. At this point therefracted ray runs along the edge of the glass, but does notcross the boundary.

So, when the angle of incidence io = co, ro= 90o

Sin c = gna Sin 90

and sin 90o = 1, so...

TRY THE WORKSHEET at the end of this section

... But What Happens Beyond the Critical Angle?

At incident angles larger than c, the ray reflects backinside the glass... this is called

TOTAL INTERNAL REFLECTION

This has one very important application in communicationtechnology...

Optical fibres are thin strands of very pure glass that cancarry communications signals in the form of laser lightbeams. The laser beams stay within the fibres because oftotal internal reflection.

Each fibre is a core strand of glass, with another layerwrapped around it. The outer layer has a lower refractiveindex than the core, so even where the fibre bends arounda corner, the laser light will generally strike the boundary atan incident angle greater than the critical angle.

Whenever the laser beam hits the boundary between the 2layers, the angle of incidence exceeds the critical angle,(io > co) so Total Internal Reflection occurs and thebeam stays totally within the fibres over long distances.

(If the fibre is kinked at a really sharp angle this wontwork... the fibres need to bend around corners inreasonably gentle curves.)

Decorative optical fibres,showing how the light stays inside due to Total Internal Reflection.

Sin c = gna = 1 1 ang

This means that the Sine ratio of thecritical angle C is equal to the reciprocalof the refractive index of the glass.

If io > cothe ray cannot get out,but reflects back insidethe glass

Core.highindex

lowerindex

Laser bounces around cornersby total internal reflection

laser be

am

Optical fibre

iiorro

1

glass

air

iiorro

2

bigger i,bigger r

iio==ccorro== 9900o

3

Critical Angle

iio>>cco

RRaayyrreefflleeccttssiinnssiiddeeggllaassss

4

Photo: Anthony Bortolloto

Worksheet 4

Part A Fill in the blanks. Check answers at the back.

When a wave meets the boundary between one mediumand another, any of 3 things can occur: the waves energycan be absorbed, or the wave can bea).......................................... or ....................................................

The Law of Reflection simply states that the angle ofb)........................... equals the angle of c)..................................The angles must be measured from the wave ray to thed)......................................, which is an imaginary line which ise).................................................. to the boundary. Concavemirrors reflect light into a f).............................. point and canproduce enlarged images, such as in a reflecting telescope.A g)........................................... mirror reflects light outwards.This produces images which are h)........................................,but have a wider field of view. A practical use for thismirror is i)......................................................................................In communications, reflection is useful for long-distanceradio reception. Some radio wavelengths reflect from thej).................................................... layer in the upperatmosphere, and are bounced around the curvature ofthe Earth. Satellite dishes and k)........................................antennas use reflection to focus wave signals into thereceiver.

Refraction occurs when waves go from one medium intoanother. The waves may change in l)......................................,and ................................. and .......................................................For example, when light goes from air into glass its speedm)......................................, and its n)..........................................gets shorter (although o)............................................ does notchange) It also changes direction, goingp)......................................... the normal.

Snells Law describes the direct relationship between theSine ratios of the angles of q)...................................... and............................................ This ratio is called ther)......................................................... It is also equal to theratio between the s)........................................ of the wave inthe 2 different mediums. The index for the wave enteringthe medium, and the index for the wave exiting the mediumare always t)............................................ of each other.

When a light ray is going from a slower medium into afaster one, the ray will refract u).........................................the normal. As the angle of incidence increases, so will theangle of refraction, until the refracted rayv).................................................................of the boundary.The angle of incidence at which this happens is called thew)....................................... angle. At angles of incidencegreater than this angle, x)..................................................................................... occurs, and the ray stays within theslower medium. This property is used in optical fibretechnology to ensure that y).............................................beams stay within the fibres.

Part B Practice Problems

Snells Law Sin i = V1 = 1n2Sin r V2

Example ProblemA beam of light goes from air into a glass block with arefractive index of 1.50. The angle of incidence is 35o.a) Find the angle of refraction.b) If light travels in air at 3.00x108 ms-1, find the velocity inthe glass.

Solution: a) Sin i = n sin 35 / sin r = 1.50Sin r

sin r = sin 35 /1.50= 0.38238...

therefore, angle of refraction, r = 22.5o

b) V1 = n 3.00x108 / V2 = 1.50

V2V2 = 3.00x10

8 / 1.50therefore, velocity in glass, V = 2.00x108 ms-1

TRY THESE

1. In an experiment, a student sent a beam of light into ablock of clear plastic. The angle of incidence was measuredas 50o. The angle of refraction was 33o.a) Find the refractive index of the plastic.b) If light travels in air at 3.0x108ms-1, find its velocity inthe plastic.

2. Light travels through a diamond at only 1.25x108ms-1.a) Find the refractive index of diamond.b) If a ray of light strikes a diamond surface at an angle of40o from the normal, find the angle of refraction as the rayenters the diamond.

3. Using a laser beam and a fish tank filled with water, therefractive index of the water was found to be 1.33.a) At what incident angle must the beam strike the water toproduce an angle of refraction of 32.5o?b) At what velocity does the laser beam travel in water?

4. Several different angles of incidence and refraction weremeasured for a light ray going from air into a high densitycrystal glass block.a) For each pair of angles, calculate a refractive index value.b) Find the average value for the refractive index.c) Use the average value to find the velocity of light in thecrystal glass.DATA Angle Angle

Incidence Refraction50.0 25.042.0 21.030.0 17.065.0 31.0

continued...

20



Worksheet 4 Part B Practice Problems (continued)

5. Window glass has a refractive index of 1.50.a) Find the velocity of light in this glass.b) If a light ray strikes the glass surface at right angles (i.e. alongthe normal line) what is the value of the angle of incidence?c) Calculate the angle of refraction for this situation.d) How do you interpret this result?

Reciprocal Indices 1n2 = 1 2n1

6. Refer to the information and answers to Q1.a) What is the refractive index for light coming out of theplastic into air?b) If a light ray in the plastic struck the boundary at anangle of incidence of 20o, at what angle of refraction will itenter the air?

7. Refer to Q3.a) What is the refractive index for light travelling from waterinto air?b) If a light ray emerged from water into air at an angle ofrefraction of 37o, what must have been the angle ofincidence?

8. In a type of lead-crystal glass, a light ray exits from theglass into air. At the interface, the angles were i = 15o, andr = 25o.a) What is the refractive index glassnair?b) What is the index airnglass?c) At what velocity does light travel in this glass?

Critical Angle Sin c = gna = 1 ang

9. a) Use the information in Q2 to find the critical anglefor light travelling inside a diamond.b) Describe what would occur (no calculation required) iflight inside a diamond hit the boundary at an angle ofincidence of:

i) 20o ii) 30o

10. a) What is the critical angle for glass with ang = 1.50?b) Describe (no calculation) what happens when light insidethe glass strikes the boundary at angle of incidence:

i) 40o ii) 41.8o iii) 45o

11. Light travelling inside a plastic block strikes theboundary at an angle of incidence = 48.6o. The refractedray is seen to run exactly along the boundary betweenplastic and air.a) What is the critical angle?b) What is the value of anp?c) At what velocity does light travel in this plastic?

FULLY WORKED SOLUTIONS AREIN THE ANSWER SECTION

21





Digital TechnologyIn the past 20-30 years our society has become more andmore digitized. Because of the speed, storage capacityand processing ability of computers, almost every aspect ofour society has gone digital.

This simply means that all information (data) whether it bea persons voice, written words, numbers, music,photos, etc. is converted into digital code for processing,storage or transmission and communication.

A simple list of some of the technologies involved is:CDs & DVDs, Mobile phones, digital cameras, theinternet, MP3 music, ATMs, computers and theirnetworks.

Increasingly, WAVES are involved in these technologies,especially when data is moved around...COMMUNICATION.

GPS is a system that allows a ship, aircraft, car or even abushwalker, to locate their exact position anywhere onEarth instantly and continuously.

The system was developed for miltary uses, but then madeavailable to anyone. The military version is thought to beaccurate to within a few metres, the civilian version towithin about 50-100 metres.

The system is based on a fleet of satellites positioned inorbit so that from anywhere on Earth, at any moment, atleast 4 satellites are in line of sight.

Each satellite constantly sends out radio signals identifyingitself, and the exact time the signal was sent.

When your portable GPS receiver picks up the signal, it cancalculate your exact distance from the satellite, from thetime delay since the signal was sent.

By doing the same for 2 other satellites, the GPS unitrapidly triangulates the signals from 3 satellites to pin-point your location on the Earths surface. (Aircraft need a4th signal to get their altitude)

Cars can now be fitted with GPS systems that show yourposition on a screen, overlaid onto a road map of the area.As you drive around, the system constantly shows yourchanging position, and can advise you where to turn toreach a designated destination.

22



5. DIGITAL COMMUNICATION & DATA STORAGE

TECHNOLOGY CASE STUDY:

GLOBAL POSITIONING SYSTEM(GPS)

DIGITALTECHNOLOGIES

Satellite orbitsSSaatteelllliittee 11

SSaatteelllliittee 22

SSaatteelllliittee 33

GGPPSS rreecceeiivveerr

Earth

Photo by Pip

Photo by John de Boer

Photo by John de Boer

Computer Hard Drive



23

CONCEPT DIAGRAM (Mind Map) OF TOPICSome students find that memorizing the OUTLINE of a topic

helps them learn and remember the concepts and important facts.Practise on this blank version.

TTHHEE WWOORRLLDDCCOOMMMMUUNNIICCAATTEESS

Practice QuestionsThese are not intended to be "HSC style" questions, but tochallenge your basic knowledge and understanding of thetopic, and remind you of what you NEED to know at theK.I.S.S. principle level.

When you have confidently mastered this level, it is stronglyrecommended you work on questions from past exampapers.

Part A Multiple Choice

1. A sound wave is best described as:A. mechanical and transverse.B. electromagnetic and transverse.C. mechanical and longitudinal.D. electromagnetic and longitudinal.

2. Which measurement inthis diagram (A,B,C or D)correctly shows theamplitude of the wave?

3. In a Transverse wave, the particles of the medium:A. vibrate perpendicular to the direction of energy flow..B. move randomly in all directions.C. vibrate parallel to the direction of energy flow..D. move with the energy from one place to another.

4. If the period of a wave is 4 seconds, then its frequencyis:A. 0.25 HzB. 0.4 HzC. 4.0 HzD. 1/16 Hz

5. The period ofthis wave is:A. 0.8sB 1.6sC. 3 mmD. 6 mm

6. If a sound wave has a velocity of 330ms-1, and itsfrequency is 660Hz, then its wavelength must be:A. 990 m B. 2.0m C. 0.5m D. 330m

7. If you heard a sound wave with small amplitude and highfrequency, you would describe it as:A. low volume and low pitch.B. low volume and high pitchC. high volume (loud) and low pitchD. high volume and high pitch.

8. Two pulses are travelling towards each other in a rope.

When they meet at point X, :A. they will interfer destructively and cancel out.B. they will reflect off each other and bounce back.C. constructive interference will increase the amplitude.D. all wave motion will stop at point X.

9. The navigation of a bat in the dark, and the depthsounding from a boat, both work on the principle of:A. ReflectionB. Refraction.C. ResonanceD. Interference

10. Compared to visible light:A. Infra-red has shorter wavelength and lower frequency.B. Ultra-violet has shorter wavelength and lower frequency.C. X-rays have longer wavelength and lower frequency.D. Microwaves have longer wavelength & lower frequency.

11. The radiation from the Sun least likely to reach theEarths surface is:A. Infra-redB. visible lightC. Ultra-violetD. radio waves.

12. The brightness of a light viewed from 40 metres,compared to viewing from 10 metres would be:A. 1/4 as bright C. 1/16 as brightB. 4 times brighter D. 16 times brighter.

13. An example of REFLECTION being helpful incommunication is:A. Radio waves bouncing off the ionosphere.B. A convex dish antenna collects satellite TV signals.C. Using a concave shaving mirror.D. A convex side mirror on a car sees into the blind spot.

14. The diagrams show a carrier wave, and the modulatedwave carrying a signal or message. The method ofmodulation used is:A. AMB. FMC. PulseD. Digital.

15. Which of the following does NOT change when a waveundergoes refraction?A. VelocityB. DirectionC. WavelengthD. Frequency

24



A

C

D

B

disp

lace

men

t (m

m)

time(s)1.0

3

-3

X

Carrier wave

Modulatedwave

16. If a light ray passed from air into one of the followingsubstances, (each at the same angle of incidence) which onewould show the least amount of refraction?A. Water (refractive index = 1.3)B. Diamond (refractive index = 2.4)C. Glass (refractive index = 1.5)D. Perspex (refractive index = 1.4)

17. Light travels in air at 3.0x108ms-1. If the refractive indexof glass = 1.5, then the velocity of light within the glass is:A. 3.0 x108ms-1.B. 2.0 x108ms-1.C. 4.5 x108ms-1.D. 1.5 x108ms-1.

18. The refractive index of water = 1.33.The Critical Angle for water would be closest to:A. 38o B. 45o C. 49o D. 53o

19. Long distance communication using laser light andoptical fibres is made practical because of:A. Refraction inside the optical fibre.B. Reflection from the ionosphereC. Total internal reflection in the optical fibre.D. Focusing of the light by a concave mirror.

20. The Global Positioning System (GPS) works on:A. laser beams carried in optical fibres.B. radio signals from local cell transmitters.C. microwave beams focused by dish antennas.D. radio signals from several satellites.

Longer Response QuestionsMark values given are suggestions only, and are to give youan idea of how detailed an answer is appropriate.

21. (3 marks) List the energy transformations that occur from when youspeak into your mobile phone to when the message isreceived at the local cell receiver.

22. (4 marks) Differentiate between:a) mechanical and EMR waves.b) transverse and longitudinal waves.

23. (5 marks) A sound wave with frequency 400Hz travels through waterat 1,500 ms-1. Show working:a) calculate the wavelength.b) calculate the waves period.

24. (5 marks) The graph describes awave in the ocean.

a) What is thefrequency of the wave?Explain your answer.

b) Given that the wave travels at 12.5ms-1, find thewavelength. Show your working.

25. (3 marks) Use the Principle of Superposition to sketch theresultant of the 3 waves shown.

26. (4 marks) With a water wave, a crest is where water has displacedupwards, and a trough where it displaced downwards, asthe wave moves through.Explain, in similar terms, what happens to air particles as asound wave passes.

27. (3 marks) Re-arrange these members of the EMR spectrum, placingthem in order from lowest to highest frequency.Radio, infra-red, gamma, light, microwaves, X-ray, ultra-violet.

28. (3 marks) Identify a method for detecting each of these EMR types:(choose a different method for each)a) visible lightb) X-rayc) infra-red

29. (3 marks) A lighthouse is viewed from 10km and its light intensity(brightness) measured to be 0.1 units. How bright would itappear if viewed from 1 km? Explain your answer.

30. (3 marks) Discuss briefly a limitation on the use of EMR forcommunication.

25



1100

-3

00

33

2200time (sec)

disp

lace

men

t (m

)

disp

lace

men

t

time



31. (3 marks) Complete each diagram to show the expected path of eachreflected light ray P, Q and R.

32. (6 marks) In an experiment, angles of incidence and refraction weremeasured as shown.a) Find the refractive index of the plastic. Show working.b) At what speed does light travel in this plastic? Showworking.

26



PPQQ

RR

air plastic



33. (4 marks) Predict the path of this light ray after it strikes theboundary. Explain your reasoning, and show any working.

34. (3 marks)Outline briefly the underlying principles used in oneapplication of physics related to waves.

airplastic.

n = 1.40



27

Answers Worksheet 1 Part B (continued)

7. 2.50 m wave: f = V / = 3x108 / 2.50 = 1.20x108 Hz

2.50 cm wave: f = V / = 3x108 / 0.0250 = 1.20x1010 Hz

comparison: The frequency of the 2.5cm wave is 100times higher than the 2.5m wave. (Makes sense: 100Xshorter wavelength --> 100X higher frequency)

8. Since 8 complete wavelengths fit in 0.96m, then = 0.12 mV= f = 384 x 0.12 = 46 ms-1

9. a) f = V / = 3.00x108 / 1.50x10-11= 2.00 x 1019 Hz

b) T = 1 / f = 1 / 2x1019 = 5.00x10-20 s.

Part C Wave Graph Exercise

wave P Q Ri) A = 0.15 m 0.10 m 0.05 mii) d= 0.1 m -0.1 m -0.05 miii)T= 0.08 s 0.16 s 0.04 siv) f= 12.5 Hz 6.25 Hz 25 Hz

v) V= f = 12.5 x 0.50 = 6.25 ms-1vii) = V / fwave Q: = 9.5 / 6.25 wave R; =9.5 / 25

= 1.5 m = 0.38 m

Worksheet 2Part Aa) mechanical b) longitudinalc) compressions d) rarefactionse) 330 f) higherg) frequency h) loudness / volumei) reflect j) navigation / hunting preyk) SONAR l) detecting fish / submarinesm) resultant n) amplitudes

Part B Superposition Graph ExerciseUsing the sum of displacements at the circled points, theresultant looks approximately like this:Note that the amplitude of the resultant starts larger andbecomes smaller. You would hear the sound volumedecrease.

Answer SectionWorksheet 1

Part Aa) energy b) matterc) medium d) sound & water wavese) vacuum f) radio, light, UV, etcg) Transverse h) at right anglesi) in the same line j) Wavelengthk) maximum displacement, from equilibrium positionl) waves / complete vibrationsm) Hertz (Hz) n) The Periodo) reciprocal p) frequencyq) wavelength r) amplitudes) period t) frequencyu) decrease v) wavelength

Part B Problems

1. a) V= f b) T = 1 / f= 280 x 5.20 = 1 / 280= 1460 m. = 0.00357 s

(3.57 x10-3 s)

2. V= f 2,500 = 0.400 x

= 2,500 / 0.400= 6,250 m (6.25 x103 m)

3. V= f 330 = 1200 x

= 330 / 1200= 0.275 m (2.75 x 10-1m)

4 a) f = 1 / T f = 1 / 16 = 6.25x10-2 Hzb) V= f

6.50= 6.25x10-2 x = 6.50 / 6.25x10-2 = 104 m

c) V= f 2.20 = 6.25x10-2 x

= 2.20 / 6.25x10-2 = 35.2 mWavelength has become a lot shorter as the wave enteredshallower water.

5. a) V= f , f = V / = 3.00x108 / 7.00x10-7

= 4.29x1014 Hzb) f = V /

= 3.00x108 / 3.00x10-7

= 1.00x1015 Hz

6.a) T= 1 / f = 1/ 53,000 =1.89x10-5 s.b) = V / f

= 3.00x108 / 53,000 =5.66x103 m.(over 5 km!)



28

Worksheet 3Part Aa) transverse b) do notc) medium d) 3 x 108e) radio f) microwavesg) infra-red h) visible lighti) ultra-violet j) X-rayk) gamma rays l) vibrate / oscillatem) resonance n) vibrateo) UV / X-rays / gamma p) ozoneq) UV r) square of distances) one quarter t) radio & microwavesu) laser beams v) opticalw) modulation x) Amplitude Modulationy) amplitude z) Frequency Modulationaa) frequency ab) Pulseac) electrical ad) radioae) microwave af) radioag) electrical

Part B Inverse Square Law Problems1. a) 12 units (1/4 as bright)

b) 3 units (1/16)c) 0.48 units (1/100)d) 192 units (4x brighter)

2. Moved to 1/4 the distance, therefore 16X more intense.

3. Intensity dropped from 32 units to 2 units.... 1/16.Therefore must be 4X further away.... answer = 8 m.

4. At double distance, intensity = 1/4.Therefore need to turn on 4 identical bulbs to get sameamount of light.

Worksheet 4Part Aa) reflected or refracted b) incidencec) reflection d) normale) perpendicular f) focalg) convex h) smaller/diminishedi) drivers side mirror j) ionospherek) microwavel) direction, wavelength & velocitym) slows down n) wavelengtho) frequency p) towardsq) incidence & refraction r) refractive indexs) velocities t) reciprocalsu) away from v) goes along the edgew) critical x) Total Internal Reflectiony) laser

Part B Snells Law Problems1.a) n = Sin i / Sin r = sin50 / sin 33 n=1.4 (no units)b) n= V1 / V2

1.4 = 3.0x108 / V2V2= 3.0x10

8 / 1.4

= 2.1 x 108 ms-1

Worksheet 4 Answers Part B (cont)2. a) n=V1 / V2 = 3.00x10

8 / 1.25x108 = 2.40b) n = Sin i / Sin r

2.40 = sin40 / Sin r Sin r = sin40 / 2.40 = 0.2678...

r = 15.5o

3. a) n = Sin i / Sin r1.33 = Sin i / sin32.5Sin i = 1.33 x sin32.5 = 0.7146...

i = 45.6o

b) n = V1 / V21.33= 3.00x108 / V2V2 = 3.00x10

8 / 1.33 = 2.26x108 ms-1.

4. a) Angle Angle RefractiveIncidence Refraction Index50.0 25.0 1.8142.0 21.0 1.8730.0 17.0 1.7165.0 31.0 1.76

b) Average =7.15/4 = 1.79c) n = V1 / V2

1.79 = 3.00x108 / V2V2 = 3.00x10

8 / 1.79 = 1.68x108 ms-1.

5. a) n = V1 / V21.50 = 3.00x108 / V2

V2 = 3.00x108 / 1.50 = 2.00x108 ms-1.

b) 0oc) n = Sin i / Sin r

1.50 = sin0 / Sin r Sin r = sin0 / 1.50 = 0

r = 0od) The ray does NOT change direction.(However, it would still change velocity and wavelength)

Reciprocal Indices Problems6. a) 1n2 = 1 = 1 / 1.4 = 0.71 (no units)

2n1b) n = Sin i / Sin r (must use the index for

0.71= sin20 / Sin r plastic --> air)Sin r = sin20 / 0.71 = 0.4817...

r = 29o

7. a) 1n2 = 1 / 2n1 = 1 / 1.33 = 0.752 (no units)b) n = Sin i / Sin r (must use the index for0.752 = Sin i / sin37 water --> air)Sin i = 0.752 x sin37 = 0.4525... So, i = 27o



29

Worksheet 4 Part B (cont)8. a) n = Sin i / Sin r

= sin15 / sin25n= 0.61 (for glass --> air)

b) n (air ->glass) = 1 / 0.61 = 1.6 (no units)c) n = V1 / V2

1.6 = 3.0x108 / V2V2 = 3.0x10

8 / 1.6

V(glass) = 1.9x108 ms-1

Critical Angle & T.I.R.

9. a) Sin c = dna = 1 and

Sin c = 1 / 2.40 = 4.1666....c = 24.6o

b) i) Would refract out into the air.ii) i > c, so ray would undergo total internal reflection

back inside the diamond.

10. a) Sin c = 1 / n = 1 / 1.50 = 0.6666...c = 41.8o

b) i) Refracts out into airii) Refracts and runs along the glass edge

iii) i > c, total internal reflection.

11. a) 48.6o (by definition of what happens at crit. angle)b) Sin c = 1 / n

n= 1 / Sin c = 1 /sin48.6 = 1/0.7501...n = 1.33

c) n = V1 / V21.33= 3.00x108 / V2

V2= 3.00x108 / 1.33 = 2.26x108 ms-1.

Practice Questions

Part A Multiple Choice1. C 5. B 9. A 13. A 17. B2. D 6. C 10. D 14. B 18. C3. A 7. B 11. C 15. D 19. C4. A 8. C 12. C 16. A 20. D

Part B In some cases there may be more than one correctanswer. The following model answers are correctbut not necessarily perfect.

21. SOUND --> ELECTRICAL --> RADIO

22.a) Mechanical waves require a medium substance totravel in. EMR waves do not, and so can travel in vacuum.b) Transverse waves vibrate at right angles to the directionof energy movement. In a longitudinal wave, the vibrationis back-and-forth in the same direction as the energy flow.

23.a) V = f

1,500 = 400 x = 1,500 / 400 = 3.75 m.

b) T = 1/f = 1/400 = 0.0025 = 2.5x10-3 Hz

24.a) from graph, T = 16 s. and f = 1/T

= 1/16 = 6.25x10-2 Hzb) V = f

12.5 = 6.25x10-2 x = 12.5 / 6.25x10-2 = 200 m.

25. summing displacements at circled points:

26.With a sound wave, a compression is where air particlesare pushed together, and a rarefaction is where they arespread apart more, as the wave moves through.

27.(lowest) Radio, microwaves, infra-red, light, ultra-violet, X-ray, gamma (highest)

28.a) human eye. b) X-ray sensitive photo film.c) receptors in human skin.

29.At 1/10 the distance it will be 100X brighter.

0.1 x 100 = 10 units.

30.There are so many radio & TV stations, mobile phonesystems and users, as well as 2-way radio for aircraft,shipping, military, taxis, etc, that the available radio bandsare becoming congested.

There is a limit to how many systems and stations canoperate without overlapping in frequencies and causinginterference to each other.

resultant(approx)



30

31.

32.a) n = Sin i / Sin r

= sin33 / sin25= 1.3 (no units)

b) n = V1 / V2 (given velocity in air =3.0x108)

1.3= 3.0x108 / V2V2= 3.0x10

8/1.3

V (in plastic)= 2.3x108 ms-1.

33.Note that the angle given is not the correct angle ofincidence. (Angle of incidence must be measured from thenormal)So, i = 60o

Next, check if io is greater than co:Sin c = 1 / n = 1 / 1.4 = 0.7142......

So, c = 46o.Therefore, i > cso ray will undergo total internal reflection, and reflect backinside the plastic block. (at angle of reflection = 60o)

34.Underlying principles of the Global Positioning System:GPS involves a small, portable receiver picking up radiosignals from a fleet of satellites in orbit. Each satellite sendsa coded message identifying itself and the precise time thatthe signal was sent.

By comparing the signals from (at least) 3 differentsatellites, the GPS receiver can triangulate to find itsposition with a high degree of accuracy... within a fewmetres in some cases.

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