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keep it simple science®

1

but first, let’s 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 energyinclude:

• HEAT• ELECTRICITY• KINETIC (energy in a moving object)• POTENTIAL (energy stored, such as

the chemical energy in petrol).

Some forms of energy move around asWAVES. A wave is a carrier of energy. Ina wave, energy moves, but matter does not.

TYPES of WAVESExamples of energy which moves around aswaves include

• SOUND

• LIGHT

• RADIO SIGNALS

• WATER WAVES

• X-RAYS

• MICROWAVES

... and many more

ENERGYCONVERSIONSEnergy can be converted from one form toanother.

In your mobile phone the SOUND WAVES ofyour voice are converted to ELECTRICALsignals then transmitted as RADIO WAVES toyour friend, whose phone converts it backagain.

SOUND ELECTRICAL RADIO

In this topic you will learnabout waves and their

properties and features, andhow they they are used for

communication.

The strings vibrate.

This causes the air tovibrate too.

Waves of vibrationspread out through

the air... sound waves.

The air vibrates, butdoes not go anywhere.

Waves Carry EnergyWithout the Transfer of Matter

Water waves carryenergy across thesurface of a pond.

The water vibrates up& down, but goes

nowhere.




2

Wavelength,Amplitude,

Frequency &Period

Light & Mirrors.Reflection in

Communication

Typesof

WavesGraphing

WavesVelocity, Pitch

& Loudness

WaveEquation Nature of

Sound Waves

Principleof

Superposition

Production,Detection &

Dangersof EM Waves

The EMSpectrum

InverseSquare

Law

Refraction.Snell’s Law,

Lenses &Total Internal

Reflection

EM Waves in Communication

THE WORLDCOMMUNICATES

The Natureof Waves

Properties ofSound Waves

ElectromagneticWaves

Reflection&

Refraction

DigitalCommunication

&Data Storage

CONCEPT DIAGRAM (“Mind Map”) OF TOPICSome students find that memorising the OUTLINE of a topic helps them learn and remember the concepts and important facts. As you proceed through the topic,

come back to this page regularly to see how each bit fits the whole. At the end of the notes you will find a blank version of this “Mind Map” to practise on.

Law ofReflection




3

Waves Carry EnergyWaves carry energy, without the transfer ofmatter.

This can occur in 1 dimension:

... or in 2 dimensions:

Ripples spreadingon the surface of apond.

...or in 3 dimensions,

such as when light radiates in all directionsfrom a glowing object.

Waves & MediumsMechanical waves are those which need a“medium” to travel through. For example, awater wave must have water to travel in. Soundwaves need air, or water, or some substance tomove in. They CANNOT travel in a vacuum.

Electromagnetic (EM) waves do NOT need amedium... they can travel through a vacuum,and in fact travel fastest in a vacuum. EM wavesinclude light, radio waves, ultra-violet and othertypes, and are studied in detail in a latersection.

Describing WavesA wave is a vibration. In a mechanical wave, the“particles” (atoms & molecules) in the mediumvibrate to transmit the wave energy. In EMwaves the vibration occurs in electric andmagnetic fields.

Consider a wave in a rope which has been givena single up-and-down “twitch”:

Energy moves along the rope, but the rope itselfdoesn’t go anywhere. Particles of the “medium”(the rope fibres) vibrate up-and-down as theenergy moves across.

This form of a wave, where the medium vibratesat right angles to the direction that the energymoves, is called a Transverse wave.

If the rope is wiggled constantly up-and-down,you get not just one pulse, but a periodic wavewith one pulse following another.

1. THE NATURE OF WAVES

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CCoommpprreesssseedd sseeccttiioonnss iinn tthhee sspprriinngg mmoovvee aalloonngg iitt lliikkee aa ““MMeexxiiccaannWWaavvee””...... eenneerrggyy iiss ttrraannssffeerrrreedd,, bbuutt tthhee ccooiillss mmeerreellyy oosscciillllaattee bbaacckk

aanndd ffoorrtthh aanndd ddoo nnoott aaccttuuaallllyy ggoo aannyywwhheerree..CCRREESSTTAA PPUULLSSEE WWAAVVEE ppaarrtt ooff tthhee rrooppee ((mmeeddiiuumm))

vviibbrraatteess uupp && ddoowwnn

TTRROOUUGGHH

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rrooppee

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TTRROOUUGGHH

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RRooppee vviibbrraatteess uupp aanndd ddoowwnn

MECHANICAL WAVESrequire a medium to travel through.

PERIODIC WAVEScontain a series of pulses, with a continuous

set of crests and troughs.

TRANSVERSE WAVESvibrate at right angles to the direction

that the energy is moving.

Energy flow

Vibration in medium




4

Longitudinal waves are when theparticles of the medium vibrate back-and-forthin the same line as the energy moves. Forexample, when a series of “compressions” and“rarefactions” are sent along a slinky spring.

LONGITUDINAL WAVES

The vibration of the medium is in the same direction

as the energy flow.

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

The Greek letter “lambda” λλis used as a symbol for wavelength.

Amplitude (a or A) = the distance that aparticle in the medium is displaced from its “restposition” at a crest or trough. i.e. the maximumdisplacement distance.

Frequency (f) = the rate at which the wave isvibrating. The number of waves that pass agiven point in 1 second, or the number ofcomplete vibrations per second.

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

Wave MeasurementsAll periodic waves,

whether Longitudinal or Transverse, Mechanical or Electromagnetic, can be described and measured by their:-

Period (T) = the time (in seconds) for onecomplete vibration to occur.

Note that there is a simple relationship betweenFrequency and Period... they are reciprocals.

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

There is a simple relationship between Velocity,Wavelength and Frequency:

Velocity = Frequency x Wavelength

THE WAVE EQUATION

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T = 1 and f = 1 f T

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rraarreeffaaccttiioonn((wwhheerree sspprriinngg iiss

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WORKSHEET at end of section

Earthquake Shock Waves




5

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

Imagine a floating cork bobbing up and down asa series of ripples move across the watersurface (i.e. a periodic wave).

If you graph the (up-down) displacement of thecork against time, the graph will look somethinglike this:

Be careful! The graph is shaped like a wave, soit’s tempting to try to read the wavelength fromthe horizontal scale... but the horizontal scale isTIME, not length.

What you CAN read from aDisplacement-Time graph:

Amplitude The vertical scale measures thedisplacement of the cork from the “equilibrium”position (i.e. the flat water surface).So,

at 0 sec, the cork was in the equilibrium position.at 0.2 sec, it was 3cm upwards... at 0.4 sec, it was back at equilibrium...

and so on.Its maximum displacement was 3cm eitherabove or below (d= -3cm) equilibrium, so theAmplitude = 3cm (0.03m)

Period Since the horizontal scale istime, you can easily read from the graph howlong it takes for one complete up-and-downcycle. 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 youcould calculate the wavelength, or vice versa.

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

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

Graphing a Longitudinal WaveYou might think these Displacement-Timegraphs wouldn’t work for a Longitudinal wavewhere the particles vibrate back-and-forth ratherthan up-and-down.

However, the graph of a longitudinal wave canbe exactly the same... you just have to realisethat the “displacement” is sidewaysdisplacement from the “equilibrium position”,instead of up-down.

Amplitude, Period and Frequency can all bedetermined in exactly the same way.

Relationship BetweenWavelength & Frequency

You may have carried out a “First HandInvestigation” in class to see how a change inFrequency (at constant velocity) affects thewavelength. Maybe you used a slinky spring, orwatched 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)

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RRiipppplleess

00..22 00..66 11..00 11..22

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To have the same speed, the shorter waves must vibrate at ahigher frequency

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TTiimmee ((ss))



Waves carry a)................................... without thetransfer of b)................................ “Mechanical”waves require a c)............................. to travel in.Examples are d)...................... and............................ “Electromagnetic” waves donot need a medium and can travel in ae)....................... Examples include f)....................and ................................

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

j)................................. is the distance from crestto crest.

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

6



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

n).............................. is the time for one completevibration. This is the o).................................... offrequency.

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

On the graph of a wave, showing Displacementv Time, 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.

Worksheet 1 The Nature of WavesFill in the blank spaces. Student Name...........................................

Example Problem 1A water wave in the ocean has a wavelength of85m, and a velocity of 4.5ms-1.a) Find the frequency. b) What is the period?Solutiona) V = f λλ

4.5 = f x 85f = 4.5 / 85

= 0.053 Hz (5.3 x 10-2 Hz)(i.e. only a small 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 of330ms-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 λλ330 = 500 x λλ

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

TRY THESE1. a) Find the velocity of a sound wave in waterif it vibrates 280 times per second and has awavelength of 5.20m.

b)What is the period of this wave?

2. An earthquake shockwave travels throughrock at a velocity of 2,500 ms-1. Its frequency is0.400 Hz. What is the wavelength?

3. What is the wavelength of a sound wave withfrequency 1200Hz? Sound travels in air at330ms-1.

4. An ocean water wave in deep water travels ata velocity of 6.50ms-1. Its period is 16.0s.a) What is the frequency?

b) Wavelength?

c) As the wave enters shallower water it keepsthe same frequency but slows down to only2.20 ms-1. What happens to the wavelength?

Worksheet 2 Practice ProblemsWave Equation 1 Student Name...........................................


1. a) Red light has a wavelength of 7.00x10-7m,and travels at 3.00x108ms-1. What is thefrequency?

b) Blue light has a wavelength of 3.00x10-7mand travels at the same speed. What is thefrequency?

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

b) What is the wavelength?

3. Compare the frequency of a radio wave 2.50mlong, with one 2.50cm long.(Assume they both travel at the speed of light)

The 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=0.03siii) What is the Period of each wave?iv) What is the Frequency of each wave?v) Given that wave “P” has a wavelength of 0.50m, calculate its velocity.vi) Waves “Q” & “R” both travel with a velocity of 9.5ms-1. Find their wavelengths.

7



Remember that for full marksin calculations, you need to show

FORMULA, NUMERICAL SUBSTITUTION,APPROPRIATE PRECISION and UNITS

4. When a guitar string is plucked, a wavevibration runs back and forth through thestring. The string is 0.96m long and it is foundthat exactly 8 complete wavelengths fit alongthe string at a time. The vibration frequency is384Hz. How fast do the waves travel throughthe string?

5. X-rays are very short wavelength EM waveswhich travel at the speed of light. If thewavelength is 1.50x10-11 metre,a) find the frequency.

b What is the period of the X-rays?

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Time (s)

Dis

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)Worksheet 3 Practice ProblemsMore Wave Equation Student Name...........................................

Worksheet 4 Practice ProblemsReading Wave Graphs Answer on reverse Student Name...........................................

Multiple Choice1. 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 measurementin this diagram (A,B,C or D) correctly shows the“amplitude” of thewave?

3. In a Transverse wave, the particles of themedium: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 itsfrequency is: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 mustbe:A. 990 m B. 2.0m C. 0.5m D. 330m

Longer Response QuestionsMark values shown are suggestions only, and are togive you an idea of how detailed an answer isappropriate. Answer on reverse if insufficient space.

7. (3 marks) List the energy transformations that occur fromwhen you speak into your mobile phone to whenthe message is received at the local “cell”receiver.

8. (4 marks) Differentiate between:a) mechanical and EM waves.

b) transverse and longitudinal waves.

9. (5 marks) A sound wave with frequency 400Hz travelsthrough water at 1,500 ms-1. Show working:a) calculate the wavelength.

b) calculate the wave’s period.

10. (5 marks) The graphdescribes awave in theocean.

a) What is thefrequency ofthe wave?Explain your answer.

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

8




Worksheet 5 Test Questions section 1 Student Name...........................................

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Sound WavesSound waves are Mechanical (they need a medium)

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

SOUND Energy movesWAVES

Particles vibrate

Instead of crests and troughs, a series of“compressions” and “rarefactions” passthrough the medium as a sound travels. Theatoms and molecules are alternately “squashedtogether” and then stretched apart as theenergy flows through.

In a compression the air pressure is higher, andlower in a rarefaction.

The back-and-forth vibration of the mediumproduces a typical wave shape if graphed.

Velocity of SoundSound travels at different speeds in differentmediums.

In air, sound travels at about 330-350 ms-1,(about 1,200 km/hr) depending on temperatureand density.

The denser the air, the slower the speed ofsound.

In liquids and solids, sound travels muchfaster...

...about 1,500ms-1 in water

...about 5,000ms-1 in most metals.

FREQUENCY = “PITCH”When you hear sounds of different “pitch” thatis the way your brain interprets sound waves ofdifferent frequency.

Low Frequency = Low PitchHigh Frequency = High Pitch

AMPLITUDE = LOUDNESS or VOLUME

Sound waves with different amplitudes areinterpreted by your brain as sounds of differentloudness or volume.

Larger Amplitude = Louder SoundSmaller Amplitude = Quieter Sound

9

2. THE PROPERTIES OF SOUND WAVES

Sound Travels

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

Some animals can send out sound waves andpick up the echoes to help locate their prey, orto navigate, in environments where they can’tsee very well, such as murky water (dolphin), orin darkness (bat).

““SSqquueeaakkss”” ooff ssoouunndd

EEcchhooeess ffrroomm iinnsseecctt

SONAR SOund Navigation And Ranging

BAT

Humans have invented SONAR technologiesfor things such as “depth sounding” anddetecting underwater objects... fish orsubmarines, it all works the same way.

USES OFSONAR The time delay between sending a

sound “ping’ and receiving the echo,gives depth and distance

AAnnttii-ssuubbmmaarriinneeWWaarrffaarree

DDeepptthhSSoouunnddiinngg




10

However, if the waves are “out of phase” (forexample, if compression coincided withrarefaction) then there is destructiveinterference... the opposite amplitudes maycancel each other out.

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

In practice, this only happens over shortdistances or time periods to give “interferencepatterns” and “beat sounds”.

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wave BDis

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t

Surfing TriviaTechnically, a breaking wave is not a wave at all.

Once it breaks, the water begins moving forward (whichallows you to catch it) and so both energy and matter are

flowing forward... this is NOT a wave!True water waves are the “swells” which you cannot catch.

Resultant

The Principle of SuperpositionAll waves have the ability to pass through otherwaves without being affected. For example, youcould shine a red spotlight across a beam of bluelight, and each colour and beam will emerge on theother side exactly the same.

However, for the instant that the 2 waves aresuperimposed upon each other, they do interactand “interfer” with each other.

Very simply, the displacement of the two wavesadd together at every point where the wavescoincide.

In this case, the waves A&B were “in phase” (crestco-incided with crest, trough with trough) so theresult was constructive interference... the resultanthas an amplitude which is the sum of A+B.

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ppooiinnttss ((cciirrcclleedd))

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“resultant” A+B

wave Awave B

IF THESE ARE SOUND WAVES, DESCRIBE WHAT YOU WOULD HEAR.

Find the resultant of these 2 waves by adding the displacements at the circled points, then join the “sum” points with an even curve.

Worksheet 6 Superposition of Waves Student Name...........................................

-vve

Dis

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+ve




11

Sound waves are a)................................. andb).................................... A sound wave consistsof a series of high pressure c)................................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 solids such asmetals.

The “pitch” of a sound is related to theg)............................... of the wave. The amplitudeof the wave determines the h).............................of the sound we hear.

Worksheet 7 Sound WavesFill in the blank spaces. Student Name...........................................

“Echoes” occur when soundsi).................................. Some animals use echoesfor j)................................ Humans use thetechnology of k)........................ for “depthsounding” and l)..............................................

When 2 or more waves coincide, they willinterfere with each other. Them)............................ wave can be found byadding together the separate waven)..............................................

COMPLETED WORKSHEETSBECOME SECTION SUMMARIES

Multiple Choice1. If you heard a sound wave with small amplitudeand high frequency, 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.

2. Two pulses are travelling towards each otherin 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.

3. The navigation of a bat in the dark, and the“depth sounding” from a boat, both work on theprinciple of:A. ReflectionB. Refraction.C. ResonanceD. Interference


4. (3 marks) Use the “Principle of Superposition” to sketchthe resultant of the 3 waves shown.

5. (4 marks) With a water wave, a “crest” is where water hasdisplaced upwards, and a “trough” where itdisplaced downwards, as the wave movesthrough.Explain, in similar terms, what happens to airparticles as a sound wave passes.


X

disp

lace

men

t

time

EM WavesElectromagnetic waves are Transverse waveswhich do NOT require a medium to travelthrough. They travel through a vacuum at3.00x108ms-1, the “speed of light”. They cantravel through many other substances atslightly slower speed. For example, light cantravel through glass or water at speeds ofaround 2.5x108ms-1. In air, the speed is so closeto the speed in a vacuum that, for simplicity,(K.I.S.S. Principle) we take it to be the same.

EM radiation does not require a medium becausethe waves propagate as vibrations of electric andmagnetic fields, not as vibrating particles.

MEMBERS OF THE EM 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 differenttypes of radiation, you must realise that they arereally all the same thing, just at differentwavelengths and frequencies.

Production of EM WavesAll EM waves are produced in basically thesame way: vibration or oscillation of electricallycharged particles.For example....Radio waves are produced by electric currentsrunning back-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” downfrom a higher to a lower orbit around an atom.

Gamma waves come from the vibrations ofcharged particles within an atomic nucleus,during a nuclear reaction in the atom.

Detection & Reception of EM Waves

Just as all EM waves are produced in the samebasic way, they are all received or detected inthe same basic way too... by a phenomenoncalled “Resonance”. When waves strikesomething and are absorbed, they may cause“sympathetic” vibrations within it.

In cartoons and the movies (not in real life) theopera singer hits a high note and all the wineglasses begin to vibrate and then shatter... afictional example of resonance.

Some real examples...

When radio waves hit asuitable aerial wire orantenna, they cause someelectrons in the metal tooscillate back-and-forth “insympathy” with the wave.

These oscillations areamplified electronically andthe signal converted tosound in the speaker,allowing you to listen to theradio.

When infra-red waves hityour skin they cause

certain molecules to beginto resonate and vibrate.

This sets off nervemessages to the brain

and you feel warmth orheat on your skin.

In a film camera the light causes resonance inchemicals in the film. Chemical reactions occurwhich permanently alter the film so that animage appears when “developed” later. Different film canbe sensitive toinfra-red, (photosin the dark) or X-rays for medicaluses.

12

3. ELECTROMAGNETIC WAVESkeep it simple science

®



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Antenna


13

A little UV gives you a suntan, but long-termexposure leads to skin damage, premature skin“ageing”, and is a major cause of deadlymelanoma skin cancer.

The Sun produces dangerous quantities of UVradiation, but luckily most of it is absorbed bythe “ozone layer” in the upper atmosphere ofthe Earth.



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Danger of High Frequency EM WavesHigh frequency EM waves (ultra-violet, X-ray & gamma) can be very dangerous to living things.

“Ozone” is a formof oxygen whichhas 3 atoms permolecule (O3)instead of thenormal 2 (O2).

The ozone moleculesresonate well at the frequency of UV and so absorb it strongly.

The Sun only produces small amounts of theeven more dangerous X-rays and gammaradiation. Once again, most is absorbed in theupper atmosphere, this time by ordinary oxygenand nitrogen gases.

Infra-red and light radiation penetrate well,(although about 30% is reflected) and whilesome radio frequencies get through, many getabsorbed or reflected.

UV RaysOxygen O22

does notAbsorb UV

Ozone O33Absorbs

UV

The Inverse Square LawAs any form of radiation spreads out from its source its intensity gets less.

For example, a sound becomes quieter if you’re further from the source, or a light is not so bright as you move further from it.

At distance “d” from the light source, some lightenergy falls on an area of x2 units. At twice thatdistance (2d) the same amount of light would fallon an area of 4x2. The brightness of the lightmust be only 1/4 as much (since the sameamount of light is falling on 4 times the area.)

So,twice the distance 1/4 as bright

3 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 brighter.at 1/3 the distance, 9 times brighter

...and so on.

Notice how the brightness (intensity) changes inproportion to the distance squared, in eachcase.

Mathematically, the relationship is that theintensity (I) (such as brightness of light) isinversely proportional to the SQUARE of thedistance (d²) from which it is viewed.

This diagram explains why:

Intensity αα 1 (distance)2

I αα 1 d2

“αα” means“proportional

to”

xx

lliigghhttssoouurrccee

ddiissttaannccee ““dd””

ddiisstaannccee ““22dd””

22xxSSqquuaarree

AArreeaa xx2

SSqquuaarree wwiitthhssiiddeess ttwwiiccee aasslloonngg..

AArreeaa == 44xx2

SSaammee aammoouunnttooff lliigghhtt ffaallllssoonn 44 ttiimmeesstthhee aarreeaa WORKSHEET at end of section

14

Radio & Microwaves carry radio and TVbroadcasts, telephone long-distance links,mobile phone networks, and satellite links fortelephone (including internet) and TV.

If you have “Satellite TV”, the “dish” on yourroof is an antenna to receive microwavesdirectly from an orbiting satellite.

Add to that, 2-way radio for military uses, CBamateurs and boating, shipping and aircraftcommunications, and you begin to realise howmany radio waves are zapping around.

Lightis being increasingly used in the form of LASERbeams carried in optical fibres for telephoneand internet communication.

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 3 common ways to modulate the wave to carry information...




What’s special about LASER LIGHT?

•It is one, pure frequency of light.

•The waves are all in phase andso they 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 it’s perfect forhigh speed digitalcommunication.

Frequency Modulation (FM)

The amplitude stays constant whilethe frequency (and wavelength)vary within a fixed range. The

information (voice, music etc) is“coded” in the variations of

frequency.

FM radio gives much better fidelityand is superior, compared to AM,

for the quality of sound (eg for music) received.

Amplitude Modulation (AM)The frequency (and wavelength) ofthe wave stays constant while the

amplitude varies.

The changing amplitude “codes for”the information being carried...

whether voice or music, or whatever.

““CCaarrrriieerrwwaavvee””

AAMMssiiggnnaall

FFMMssiiggnnaall

DDiiggiittaallssiiggnnaall

DDiiggiittaall 11 00 11 11 00 11ddaattaa

NNooiinnffoorrmmaattiioonnccaarrrriieedd

AAmmpplliittuuddeecchhaannggeess..FFrreeqquueennccyyccoonnssttaanntt

FFrreeqq.. cchhaannggeess..AAmmpp..ccoonnssttaanntt

WWaavvee ppuullsseessoonn aanndd ooffff

This diagram compares the effect of AM, FM & Digital Modulation

on the same “carrier wave”

WAVEMODULATION

EM Waves & CommunicationHumans rely on sound waves for communicating by direct speech, but all our modern communication technologies rely on EM waves.

Pulse Modulation(Digital)

To carry information indigital form the wave mustswitch rapidly between 2

different states,representing the “1” and“0” of digital codes. The

wave can be switchedrapidly on and off (as inthe diagram) or switchedback-and-forth between

different “phase states”...phase modulation.

OpticalFibres

carry PulseModulated

laserbeams




15

Case Study: MOBILE (CELL) PHONESWhen you use a mobile phone, the sound of your

voice goes into a microphone and almost instantly pops out the other end

into your friend’s ear. What happens in between?

In the future we will need to switch morecommunications to use the laser light / opticalfibre method wherever possible, and to makebetter use of the RF bands. For example, it ispossible to use the same frequency “channel”for several different purposes as long as thedifferent signals are modulated differently andas long as the radio receivers aresophisticated enough to pick out only thedesired signal and ignore the others.

One thing is for sure... humans will keepcommunicating and the need for new serviceswill keep expanding. So far, our technologyhas always managed to keep up, and it willprobably continue to do so.

Modern communication systems havedeveloped rapidly and new features andcapabilities seem to come out every day. Itseems that the entire system is unlimited andthat it can continue to expand and improveforever.

Well perhaps it can, but NOT while continuingto use the radio end of the EMR spectrum.Each “station” or channel must operate on adifferent frequency or else signals can “jam”or “interfere” with each other.

The simple fact is that there are now so manyradio & TV stations, mobile phone networks,aircraft and shipping channels, military, policeand emergency service channels, etc. etc. allusing the RF (Radio Frequency) part of theEMR spectrum, that it is becoming difficult tokeep expanding services without interferingwith existing channels.

1. The SOUND energy ofyour voice is converted toELECTRICAL signals by

the microphone. Theelectrical signal is used to

digitally modulate aRADIO wave.

2. The digital RADIOsignal is

transmitted by yourphone and receivedby the local “cell”

antenna.

3. If your call is goingto a person in another

location (a different“cell”) the signal is

converted into amodulated

MICROWAVE andbeamed, via hilltoprelay towers, to the

correct area.(Alternatively, it might

be sent as amodulated Laser

LIGHT beam throughoptical fibres).

4. In the othercell area, the

signal isconvertedback to a

modulatedRADIO signal

andtransmitted.

5. Your friend’sphone receives the

RADIO signal,amplifies it as anELECTRICAL

signal and this isconverted to

SOUND waves intheir earphone.

SOUND ELECTRICITY RADIO MICROWAVE RADIO SOUND(or LASER LIGHT)

ENERGYY CHANGES

Discussion:LIMITATIONS OF COMMUNICATION CHANNELS

Electromagnetic waves area)................................... waves which b)...............(do/do not) require a c)............................ to travelin. They all move at the “speed of light”, whichis d)................... ms-1 in a vacuum.

The members of the EM spectrum (in order ofincreasing frequency) are:e)......................, f)......................., g)......................,h)..................., i)......................., j)..........................., and k).........................

All EM waves are produced when electricalcharges l)......................... They are alldetected/received by the process ofm).................................. This is when the wave isabsorbed by a substance and causes electronsor molecules to n)............................. “insympathy” with the wave.

High frequency EM waves, such aso).............................. is dangerous to life. Luckily,the p).............................. layer of the atmosphereabsorbs most of the dangerousq)........................... rays from the Sun.

All forms of radiation decrease in intensity inproportion to the r)..........................................from the source, so if distance is doubled, theintensity will drop to s)...................... (fraction)

EM waves are very important in moderncommunications. The wave types used aremainly t)............................. and .............................,but light is being used more and more in theform of u)................................ carried insidev)............................. fibres.

Waves carry information by the process ofw).................................. This can be done in 3ways:• “AM” stands for x)................................................................., in which the information iscarried as fluctuations in the y).............................of the wave.• “FM” stands for z).............................................................................., in which the signal iscarried by variations in aa)................................. ofthe wave.• Digital signals are carried byab)................................ modulation in which thecarrier wave rapidly switches between 2 states(e.g. on and off).

The energy changes occurring in a mobilephone call are as follows: Sound waves of yourvoice are converted to an ac)................................signal. This is used to modulate aad)..................................... wave transmitted tothe local “cell” antenna. Next, the signal is sentvia ae)......................................... link, or LASERbeam to another “cell” station. Here it istransmitted again as a af)......................................signal. The receiving phone converts this to anag)................................... signal and finally it isconverted back to sound waves.

16





Example Problem:At a distance of 5m the brightness of a light ismeasured to be 36 units. How bright would itbe if viewed from 15m?

Answer: Since the distance is 3x further, then intensitywill be 1/9. So new brightness = 36/9 = 4units.

TRY THESE:1. At a distance of 10m from a light, thebrightness (intensity) is 48 units. What intensitywould it have at distance:a) 20m? b) 40m?

c) 100m? d) 5m?

Worksheet 9 EM WavesFill in the blank spaces. Student Name...........................................

Worksheet 10 Practice ProblemsInverse Square Law Student Name...........................................

2.How much stronger would a radio signal be ifyou moved from 100km, to 25km distance fromthe transmitter?

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

4. One light bulb (at a certain distance) gives “I”units of light intensity. To get the same lightintensity at double the distance, how manyidentical bulbs need to be switched on?

Usage & copying is permitted according to the Site Licence Conditions only17



Multiple Choice1. Compared to visible light:A. Infra-red has shorter wavelength &

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.

2. The radiation from the Sun least likely to reachthe Earth’s surface is:A. Infra-redB. visible lightC. Ultra-violetD. radio waves.

3. The brightness of a light viewed from 40 metres,compared to viewing from 10 metres would be:A. 1/4 as brightB. 4 times brighterC. 1/16 as brightD. 16 times brighter.

4. The diagrams show a “carrier wave”, and themodulated wave carrying a signal or message.

The method of modulation used is:A. AMB. FMC. PulseD. Digital.


5. (3 marks) Re-arrange these members of the EM spectrum,placing them in order from lowest to highestfrequency.Radio, infra-red, gamma, light, microwaves, X-ray, ultra-violet.

6. (3 marks) Identify a method for detecting each of these EMtypes: (choose a different method for each)a) visible light

b) X-ray

c) infra-red

7. (3 marks) A lighthouse is viewed from 10km and its lightintensity (brightness) measured to be 0.1 units.How bright would it appear if viewed from 1 km?Explain your answer.

8. (3 marks) Discuss briefly a limitation on the use of EMRfor communication.


CCaarrrriieerrwwaavvee

MMoodduullaatteeddwwaavvee

When a Wave Hits a BoundaryWhen a wave is travelling through one mediumand then strikes a different medium, one of 3things can happen at the boundary:

It is quite possible that all 3 things can happenat once. For example, if a beam of light istravelling through air, and then strikes a glasswindow:• 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 will bounce 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 must be measured between the rays andthe “NORMAL”... an imaginary line at rightangles to the surface.

What if the Surface Isn’t Flat?The Law of Reflection is still obeyed, as shown:

Reflection of Light from Curved Mirrors

“Concave” mirrors (“go in like a CAVE”)reflect light to a “Focus”, or “focalpoint”.

Concave mirrors can give ENLARGEDimages if viewed from the rightdistance, such as a household shavingmirror or make-up mirror, which gives amagnified reflection of your face. This isalso the basis of a “reflectingtelescope.”

“Convex” mirrors reflect light so therays diverge outwards, as if comingfrom a focus behind the mirror.

Convex mirrors produce smaller

(“diminished”) images, but give a wider-angle view. An example of use is theside mirrors on a car which give you awide-angle view into the driver’s “blind-spot”. (BUT things look smaller. This can confuse adriver into thinking that other cars are further away.)

18

4. REFLECTION & REFRACTIONkeep it simple science

®



TThhee IInncciiddeenntt rraayyss PP,,QQ &&RR aarree ppaarraalllleell..

EEaacchh oobbeeyyss tthhee LLaaww ooffRReefflleeccttiioonn,, bbuutt tthheerreefflleecctteedd rraayyss ggoo iinnddiiffffeerreenntt ddiirreeccttiioonnss..

TThhee ““NNoorrmmaall”” ffoorr eeaacchhrraayy iiss sshhoowwnn aass aa ddootttteedd

lliinnee

PP QQ

RR

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

AAbbssoorrbbeedd eenneerrggyybbeeccoommeess hheeaatt

AABBSSOORRPPTTIIOONNooff tthhee eenneerrggyy

RREEFFLLEECCTTIIOONN((bboouunncceess ooffff))

TTRRAANNSSMMIISSSSIIOONNiinnttoo tthhee

nneeww mmeeddiiuumm,,wwiitthh ppoossssiibblleeRREEFFRRAACCTTIIOONN((cchhaannggee ooffddiirreeccttiioonn))

““NNoorrmmaall””lliinnee

IInncciiddeenntt rraayy

RReefflleecttee

dd

rrayy

ioo

roo

RReefflleeccttiivveessuurrffaaccee

ssuucchh aass aammiirrrroorr

FFooccuuss

““VViirrttuuaall””FFooccuuss

This is why uneven, rough surfaces don’t give “shiny” reflections.

Light is scattered in all directions.


19

Wave reflection from the ionosphere can helpwith long distance radio communications. Itworks best with the longer wavelength AMsignals.

The Ionosphere is a zone in the upperatmosphere where the air molecules are partlyionised (electrically charged) by radiations from the Sun. The ionised gases actas a reflective surface to radio waves of certainwavelengths.

TV signals and FM (shorter wavelengths) radiodo not reflect so well and generally you need tobe in “line of sight” from the transmitter to getgood reception.

When going from a more dense, to a less densemedium the opposite changes occur.

• The velocity increases: wave speeds up• The wavelength gets longer.• Wave refracts away from the normal.

In this case,

io < ro



TTrraannssmmiitttteerr

IIoonnoosspphheerreellaayyeerr

RReecceeiivveerr

ioro

Incident Ray

Glass Air

Refracted Ray

nnoorrmmaall

EEAARRTTHH

Reflections in CommunicationsAnother example involves howMicrowaves are transmitted andreceived. Microwaves are usedto relay TV programs to regionaltransmitters and to relay longdistance phone calls (includinginternet) from city to city.

At the transmission end, acurved reflector keeps thewaves in a tight beam aimed atthe next relay station. Thereceiver has a similar dish tofocus the waves into thereceiving antenna.

MMiiccrroowwaavveebbeeaamm ttrraavveellssbbeettwweeeenn rreellaayy

ssttaattiioonnss

YYoouurr ssaatteelllliittee TTVV ddiisshh iiss aa rreefflleeccttoorr ttoooo

Microwave Reflector Dishes

TTrraannssmmiitttteerrddiisshh

RReecceeiivveerrddiisshh

RefractionRefraction occurs when waves enter a new medium. The waves change their speed and their

wavelength and, depending on the angle of incidence, may change direction.All waves can undergo refraction, but here we will concentrate entirely on light waves.

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

• The velocity slows.• The wavelength gets shorter.• The beam changes direction

towards the normal.

io > roio

ro

Incident Ray

Refracted Ray

Air Glass

nnoorrmmaall

Angle ofRefraction

Angle ofIncidence

When a light ray refracts, its wavelength changes, but frequency stays the same.

Since COLOUR is determined by frequency, there is no colour change during refraction.

20

Snell’s LawYou may have carried out an investigation inclass using a “Ray Box Kit” to measure anglesof incidence and angles of refraction.

When you graph the angles the result is a curve.

This is not much use for defining anyrelationship that may exist.

In 1621, Willebrord Snell discovered that if yougraph the Sine ratios of the angles, the points liein a straight line. You may have done the samewith your experimental data.

The fact that it’s a straight line means there is adirect relationship between Sin i and Sin r.

The gradient of the line is not only the ratiobetween the Sine of the angles, but is also equalto the ratio of velocities of the wave in the 2mediums involved.

This special ratio is known as the“REFRACTIVE INDEX” (n)

This is now called Snell’s Law:

Refractive IndexWhen waves enter a new medium, and then exit itagain, the refractions that occur on the way in, arethe opposite of what happens on the way out.

For example, this light ray goes from air, intoglass and out into air again.

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

and

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

These 2 values are RECIPROCALS !! ...and this willalways be the case... the index of refraction goingin is the reciprocal of the index coming out.




Angle of refraction, roo

Angl

e of

inc

iden

ce, i

oo

Sin roo

Sin

ioo

GGrraaddiienn

tt == rriiss

e == SSiinn

ii

rruunn SSiinn

rr

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

Sin i = V1 = 1n2Sin r V2

1n2 = 1 2n1

The spoonappears “broken”at the surface ofthe tea due to

refraction of thelight by which we

see it.

4455o

4455o

2288o

2288o

RReeffrraaccttiioonnaaiirr ->> ggllaassss

RReeffrraaccttiioonnggllaassss ->> aaiirr

nnoorrmmaall

glass


3 beams of light being refractedthrough a perspex block.

21

Total Internal Reflection & the Critical Angle

Consider the situation when waves are goingfrom a more dense medium into a less densemedium, such as light going from glass into air.

The waves refract away from the normal.

Now think about increasing the incident angleas shown in this series of diagrams.

There comes an angle (called the “CriticalAngle”, (c)) where the angle of refraction = 90o.At this point the refracted ray runs along theedge of the glass, but does not cross theboundary.

So, when the angle of incidence equals the“critical angle”, the angle of refraction is a rightangle.

If io = co, then ro= 90o

Remember that Sin i = gnaSin r so at the critical angle Sin c = gna

Sin 90

and sin 90o = 1, so...

... But What Happens Beyond the Critical Angle?At incident angles larger than “c”, the rayreflects back inside the glass... this is called

“TOTAL INTERNAL REFLECTION”

This has one very important application incommunication technology...

Optical fibres are thin strands of very pure glassthat can carry communications signals in theform of laser light beams. The laser beams staywithin the fibres because of total internalreflection.

Each fibre is a core strand of glass, with anotherlayer wrapped around it. The outer layer has alower refractive index than the core, so evenwhere the fibre bends around a corner, the laserlight will generally strike the boundary at anincident angle greater than the critical angle.

Whenever the laser beam hits the boundarybetween the 2 layers, the angle of incidenceexceeds the critical angle, (io > co) so TotalInternal Reflection occurs and the beam staystotally within the fibres over long distances.




Sin c = gna = 1 1 ang

This means that the Sine ratio of thecritical angle “C” is equal to

the reciprocal of the refractive index of the glass.

If ioo > coo

the ray cannot get out,but reflects back insidethe glass

CCoorree..hhiigghhiinnddeexx

lloowweerr iinnddeexxoouutteerr llaayyeerr

TThhee llaasseerr lliigghhtt ““bboouunncceess”” aarroouunndd ccoorrnneerrssbbyy ttoottaall iinntteerrnnaall rreefflleeccttiioonn

OOppttiiccaall ffiibbrree

iiorro

11

ggllaassss

aaiirr

iiorro

22

bbiiggggeerr ii,,bbiiggggeerr rr

iio==cco

rro== 9900o

33

CCrriittiiccaall AAnnggllee

iio>>cco

RRaayyrreefflleeccttssiinnssiiddeeggllaassss

4

llaasseerr bbeeaamm



When a wave meets the boundarybetween one medium and another, anyof 3 things can occur: the wave’s energycan be absorbed, or the wave can bea)............................... or ............................

The Law of Reflection simply states thatthe angle of b)..................... equals theangle of c)........................... The anglesmust be measured from the wave “ray”to the d)................................. This is animaginary line which ise)................................. to the boundary.

Concave mirrors reflect light into af)........................ point and can produceenlarged images, such as in a reflectingtelescope.

A g)........................ mirror reflects lightoutwards. This produces images whichare h)..........................., but have a widerfield of view. A practical use for thismirror is i)..................................................

In communications, reflection is usefulfor long-distance radio reception. Someradio wavelengths reflect from thej)................................ layer in the upperatmosphere, and are “bounced” aroundthe curvature of the Earth. Satellite“dishes” and k)....................................antennas use reflection to focus wavesignals into the receiver.

Refraction occurs when waves go fromone medium into another. The wavesmay change in l)........................, and........................... and .................................

22



Worksheet 12 Reflection & RefractionFill in the blank spaces. Student Name...........................................

For example, when light goes from airinto glass its speed m)......................, andits n)............................... gets shorter(although o)................................... doesnot change). It also changes direction,going p)............................. the normal.

Snell’s Law describes the directrelationship between the Sine ratios ofthe angles of q).......................... and................................... This ratio is calledthe r).......................... ...................... It isalso equal to the ratio between thes)............................. of the wave in the 2different mediums. The index for thewave entering the medium, and theindex for the wave exiting the mediumare always t)..................................... ofeach other.

When a light ray is going from a“slower” medium into a “faster” one, theray will refract u)............................ thenormal. As the angle of incidenceincreases, so will the angle of refraction,until the refracted rayv).............................................. of theboundary. The angle of incidence atwhich this happens is called thew)............................... angle. At angles ofincidence greater than this angle,x).............................................. ..............occurs, and the ray stays within the“slower” medium. This property is usedin optical fibre technology to ensurethat y).................................. beams staywithin the fibres.


Test Questions for this section areat the end of section 5


Snell’s Law Sin i = V1 = 1n2Sin r V2

Example ProblemA beam of light goes from air into a glass blockwith a refractive index of 1.50. The angle ofincidence is 35o.a) Find the angle of refraction.b) If light travels in air at 3.00x108 ms-1, find thevelocity in the glass.

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

Sin r sin r = sin 35 /1.50 = 0.38238

therefore, angle of refraction, r = 22.5o

b) V1 = n 3.00x108 / V2 = 1.50V2

V2 = 3.00x108 / 1.50therefore, velocity in glass, V = 2.00x108 ms-1

TRY THESE1. In an experiment, a student sent a beam oflight into a block of clear plastic. The angle ofincidence was measured as 50o. The angle ofrefraction was 33o.a) Find the refractive index of the plastic. b) If light travels in air at 3.0x108ms-1, find itsvelocity in the plastic.

2. Light travels through a diamond at only1.25x108ms-1.a) Find the refractive index of diamond.b) If a ray of light strikes a diamond surface at anangle of 40o from the normal, find the angle ofrefraction as the ray enters the diamond.

3. Using a laser beam and a fish tank filled withwater, the refractive index of the water wasfound to be 1.33.a) At what incident angle must the beam strikethe water to produce an angle of refraction of32.5o?b) At what velocity does the laser beam travel inwater?

4. Several different angles of incidence andrefraction were measured for a light ray goingfrom air into a high density crystal glass block. a) For each pair of angles, calculate a refractiveindex value.b) Find the average value for the refractiveindex.c) Use the average value to find the velocity oflight in the crystal glass.

DATA Angle of Angle ofIncidence Refraction50.0 25.042.0 21.030.0 17.065.0 31.0

23



Worksheet 13 Practice ProblemsRefraction Student Name...........................................

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. along the normal line) what is the value of theangle 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 comingout of the plastic into air?b) If a light ray in the plastic struck the boundaryat an angle of incidence of 20o, at what angle ofrefraction will it enter the air?

7. Refer to Q3. a) What is the refractive index for light travellingfrom water into air?b) If a light ray emerged from water into air at anangle of refraction of 37o, what must have beenthe angle of incidence?

8. In a type of lead-crystal glass, a light ray exitsfrom the glass into air. At the interface, theangles were i = 15o, and r = 25o. a) What is the refractive index glassnair?b) What is the index airnglass?c) At what velocity does light travel in thisglass?

Critical Angle Sin c = gna = 1 ang

9. a) Use the information in Q2 to find the“critical angle” for light travelling inside adiamond.b) Describe what would occur (no calculationrequired) if light inside a diamond hit theboundary at an angle of incidence of:

i) 20o ii) 30o

10. a) What is the critical angle for glass with ang = 1.50?

b) Describe (no calculation) what happens whenlight inside the glass strikes the boundary atangle of incidence:

i) 40o ii) 41.8oiii) 45o

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

Digital TechnologyIn the past 20-30 years our society hasbecome more and more “digitised”.Because of the speed, storage capacityand processing ability of computers,almost every aspect of our society has“gone digital”.

This simply means that all information(data) whether it be a person’s voice,written words, numbers, music, photos,etc. is converted into digital code forprocessing, storage or transmissionand communication.

A simple list of some of thetechnologies involved is:

CD’s & DVD’s,Mobile phones, Digital cameras,

Computers & Internet, MP3 music,

ATM’sGPS

Increasingly, WAVES are involved inthese technologies, especially whendata is moved around...

COMMUNICATION.

GPS is a system that allows a ship, aircraft, caror bushwalker, to locate their exact positionanywhere on Earth instantly and continuously.

The system was developed for miltary uses, butthen made available to anyone. The militaryversion is thought to be accurate to within ametre, the civilian version to within about 10 m.

The system is based on a fleet of 32 satellites(controlled by the US Air Force) positioned inorbit so that from anywhere on Earth, at anymoment, several satellites are in “line of sight”.

Each satellite constantly sends out microwavesignals identifying itself, its orbit details and theprecise time the signal was sent. When yourportable GPS receiver picks up the signal, it cancalculate your exact distance from the satellite,from the time delay since the signal was sent.

By doing the same for 2 other satellites, the GPSunit rapidly “triangulates” the signals from 3satellites to pin-point your location on theEarth’s surface. (Aircraft need a 4th signal toget their altitude)

GPS systems for cars show your position on ascreen overlaid onto a road map of the area. Asyou drive around, the system constantly showsyour changing position, and can advise youwhere to turn to reach your destination.

24

5. DIGITAL COMMUNICATION & DATA STORAGEkeep it simple science

®



TECHNOLOGY CASE STUDY:

GLOBAL POSITIONING SYSTEM(GPS)

Satellite orbitsSSaatteelllliittee 11

SSaatteelllliittee 22

SSaatteelllliittee 33

GGPPSS rreecceeiivveerr

Earth

GPS

Multiple Choice1. 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”.

2. Which of the following does NOT change when awave undergoes refraction?A. VelocityB. DirectionC. WavelengthD. Frequency

3. If a light ray passed from air into one of thefollowing substances, (each at the same angle ofincidence) which one would show the leastamount 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)

4. Light travels in air at 3.0x108ms-1. If therefractive index of glass = 1.5, then the velocityof 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.

5. The refractive index of water = 1.33.The “CriticalAngle” for water would be closest to:A. 38o B. 45o C. 49o D. 53o

6. Long distance communication using laser lightand optical 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.

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


8. (3 marks) Complete each diagram to show the expectedpath of each reflected light ray P, Q and R.

9. (6 marks) In an experiment, angles ofincidence and refractionwere measured as shown.

a) Find the refractive index of theplastic. Show working.

b) At what speed does light travel in this plastic?Show working.

10. (4 marks) Predict the path of this light rayafter it strikes the boundary.Explain your reasoning, andshow any working.

11. (3 marks)Outline briefly the underlying principles used inone application of physics related to wavesused in communication.

25




Worksheet 14 Test Questions sections 4&5 Student Name...........................................

PP

RR

aaiirr

Remember that for full marksin calculations, you need to show

FORMULA, NUMERICAL SUBSTITUTION,APPROPRIATE PRECISION and UNITS

aaiirr

QQ

ppllaassttiicc

3333o

2255o

3300o

plastic.n = 1.40




26

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

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

THE WORLDCOMMUNICATES




27

Worksheet 4wave P Q R

i) 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 λλ = 12.5 x 0.50 = 6.25 ms-1

vii) λλ = V / fwave Q: = 9.5 / 6.25 wave R; =9.5 / 25

= 1.5 m = 0.38 m

Worksheet 51. C 2. D 3. A 4. A 5. A 6. C

7. SOUND --> ELECTRICAL --> RADIO

8.a) Mechanical waves require a “medium” substanceto travel in. EM waves do not, and so can travel invacuum.b) Transverse waves vibrate at right angles to thedirection of energy movement. In a longitudinal wave,the vibration is back-and-forth in the same directionas the energy flow.

9.a) V = f λλ

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

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

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

= 1/16 = 6.25x10-2 Hzb) V = f λλ

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

Worksheet 6Using the sum of displacements at the circled points,the resultant looks approximately like the solid curveshown.

Note that the amplitude of the resultant starts largeand becomes smaller. You would hear the sound volume decrease.

Answer SectionWorksheet 1a) 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

Worksheet 21. a) V= f λλ b) T = 1 / f

= 280 x 5.20 = 1 / 280= 1460 m. = 0.00357 s

(3.57 x10-3 s)2. V= f λλ

2,500 = 0.400 x λλl= 2,500 / 0.400

= 6,250 m (6.25 x103 m)

3. V= f λλ330 = 1200 x λλ

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

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

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

c) V= f λλ2.20 = 6.25x10-2 x λλ

λλ= 2.20 / 6.25x10-2 = 35.2 mWavelength has become a lot shorter as the waveentered shallower water.

Worksheet 31. a) V= f λλ, f = V / λλ = 3.00x108 / 7.00x10-7

= 4.29x1014 Hzb) f = V / λλ

= 3.00x108 / 3.00x10-7

= 1.00x1015 Hz

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

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

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

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

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

4. Since 8 complete wavelengths fit in 0.96m, thenλλ = 0.12 m

V= f λλ = 384 x 0.12 = 46 ms-1

5. a) f = V / λλ = 3.00x108 / 1.50x10-11

= 2.00 x 1019 Hzb) T = 1 / f = 1 / 2x1019 = 5.00x10-20 s.

28

Worksheet 7Part Aa) mechanical b) longitudinalc) compressions d) rarefactionse) 330 f) higherg) frequency h) loudness / volumei) reflect j) navigation / hunting preyk) SONARl) detecting fish / submarinesm) resultant n) amplitudes

Worksheet 81. B 2. C 3. A

4. summing displacements at circled points:

5. With a sound wave, a “compression” is where airparticles are pushed together, and a “rarefaction” iswhere they are spread apart more, as the wave movesthrough.

Worksheet 9a) transverse b) do notc) medium d) 3 x 108

e) 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

Worksheet 101. 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 moreintense.

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 getsame amount of light.

Worksheet 111. D 2. C 3. C 4. B

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

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

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

0.1 x 100 = 10 units.

8. 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“bands” are becoming congested.

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

Worksheet 12a) reflected or refracted b) incidencec) reflection d) normale) perpendicular f) focalg) convex h) smaller/diminishedi) driver’s 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




disp

lace

men

t

time

rreessuullttaanntt((aapppprrooxx))

29

Worksheet 13Snell’s 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= 3.0x108 / 1.4

= 2.1 x 108 ms-1

2. a) n=V1 / V2 = 3.00x108 / 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 = 3.00x108 / 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 = 3.00x108 / 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) 0o

c) n = Sin i / Sin r1.50 = sin0 / Sin r Sin r = sin0 / 1.50 = 0

r = 0o

d) The ray does NOT change direction.(However, it would still change velocity andwavelength)

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

2n1

b) 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

8. a) n = Sin i / Sin r= sin15 / sin25

n= 0.61 (for glass --> air)b) n (air ->glass) = 1 / 0.61 = 1.6 (no units)

8. c) n = V1 / V21.6 = 3.0x108 / V2V2 = 3.0x108 / 1.6

V(glass) = 1.9x108 ms-1

Critical Angle & T.I.R.9. a) Sin c = dna = 1

andSin 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.

Worksheet 141. A 2. D 3. A 4. B 5. C 6. C 7. D8.

9.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= 3.0x108/1.3

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

10. Note that the angle given is not the correct angleof incidence. (Angle of incidence must be measuredfrom the normal)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 > c so ray will undergo total internalreflection, and reflect back inside the plastic block.(at angle of reflection = 60o)

11.Underlying principles of GPS:GPS involves a small, portable receiver picking upmicrowave signals from a fleet of satellites in orbit.Each satellite sends a coded message identifyingitself and the precise time that the signal was sent.

By comparing the signals from (at least) 3 differentsatellites, the GPS receiver can “triangulate” to findits position with a high degree of accuracy... within afew metres in some cases.




PP

RR

QQ

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