KISS Notes The World Communicates

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Preliminary Physics Topic 1

THE WORLD COMMUNICATES What 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 WAVES 3. ELECTROMAGNETIC WAVES 4. REFLECTION & REFRACTION 5. DIGITAL COMMUNICATION & DATA STORAGE ...in the context of communications

but first, let’s revise... ENERGY

TYPES of WAVES

Energy 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).

Examples of energy which moves around as waves include • SOUND • LIGHT • RADIO SIGNALS • WATER WAVES • X-RAYS

Some forms of energy move around as WAVES . A wave is a carrier of energy. In a wave, energy moves, but matter does not.

The strings vibrate.

• MICROWAVES ... and many more

This causes the air to vibrate too. Waves of vibration spread out through the air... sound waves.

ENERGY CONVERSIONS Energy can be converted from one form to another.

The air vibrates, but does not go anywhere.

In your mobile phone the SOUND WAVES of your voice are converted to ELECTRICAL signals then transmitted as RADIO WAVES to your friend, whose phone converts it back again.

Water waves carry energy across the surface of a pond. The water vibrates up & down, but goes nowhere.

SOUND

RADIO

In this topic you will learn about waves and their properties and features, and how they they are used for communication.

Waves Carry Energy Without the Transfer of Matter Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

ELECTRICAL

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CONCEPT DIAGRAM (“Mind Map”) OF TOPIC Some 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.

Types of Waves

Wavelength, Amplitude, Frequency & Period

Wave Equation

Nature of Sound Waves

Graphing Waves

Velocity, Pitch & Loudness

The Nature of Waves

Principle of Superposition

Properties of Sound Waves

Electromagnetic Waves

THE WORLD COMMUNICATES

Inverse Square Law

Digital Communication & Data Storage

Production, Detection & Dangers of EM Waves EM Waves in Communication

Reflection & Refraction Law of Reflection Refraction. Snell’s Law, Lenses & Total Internal Reflection

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The EM Spectrum

2

Light & Mirrors. Reflection in Communication

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1. THE NATURE OF WAVES Waves Carry Energy

Describing Waves

Waves carry energy, without the transfer of matter.

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

This can occur in 1 dimension: Pulses moving along a slinky spring

Consider a wave in a rope which has been given a single up-and-down “twitch”: Compressed sections in the spring move along it like a “Mexican Wave”... energy is transferred, but the coils merely oscillate back and forth and do not actually go anywhere.

A PULSE WAVE

CREST

Energy moves along the rope

... or in 2 dimensions: Ripples spreading on the surface of a pond.

part of the rope (medium) vibrates up & down rope TROUGH

Energy moves along the rope, but the rope itself doesn’t go anywhere. 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 right angles to the direction that the energy moves, is called a Transverse wave.

...or in 3 dimensions,

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

such as when light radiates in all directions from a glowing object.

A PERIODIC, TRANSVERSE WAVE Energy moves

Rope vibrates up and down

CREST

TROUGH

MECHANICAL WAVES require a medium to travel through.

Waves & Mediums

PERIODIC WAVES contain a series of pulses, with a continuous set of crests and troughs.

Mechanical waves are those which need a “medium” to travel through. For example, a water wave must have water to travel in. Sound waves need air, or water, or some substance to move in. They CANNOT travel in a vacuum.

TRANSVERSE WAVES vibrate at right angles to the direction that the energy is moving.

Electromagnetic (EM) waves do NOT need a medium... they can travel through a vacuum, and in fact travel fastest in a vacuum. EM waves include light, radio waves, ultra-violet and other types, and are studied in detail in a later section. Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

Energy flow Vibration in medium 3

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Longitudinal waves

are when the particles of the medium vibrate back-and-forth in the same line as the energy moves. For example, 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.

LONGITUDINAL WAVE IN A SPRING Spring vibrates

Energy moves

Earthquake Shock Waves compression in spring

rarefaction (where spring is stretched)

Wave Measurements All periodic waves, whether Longitudinal or Transverse, Mechanical or Electromagnetic, can be described and measured by their:-

Wavelength = the distance from one crest to the next. (or from one trough to the next, or from one compression to the next) The S.I. unit is the metre (m). The Greek letter “lambda”

Period

(T) = the time (in seconds) for one complete vibration to occur. Note that there is a simple relationship between Frequency and Period... they are reciprocals.

λ

T= 1 f

is used as a symbol for wavelength.

Amplitude

and f = 1 T

Velocity

(a or A) = the distance that a particle in the medium is displaced from its “rest position” at a crest or trough. i.e. the maximum displacement distance.

(v) = the speed of the wave, in metres/sec.(ms-1) There is a simple relationship between Velocity, Wavelength and Frequency:

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.

THE WAVE EQUATION Velocity = Frequency x Wavelength

S.I. unit is the “hertz” (Hz)

1 Hz = 1 wave/sec.

V = fλ

WAVELENGTH

Wave cycles per second is FREQUENCY

AMPLITUDE

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WORKSHEET at end of section

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Graphing Waves

Graphing a Longitudinal Wave

A good way to represent a wave is by using a graph.

You might think these Displacement-Time graphs wouldn’t work for a Longitudinal wave where the particles vibrate back-and-forth rather than up-and-down.

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

However, the graph of a longitudinal be exactly the same... you just have that the “displacement” is displacement from the “equilibrium instead of up-down.

Ripples

Cork bobs up and down

+3

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

0

One period = 0.8 s 0.2

0.4

0.6

Relationship Between Wavelength & Frequency

Time (s) 0.8

1.0

You may have carried out a “First Hand Investigation” in class to see how a change in Frequency (at constant velocity) affects the wavelength. Maybe you used a slinky spring, or watched the water waves in a “ripple tank”.

1.2

-3 3

Displacement (cm)

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

wave can to realise sideways position”,

Be careful! The graph is shaped like a wave, so it’s tempting to try to read the wavelength from the horizontal scale... but the horizontal scale is TIME, not length.

You would have found... INCREASING the FREQUENCY

What you CAN read from a Displacement-Time graph:

DECREASE in WAVELENGTH and

Amplitude The vertical scale measures the displacement 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 either above or below (d= -3cm) equilibrium, so the Amplitude = 3cm (0.03m)

DECREASING the FREQUENCY

INCREASE in WAVELENGTH

(If VELOCITY is the same)

Longer Wavelength

Lower Frequency

Period

Since the horizontal scale is time, you can easily read 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 could calculate the wavelength, or vice versa.

Shorter Wavelength

e.g. if the ripples are 0.45m apart: (i.e. λ = 0.45m)

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

V=fxλ

So, velocity

= 1.25 x 0.45 = 0.56 ms-1

WORKSHEET at end of section Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

Higher Frequency

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Worksheet 1

The Nature of Waves

Fill in the blank spaces.

Student Name...........................................

Waves carry a)................................... without the transfer of b)................................ “Mechanical” waves require a c)............................. to travel in. Examples are d)...................... and ............................ “Electromagnetic” waves do not need a medium and can travel in a e)....................... Examples include f).................... and ................................

Frequency is the number of l)................................ per second. The SI unit is the m)................. (........) n).............................. is the time for one complete vibration. This is the o).................................... of frequency. Velocity is the speed of the wave and is equal to p)....................... multiplied by q)...........................

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

On the graph of a wave, showing Displacement v Time, the vertical scale shows the r)..................................... of the wave, while the horizontal allows you to read the value of the s).................................. and then easily calculate the t)......................................................

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

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

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

Worksheet 2 Practice Problems Wave Equation 1

Student Name........................................... TRY THESE 1. a) Find the velocity of a sound wave in water if it vibrates 280 times per second and has a wavelength of 5.20m.

Example Problem 1 A water wave in the ocean has a wavelength of 85m, and a velocity of 4.5ms-1. a) Find the frequency. b) What is the period? Solution a) V= fλ 4.5 = f x 85 f = 4.5 / 85 = 0.053 Hz (5.3 x 10-2 Hz)

b)What is the period of this wave? 2. An earthquake shockwave travels through rock at a velocity of 2,500 ms-1. Its frequency is 0.400 Hz. What is the wavelength?

(i.e. only a small fraction of a wave passes by each second.)

b)

T=1/f = 1 / 0.053 = 19 s 3. What is the wavelength of a sound wave with frequency 1200Hz? Sound travels in air at 330ms-1.

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

Example Problem 2 A 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.

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

Solution a) f =1/T = 1 / 0.002 = 500Hz (i.e. 500 vibrations per sec.) b)

b) Wavelength?

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

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c) As the wave enters shallower water it keeps the same frequency but slows down to only 2.20 ms-1. What happens to the wavelength? 6

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Worksheet 3 Practice Problems More Wave Equation

Student Name...........................................

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

4. When a guitar string is plucked, a wave vibration runs back and forth through the string. The string is 0.96m long and it is found that exactly 8 complete wavelengths fit along the string at a time. The vibration frequency is 384Hz. How fast do the waves travel through the string?

b) Blue light has a wavelength of 3.00x10-7m and travels at the same speed. What is the frequency? 2. Radio signals travel at the speed of light. (3.00x108ms-1) A radio station has a frequency of 530 kHz (=530,000Hz). a) What is the period of the waves?

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

b) What is the wavelength? b What is the period of the X-rays? 3. Compare the frequency of a radio wave 2.50m long, with one 2.50cm long. (Assume they both travel at the speed of light)

Remember that for full marks in calculations, you need to show FORMULA, NUMERICAL SUBSTITUTION, APPROPRIATE PRECISION and UNITS

Worksheet 4 Practice Problems Reading Wave Graphs Answer on reverse

Student Name...........................................

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=0.03s iii) 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.

P

0

Time (s) 0.05

0.1

R

-0 0.1

Displacement

(m)

0.1

Q

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Worksheet 5 Test Questions

section 1

Multiple Choice

Longer Response Questions Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. Answer on reverse if insufficient space.

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 in this diagram (A,B,C or D) correctly shows the “amplitude” of the wave?

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

B D

A

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

C

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.

b) transverse and longitudinal waves.

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

4. If the period of a wave is 4 seconds, then its frequency is: A. 0.25 Hz B. 0.4 Hz C. 4.0 Hz D. 1/16 Hz

b) calculate the wave’s period.

1.0

10. (5 marks) The graph describes a wave in the ocean.

time (s)

-3

3 displacement (m)

3 displacement (mm)

5. The period of this wave is: A. 0.8s B 1.6s C. 3 mm D. 6 mm

Student Name...........................................

time (s) 10

20

a) What is the -3 3 frequency of the wave? Explain your answer.

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

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

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2. THE PROPERTIES OF SOUND WAVES Sound Waves

Velocity of Sound

Sound waves are Mechanical (they need a medium) and Longitudinal (vibrate back-and-forth

Sound travels at different speeds in different mediums.

in the line of the energy flow)

SOUND WAVES

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

Energy moves Particles vibrate

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

Instead of crests and troughs, a series of “compressions” and “rarefactions” pass through the medium as a sound travels. The atoms and molecules are alternately “squashed together” and then stretched apart as the energy flows through.

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

Sound Travels

FREQUENCY = “PITCH” Compression

Rarefaction

Compression

When you hear sounds of different “pitch” that is the way your brain interprets sound waves of different frequency.

Rarefaction

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

Low Frequency = Low Pitch High Frequency = High Pitch

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

AMPLITUDE = LOUDNESS or VOLUME

Displacement from the equilibrium

Compressions.

Higher air pressure

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

Time

Rarefaction.

Larger Amplitude = Louder Sound Smaller Amplitude = Quieter Sound

Lower pressure

USES OF SONAR

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.

BAT

The time delay between sending a sound “ping’ and receiving the echo, gives depth and distance

SONAR SOund Navigation And Ranging

Anti-s submarine Warfare

Echoes from i nsect

Depth Sounding

“Squeaks” of sound

Some animals can send out sound waves and pick up the echoes to help locate their prey, or to navigate, in environments where they can’t see very well, such as murky water (dolphin), or in darkness (bat). Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

Humans have invented SONAR technologies for things such as “depth sounding” and detecting underwater objects... fish or submarines, it all works the same way. 9

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The Principle of Superposition All waves have the ability to pass through other waves without being affected. For example, you could shine a red spotlight across a beam of blue light, and each colour and beam will emerge on the other side exactly the same.

However, if the waves are “out of phase” (for example, if compression coincided with rarefaction) then there is destructive interference... the opposite amplitudes may cancel each other out. Add positive & negative displacements at the circled points

However, for the instant that the 2 waves are superimposed upon each other, they do interact and “interfer” with each other. Displacement

Displacement

“resultant” A+B wave A wave B To find a “resultant”, add the displacements of A&B at convenient points (circled)

wave A Resultant wave B

Theoretically, if 2 sound waves had the same amplitude and were 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!)

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

In practice, this only happens over short distances or time periods to give “interference patterns” and “beat sounds”.

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

Worksheet 6 Superposition of Waves

Student Name...........................................

-v ve

Displacement

+ve

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

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

Surfing Trivia Technically, a breaking wave is not a wave at all. Once it breaks, the water begins moving forward (which allows 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. Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

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Worksheet 7 Sound Waves Fill in the blank spaces.

Student Name...........................................

Sound waves are a)................................. and b).................................... A sound wave consists of a series of high pressure c)................................ and lower pressure d)........................................ travelling through the medium.

“Echoes” occur when sounds i).................................. Some animals use echoes for j)................................ Humans use the technology of k)........................ for “depth sounding” and l)..............................................

In air, the speed of sound is about e).......... ms-1, but it is much f)............................... (higher/lower) in water or in solids such as metals.

When 2 or more waves coincide, they will interfere with each other. The m)............................ wave can be found by adding together the separate wave n)..............................................

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

COMPLETED WORKSHEETS BECOME SECTION SUMMARIES

Worksheet 8 Test Questions Multiple Choice

Student Name...........................................

section 2

Longer Response Questions Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. Answer on reverse if insufficient space.

1.

If you heard a sound wave with small amplitude and high frequency, you would describe it as: A. low volume and low pitch. B. low volume and high pitch C. high volume (loud) and low pitch D. high volume and high pitch.

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

displacement

2. Two pulses are travelling towards each other in a rope. X

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.

5. (4 marks) With a water wave, a “crest” is where water has displaced upwards, and a “trough” where it displaced downwards, as the wave moves through. Explain, in similar terms, what happens to air particles as a sound wave passes.

3. The navigation of a bat in the dark, and the “depth sounding” from a boat, both work on the principle of: A. Reflection B. Refraction. C. Resonance D. Interference

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time

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3. ELECTROMAGNETIC WAVES Detection & Reception of EM Waves

EM Waves Electromagnetic waves are Transverse waves which do NOT require a medium to travel through. They travel through a vacuum at 3.00x108ms-1, the “speed of light”. They can travel through many other substances at slightly slower speed. For example, light can travel through glass or water at speeds of around 2.5x108ms-1. In air, the speed is so close to the speed in a vacuum that, for simplicity, (K.I.S.S. Principle) we take it to be the same.

Just 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 strike something and are absorbed, they may cause “sympathetic” vibrations within it.

EM radiation does not require a medium because the waves propagate as vibrations of electric and magnetic fields, not as vibrating particles. When the fat lady sings...

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

low

v.long

MEMBERS OF THE EM SPECTRUM

Radio (and TV) waves

visible LIGHT ultra-violet very short

X-rays Gamma rays

Frequency increasing

infra-red (heat radiation)

Some real examples...

These oscillations are amplified electronically and the signal converted to sound in the speaker, allowing you to listen to the radio.

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

When infra-red waves hit your skin they cause certain molecules to begin to resonate and vibrate. This sets off nerve messages to the brain and you feel warmth or heat on your skin.

Production of EM Waves All 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 running back-and-forth in a conducting wire.

In a film camera the light causes resonance in chemicals in the film. Chemical reactions occur which permanently alter the film so that an image appears when “developed” later. Different film can be sensitive to infra-red, (photos in the dark) or Xrays for medical uses.

Infra-red waves are made by molecules vibrating rapidly because of the heat energy they contain. Light is emitted when electrons rapidly “jump” down from a higher to a lower orbit around an atom.

Gamma waves come from the vibrations of charged particles within an atomic nucleus, during a nuclear reaction in the atom. Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

Antenna

When radio waves hit a suitable aerial wire or antenna, they cause some electrons in the metal to oscillate back-and-forth “in sympathy” with the wave.

very high

Wavelength decreasing

microwaves

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

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

Sun

X-rray & gamma

UV rad io

some reflected

e on inf oz rar ed &l igh t

UV Rays

“Ozone” is a form Oxygen O2 does not of oxygen which Absorb UV has 3 atoms per molecule (O3) instead of the normal 2 (O2).

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

Ozone O3 Absorbs UV

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

The Sun only produces small amounts of the even more dangerous X-rays and gamma radiation. Once again, most is absorbed in the upper atmosphere, this time by ordinary oxygen and nitrogen gases.

r ye la

Earth’s surface

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

upper atmosphere

The Inverse Square Law As 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. Mathematically, the relationship is that the intensity (I) (such as brightness of light) is inversely proportional to the SQUARE of the distance (d²) from which it is viewed.

Intensity α

1 (distance)2

I α

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

“α” means “proportional to”

1 d2

This diagram explains why:

d” e “2 anc dist x light source

dista nce “d”

Square Area x2

3 times the distance

1/9

10 times the distance

1/100

bright

as bright as bright

Notice how the brightness (intensity) changes in proportion to the distance squared, in each case.

Area = 4x2 Same amount of light falls on 4 times the area

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1/4 as

...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.

Square with sides twice as long. 2x

So, twice the distance

WORKSHEET at end of section 13

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

Radio & Microwaves carry radio and TV

What’s special about LASER LIGHT?

broadcasts, telephone long-distance links, mobile phone networks, and satellite links for telephone (including internet) and TV.

•It is one, pure frequency of light.

If you have “Satellite TV”, the “dish” on your roof is an antenna to receive microwaves directly from an orbiting satellite.

•The waves are all in phase and so they interfere constructively to form a very intense, tight beam.

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

•A laser beam will stay inside an optical fibre and not “leak” out or dissipate for long distances. •A laser can be turned on & off very rapidly, so it’s perfect for high speed digital communication.

Light is being increasingly used in the form of LASER beams carried in optical fibres for telephone and internet communication.

How a Wave Carries Information How 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...

Frequency Modulation (FM)

WAVE MODULATION

The amplitude stays constant while the frequency (and wavelength) vary within a fixed range. The information (voice, music etc) is “coded” in the variations of frequency.

Amplitude Modulation (AM) The frequency (and wavelength) of the wave stays constant while the amplitude varies. The changing amplitude “codes for” the information being carried... whether voice or music, or whatever.

FM radio gives much better fidelity and is superior, compared to AM, for the quality of sound (eg for music) received. This diagram compares the effect of

AM, FM & Digital Modulation on the same “carrier wave”

Pulse Modulation (Digital)

Optical Fibres carry Pulse Modulated laser beams

To carry information in digital form the wave must switch rapidly between 2 different states, representing the “1” and “0” of digital codes. The wave can be switched rapidly on and off (as in the diagram) or switched back-and-forth between different “phase states”... phase modulation.

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“Carrier wave”

No information carried

AM signal

Amplitude changes. Frequency constant

FM signal

Freq. changes. Amp.constant

Wave pulses on and off

Digital signal Digital data

1

0

1

1

0

1

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Case Study: MOBILE (CELL) PHONES

2. The digital RADIO signal is transmitted by your phone and received by the local “cell” antenna.

When 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? 1. The SOUND energy of your voice is converted to ELECTRICAL signals by the microphone. The electrical signal is used to digitally modulate a RADIO wave.

5. Your friend’s phone receives the RADIO signal, amplifies it as an ELECTRICAL signal and this is converted to SOUND waves in their earphone.

4. In the other cell area, the signal is converted back to a modulated RADIO signal and transmitted.

3. If your call is going to a person in another location (a different “cell”) the signal is converted into a modulated MICROWAVE and beamed, via hilltop relay towers, to the correct area. (Alternatively, it might be sent as a modulated Laser LIGHT beam through optical fibres).

ENERGY Y CHANGES

SOUND

ELECTRICITY

RADIO

MICROWAVE (or LASER LIGHT)

RADIO

SOUND

Discussion: LIMITATIONS OF COMMUNICATION CHANNELS Modern communication systems have developed rapidly and new features and capabilities seem to come out every day. It seems that the entire system is unlimited and that it can continue to expand and improve forever.

In the future we will need to switch more communications to use the laser light / optical fibre method wherever possible, and to make better use of the RF bands. For example, it is possible to use the same frequency “channel” for several different purposes as long as the different signals are modulated differently and as long as the radio receivers are sophisticated enough to pick out only the desired signal and ignore the others.

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

One thing is for sure... humans will keep communicating and the need for new services will keep expanding. So far, our technology has always managed to keep up, and it will probably continue to do so.

The simple fact is that there are now so many radio & TV stations, mobile phone networks, aircraft and shipping channels, military, police and emergency service channels, etc. etc. all using the RF (Radio Frequency) part of the EMR spectrum, that it is becoming difficult to keep expanding services without interfering with existing channels. Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

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Worksheet 9

EM Waves

Fill in the blank spaces.

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

Electromagnetic waves are a)................................... waves which b)............... (do/do not) require a c)............................ to travel in. They all move at the “speed of light”, which is d)................... ms-1 in a vacuum. The members of the EM spectrum (in order of increasing frequency) are: e)......................, f)......................., g)......................, h)..................., i)......................., j)..........................., and k).........................

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

All EM waves are produced when electrical charges l)......................... They are all detected/received by the process of m).................................. This is when the wave is absorbed by a substance and causes electrons or molecules to n)............................. “in sympathy” with the wave. High frequency EM waves, such as o).............................. is dangerous to life. Luckily, the p).............................. layer of the atmosphere absorbs most of the dangerous q)........................... rays from the Sun.

The energy changes occurring in a mobile phone call are as follows: Sound waves of your voice are converted to an ac)................................ signal. This is used to modulate a ad)..................................... wave transmitted to the local “cell” antenna. Next, the signal is sent via ae)......................................... link, or LASER beam to another “cell” station. Here it is transmitted again as a af)...................................... signal. The receiving phone converts this to an ag)................................... signal and finally it is converted back to sound waves.

All forms of radiation decrease in intensity in proportion to the r).......................................... from the source, so if distance is doubled, the intensity will drop to s)...................... (fraction) COMPLETED WORKSHEETS BECOME SECTION SUMMARIES

Worksheet 10 Practice Problems Inverse Square Law

Student Name...........................................

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

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

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

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

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

c) 100m?

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

d) 5m?

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Worksheet 11 Test Questions Multiple Choice

section 3

Student Name...........................................

Longer Response Questions Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. Answer on reverse if insufficient space.

1.

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.

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

2. The radiation from the Sun least likely to reach the Earth’s surface is: A. Infra-red B. visible light C. Ultra-violet D. radio waves.

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

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

b) X-ray c) infra-red

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

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

Modulated wave

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

The method of modulation used is: A. AM B. FM C. Pulse D. Digital.

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4. REFLECTION & REFRACTION When a Wave Hits a Boundary

Reflection of Light from Curved Mirrors

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

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

Example: Light waves travelling in air, then hitting glass. ABSORPTION of the energy Absorbed energy becomes heat

REFLECTION (bounces off) TRANSMISSION into the new medium, with possible REFRACTION (change of direction)

Focus

It is quite possible that all 3 things can happen at once. For example, if a beam of light is travelling through air, and then strikes a glass window: • the glass ABSORBS some of the light. • some REFLECTS off the glass • some is TRANSMITTED through the glass.

Concave mirrors can give ENLARGED images if viewed from the right distance, such as a household shaving mirror or make-up mirror, which gives a magnified reflection of your face. This is also the basis of a “reflecting telescope.”

Reflection The “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 of Incidence Reflection

io

=

In cid en tr ay

io

“Normal” line

ro

ro

ed ct fle y e R ra

“Convex” mirrors reflect light so the rays diverge outwards, as if coming from a focus behind the mirror.

Reflective surface such as a mirror

The trickiest bit is how the angles are measured. They must be measured between the rays and the “NORMAL”... an imaginary line at right angles to the surface.

“Virtual” Focus

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

The Incident rays P,Q & R are parallel. Each obeys the Law of Reflection, but the reflected rays go in different directions. The “Normal” for each ray is shown as a dotted line

smaller Convex mirrors produce (“diminished”) images, but give a widerangle view. An example of use is the side mirrors on a car which give you a wide-angle view into the driver’s “blindThis is why uneven, rough surfaces spot”. (BUT things look smaller. This can confuse a P

Q

don’t give “shiny” reflections. Light is scattered in all directions.

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Reflections in Communications Wave reflection from the ionosphere can help with long distance radio communications. It works best with the longer wavelength AM signals.

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

Ion os p lay here er

At the transmission end, a curved reflector keeps the waves in a tight beam aimed at the next relay station. The receiver has a similar dish to focus the waves into the receiving antenna.

EARTH Receiver Transmitter

Microwave Reflector Dishes

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

Microwave beam travels between relay stations

TV signals and FM (shorter wavelengths) radio do not reflect so well and generally you need to be in “line of sight” from the transmitter to get good reception.

Transmitter dish

Your satellite TV dish is a reflector too

Receiver dish

Refraction Refraction 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)

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

• The velocity slows. • The wavelength gets shorter. • The beam changes direction towards the normal.

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

Incident Ray io

normal

io

>

Angle of Incidence

ro

io

ro

Incident Ray

Refracted Ray Air

Glass

ro

Glass

normal

Air

In this case,

Angle of Refraction

io < ro

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. Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

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Snell’s Law

Refractive Index

You may have carried out an investigation in class using a “Ray Box Kit” to measure angles of incidence and angles of refraction.

When waves enter a new medium, and then exit it again, the refractions that occur on the way in, are the opposite of what happens on the way out.

When you graph the angles the result is a curve. Angle of incidence, io

For example, this light ray goes from air, into glass and out into air again. glass 45o Refraction air -> > glass

Angle of refraction,

28o

Refraction glass -> > air 28o

45o

ro

This is not much use for defining any relationship that may exist.

Refractive index (air -> glass)

In 1621, Willebrord Snell discovered that if you graph the Sine ratios of the angles, the points lie in a straight line. You may have done the same with your experimental data.

and

Sin io

normal

ng = sin45 / sin28 = 1.5

a

Refractive index (glass -> air)

na = sin28 / sin45 = 0.66

g

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

i Sin r = Sin e ris = run t ien ad Gr

n2

1

=1

n1

2

WORKSHEET at end of section Sin ro

The fact that it’s a straight line means there is a direct relationship between Sin i and Sin r. The gradient of the line is not only the ratio between the Sine of the angles, but is also equal to the ratio of velocities of the wave in the 2 mediums involved.

3 beams of light being refracted through a perspex block.

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

The spoon appears “broken” at the surface of the tea due to refraction of the light by which we see it.

This is now called Snell’s Law: Sine (angle incidence) = velocity (medium 1) = n Sine (angle refraction) velocity(medium 2) Sin i = V1 = Sin r V2

1n2

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... But What Happens Beyond the Critical Angle?

Total Internal Reflection & the Critical Angle

At incident angles larger than “c”, the ray reflects back inside the glass... this is called “TOTAL INTERNAL REFLECTION”

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

4

The waves refract away from the normal.

Ray reflects inside glass

Now think about increasing the incident angle as shown in this series of diagrams. 1

air

io>co

glass

io

ro

If io > co the ray cannot get out, but reflects back inside the glass

This has one very important application in communication technology...

2 bigger i, bigger r ro

Optical fibres are thin strands of very pure glass that can carry communications signals in the form of laser light beams. The laser beams stay within the fibres because of total internal reflection.

3

io

ro= 90o

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

io=co

Critical Angle

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

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

So, when the angle of incidence equals the “critical angle”, the angle of refraction is a right angle. If io = co, then ro= 90o Remember that

Sin i = Sin r

so at the critical angle

Sin c = Sin 90

Optical fibre laser b eam

gna

g

Core. high index

na

lower index outer layer

and sin 90o = 1, so...

Sin c = 1

gna

The laser light “bounces” around corners by total internal reflection

= 1 ang

This means that the Sine ratio of the critical angle “C” is equal to the reciprocal of the refractive index of the glass. WORKSHEET at end of section Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

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Worksheet 12

Reflection & Refraction

Fill in the blank spaces.

Student Name...........................................

When a wave meets the boundary between one medium and another, any of 3 things can occur: the wave’s energy can be absorbed, or the wave can be a)............................... or ............................ The Law of Reflection simply states that the angle of b)..................... equals the angle of c)........................... The angles must be measured from the wave “ray” to the d)................................. This is an imaginary line which is e)................................. to the boundary. Concave mirrors reflect light into a f)........................ point and can produce 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 this mirror is i).................................................. In communications, reflection is useful for long-distance radio reception. Some radio wavelengths reflect from the j)................................ layer in the upper atmosphere, and are “bounced” around the curvature of the Earth. Satellite “dishes” and k).................................... antennas use reflection to focus wave signals into the receiver. Refraction occurs when waves go from one medium into another. The waves may change in l)........................, and ........................... and ................................. Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

COMPLETED WORKSHEETS BECOME SECTION SUMMARIES

22

For example, when light goes from air into glass its speed m)......................, and its n)............................... gets shorter (although o)................................... does not change). It also changes direction, going p)............................. the normal. Snell’s Law describes the direct relationship between the Sine ratios of the angles of q).......................... and ................................... This ratio is called the r).......................... ...................... It is also equal to the ratio between the s)............................. of the wave in the 2 different mediums. The index for the wave entering the medium, and the index for the wave exiting the medium are always t)..................................... of each other. When a light ray is going from a “slower” medium into a “faster” one, the ray will refract u)............................ the normal. As the angle of incidence increases, so will the angle of refraction, until the refracted ray v).............................................. of the boundary. The angle of incidence at which this happens is called the w)............................... angle. At angles of incidence greater than this angle, x).............................................. .............. occurs, and the ray stays within the “slower” medium. This property is used in optical fibre technology to ensure that y).................................. beams stay within the fibres.

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Test Questions for this section are at the end of section 5

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Worksheet 13 Refraction

Practice Problems 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 the angle of incidence? c) Calculate the angle of refraction for this situation. d) How do you interpret this result?

Snell’s Law

Sin i = V1 = 1n2 Sin r V2 Example Problem A beam of light goes from air into a glass block with a refractive 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 in the glass.

Reciprocal Indices

n2

1

=1 2

n1

Solution: a) Sin i = n Sin r

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

sin 35 / sin r = 1.50

sin r = sin 35 /1.50 = 0.38238 therefore, angle of refraction, r = 22.5o b)

V1 = n V2

3.00x108 / V2 = 1.50

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

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

TRY THESE 1. In an experiment, a student sent a beam of light into a block of clear plastic. The angle of incidence was measured as 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 in the plastic.

8. In a type of lead-crystal glass, a light ray exits from the glass into air. At the interface, the angles 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 this glass?

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 of 40o from the normal, find the angle of refraction as the ray enters the diamond.

Critical Angle Sin c =

g

na =

1

a

ng

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

9. a) Use the information in Q2 to find the “critical angle” for light travelling inside a diamond. b) Describe what would occur (no calculation required) if light inside a diamond hit the boundary at an angle of incidence of: i) 20o ii) 30o

4. Several different angles of incidence and refraction were measured for a light ray going from air into a high density crystal 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 the crystal glass.

10. a) What is the critical angle for glass with ang = 1.50? b) Describe (no calculation) what happens when light inside the glass strikes the boundary at angle of incidence: i) 40o ii) 41.8oiii) 45o

DATA Angle of Incidence 50.0 42.0 30.0 65.0

11. Light travelling inside a plastic block strikes the boundary at an angle of incidence = 48.6o. The refracted ray is seen to run exactly along the boundary 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 this plastic?

Angle of Refraction 25.0 21.0 17.0 31.0

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5. DIGITAL COMMUNICATION & DATA STORAGE Digital Technology

TECHNOLOGY CASE STUDY:

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

GLOBAL POSITIONING SYSTEM (GPS) GPS is a system that allows a ship, aircraft, car or bushwalker, to locate their exact position anywhere on Earth instantly and continuously. The system was developed for miltary uses, but then made available to anyone. The military version is thought to be accurate to within a metre, the civilian version to within about 10 m.

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 for processing, storage or transmission and communication. A simple list of some technologies involved is:

of

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

the

CD’s & DVD’s, Mobile phones, Digital cameras, Computers & Internet, MP3 music, ATM’s GPS

Satellite orbits

Satellite 1

GPS receiver

Earth Satellite 3

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

COMMUNICATION.

Satellite 2

GPS

Each satellite constantly sends out microwave signals identifying itself, its orbit details and the precise time the signal was sent. When your portable GPS receiver picks up the signal, it can calculate your exact distance from the satellite, from the time delay since the signal was sent. By doing the same for 2 other satellites, the GPS unit rapidly “triangulates” the signals from 3 satellites to pin-point your location on the Earth’s surface. (Aircraft need a 4th signal to get their altitude) GPS systems for cars show your position on a screen overlaid onto a road map of the area. As you drive around, the system constantly shows your changing position, and can advise you where to turn to reach your destination.

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Worksheet 14 Test Questions

sections 4&5

Multiple Choice

Longer Response Questions Mark values shown are suggestions only, and are to give you an idea of how detailed an answer is appropriate. Answer on reverse if insufficient space.

1. An example of REFLECTION being helpful in communication 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”.

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

9. (6 marks) In an experiment, angles of incidence and refraction were measured as shown.

33o

a) Find the refractive index of the plastic. air Show working.

3. If a light ray passed from air into one of the following substances, (each at the same angle of incidence) which one would 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)

25o

plastic

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

4. Light travels in air at 3.0x108ms-1. If the refractive index of 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.

10. (4 marks) plastic. Predict the path of this light ray n = 1.40 after it strikes the boundary. Explain your reasoning, and show any working.

30o air

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

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

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

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Q R

2. Which of the following does NOT change when a wave undergoes refraction? A. Velocity B. Direction C. Wavelength D. Frequency

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.

Student Name...........................................

Remember that for full marks in calculations, you need to show FORMULA, NUMERICAL SUBSTITUTION, APPROPRIATE PRECISION and UNITS

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CONCEPT DIAGRAM (“Mind Map”) OF TOPIC Some 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 WORLD COMMUNICATES

Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

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keep it simple science

Worksheet 4

Answer Section Worksheet 1

i) A = ii) d= iii)T= iv) f=

a) energy b) matter c) medium d) sound & water waves e) vacuum f) radio, light, UV, etc g) Transverse h) at right angles i) in the same line j) Wavelength k) maximum displacement, from equilibrium position l) waves / complete vibrations m) Hertz (Hz) n) The Period o) reciprocal p) frequency q) wavelength r) amplitude s) period t) frequency u) decrease v) wavelength

330

V= f λ = 280 x 5.20 = 1460 m.

Worksheet 5 1. C

b) T = 1 / f = 1 / 280 = 0.00357 s (3.57 x10-3 s)

3. A

4. A

5. A

6. C

9. a)

V=f λ 1,500 = 400 x λ λ= 1,500 / 400 = 3.75 m. b) T = 1/f = 1/400 = 0.0025 = 2.5x10-3 Hz

4 a) f = 1 / T f = 1 / 16 = 6.25x10-2 Hz b) V= f λ 6.50= 6.25x10-2 x λl λ= 6.50 / 6.25x10-2 = 104 m c) V= f λ 2.20 = 6.25x10-2 x λ λ= 2.20 / 6.25x10-2 = 35.2 m Wavelength has become a lot shorter as the wave entered shallower water. 1. a) V= f λ,

2. D

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

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

Worksheet 3

R 0.05 m -0.05 m 0.04 s 25 Hz

7. SOUND --> ELECTRICAL --> RADIO

2. V= f λ 2,500 = 0.400 x λ l= 2,500 / 0.400 = 6,250 m (6.25 x103 m) 3.

Q 0.10 m -0.1 m 0.16 s 6.25 Hz

v) V= f λ = 12.5 x 0.50 = 6.25 ms-1 vii) λ = V / f wave Q: = 9.5 / 6.25 wave R; =9.5 / 25 = 1.5 m = 0.38 m

Worksheet 2 1. a)

wave P 0.15 m 0.1 m 0.08 s 12.5 Hz

10. a) from graph, T = 16 s. and f = 1/T = 1/16 = 6.25x10-2 Hz b) V = f λ 12.5 = 6.25x10-2 x λ λ = 12.5 / 6.25x10-2 = 200 m.

f = V / λ = 3.00x108 / 7.00x10-7 = 4.29x1014 Hz

Worksheet 6

b) f = V / λ = 3.00x108 / 3.00x10-7 = 1.00x1015 Hz

Using the sum of displacements at the circled points, the resultant looks approximately like the solid curve shown.

2.a) T= 1 / f = 1/ 53,000 =1.89x10-5 s. b) λ= V / f = 3.00x108 / 53,000 =5.66x103 m. (over 5 km!) 3. 2.50 m wave: f = V / λ = 3x108 / 2.50 = 1.20x108 Hz 2.50 cm wave: f = V / λ = 3x108 / 0.0250 = 1.20x1010 Hz comparison: The frequency of the 2.5cm wave is 100 times higher than the 2.5m wave. (Makes sense: 100X shorter wavelength --> 100X higher frequency)

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

4. Since 8 complete wavelengths fit in 0.96m, then λ = 0.12 m V= f λ = 384 x 0.12 = 46 ms-1 5. a) f = V / λ

= 3.00x108 / 1.50x10-11 = 2.00 x 1019 Hz b) T = 1 / f = 1 / 2x1019 = 5.00x10-20 s.

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Worksheet 7

Worksheet 10

Part A a) mechanical b) longitudinal c) compressions d) rarefactions e) 330 f) higher g) frequency h) loudness / volume i) reflect j) navigation / hunting prey k) SONAR l) detecting fish / submarines m) resultant n) amplitudes

1.

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 same amount of light.

Worksheet 8 1. B

2. C

3. A

4. summing displacements at circled points:

Worksheet 11 1. D

2. C

3. C

4. B

displacement

5. (lowest) Radio, microwaves, infra-red, light, ultraviolet, X-ray, gamma (highest) 6. a) human eye. b) X-ray sensitive photo film. c) receptors in human skin.

time

7. At 1/10 the distance it will be 100X brighter. 0.1 x 100 = 10 units. resultant (approx)

8. There are so many radio & TV stations, mobile phone systems and users, as well as 2-way radio for aircraft, shipping, military, taxis, etc, that the available radio “bands” are becoming congested.

5. With a sound wave, a “compression” is where air particles are pushed together, and a “rarefaction” is where they are spread apart more, as the wave moves through.

There is a limit to how many systems and stations can operate without overlapping in frequencies and causing interference to each other.

Worksheet 9 a) transverse c) medium e) radio g) infra-red i) ultra-violet k) gamma rays m) resonance o) UV / X-rays / gamma q) UV s) one quarter u) laser beams w) modulation y) amplitude aa) frequency ac) electrical ae) microwave ag) electrical

b) do not d) 3 x 108 f) microwaves h) visible light j) X-ray l) vibrate / oscillate n) vibrate p) ozone r) square of distance t) radio & microwaves v) optical x) Amplitude Modulation z) Frequency Modulation ab) Pulse ad) radio af) radio

Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

Worksheet 12 a) reflected or refracted b) incidence c) reflection d) normal e) perpendicular f) focal g) convex h) smaller/diminished i) driver’s side mirror j) ionosphere k) microwave l) direction, wavelength & velocity m) slows down n) wavelength o) frequency p) towards q) incidence & refraction r) refractive index s) velocities t) reciprocals u) away from v) goes along the edge w) critical x) Total Internal Reflection y) laser

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keep it simple science 8. c)

n = V1 / V2 1.6 = 3.0x108 / V2 V2 = 3.0x108 / 1.6 V(glass) = 1.9x108 ms-1

Worksheet 13

Snell’s Law Problems 1.a) n = Sin i / Sin r = sin50 / sin 33 n=1.4 (no units) b) n= V1 / V2 1.4 = 3.0x108 / V2 V2= 3.0x108 / 1.4 = 2.1 x 108 ms-1 2. a) n=V1 / V2 = 3.00x108 / 1.25x108 = 2.40 b) n = Sin i / Sin r 2.40 = sin40 / Sin r Sin r = sin40 / 2.40 = 0.2678... r = 15.5o

Critical Angle & T.I.R. 9. a)

= 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.

3. a)

n = Sin i / Sin r 1.33 = Sin i / sin32.5 Sin i = 1.33 x sin32.5 = 0.7146... i = 45.6o b) n = V1 / V2 1.33= 3.00x108 / V2 V2 = 3.00x108 / 1.33 = 2.26x108 ms-1. 4. a)

Angle Angle Incidence Refraction 50.0 25.0 42.0 21.0 30.0 17.0 65.0 31.0 b) Average =7.15/4 = 1.79 c) n = V1 / V2 1.79 = 3.00x108 / V2 V2 = 3.00x108 / 1.79 = 1.68x108 ms-1.

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

Refractive Index 1.81 1.87 1.71 1.76

Worksheet 14 1. A 8.

b)

3. A P

4. B

5. C

6. C

7. D

Q

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 / V2 V2= 3.0x108/1.3 V (in plastic)= 2.3x108 ms-1. 10. Note that the angle given is not the correct angle of incidence. (Angle of incidence must be measured from 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 internal reflection, and reflect back inside the plastic block. (at angle of reflection = 60o)

= 1 / 1.4 = 0.71 (no units)

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

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

7. a) 1n2 = 1 / 2n1 = 1 / 1.33 = 0.752 (no units) b) n = Sin i / Sin r (must use the index for 0.752 = 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)

Preliminary Physics Topic 1 “World Communicates” Copyright © 2005-2 2009 keep it simple science www.keepitsimplescience.com.au

2. D

R

Reciprocal Indices Problems 1n2 = 1 2n1

dna

10. a) Sin c = 1 / n = 1 / 1.50 = 0.6666... c = 41.8o b) i) Refracts out into air ii) Refracts and runs along the glass edge iii) i > c, total internal reflection.

5. a) n = V1 / V2 1.50 = 3.00x108 / V2 V2 = 3.00x108 / 1.50 = 2.00x108 ms-1. b) 0o c) n = Sin i / Sin r 1.50 = sin0 / Sin r Sin r = sin0 / 1.50 = 0 r = 0o d) The ray does NOT change direction. (However, it would still change velocity and wavelength)

6. a)

Sin c =

By comparing the signals from (at least) 3 different satellites, the GPS receiver can “triangulate” to find its position with a high degree of accuracy... within a few metres in some cases.

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