Page 1 of 92

Unit 2

Page 2 of 92

Syllabus Topic Pages

28 - 35 Mechanical & Electromagnetic Waves 3 - 21

36 - 38 Refraction 22 - 26

39 - 40 Polarisation 27 - 29

41 Diffraction 30 - 32

42 - 43 Wave Nature of Electrons 33 - 34

44 – 49 Pulse Echo Techniques 35 - 40

50 Charge and Current 41 – 44

51- 56 Voltage, Current, Resistance & Power 45 – 58

57 Resistivity 59 – 61

59 EMF & Internal Resistance 62 - 65

58, 60 - 62 Potential Divider 66 – 72

63 -67 Wave/Particle nature of light 73 – 81

68 Spectra & Energy Levels 82 – 88

69 – 70 Radiation Flux 89 - 92

Page 3 of 92

Waves

Mechanical waves

These would be set up within a solid, a liquid or a gas due to the vibration of the molecules there. They are transmitted by both intermolecular forces and by collisions

between the molecules. In solids they could be either longitudinal or transverse vibrations while in a liquid or a gas only longitudinal vibrations are really possible.

Examples of these waves would be the P and S waves in the Earth's crust due to an earthquake and sound waves in air.

Earthquake waves.SWF

Electromagnetic waves

They all travel at the same speed 3.00 x 108 m s-1.

They do not require a medium through which to pass.

They are generated by accelerating charged particles.

Electromagnetic waves.SWF

Gamma X-rays U-V Visible I-R Microwaves Radio waves

10-15 10-9 4x10-7 4x10-7- 7x10-7 10-6 10-4 - 10-1 0.1 – 103 m

Electromagnetic waves share many of the general wave properties of mechanical waves:-

they transfer energy from place to place can be reflected can be refracted

can be superposed can be diffracted

ripple.exe

Refraction.SWF

Page 4 of 92

Wave motion

A wave motion is the transmission of energy from one place to another through a

material or a vacuum. Wave motion may occur in many forms such as water waves, sound waves, radio waves and light waves, but the waves are basically of only two types:

Sound in Helium.avi

(a) transverse waves - the oscillation is at right angles to the direction of

propagation of the wave (Figure 1(a)). Examples of this type are water waves and most electromagnetic waves.

Transverse wave generation.SWF

Waves, peaks & troughs.SWF

http://www.acoustics.salford.ac.uk/feschools/waves/waves.htm

(b) longitudinal waves - the oscillation is along the direction of propagation of the

wave (Figure 1(b)). An example of this type is sound waves in air.

Longitudinal wave creation.SWF

http://www.acoustics.salford.ac.uk/feschools/waves/waves.htm

Wavelength ()

Wavelength ()

y0

y0

Figure 1(a)

wavelength

wavelength

wavelength

Figure 1(b)

Page 5 of 92

Basic definitions: Wavelength: the distance between any two successive corresponding points that are

vibrating in phase

Displacement: the distance from the mean, central, undisturbed position at any point

on the wave (y)

Amplitude: the maximum displacement (y0)

Frequency: the number of vibrations per second made by each particle/wave

Period: the time taken for one complete oscillation (T= 1/f)

Phase: the „delay‟ between the oscillations of neighbours

Wave terms.SWF Amplide, wavelength & phase.SWF

Wave Speed

The number of oscillations per second of each part of the medium is the

FREQUENCY f. It also equals the number of complete waves passing any place in one second.

If f waves per second go past a place, and the wavelength is , then the distance travelled by the waves per second (i.e. the WAVE SPEED, c) is given by the

equation

fc

Sound in Helium.avi

Graphical representation of longitudinal waves

Longitudinal wave particle vibrations.GIF

Longitudinal waves.GIF

Longitudinal wave pause.SWF

Longitudinal wave.SWF

1 At large sports meetings the crowds sometimes produce a „Mexican wave‟.

(a) What would be a better name for this manoeuvre?

(b) Describe what the crowd would need to do to produce such a wave and suggest values for its frequency and amplitude.

Page 6 of 92

2 The diagram shows a transverse wave on a rope. It is moving to the right. Several particles on the rope have been labelled.

(a) On the diagram draw arrows to show the direction in which particles P,

R and T are moving. (b) What can you say about the motion of Q and S?

(c) Mark on your diagram two particles that are (i) in phase, i.e. moving together - call them P and P‟ (ii) in antiphase, i.e. moving oppositely- call them A and A '.

3 Describe, with the aid of a sketch, an electromagnetic wave.

4 Who will hear the singer first - a person who is 45 m from the stage, or a person watching the concert on TV at home 2400 km away?

Assume that the microphone is very close to the singer and that the viewer is sitting close to the TV.

Speed of electromagnetic waves = 3.00 x 108 m s-1

Speed of sound in air = 340 m s-1

5 Diagram (i) represents part of a stretched spring. Diagram (ii) represents the

same section of the spring at one instant of time when a sinusoidal longitudinal wave is travelling along it.

(a) Use the diagram (ii) to determine the wavelength of the longitudinal wave.

(b) The wave speed is 2.00 m s-1. Calculate the frequency of this wave.

(c) Describe qualitatively the motion of an individual coil of this spring as the

longitudinal wave travels along the spring.

Page 7 of 92

6 The diagram shows the shape of a wave on a stretched rope at one instant of time. The wave is travelling to the right.

(a) Determine the wavelength of the wave.

(b) Mark on the diagram a point on the rope whose motion is exactly out of phase with the motion at point A. Label this point X.

(c) Mark on the diagram a point on the rope which is at rest at the instant shown. Label this point Y.

(d) Draw an arrow on the diagram at point C to show the direction in which

the rope at C is moving at the instant shown. (e) The wave speed is 3.2 m s

-1. After how long will the rope next appear

exactly the same as in the diagram above?

7 A microwave generator produces plane polarised electromagnetic waves of wavelength 29 mm.

(a) (i) Calculate the frequency of this radiation. (ii) Complete the diagram of the electromagnetic spectrum below

by adding the names of the parts of the electromagnetic

spectrum.

(iii) State a typical value for the wavelength of radiation at boundary X.

Page 8 of 92

8 The table below summarises some features of the electromagnetic spectrum.

Complete the table by filling in the missing types of radiation, wavelengths and sources.

Radiation Typical

wavelength Source

Visible light Very hot objects

Gamma

100 m High frequency

oscillator

10-6

m

Page 9 of 92

Standing waves

A stationary or standing wave is one in which the amplitude varies from place to

place along the wave. Figure 1 is a diagram of a stationary wave. There are places where the amplitude is zero and, halfway between, places where the amplitude is a maximum; these are known as nodes and antinodes respectively.

Any stationary wave can be formed by the addition of two travelling waves moving in opposite directions. http://www.ngsir.netfirms.com/englishhtm/TwaveStatA.htm

(Transverse Standing wave)

string1.swf

string2.swf

Standing waves 1.SWF

string3.swf

string4.swf

string5.swf

string6.swf

string7.swf

Stationary waves 2.SWF

string8.swf

string9.swf

Tacoma Narrows.SWF

http://www.enm.bris.ac.uk/anm/tacoma/tacoma.html#mpeg (Tacoma Narrows Bridge)

A node is a place of zero amplitude An antinode is a place of maximum amplitude

Fig 1

X

N N N N N N

A A A A A A

a1 a3 a2 Y

N

Page 10 of 92

9 The diagram shows a wire with a mass of 1.30 kg at one end and a vibrator at the other end.

(a) How many wavelengths does the diagram show?

(b) Use your answer to calculate the wavelength of the stationary wave.

(c) The frequency of the vibrator is 250 Hz. Calculate the speed of the waves.

(d) Point N in the diagram is a node. On the diagram, mark with an A an

antinode. (e) State one difference between a node and an antinode.

(f) Explain how a node is formed from two progressive waves. (g) In energy terms, what is the difference between a standing wave and a progressive wave?

10 (a)

Sketch three stationary patterns that could be formed on the cord in the

diagram. Mark the nodes and antinodes in each sketch.

(b) If in one of your sketches the distance between nodes is 0.55 m when the signal generator frequency is 40 Hz, what is the speed of the waves on the string?

11 The natural frequencies f of vibration of the string in Q2 can be found by

putting n = 1,2,3, etc. in the formula

Tnf

2

where is the length of the string, T the tension in the string and its mass per

unit length. (a) Show that the units of the right hand side of the formula are s-1 or Hz.

(b) Using the arrangement in the diagram explain how you would show

experimentally that f was inversely proportional to for a fixed value of n, e.g. n = 2.

0.90 m

Vibrator

N

1.30 kg

Page 11 of 92

12 (a) State two differences between a stationary wave and a progressive wave.

Difference 1 Difference 2

(b) Spiders are almost completely dependent on vibrations transmitted through their webs for receiving information about the location of their prey. The threads of the web are under tension. When the threads are

disturbed by trapped prey, progressive transverse waves are transmitted along the sections of thread and stationary waves are formed.

Early in the morning droplets of moisture are seen evenly spaced along the thread when prey has been trapped.

(i) Explain why droplets form only at these points.

(ii) The speed of a progressive transverse wave sent by trapped prey along a thread is 9.8 cm s-1. Use the diagram to help you

determine the frequency of the stationary wave.

13 A stationary wave is produced on a stretched string by a vibration generator attached to one end. The graph shows part of the wave. The two full lines represent the extreme positions of the string.

(a) State the wavelength of this wave (b) Mark a letter A on the graph to label an antinode.

(c) The stationary wave is formed by the superposition of two waves travelling along the string in opposite directions. The frequency of the

vibrator is 36.0 Hz. Calculate the speed of the travelling waves. (d) State the phase relationship between the two travelling waves at an

antinode.

(e) Determine the amplitude of each of the travelling waves.

Page 12 of 92

14 (a) Describe, with the aid of a diagram, an experiment to demonstrate stationary waves using microwaves.

(b) Using the idea of wave superposition, explain what is observed in your experiment.

(c) Describe how you could use the experiment to measure the wavelength of microwaves.

15 A piece of string is connected to a variable frequency vibration generator. The fundamental frequency of this system is 60 Hz.

(a) Complete the table to show what would be observed as the frequency is gradually increased from 40 Hz to 180 Hz.

16 The cello is a stringed musical instrument that may be played either by

stroking the strings with a bow or by plucking the strings with the fingers.

Page 13 of 92

(a) One of the attached strings on the cello has a vibrating length of 0.80 m. The string is made to oscillate as a stationary wave by means

of a bow and the following pattern of oscillations is seen. The position of the string at two different times is shown.

(i) Explain how the movement of the bow causes this wave

pattern.

(ii) Using the diagram calculate the wavelength of the wave. (iii) State two differences between the wave on the string and the

sound wave it produces.

(b) The cello string is then plucked and the waveform of the resulting

sound is analysed by an oscilloscope. It is found to consist of two frequencies of different amplitudes.

The frequency spectrum is shown below,

The waveform of the 200 Hz wave has been drawn on the axes below. On the

same axes sketch the waveform of the 1000Hz wave.

Page 14 of 92

Superposition

When two groups of waves (called wave trains) meet and overlap they „interfere‟ with each other. The resulting amplitude will depend on the amplitudes of both the waves

at that point.

If the crest of one wave meets the crest of the other the waves are said to be in phase and the resulting intensity will be large. This is known as constructive interference. If the crest of one wave meets the trough of the other they are said to be out of phase

by 21 then the resulting intensity will be less/zero (if the waves have equal

amplitudes). This is known as destructive interference. http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=19 (Superposition Principle)

Superposition principle.GIF

This phase difference may be produced by allowing the two sets of waves to travel

different distances - this difference in distance of travel is called the path difference between the two waves.

doubleslit.exe

The diagrams in Figure 1 below show two waves of equal amplitudes with different

phase and path differences between them. The first pair has a phase difference of 21

or 180o and a path difference of an odd number of half-wavelengths. The second pair have a phase difference of zero and a path difference of a whole number of

wavelengths, including zero.

Figure 1 destructive interference constructiv e interf erence

+ = + =

Page 15 of 92

Maximum – crest meets crest

Minimum – crest meets trough

Maximum – crest meets crest

Maximum – crest meets crest

Minimum – crest meets trough

Minimum – crest meets trough

Minimum – crest meets trough

Figure 3

S1

S2

To obtain a static interference pattern at a point (that is, one that is constant with time) we must have

two sources of the same wavelength, and

two sources which have a constant phase difference between them.

Sources with synchronised phase changes between are called coherent sources and

those with random phase changes are called incoherent sources. This condition is met by two speakers connected to a signal generator because the

sound waves that they emit are continuous – there are no breaks in the waves. However two separate light sources cannot be used as sources for a static interference

pattern because although they may be monochromatic the light from them is emitted in a random series of pulses of around 10-8 s duration. The phase difference that may

exist between one pair of pulses emitted from the source may well be quite different from that between the next pair of pulses (Figure 2).

Therefore although an interference pattern still occurs, it changes so rapidly that you get the impression of uniform illumination. Another problem is that the atoms

emitting the light may collide with each other so producing phase changes within one individual photon. We must therefore use one light source and split the waves from it into two.

Diagrams in Figure 3-7 show two sources S1 and S2

emitting waves - they could

be light, sound or microwaves.

The plan view of the waves in Figure 3 shows waves

coming from two slits and „interfering‟ with each other.

The lines along which the path differences will give maxima or minima.

Pulses from source A

Pulses from source B

Figure 2

Page 16 of 92

maximum

maximum

maximum minimum

minimum

Figure 5

Figure 4 (a)

Figure 4 (b)

This type of arrangement is like that produced in a ripple tank or in the double slits experiment with light (see later).

Figure 4(a) shows light

interfering as it passes through two slits. In Figure 4(a) the appearance of the interference

pattern on a screen placed in the path of the beam is shown.

You can see the maxima and minima and the way in which

the intensity changes from one to the other.

Changing the wavelength of the light (its wavelength), the

separation of the slits or the distance of the slits from the

screen will all give changes in the separation of the maxima in the interference pattern.

Two dippers Ripple tank.GIF

Two source Max & Min.SWF

Young's slits Max & Min.SWF

Figure 5 shows the interference

effects of two speakers. The sound waves spread out all round the

speakers and a static interference pattern is formed. (Not all the maxima and minima are labelled).

You can hear this by setting up two speakers in the lab connected to one

signal generator and then simply walking round the room. You will hear the sound go from loud to soft

as you pass from maximum to minimum.

Page 17 of 92

shallow deep

In Figures 6 and 7 you can see that at the different points on the screen the waves from S1 have travelled a different distance from those from S2. In Figure 6 the path

difference is zero, in Figure 7 it is half a wavelength

Interference

The effect of amplitude

If two sources have very different amplitudes, then you will not be able to observe superposition patterns.

The wave with the larger amplitude will dominate, because it makes little difference to the total whether the wave with the smaller amplitude is in phase or out of phase with it.

Wavefronts

All points on a wavefront vibrate in phase. If you dip

your finger in the water of a ripple tank you will notice that circular ripples spread out.

This is because waves travel at the same speed in all

directions.

If we tilt the tank so that the depth of the

water now varies:

The wavefront has an oval shape because waves travel faster in deep water than in shallow water.

Figure 7

S1

S2 screen

Path difference = /2

Figure 6

Path difference = 0

screen

S1

S2

Page 18 of 92

Phase Differences

Places marked and oscillate with a phase difference

2x

http://www.acoustics.salford.ac.uk/feschools/waves/super.htm

Phase difference.SWF

speed of light as = 3.00 x 108 ms-1 speed of sound in air = 330 ms-1.

17 (a) What is meant by the principle of superposition of waves?

(b) Are there any types of wave which do not obey the principle or any circumstances in which it does not apply?

18 The diagram shows an experimental arrangement for investigating the superposition of sound waves from two sources S1 and S2. The sources are in phase and produce sound of wavelength 80 mm.

(a) When S1M is 800 mm the trace on the oscilloscope is a maximum.

Suggest three possible values for S2M.

Without moving any of the apparatus the leads to one of the speakers are reversed, i.e. the sources are now in antiphase.

(b) Explain the meanings of in phase and in antiphase and describe how the trace on the oscilloscope changes when the leads are reversed.

19 Why can‟t you get a static interference pattern with two light bulbs while it is

possible with two loudspeakers.

x

Page 19 of 92

20 A motorist drives along a motorway at a steady speed of 30 ms-1. There are radio transmitters at each end of the motorway. She is listening to the car

radio and as she travels along she notices that the radio signal varies in strength, 5s elapsing between successive maxima. Explain this effect and

calculate the wavelength of the radio signal that she is tuned to.

21 The diagram shows an arrangement with sound waves.

A loudspeaker connected to a signal generator is mounted, pointing

downwards, above a horizontal bench. The sound is detected by a microphone connected to an oscilloscope. The height of the trace on the

oscilloscope is proportional to the amplitude of the sound waves at the microphone.

When the vertical distance x between the microphone and the bench is varied,

the amplitude of the sound waves is found to vary as shown on the graph.,

(a) Explain why the amplitude of the sound has a number of maxima and

minima.

(b) The frequency of the sound waves is 3.20 kHz. Use this, together with

information from the graph, to determine a value for the speed of sound in air.

(c) The contrast between the maxima and minima becomes less pronounced

as the microphone is raised further from the surface of the bench. Suggest

an explanation for this.

Page 20 of 92

22 The diagram is a plan view of an experiment to measure the wavelength of microwaves. The diagram is to scale but one third of full size.

(a) As a microwave detector is moved around the arc from A to B, alternate

maxima and minima of intensity are observed. Explain why.

Page 21 of 92

(b) A maximum is observed at point 0, and the next maximum at point X. By means of suitable measurements on the diagram, determine the wavelength of

the microwaves.

(c) A teacher demonstrating this experiment finds that, even at the maxima, the wave intensity is small. A student suggests making the slits wider to let more energy through. Explain why this might not be a good idea.

(c) For an interference pattern to be observed between waves from two sources,

the sources must be coherent. Explain what is meant by coherent, and what makes the two sources in this experiment coherent.

Page 22 of 92

Refraction

Change of Speed For all refracted waves the path is deviated away from the normal when the speed

increases and towards the normal when the wave is slowed down.

Light moves slower in medium 1 than in medium 2.

2

1

2

121

sin

sin

c

c

Since the path of the light is reversible

1

2

1

212

sin

sin

c

c

Refraction occurs for all waves. Sound can be deviated as it passes from warm air to cooler air and microwaves can be refracted by wax.

2

1

Medium 1

Medium 2

2

1

Speed c2

Speed c1

Page 23 of 92

Refraction – wave fronts

Refraction.SWF

Wavefronts animation

The following diagram shows the refraction of a plane wave at a plane interface. The

position of the refracted wave is formed using the idea of secondary wavelets

One side of the wave moves from A to C (a distance vgt) in glass in the same time that

the other side of the wave front moves the form B to D (a distance vat) in the same time in air. The wave front recombines at CD.

g

a

g

a

g

aga

v

v

AD

tvAD

tv

sin

sin

vat

vgt

air

glass

a

g

A

D

C

B

Page 24 of 92

Total internal reflection and critical angle

When light passes from a material such as water into one of lower refractive index

such as air it is found that there is a maximum angle of incidence in the water that will give a refracted beam in the air, that is, the angle of refraction is 90o. The angle of

incidence in the denser medium corresponding to an angle of refraction of 90o in the less dense medium is known as the critical angle (c) (Figure 1). The reason for this is clear if we consider the formulae. For an angle of refraction of 90o we have:

211

212

1sin

90sin

sin

sin

sin

c

c

Example problem

The refractive indices from air to glass and from air to water are 1.50 and 1.33

respectively. Calculate the critical angle for a water-glass surface.

13.150.1

1.

1.

sin

sin.

sin

sin

sin

sin

sin

sin,

sin

sin

wa

ga

waag

w

a

a

g

w

g

wg

w

awa

g

aga

Total internal ref lection

>c c

medium one

medium tw o

Figure 1

<c

a

g

w w

g

a

Air

Water

Glass

Air

Page 25 of 92

Therefore the critical angle for light passing from glass to water is

5.62

50.1

13.1

90sin

sin

c

cwg

For an air-glass boundary

42

50.1

11sin

c

cga

And for an air-water boundary

5.48

13.1

11sin

c

cwa

For angles of incidence greater than the critical angle all the light is reflected back

into the optically more dense material, that is, the one with the greater refrac tive index. This is known as total internal reflection and the normal laws of reflection are

obeyed. Total internal reflection explains the shiny appear-

ance of the water surface of a swimming pool when viewed at an angle from below. The phenomenon is

used in prismatic binoculars (Mirages are caused by continuous internal reflection.)

1 Blue light is deviated more than red light when it enters a glass block because:

A it has a longer wavelength B it has a lower frequency

C it travels at a greater speed in glass than red light D it travels at a lower speed in glass than red light

2 The diagram shows how a narrow beam of light strikes a layer of oil on the surface of a tank of water at an angle of 58.0.

Refractive index of oil = 1.28 Refractive index of water = 1.34

Calculate

(a) The angle of refraction in the oil (b) The angle of refraction in the water (c) The angle of refraction in the water if the layer of oil was removed.

Figure 2

glass

58

Page 26 of 92

3 Draw a diagram showing wavefronts when light passes from air into glass.

50.1ga at incident angle of 45.

4 (a) A ray of light enters one side of a rectangular glass block at incident

angle of 40. Calculate the angle of refraction 55.1ga .

(b) The opposite side of the block is immersed in a clear liquid. The angle

of refraction is 28 when the ray passes into the liquid. Calculate the

refractive index of the liquid a .

5 Calculate the critical angle for the interface between the core of an optical

fibre with refractive index 1.60 and the cladding with refractive index 1.52.

6 Rainbows are caused when sunlight is dispersed by raindrops. The

different colours follow separate paths.

The diagram shows some of the rays of light passing through a raindrop.

(a) Name the process which occurs at A.

(b) The ray at B is actually only partially reflected at the surface of the water. Continue the ray to show the path of the red light which is not

reflected.

(c) Explain the condition that would be required to prevent the red light from emerging at B.

Light changes its direction at A because of a change of speed on entering the water.

(d) Red light has a frequency of 4.2 × 1014 Hz. Calculate its wavelength in a raindrop.

Speed of red light in water = 2.2 × 108 m s–1.

Sunlight

Violet

Red

Raindrop

A

B

Page 27 of 92

Polarisation If you get hold of one end of a rubber rope, tie the o ther end to a post, stretch it and then send a series of pulses down the rope the vibration travels down the rope.

Although each successive pulse may be sent in a different plane each pulse only vibrates in one direction.

A wave in which the plane of vibration is constantly changing is called an unpolarised wave.

However if the vibrations of a transverse wave are in one plane only then the wave is said to be plane polarised.

Light is plane-polarised when the vibrations are made to occur in one plane only.

Light is a transverse electromagnetic wave with the vibrations of an electric and a magnetic field occurring at right angles to each other and in any plane at right angles

to the direction of travel of the light. Polarisation is easily observed with the rubber rope experiment described above but it

can also be shown with electromagnetic waves such as microwaves, TV, radio and light.

Electric field (E)

Magnetic field (B)

Unpolarised radiation Plane polarised radiation

Figure 1

It is important to realise that transverse waves can be polarised

while longitudinal waves cannot.

Electric field

Magnetic field

e.m. radiation

Figure 2

Page 28 of 92

Effects of polarisation with light.

Sunglasses Car windscreens Polarisation by reflection - glare/shine off roads Optical activity Polarisation of scattered sunlight Stresses in materials

The effect of a polariser and an analyser is shown in the following diagrams.

Polarisation of transverse waves.SWF

Polarisation of microwaves.SWF

Sensation of light polarization

Our own eyes are poor at sensing light polarization but very good at sensing its colour

and brightness; however, sensation of light polarization is not at all unusual in the animal kingdom.

It is found in some insects, crustaceans, fish, birds, and particularly in cephalopods, a class of molluscs that includes squids, octopuses and cuttlefish. A key requirement for polarization sensitivity is the excitation of two or more classes

of visual pigments that have different alignments in the eye (technically, different axes of maximal excitation). Comparison of the neural inputs from these two rows

allows the animal to sense polarized light.

Stomatopod1.JPG

Stomatopod2.GIF

The images in the animation were taken using a polarizing filter - we wouldn‟t see the

flashing red and white signals with only the naked eye.

Light reflected off shiny surfaces may be polarised. http://www.colorado.edu/physics/2000/applets/polarized.html

Photoelastic stress analysis

When polarised light passes through some transparent materials, the plane of polarisation is rotated. If the material is put under stress, the amount of stress affects the

degree of polarisation. If the incident polarised light is white, each of its component

colours is rotated by a different amount, creating a pattern of coloured fringes when viewed through the analyser polaroid. Engineers make models of structural components out of materials such as Perspex.

The model is then stressed, and the pattern of coloured fringes is examined in order to identify regions of high stress.

Page 29 of 92

1 What is meant by plane polarized light?

2 Explain whether the human eye can distinguish polarized light from

unpolarised light.

3 How could you tell if a television signal was plane polarized and also find out the direction of polarization?

4 Why do yachtsmen wear Polaroid sunglasses?

5 (a) Explain with the aid of a diagram why transverse waves can be plane polarised but longitudinal waves cannot be plane polarised.

(b) (i) A filament lamp is observed directly and then through a sheet of Polaroid. Describe and explain the effect of the sheet of

Polaroid on the intensity of the light seen.

(ii) The sheet of Polaroid is now rotated in a plane perpendicular to

the direction of travel of the light. What effect, if any, will this have on the intensity of the light seen?

6 Describe, with the aid of a diagram, how you would demonstrate that these

microwaves were plane polarised.

7 (a) Describe how you would demonstrate experimentally that light waves

can be polarised, using either light or microwaves. Include a diagram of the apparatus you would use.

(b) What does the experiment tell you about the nature of electromagnetic waves?

Page 30 of 92

Diffraction When a wave hits an obstacle it does not simply go straight past, it bends round the obstacle. The same type

of effect occurs at a hole - the waves spread out the other side of the hole. This phenomenon is known as

diffraction and examples of the diffraction of plane waves are shown in the diagram.

The effects of diffraction are much more noticeable if the size of the obstacle is small (a few wavelengths across),

while a given size of obstacle will diffract a wave of long wavelength more than a shorter one.

Diffraction can be easily demonstrated with sound waves or microwaves. It is quite easy to hear a sound even if

there is an obstacle in the direct line between the source and your ears. By using the 2.8 cm microwave apparatus very good diffraction effects may be observed with

obstacles a few centimetres across.

One of the most powerful pieces of evidence for light being some form of wave motion is that it also shows diffraction. The problem with light and that which led

Newton to reject the wave theory is that the wavelength is very small and therefore diffraction effects are hard to

observe. You can observe the diffraction of light, however, if you know just where to look.

The coloured rings round a street light in frosty weather, the coloured bands viewed by reflection from a record

and the spreading of light round your eyelashes are all diffraction effects. Looking through the material of a stretched pair of tights at a small torch bulb will also

show very good diffraction. A laser will also show good diffraction effects over large distances because of the

coherence of laser light. Diffraction is essentially the effect of removing some of

the information from a wave front; the new wave front will be altered by the obstacle or aperture. Huygens'

theory explained this satisfactorily.

.

small gap

diffraction round an obstacle

diffraction at an edge

small wavelength

large gap

Page 31 of 92

http://www.launc.tased.edu.au/online/sciences/physics/diffrac.html

http://www.ngsir.netfirms.com/englishhtm/Diffraction.htm

http://lectureonline.cl.msu.edu/~mmp/kap27/Gary-Diffraction/app.htm

Huygens' principle1.SWF

Diffraction Intro.SWF

Diffraction&Huygens.SWF

Diffraction through a slit.SWF

Diffraction variable slit.SWF

Diffraction around an object.SWF

Diffraction of radio waves.SWF

Diffraction & resolving power.SWF

Page 32 of 92

1 Red monochromatic light falls on a narrow slit. Describe what happens to the diffraction pattern as the slit is slowly opened.

2 How would the diffraction pattern in Q1 be affected if blue light were used

instead of red? 3 Which would be easier to receive in hilly areas and why, television or radio?

4 Why is the diffraction of light much more difficult to observe than the

diffraction of microwaves?

5 Each of the diagrams below shows a series of wavefronts, one wavelength apart, approaching a gap between two barriers in a ripple tank.

(a) What is a wavefront?

(b) Add further wavefronts to each diagram to show what happens as the

waves pass through each gap.

(c) The station BBC Radio 4 broadcasts both on the Long Wave band at

198 kHz and on VHF at approximately 94 MHz. In mountainous parts of the country, reception is better on Long Wave than on VHF. Suggest why.

Page 33 of 92

Wave Particle Duality for Electrons

Electrons scattered by a particle.SWF

The electron beam is accelerated through a p.d. of a few kV and it hits a thin film of graphite in an evacuated tube.

Electrons scattered by a crystal.SWF

Many of the electrons arrive near the centre of the screen but others arrive at other distances from the centre of the screen. So in addition to a bright spot in the centre,

two bright concentric rings are formed as well as other faint rings. http://cst-www.nrl.navy.mil/lattice/struk.jmol/a9.html

The electrons appear to be behaving like waves (diffracting)

If the accelerating voltage of the gun is increased the rings become smaller.

Fast moving electrons (when regarded as particles) appear to have a short wavelength (when regarded as waves). After the discovery of the photoelectric effect it was realised that waves can possess

particle- like properties, and a search was made to see if the reverse was true, could particles behave like waves.

http://www.colorado.edu/UCB/AcademicAffairs/ArtsSciences/physics/PhysicsInitiative/Physics2000/applets/twoslitsb.html

http://www.launc.tased.edu.au/online/sciences/physics/debroglie.html (But don‟t bother about the formulae!)

What are electrons?

When we are dealing with the forces on particles like electrons, and their

energy changes, we can treat them as particles.

When we want to know where they are, we have to use wave ideas.

The phrase wave-particle duality is used to describe the two-sided nature of

electromagnetic radiation, and the two-sided nature of particles.

screen

electron gun graphite

Page 34 of 92

http://chaos.nus.edu.sg/simulations/Modern%20Physics/Interference/interference.html

Two slit interference electrons & light.SWF

..\..\..\Mult imedia\Movies\Electron Waves.mov

The location of an electron inside an atom may be described in terms of a wave whose amplitude at any place determines the probability of finding the electron at that position.

Particle or wave.SWF

Electrons in atoms

The amplitude of the „electron wave‟ at a place determines the probability of finding the electron at that place.

Only particular standing waves are possible and these depend on the energy state of the electron. Animations\Schrodinger.xls

Quantised orbits

The simple Rutherford model of the atom had one serious disadvantage concerning the stability of the orbits. Bohr showed that in such a model the electrons would spiral into the nucleus in about 10-10 s, due to electrostatic attraction. He therefore

proposed that the angular momentum of the electron should be quantised, in line with Planck's quantum theory of radiation.

1 Which experiment shows particles behaving like waves?

2 Write down one device that uses the idea that particles have wave properties.

Scanning tunneling electron microscope image of graphite.GIF

Page 35 of 92

Pulse-echo techniques One method of finding the speed of sound, v, in air is to bang a drum while standing a measured distance, d, (at least 100 m) from the wall of a large building and measuring the time, t, between striking the drum and hearing the echo.

t

dv

2

Sonar and radar are methods during the Second World War that are still widely used to gauge the position of ships and aircraft. They achieve this by sending out pulses of

sound and radio waves and noting the time and direction of the reflected pulses. Bats and dolphins are examples of animals that emit and receive high-frequency sounds.

What is Ultrasound?

Ultrasound is the name given to high frequency sound - defined as sound with a

frequency over 20 000Hz. Sounds with this frequency are too high in pitch to be heard by the human ear. These waves can be transmitted in beams (like light) and are

used to produce live 2-D images of the internal organs. Recently it has become possible to generate 3-D images by means of ultrasound. The ultrasound pulse travels through the body and echoes off the internal organs. These ultrasound echoes are then

recorded and displayed as a live image. It is used across a wide range of medical specialties including obstetrics, gynaecology, cardiology, surgery, and

gastroenterology. Ultrasound is favoured in these areas as it is a safe and relatively inexpensive imaging method.

For medical diagnostic purposes, frequencies used in Ultrasound scanning are in the range of 2.5-10MHz (2.5 to 10 million Hz). Very short bursts of sound lasting around

one millionth of a second are transmitted into the patient approximately 500 - 1000 times a second. As the sound travels in the body, it is reflected at the junction of

different the tissues, to produce echoes which are picked up by the transducer.

The sound is reflected because different tissues carry sound at different rates. The fraction of sound that is

reflected depends on the difference in a property known as the acoustic impedance of the tissue on

each side of the interface. The acoustic impedance depends on the density of the medium so a much bigger reflection occurs at a tissue-bone boundary

than a tissue-muscle interface.

These echoes need to be electronically amplified in the scanner. The echoes that come from deep within the body are more attenuated (energy is absorbed or scattered

d

Page 36 of 92

within the body) than those from more superficial (or shallower) parts and therefore require more amplification. When the echoes return to the transducer, it is possible to

reconstruct a 2-D map of all the tissues that have been exposed to the ultrasound pulse. The information is stored in a computer and then displayed as an image on a

television monitor. Stronger echoes appear as brighter dots on the screen.

Ultrasound waves are produced by a piezoelectric transducer which is capable of changing electrical signals into mechanical waves (ultrasound). The same transducer can also receive the reflected ultrasound and change it back into electric signals. This

effect - known as the piezoelectric effect - is displayed by a number of naturally occurring materials

Ultrasound is an extremely useful means of diagnostic imaging. It is widely used in

obstetrics, for foetal imaging and to guide diagnostic procedures, as well as gynaecology. As fluid is a good conductor of sound, ultrasound is particularly good

for differentiating between cysts and solid structures and viewing fluid-filled structures such as the bladder, or the foetus in the sac of amniotic fluid.

Doppler Ultrasound

When ultrasound is transmitted towards a stationary reflector, the reflected waves will

be of the same frequency as those originally transmitted. However, if the reflector is moving towards the transmitter, the reflected frequency will be higher than the

transmitted frequency. On the other hand, if the reflector is moving away from the transmitter, the reflected frequency will be lower than the transmitted frequency. This phenomenon is called the Doppler Effect, after its discoverer, Christian Doppler.

The difference between the frequencies is called the Doppler shift. This may sound very complicated, but it's a surprisingly everyday effect, most commonly illustrated by a siren on an ambulance or police car. As the car approaches, the sound appears

higher and higher pitched - until the moment where it passes and the tone is heard to drop sharply. In medical practice, this is used in particular to measure the flow of

blood through vessels and within the heart.

As ultrasound is non- invasive and involves no ionising radiation. It is an extremely safe method of medical imaging. At the energies and doses currently used in

diagnostic ultrasound, no harmful effects on any tissues have ever been demonstrated, over a very long period of use. This is another reason why it is ideal for foetal imaging, and indeed, is usually considered the only scan safe for pregnant women.

Apart from its safety, ultrasound is also the method of choice for seeing many internal organs. Ultrasound provides high quality images of internal organs that some other

scans such as MRI's cannot achieve. Finally, compared to many other techniques, it is relatively inexpensive - meaning it is very widely available.

Microsoft Office PowerPoint 97-2003 Presentation

http://www.virtualcancercentre.com/investigations.asp?sid=8 The principles of medical Ultrasound http://www.mrcophth.com/commonultrasoundcases/principlesofultrasound.html#acoustic

Page 37 of 92

Thickness measurement (includes a good animation)

http://www.ndt-ed.org/EducationResources/HighSchool/Sound/ultrasound.htm

The amount of detail in a scan is affected by diffraction effects.

Microsoft Office Excel 97-2003 Worksheet

The smaller the wavelength of sound used in an ultrasound scan, the smaller the finest detail that can be distinguished. However there is a contrary trend that the shorter the

wavelength the sooner a wave pulse will be absorbed. Medical ultrasound scans generally compromise these competing factors using wavelengths in the range 0.075

mm to 1.5 mm. As a general guide the smallest detail that can be resolved will be about the size of a wavelength. To make out a foetal thumb that is 0.5 mm wide would require the use of ultrasound of wavelength

0.5 mm or less.

There is an additional constraint that the resolution will be half the length of a pulse. A pulse may be a few wavelengths and its size may be calculated thus:

on is pulse time speed length

Example

An ultrasound system for examining the eye sends out a pulse of ultrasound waves

with a frequency of 6 MHz. The pulse duration is 0.6 s. The speed of sound in the

eye averages 1510 m s-1. What is the smallest detail that can distinguished?

1 Wavelength method mmms

sm

f

v251.01051.2

106

1510 4

16

1

2 Pulse length method

mmlengthpulsehalfresolution

mssmlengthpulse

453.0

10906.0106.01510 361

The worse resolution is 0.453 mm and finer details than this could not be seen in the image produced.

1 Ultrasound is preferred to X-rays for some diagnostic images because:

A it gives a more detailed image

B it is a longitudinal wave C it is less harmful to the patient D it penetrates the body more easily

2 A firework accelerates upwards and emits a constant high-pitched sound. An

observer will hear: A a constant higher pitched sound B a constant lower pitched sound

C a continually decreasing pitch D a continually increasing pitch

Page 38 of 92

3 A recorder at the finish line of a 100 m race sees the flash of the starting pistol and starts her stopwatch. A second recorder fails to see the flash and starts his

watch on hearing the bang. The winner‟s time differs by 0.3 s on the two watches. Explain the likely reason for this and use the difference to estimate

the speed of sound in air.

4 A trawlerman uses sonar to detect shoals of fish. A strongly reflected pulse is

received 1.60 s after it was transmitted. If the speed of sound in water is 1500 m s

-1, how far from the boat are the fish?

5 Give on advantage and on disadvantage for using high frequency ultrasound

for diagnostic images in medicine.

6 A space probe sends radio signals back to Earth. Why may the radio receiver

at ground control have to be retuned after the launch of the probe? 7 A radar measurement of the distance to the Moon gives a round-trip time of

2.57 s. Calculate how far away the Moon is.

(Speed of radio waves = 3.00 x 108 m s-1).

Page 39 of 92

Doppler Effect

The Doppler Effect is the apparent change of frequency and wavelength when a

source of waves and an observer move relative to each other. These effects were first explained by Doppler in 1842 as a bunching up and a spreading out of waves.

Looking at a duck swimming in a pond would show you that the waves it generates in the direction it is swimming are bunched while those behind it are spread out. (Figure 1)

To demonstrate his theory he persuaded a group of trumpeters to stand and play in an open railway carriage while the carriage travelled across the Dutch countryside.

Observers on the ground heard a change of pitch as the truck passed them. One of the most important applications of the Doppler Effect is in the study of the expansion of the Universe. Galaxies have their light shifted towards the red due to

their speed of recession and when we receive the light at the Earth we describe it as Red Shifted.

http://lectureonline.cl.msu.edu/~mmp/applist/doppler/d.htm

Red shift.SWF

When a source of light is moving toward someone, the light will appear „bluer‟.

If a source of light is moving away, the light will appear „redder‟.

These two frequency shifts are called blue

shift and red shift.

Doppler effect.SWF

Blue shift and red shift are used to measure the velocity and rotation of stars and

galaxies.

Spiral Galaxy Red Blue Shift.MOV

Red shifts and blue shifts have been used to measure the orbital velocity of the Earth,

to detect stars and quasars, and to detect the rotation of other galaxies. They have also been used to determine the speed of dust clouds in the Milky Way, and have helped to prove that our galaxy is rotating.

Astronomers have discovered that all the distant galaxies are moving away from us, and by measuring their red shifts their speeds may be estimated.

Figure 1

Waves bunched together – wavelength shortened Waves spread out –

wavelength increased

Page 40 of 92

The furthermost galaxies have been estimated to have speeds approaching the speed of light.

Hubble photographed the spectra of galaxies and detected that certain characteristic absorption lines were shifted towards the red end of the spectrum.

http://zebu.uoregon.edu/nsf/hub.html The effect can be also be observed in the following uses and applications of the Doppler Effect

(a) Change in the pitch of a buzzer when it is whirled around your head

(b) Change in pitch of a train hooter or whistle as it passes through a station (c) Shift of the frequency of the light from the two sides of the solar disc due to

the Sun's rotation

(d) Variation in the frequency of the light from spectroscopic binaries (e) Police radar speed traps

(f) Doppler broadening of spectral lines in high temperature plasmas (g) Measurement of the speed of the blood in a vein or artery

The Doppler Effect is commonly used in medicine. An ultrasound transducer coupled

to the body near an artery emits pulses that are reflected by moving blood cells within the artery. A blockage due to a blood clot (thrombosis) or a constriction caused by a thickening of the arterial wall is detected by a sudden change in the shift frequency.

8 What is meant by the Doppler Effect?

9 Write down four uses or effects of the Doppler Effect.

10 If the light from a star is observed to be blue shifted as seen from the Earth

what does this tell you about the motion of the star?

Page 41 of 92

Electron flow in a conductor

Electric current

Current & electron flow 1.SWF

When you turn on an electric light an electric current flows in the wire. Do not think

of it like water coming from a tap – the electricity current does not flow out from the switch – electric charge is already in the wire – connecting the lamp to the power supply via the switch simply gives the charged particles the energy to flow.

This energy can come from a variety of sources – kinetic as in a dynamo, a chemical

reaction in a cell, light falling on a photoelectric cell, heating the junction of two metals in a thermocouple, sound in a microphone or mechanical stress in a piezo- electric crystal.

When an electric current flows electrical energy is converted to other forms of energy such as „heat‟, light, chemical, magnetic and so on.

Consider a piece of metal wire - a very much enlarged view of which is shown in

Figure 1 .

A piece of wire is made of a huge number of atoms and each one of these has its own cloud of electrons. However in a metal there are a large number of electrons that are not held around particular nuclei but are free to move at high speed and in a random

way through the metal. These are known are „free‟ electrons and in a metal there are always large numbers of these. It is when these free electrons are all made to move in

a certain direction by the application of a voltage across the metal that we have an electric current. The difference between a metal (a large and constant number of free electrons), a

semiconductor (a few free electrons, the number of which varies with temperature) and an insulator (no/very few free electrons) is shown in Figure 2.

Each electron has a very small amount of electric charge, and it is more convenient to use a larger unit when measuring charge. This unit is the coulomb.

electron

atom/ion

Figure 1

Figure 2 semiconduct

or insulator conductor

Page 42 of 92

The charge on one electron is -1.6 x 10-19 C. This is usually written as e. You would need about 6x1018 electrons to have a charge of one coulomb!

The electrical charge passing any one point in a circuit in one second is called the

electric current, and it is measured in Amperes (A).

The Amp can be defined in the following way:

charge of flow of Rate Current

Rate of flow of charge.SWF

Velocity of free electrons in a wire

The free electrons in a metal have three distinct velocities associated with them:

a random velocity ( about 105 ms-1)

a velocity with which electrical energy is transferred along the wire (about 108ms-1)

a drift velocity of the electrons as a whole when a current flows through the wire (this depends on the applied voltage but is usually a few mm s

-1 for

currents of a few amps in normal connecting leads).

Electron drift velocity

Microsoft Office PowerPoint 97-2003 Presentation

nAQvI

Current & electron flow 2.SWF

A current of 1A flows in a wire if a charge of 1C passes any point

in the wire each second.

Page 43 of 92

The table below shows some free electron concentrations

Metal Free electron concentration (m-3) (at 300 K)

Lithium 4.7x1028

Sodium 2.7x1028

Silver 5.9x1028

Copper 8.5x1028

Distinction between metals, semiconductors and insulators.

Metals are good conductors because the charge is carried by free electrons.

For a good conductor like copper the number of free electrons per unit volume is

n 1029 m-3.

n is VERY BIG.

Semi-conductors, like silicon, have fewer charge carriers per unit volume compared with metals.

n is medium size.

Insulators contain hardly any charge carriers at all (n is very small)

1 What is meant by the free electron concentration in a metal?

2 Why might a good electrical conductor also be a good thermal conductor? 3 Calculate the current flowing in a copper wire of cross-sectional area 2x10-7m2

if the drift velocity is 3.5x10-4 ms-1.

4 Calculate the drift velocity in a silver wire of diameter 0.26 mm if a current of 20 mA flows through it.

5 A table tennis ball oscillates between two charged vertical metal plates. A

sensitive ammeter connected between the plates records a current of 3 A

when the period of oscillation of the ball is 2 s. Calculate: (a) the charge carried between the plates by the ball in each oscillation

(b) the number of electrons carried between the plates by the ball in each oscillation

6 In a television tube 0.25 m long the electron velocity is 5x107 ms-1. If the current in the tube is 1.5 mA calculate:

(a) the number of electrons reaching the screen every second (b) the number of electrons in 1 cm of the beam

Page 44 of 92

7 Diagram 2 shows the results of an experiment to find the velocity of charged ions.

It shows their movement during 50 s.

Calculate:

(a) the ion drift velocity

(b) the charge reaching the positive electrode per second if the current is 1.25 mA

(c) the number of ions reaching the positive

electrode per second if there are each

doubly charged

microscope slide

f ilter paper

crocodile clip

crystal

0 V

+75V

Diagram 2

mm

10

20

30

Page 45 of 92

Potential and Potential Difference

As a charge moves round a circuit from the positive to the negative it loses energy.

There is a problem here. As you know an electric current is a flow of negatively charged electrons and these flow away from the negative terminal of a supply, round

the circuit and back to the positive terminal. However the „traditional‟ view of current flow is from positive to negative and we will take that view when looking at the energy of electrical charge.

We define the amount of electrical potential energy that a unit charge has as:

The electrical potential energy of a unit charge at a point in a circuit is called the potential at that point.

The next set of diagrams (Figure 2) show how the potential varies round some basic circuits. To simplify the treatment we are going to assume that the energy lost in the connecting wires is neglibgible and we are going to ignore it. This means that the energy of the charge at one end of a connecting wire is the same as that at the other end. The bigger the energy change the bigger the difference in potential. We call the difference in electrical potential between two points in the circuit the potential difference between those two places.

The potential difference between two points is defined as: Potential difference between two points in a circuit is the work done in moving unit charge (i.e. one coulomb) from one point to the other

http://regentsprep.org/Regents/physics/phys03/apotdif/default.htm

http://www.rkm.com.au/ANIMATIONS/animation-electrical-circuit.html

The units for potential difference are therefore Joules per coulomb, or volts. (1 volt = 1 Joule/coulomb).

So if a charge Q moves between two points in a circuit that have a potential difference

of V volts between them the energy gained (or lost) by the charge is given by the formula:

Electrical energy = Charge x Potential difference(Voltage)

Joules = Coulombs x Volts = Amps x Time x Volts

Electrical energy = ItV

Energy = VIt.SWF

Electron flow direction Traditional current direction

- to + + to -

Figure 1

Page 46 of 92

8 Calculate how much electrical energy is supplied by a 1.5V battery when: (a) a charge of 1000C passes through it

(b) a current of 2.5A flows from it for 2s

9 How much energy is drawn from a 12V car battery if it is used to supply 200A for 1.5s to the starter motor.

10 10 identical torch bulbs are connected in series across a 12 V d.c supply. (a) What is the p.d across each bulb

(b) What is the potential at the join of the second and third bulbs from the negative terminal?

11 (a) Write the word equation that defines potential difference.

(b) The unit of potential difference is the volt. Express the volt in terms of

base units only. (c) A 6.0 V battery of negligible internal resistance is connected to a

filament lamp. The current in the lamp is 2.0 A.

Calculate how much energy is transferred in the filament when the battery is connected for 2.0 minutes.

12 (a) State the word equation that is used to define charge.

(b) A 9.0 V battery of negligible internal resistance is connected to a light bulb.

Calculate the energy transferred in the light bulb when 20 C of charge flows through it.

13 A thick wire is connected in series with a thin wire of the same material and a battery

(a) In which wire do the electrons have the greater drift velocity? Explain your answer.

(b) A battery is connected across a large resistor and a small resistor is connected in parallel. The currents through the resistors are different.

Which resistor has the higher dissipation of power? Explain your answer.

Page 47 of 92

Variation of current with applied voltage

There are several ways in which the current through a device can be altered. Elastic strain, temperature and light are examples of these. Figure 1 shows examples of

current-voltage curves for a number of different situations.

Voltage-current characteristics

The p.d. applied to the device is varied over a suitable

range and the current is measured. The device is turned round to see if it behaves the same in both directions.

Graphs of current against voltage are plotted.

Filament lamp

Resistor

Semiconductor Diode

IV Graphs.SWF

http://micro.magnet.fsu.edu/electromag/java/filamentresistance/ (Resistance at molecular level)

A

Device

V

I

V

I V

R

V

R increases with temperature

I

V

I V

R

V

R is constant

I

V

The diode allows current to flow freely in one direction only.

The current increases rapidly for

‘forward’ voltages greater than

0.5 V.

Page 48 of 92

Resistance and temperature

When a material is heated its resistance may change. This is due to the thermal

motion of the atoms within the specimen. For a metal the temperature coefficient of resistance is positive - in other words and increase in the temperature gives an increase in resistance. This can be explained by

the motion of the atoms and free electrons within the solid. At low temperatures the thermal vibration is small and electrons can move easily within the lattice but at high

temperatures the motion increases giving a much greater chance of collisions between the conduction electrons and the lattice and so impeding their motion. In a light bulb the filament is at about 2700 oC when it is working and its resistance when hot is

about ten times that when cold. (For a typical domestic light bulb the resistance measured at room temperature was 32 Ω and this rose to 324 Ω at its working

temperature). However in non-metals such as semiconductors an increase in temperature leads to a drop in resistance. This can be explained by electrons gaining energy and moving

into the conduction band - in fact changing from being bound to a particular atom to being able to move freely - an increase in the number of free electrons. The

temperature coefficient of resistance and also that of the temperature coefficient of resistivity is therefore negative.

14 Why does the resistance of a metal increase as its temperature is raised?

15 Why does the resistance of a semiconductor decrease as its temperature is

raised?

16 Explain in terms of its „secret‟ composition, how the resistance of a fixed resistor may remain constant as its temperature is raised?

Page 49 of 92

Figure 1 electron metal atom

Resistance The free electrons in a metal are in constant random motion. As they

move about they collide with each other and with the atoms of the

metal. If a potential difference is now applied across the metal the electrons tend to move towards the

positive connection. As they do so their progress is interrupted by

collisions. These collisions impede their movement and this property of the material is ca lled its resistance. If the temperature of the metal is raised the atoms vibrate more strongly

and the electrons make more violent collisions with them and so the resistance o f the metal increase. The electron drift velocity v (in the equation I = nAQv) decreases.

The resistance of any conducting material depends on the following factors:

the material itself (actually how many free electrons there are per m3)

its length

its cross-sectional area and

its temperature

Resistivity of warm wire.GIF

http://regentsprep.org/Regents/physics/phys03/bresist/default.htm

The resistance of a given piece of material is related to the current flowing through it and the potential difference between its ends by the equation:

I

VR

The units of resistance are ohms (Ω).

If the ratio of p.d to current remains constant for a series of different p.d.s the material is said to obey Ohm's Law. This is true for a metallic conductor at a constant

temperature. This means that although we can always work out the resistance of a specimen knowing the current through it and the p.d across it. However if these quantities are

altered we can only PREDICT how it will behave under these new conditions if it obeys Ohm's law.

It is also vital to realise that the resistance is simply the ratio of the voltage and current at a particular point and NOT generally the gradient of the V I curve.

http://micro.magnet.fsu.edu/electromag/java/filamentresistance/ (Resistance at molecular level)

Page 50 of 92

Voltage

(V)

Current (I)

Figure 3

75oC

15oC

It is important to realise that Ohm‟s Law only holds for a metallic conductor if the temperature is constant.

This means that if the temperature of the metal is held steady at say 15o C the variation of

current and voltage will be linear. However if the temperature of the metal changes (as in the filament of a light bulb) then the resistance will

also change. The collisions between the electrons and the atoms will occur more often

and be more violent. So if the wire is raised to 75o C a second set of readings can be taken – they will still be linear

but the resistance of the wire (the ratio of V to I) will be greater (see Figure 3).

It is worth having a look at two graphs that show how the resistance of two types of material change when their temperature is changed.

The first is a metal wire (Figure 4(a)), and the second is a (negative temperature coefficient) thermistor (Figure 4 (b).

In the case of the metal wire the resistance increases as the temperature increases, you can see this because the ratio of pairs of points on the V-I graph increases at high

currents (hot wires). In the case of the thermistor the resistance decreases as the temperature increases, you can see because the ratio of pairs of points on the V-I

graph decreases at high currents (high temperatures.).

Although the gradients of the graphs suggest a change in resistance do not be tempted

to use the gradient to work out the resistance.

You must still deal with the voltage/current ratio only.

The reason that the thermistor decreases is because the thermistor is a semiconductor and more free electrons are produced as the temperature is raised. The number n (in the equation I = nAQv) increases.

(In fact more electrons are raised to the conduction band of the material.)

Volta

ge

Current

Volta

ge

Current

Figure 4 (a) Figure 4 (b)

Page 51 of 92

17 Calculate the current through the following resistors:

(a) 120 Ω connected to 240V (b) 4700 Ω connected to 12V (c) 10kΩ connected to 6V

(d) 2.5MΩ connected to 25V

18 What is the resistance of the following?

(a) a lamp that draws 2A from a 12V supply

(b) a kettle that draws 4A from a 240V supply

19 The two graphs below show a specimen of metal wire at two different

temperatures.

(a) Calculate the resistance of the wire at

each temperature (b) Which graph shows the higher

temperature?

(c) Does the material disobey Ohm's Law? Explain your answer.

20 Why is it more likely for the filament in a light bulb to break when you switch it on rather than when it has been on for some time? Explain your answer.

21 The circuit shows a battery of negligible internal resistance connected to three resistors.

(a) Calculate the potential difference across the 15 resistor.

(b) Calculate the current I1 in the 4.0 resistor.

(c) Calculate the current 12 and the resistance R.

I (A)

V (V)

10

5

20

Page 52 of 92

22 The circuit diagram shows a 12 V d.c. supply of negligible internal resistance connected to an arrangement of resistors. The current at three places in the

circuit and the resistance of two of the resistors are given on the diagram.

(a) Calculate the potential difference across the 4.0 resistor.

(b) Calculate the resistance of resistor R2. (c) Calculate the resistance of resistor R1.

23 Two filament lamps are designed to work from a 9.0 V supply but they have

different characteristics. The graph shows the current-potential difference

relationship for each lamp.

(a) The lamps are connected in parallel with a 9.0 V supply as shown.

(i) Which lamp is brighter? Give a reason for your answer.

(ii) Determine the current in the supply. (iii) Calculate the total resistance of the two lamps when they are connected in parallel.

Page 53 of 92

(b) The lamps are now connected in series to a variable supply which is adjusted until the current is 0.8 A.

Compare and comment on the brightness of the lamps in this circuit.

24 A student connects the circuit as shown in the diagram.

(a) The reading on the voltmeter is 1.8 V. Calculate the current in the resistor.

(b) Calculate the resistance of the thermistor.

The graph shows how the resistance of the thermistor depends on its

temperature

(c) Determine the temperature of the thermistor.

(d) If the e.m.f, of the supply were doubled, would the reading on the voltmeter double? Explain your answer.

Page 54 of 92

25 Use the axes to draw the current-voltage characteristics of a diode and a

filament lamp

Diode Filament lamp

Electrical Power

Power is the rate at which work is done or energy changed from one form to another

and so:

tIV

IVt

QV

t

QV

timePower Energy

time

EnergyPower Electrical

Example problems

1. Calculate the power of a 12V light bulb using 2.5 A. Power = VI = 12V x 2.5A = 30 W

2. Calculate the current used by a 12V immersion heater that is designed to

deliver 30000J in 5 minutes Energy = Power x Time = 30000

30000 = Power x 300

Power = 100W. Current I = P/V = 100W/12V = 8.3A

3. Calculate the energy transformed by a 12V car battery that delivers a

current of 200A for 3 s. Energy = Power x time = VIt = 12 x 200 x 3 = 72 000 J

Two alternative formulae for electrical power

We can combine the formula I

VR with the basic formula for electrical power to

give two alternative formulae:

Since power = VI, we can write

R

VRIIV

22Power

I

V

I

V

Page 55 of 92

Example problems

1 Calculate the resistance of a 100W light bulb if it takes a current of 0.8 A

1568.0

10022

2 A

W

I

PRRIP

2 Calculate the power of a 12V immersion heater with a resistance of 10

WR

VP 4.14

10

1222

http://www.ukpower.co.uk/running-costs-elec.asp (Electricity Running Costs Calculator)

26 Calculate the power loss in an electrical transmission cable, 15 km long, carrying a current of 100A at a potential of 200 kV. The resistance per km of

the cable is 0.2 Ω. 27 What power is supplied to the heater of an electric bar fire with a resistance of

50 Ω connected to the mains 240V supply?

28 What is the power loss down a copper connecting lead 50cm long with a resistance of 0.005 Ω per metre when it carries a current of 1.5A?

29 A 240 V mains lamp draws a current of 2 A from the supply when operating normally.

Calculate: (a) the resistance of the lamp when hot

(b) the power of the lamp when operating normally (c) the number of electrons passing through the lamp filament each second (d) the energy transferred to each coulomb by the supply

(e) the energy transferred to each electron by the supply

30 What is the power of an electrical heater operating from 110 V if the

resistance of the heater is 15 ? 31 What is the power loss down a copper connecting lead 75cm long with a

resistance of 0.13 per metre when a current of 4.5 A flows through it?

Page 56 of 92

Two resistors in series (Figure 1)

Series circuit current.SWF

Series circuit voltage.SWF

The current (I) flowing through R1 and R2 is the

same and so the potential differences across them are V1 = IR1 and V2 = IR2

21 VVV

R is the effective series resistance of the two resistors.

So:

21

21

RRR

IRIRIRV

Two resistors in parallel (Figure 2)

Parallel circuit current.SWF

Parallel circuit voltage.SWF

The potential difference (V) across each of the two

resistors is the same and the current (I) flowing into junction A is equal to the sum of the currents in the two branches, therefore:

21 III

21

21

R

V

R

V

R

VI

IRIRV

21

111

RRR

where R is the effective resistance of the two resistors in parallel.

Notice that

two resistors in series always have a larger effective resistance than either of the two resistors on their own

two in parallel always have a lower resistance than either of the two resistors on their own

This means that connecting two or more resistors in parallel, such as the use of a mains adaptor, will increase the current drawn from a supply.

Resistances in parallel.mdl

http://schools.matter.org.uk/Content/Resistors/Default.htm

V1 V2

Figure 1

R1 R2

I

Figure 2

A

R1

R2

V I

I2

I1

Page 57 of 92

Examples

1 Calculate the resistance of the following combinations:

(a) 100 Ω and 50 Ω in series

(b) 100 Ω and 50 Ω in parallel

(a) R = R1 + R2 = 100 + 50 = 150

(b)

33

50

1

100

11

R

R

2 Calculate the current flowing through the following when a p.d of 12V is applied across the ends:

(a) 200 Ω and 1000 Ω in series (b) 200 Ω and 1000 Ω in parallel

(a) R = 1200 I = V/R = 12/1200 = 0.01 A = 10 mA

(b) R = 167 I = V/R = 12/167 = 0.072 A = 72 mA

3 You are given one 100 Ω resistor and two 50 Ω resistors. How would you

connect any combination of them to give a combined resistance of: (a) 200 Ω, (b) 125 Ω

(a) 100 in series with both the 50

(b) the two 50 in parallel and this in series with the 100

32 Four 10 resistors are connected as shown in the diagram

(a) Calculate the total resistance of the combination. (b) Comment on your answer and suggest why such a combination of resistors

might be used.

33 Complete each of the following statements in words:

(a) The resistance of an ammeter is assumed to be (b) The resistance of a voltmeter is assumed to be

(c) Calculate the total resistance of four 5.0 resistors connected in

parallel.

Page 58 of 92

34 An electric room heater consists of three heating elements connected in parallel across a power supply.

Each element is made from a metal wire of resistivity () 5.5 x 10-5 m at room

temperature. The wire has a cross-sectional area (A) 8.0 x 10-7 m2 and length 0.65 m. The heater is controlled by two switches, X and Y.

(a) Show that the resistance of one heating element at room temperature is approximately 45 .

AR

(b) Calculate the total resistance of the heater for the following

combinations of switches at the moment the switches are closed.

Switch X Switch Y Resistance of

heater/

Open Closed

Closed Open

Closed Closed

(c) Calculate the maximum power output from the heater immediately it is

connected to a 230 V supply. (d) After being connected to the supply for a few minutes the power output

falls to a lower steady value. Explain why this happens.

Page 59 of 92

Resistivity

There are three factors that affect the resistance of a specimen of material:

the temperature

the dimensions of the specimen - the smaller the cross sectional area and the

longer the specimen the larger the resistance

the material from which the specimen is made

The property of the material that affects its resistance is called the resistivity of the

material and is given the symbol .

Resistivity is related to resistance of a specimen of length l and cross sectional area A by the formula:

RA

AR The units for resistivity are Ωm

Resistivity & Area.GIF

http://regentsprep.org/Regents/physics/phys03/bresist/default.htm

The following table gives the resistivities of a number of common materials.

Material Resistivity (x 10-8 Ωm) Material Resistivity (Ωm)

Copper 1.69 Non metals 104

Nichrome 130 Insulators 1013 - 1016

Aluminium 3.21 Germanium 0.65

Eureka 49 Silicon 2.3x10-5

Lead 20.8 Carbon 33 - 185x10-8

Manganin 44 Silver 1.6x10-8

The resistivities of solutions cannot be quoted generally because they depend on the

concentrations and are therefore variable quantities. As an example however the

resistivity of pure water is about 2.5 x105 m and that of a saturated solution of

sodium chloride about 0.04 m at 20oC.

The reciprocal of resistivity is known as the conductivity of the material ()

1. Calculate the resistance of a 1.5 m long piece of eureka wire of diameter 0.5 mm

7.3

1025.0

5.1104923

8

m

mm

AR

2. A piece of wire needed for a heater is to be made of Manganin. It is to have a cross sectional area of 1.5x10

-7 m

2 and a resistance of 5 Ω. How long must it be?

mm

mRA7.1

1044

105.158

27

A

Page 60 of 92

Data:

Resistivity of copper 1.7 x 10-8 m

Resistivity of constantan 47 x 10-8 m

Resistivity of germanium 0.65 m

35 Calculate the length of a copper wire of cross sectional area 0.65 x10-7 m2 that has a resistance of 2Ω

36 Calculate the resistivity of a material if a 2.6 m length of wire of that material

with a diameter of 0.56 mm has a resistance of 3Ω. 37 Calculate the resistance between the large faces of a slab of germanium of

thickness 1 mm and area 1.5 mm2.

38 Calculate the length of a section of constantan wire of diameter 0.32 mm that has a resistance of 10 Ω

39 (a) A student is asked to carry out an experiment to find the resistivity of the material of a length of resistance wire. Draw an appropriate circuit

diagram. (b) List all the measurements the student should take to find the resistivity. (c) How should these measurements be used to find the resistivity?

(d) Suggest two precautions the student should take to ensure an accurate result.

40 Lord Kelvin discovered that the electrical resistance of iron wire changed when the wire was stretched or compressed. This is the principle on which a

resistance strain gauge is based. Such a gauge consists of a length of very fine iron wire cemented between two very thin sheets of paper.

(a) The cross-sectional area of the wire is 1.1 x 10-7 m2 and the gauge length

as shown in the diagram is 2.4 x 10-2 m. The resistivity of iron is

9.9 x 10-8 m. Calculate the resistance of the strain gauge.

(b) When this gauge is stretched its length is increased by 0.1% but its cross-

sectional area remains the same. What is the change in the resistance of the gauge?

(c) Explain the effect that stretching the wire will have on the drift velocity of electrons in the wire. Assume that the other physical dimensions of the wire remain unchanged and that there is a constant potential difference

across the wire.

Page 61 of 92

41 The diagram shows a type of resistor commonly used in electronic circuits.

It consists of a thin film of carbon wrapped around a cylindrical insulator. The

diagram below (not to scale) shows a typical film of carbon, resistivity p, before it is wrapped round the insulator.

(a) Show that the resistance R of the carbon film is given by

twR

(b) This film has length l = 8.0 mm, width w = 3.0 mm and thickness t = 0.0010 mm (i.e. t = 1.0 x 10-6 m). If the resistivity of carbon is 6.0 x

10-5 m, calculate the resistance of the carbon film.

(c) Show that the resistance of a square piece of carbon film of uniform thickness is independent of the length of the sides of the square.

Page 62 of 92

Electromotive force and internal resistance

When current flows round a circuit energy is transformed in both the external resistor but also in the cell itself. All cells have a resistance of their own and we

call this the internal resistance of the cell. The voltage produced by the cell is called the electromotive force or e.m.f for short and this produces a p.d across the cell and across the external resistor.

The e.m.f of the cell can be defined as the maximum

p.d that the cell can produce across its terminals or the open circuit p.d since when no current flows from the cell then no electrical energy can be lost within it.

Consider the circuit shown in Figure 1. If the e.m.f of

the cell is E and the internal resistance is r and the cell is connected to an external resistance R then:

IrIRIrVE The quantity of useful electrical energy available

outside the cell is IR and Ir is the energy transformed to other forms within the cell itself.

We usually require the internal resistance of a cell to be small to reduce the energy transformed within the cell; however it is sometimes helpful to have a

rather larger internal resistance to prevent large currents from flowing if the cell terminals are shorted.

The more correct definition of emf is

IrV

IrIR

It

rtIRtI

22

charge

circuit complete in the nsformedenergy tra

Hence IrEV , and E = V when I = 0, i.e. when the load R is infinite (VERY BIG). We say that the battery (cell) is on open circuit.

R r

Ir

IR E

Figure 1

R

r

E

Page 63 of 92

Example problem

A cell of e.m.f 12V and internal resistance 0.1Ω is used in two circuits. Calculate the p.d between its terminals when it is connected to (a) 10 Ω and (b) 0.2 Ω .

(a) Total resistance = 10 + 0.1 = 10.1 Ω

Therefore current = AV

19.11.10

12

Loss of energy per coulomb in the cell = 1.19 x 0.1 = 0.119V P.d between terminals = 12 - 0.119 = 11.88V = 11.9 V

(b) Total resistance = 0.2 + 0.1 = 0.3 Ω

Therefore current = AV

403.0

12

Loss of energy per coulomb in the cell = 40 x 0.1 = 4V P.d between terminals = 12 - 4 = 8V

Short circuit current

When a power supply is short-circuited by connecting the terminals together with a low resistance (R), the total resistance of the circuit is almost entirely due

to the internal resistance (r) of the power supply. Car batteries need to supply currents up to 200 A, so they are designed to have

very low internal resistance. High-voltage power supplies are designed to have large internal resistance so as

to prevent them from supplying dangerously high currents.

Experiment to measure internal resistance R may be varied and a range of values of V and I

are measured. Keep a record of values of R for later.

ErI

IrEV

IRV

IrIRE

A graph of V against I is plotted.

The graph is a straight line

which has a gradient -r

and a „y intercept‟ E

V r

E

R

V

I

E Gradient = -r

Page 64 of 92

The power transformed in the load R is VIP

How can we maximise the power? Make V and I as large as possible.

What is the largest possible value of V? E

How big is the current when V E? Zero!

What happens to V if we try to increase the current? V decreases.

Maximum power is delivered to the load when the load resistance R equals the internal resistance r of the supply.

Microsoft Office Excel 97-2003 Worksheet

42 What is meant by:

(a) the E.M.F. of a cell (b) the internal resistance of a cell?

43 How does the internal resistance of a cell affect the current drawn from it?

44 A cell of e.m.f 3.0 V is connected to a resistor of 2700 and when a voltmeter

of very high resistance is connected across its terminals the voltmeter reads 2.8 V

(a) Explain the difference between these two voltages (b) Calculate the internal resistance of the cell

(c) Calculate the new reading of the voltmeter if the voltmeter has a

resistance of 1500 .

45 A student wants to measure the internal resistance of a cell and he connects it

in series with an ammeter and a variable resistor. The student then connects a

voltmeter across the variable resistor and, for different settings of the variable resistor, obtains the following readings:

V/V 1.43 1.41 1.39 1.33 1.20

/mA 143 176 231 333 600

Plot a graph of V against and deduce the internal resistance and e.m.f. of the cell.

(V = E - r or V = -r + E )

Page 65 of 92

46 A resistance substitution box R is connected to a power supply. For various

values of R the current is measured.

R/ 8.0 6.0 4.0 2.0 1.0 0.80 0.60

/mA 167 214 300 500 750 833 938

P/mW

(a) Calculate the power of the resistor for each value of R.

(b) Predict the power of the resistor when R is zero and when R is infinitely large.

(c) Plot a graph of the power, P (dissipated in R) against resistance R.

(d) For what value of R is the power a maximum?

(e) The internal resistance of the power supply is known to be 1.0 .

Comment.

47 Why are car batteries designed to have negligible internal resistance whereas high voltage (EHT) laboratory supplies are manufactured with a very large

internal resistance?

Page 66 of 92

Potential divider

The basic circuit is shown in the first circuit diagram. The output voltage across AB is given by:

VRR

RR

RR

VRIV

21

22

2122

Note that the input voltage (V) in this case

supplied by the battery is constant. The current flowing through both resistors is the same (series circuit) and so the output

voltage across one of them depends simply on the two resistance values and the input

voltage.

Measuring the output

If we now attempt to actually measure the output voltage things may change.

(a) Firstly consider using a digital voltmeter with very high (if not virtually

infinite) resistance (RV). The resistor R2 and the voltmeter are connected in parallel and so their combined resistance (R) is given by the equation;

VRRR

111

2

RV is huge – almost

infinite and so 01

VR

and

can be ignored. This

means that 2

11

RR and

so R R2.

The output voltage (V0) measured by the meter really is that across R2, in other

words V2.

V2

R1

R2

A

B

V

V2

R1

R2 Digital

voltmeter

VO

V

Page 67 of 92

(b) A moving coil meter. These meters have a

much lower resistance than a digital meter,

usually some tens of k.

This means that the combined resistance of

R2 and RV is affected by the resistance of the

voltmeter and is actually lower than R2.

The proportion of the input voltage (V) dropped across R2 therefore falls and so the

output voltage (V0) is less than that measured with a digital meter.

Replacing R1 with a Light dependent resistor (LDR)

The LDR is a component that has a resistance that changes when light falls on

it. As the intensity of the light is increased so the resistance of the LDR falls. If the LDR is connected as part of a

potential divider as shown in the diagram then as the light level is increased its

resistance falls and the proportion of the input voltage dropped across it will also fall.

So in the light V2 is low and in the dark

V2 is high.

Replacing R1 with a thermistor

Something very similar happens if R2 is replaced by a thermistor. As the

temperature of the thermistor rises its resistance falls and so the voltage dropped across it falls.

When the thermistor is hot V2 is low and

when the thermistor is cold V2 is high.

http://www.crocodile-clips.com/absorb/AP5/sample/020201.html (Practice questions)

V2

R1

LDR V

V2

R1

thermistor

V

V2

R1

R2 Moving

coil

voltmeter

V

VO

Page 68 of 92

48 Complete the following table showing the readings of a digital voltmeter of infinite resistance for the output voltage (V) for a series of different resistances

and input voltages.

Input voltage (Vo)

R1/ R2/ Output voltage (V)

12 100 k 200k

6 25 k 10 k

24 5 k 20 k

6 250 100

49 Repeat question one but this time assume that the meter connected to measure

the output voltage is an analogue meter with a resistance of 200 k.

Input voltage (V0) R1/ R2/ Output voltage (V)

12 100 k 200k

6 25 k 10 k

24 5 k 20 k

6 250 100

50 Now assume that resistor R1 is

replaced by a thermistor T1 (one where the resistance decreases as the

temperature rises). If the values of the resistance R2 and the thermistor are equal at the start

what will happen to the output potential difference (V) as the

thermistor is cooled?

51 The thermistor is now replaced by an LDR. What happens to the output

potential difference (V) as the light intensity falling on the LDR is increased? 52 Assuming that the voltmeter used to measure V in Q51 has an almost infinite

resistance what happens to the current through R2 as the light intensity falling on the LDR is decreased?

R1

R2

Vo

V

T1

R2

Vo

V

Page 69 of 92

53 A student connects a 9.0 V battery in series with a resistor R, a thermistor and a milliammeter. He connects a voltmeter in parallel with the resistor. The

reading on the voltmeter is 2.8 V and the reading on the milliammeter is 0.74 mA.

(a) (i) Show that the resistance of R is approximately 4000 .

(ii) Calculate the resistance of the thermistor

(b) The thermistor is mounted on a plastic base that has steel sprung clips

for secure connection in a circuit board.

Another student is using an identical circuit except that the bare metal pins of his thermistor are twisted together.

(c) Suggest an explanation for how the reading on this student's milliammeter will compare with that of the first student.

Page 70 of 92

54 (a) A student sets up a circuit and accidentally uses two voltmeters V1 and V2 instead of an ammeter and a voltmeter. The circuit is shown below.

(i) Circle the voltmeter which should be an ammeter.

(ii) Both voltmeters have a resistance of 10 M. The student sees that the reading on V2 is 0 V. Explain why the potential difference across the

100 resistor is effectively zero.

(b) The student replaces the 100 resistor with another resistor of resistance R.

The reading on V2 then becomes 3.0 V.

(i) Complete the circuit diagram

below to show the equivalent resistor network following this change.

Label the resistor R.

(ii) Calculate the value of R.

Page 71 of 92

55 A light-emitting diode (LED) is a diode that emits light when it conducts. Its circuit symbol is

A student connects the circuit shown below.

She notices that the reading on the high resistance voltmeter remains at 0 Vas she slides the contact between terminals A and B.

(a) Explain this observation as fully as you can.

(b) The student then disconnects the LED and reconnects the circuit as shown below. She intends to vary the intensity of the light emitted by the LED by

sliding the contact between terminals A and B.

The student cannot detect any light emitted by the LED. Briefly explain why

the LED is so dim.

(c) Draw the circuit that the student should have connected using this apparatus in order to vary the brightness of the LED and measure the

potential difference across it.

Page 72 of 92

56 The following circuit can be used as a light meter.

(a) The maximum value of resistance of the light-dependent resistor (LDR) is 950 k.

What is the reading on the voltmeter for this resistance? (b) The minimum value of resistance of the LDR is 1.0 k. What is the

reading on the voltmeter for this resistance?

(c) For this light meter the voltmeter is connected across the 10 k resistor, rather than the LDR. Explain how the readings on the voltmeter enable this circuit to be used as a light meter.

57 Two resistors of resistance 2.0 M and 4.0 are connected in series across a

supply voltage of 6.0 V. Together they form a simple potential divider circuit.

State the potential difference across each resistor.

P.d. across the 2.0 M resistor =

P.d. across the 4.0 resistor =

A second potential divider circuit uses a resistor and a diode connected in

series with the same supply. Calculate the potential difference across each

component when the resistance of the resistor and diode are 45 and 5.0

respectively.

P.d. across the 45 resistor =

P.d. across the diode =

Page 73 of 92

or

no effect no effect Leaf falls

immediately

Nature of light

The Photoelectric Effect

When is a wave not a wave - when it's a particle! In 1887 Heinrich Hertz noticed that sparks would jump between two spheres when their surfaces were illuminated by light from another spark. This effect was studied

more carefully in the following years by Hallwachs and Lenard. They called the effect photoelectric emission and a very simple experiment can be used to investigate

it.

In the diagram shown above a clean zinc plate is fitted to the top of a gold leaf

electroscope and then given a positive charge (you can do this either with a charged glass rod or an EHT supply. The next thing is to shine some radiation on it, using an ordinary lamp, a helium-neon laser (giving out intense red light) or an ultra violet

light has absolutely no effect. The electroscope stays charged and the leaf stays up. However if the plate is given a negative charge to start with (using say a charged

polythene rod) there is a difference. Using the lamp and even the laser has no effect, but when ultra violet light is shone on the plate the leaf falls immediately: the electroscope has been discharged. (Doing the experiment in a vacuum proves that it is

not ions in the air that are causing the discharge.) No effect can be produced with radiation of longer wavelength (lower frequency) no

matter how long the radiation is shone on the plate. The plate was emitting electrons when the ultra violet radiation fell on it and this explained why the leaf only fell when it had an initial negative charge - when it was

positive the electrons were attracted back to the plate.

The researchers found five important facts about the experiment:

no electrons were emitted from the plate if it was positive

the number of electrons emitted per second depended on the intensity of the incident radiation

the energy of the electrons depended on the frequency of the incident radiation

there was a minimum frequency (f0) below which no electrons were emitted no

matter how long radiation fell on the surface

If electrons are emitted this occurs immediately

Page 74 of 92

This minimum frequency is called the threshold frequency for that material. Photons with a lower frequency will never cause electron emission. This can be explained as follows

The free electrons are held in the metal in a "hole" in the electric field, this is called a potential well. Energy has to be supplied to them to enable them to escape from the surface. Think of a person down a hole with very smooth sides. They can only escape if they can jump out of the hole in one go. They cannot get half way up and then have a rest - it's all or nothing! This is just like the electrons. The deeper the "hole" the more tightly bound the electrons are and the greater energy and therefore the greater is the frequency of radiation that is needed to release them.

The quantum theory of Max Planck is needed to explain the photoelectric effect. In trying to explain the variation of energy with wavelength for the radiation emitted by

hot objects he came to the conclusion that all radiation is emitted in quanta and the

energy of one quantum or photon is given by the equation:

hfEnergyPhoton

The amount of energy needed to just release a photoelectron is known as the work function for the metal. This can also be expressed in terms of the minimum frequency that will cause photoelectric emission.

0Function Work hf

The table below gives the work function for a number of surfaces - both in joules and in electron volts. The threshold frequency for each surface is also included.

Another way of looking at it is to think of a fairground coconut shy. A brother and sister are trying to knock the coconuts off their stands. The boy has a large box of table tennis balls which he is throwing at the coconuts, with little effect. No matter how many of the table tennis balls he throws at a coconut it will still stay in place – the table tennis balls represent the “red” quanta. However his sister has a pistol! This represents the violet quanta. A single shot from the pistol will knock off a coconut and it will do it immediately. As we saw in the previous experiment we could illuminate the zinc plate all day with a high powered laser and the leaf of the electroscope would not fall. However as soon as we shone the ultra violet light on the plate the leaf dropped. This is because the ultra violet light has a high enough frequency and therefore each quantum of ultra violet has sufficient energy. One quantum has enough energy to kick out an electron in one go. The photoelectric effect is therefore very good evidence for the particulate nature of light.

Work function /10-19

J /eV Frequency Wavelength

/1014

Hz /nm

Sodium 3.8 2.40 5.8 520

Caesium 3.0 1.88 4.5 666

Lithium 3.7 2.31 5.6 560

Calcium 4.3 2.69 6.5 462

Magnesium 5.9 3.69 8.9 337

Silver 7.6 4.75 11.4 263

Platinum 10.0 6.75 15.1 199

Page 75 of 92

Stopping

potential

Frequency

fo

Metal 1

Metal 2

Measuring the Planck Constant (h = 6.6 x 10-34 J s)

For each coloured filter the positive („stopping‟) potential on the cathode is increased until the ammeter reading is zero

- this means that no photoelectrons are reaching the detector.

A graph can then be plotted of stopping potential

(electron energy) against frequency.

Theory

If hf (or h

f

), f is the threshold frequency for that metal, and electrons are

just able to be removed. However they will have no K.E. after they have been emitted.

If hf , the surplus photon energy hf is given to the electron as kinetic

energy. A stopping p.d. is applied so that the collector is negative with respect to the emitter and this makes it difficult for the electrons to reach the collector.

As this reverse voltage V is gradually increased the ammeter reading is reduced.

Eventually when maxkeeV the ammeter reading will be ZERO and electrons will

no longer be able to reach the collector.

ef

e

hV

hfeV

A graph of V v f is plotted, and

gradient = e

h y intercept =

e

x intercept =

h

collector

V

A

sodium

emitter

filter

lens * lamp

electron

A

V

Page 76 of 92

If we change the sensitive surface to another metal (metal 2) with a higher work

function then since the gradient of the line is constant

e

h and we simply get a

second line parallel to the first but shifted to the right.

This simple experiment is very good evidence that light sometimes behaves like a stream of particles.

http://www.launc.tased.edu.au/online/sciences/physics/photo-elec.html http://lectureonline.cl.msu.edu/~mmp/kap28/PhotoEffect/photo.htm

Charge on the electron = (-)1.6 x 10-19 C

Planck‟s constant = 6.63 x 10-34 Js Speed of light in free space = 3.00 x 108 ms-1

Mass of an electron = 9.11 x 10-31 kg

1 What is meant by a quantum or photon?

2 Write down Planck‟s equation for the energy of a quantum of radiation.

3 Which has the greater energy, a quantum of yellow light or a quantum of violet light?

4 Which experiment shows waves behaving like particles?

5 What is meant by the “stopping potential” in the photoelectric effect?

6 What is meant by the “work function” in the photoelectric effect? 7 What is the energy of a quantum of radiation that has a wavelength of 500 nm?

8 What is meant by an “electron volt”?

9 If the frequency of the incident radiation that falls on a metal surface is

increased what happens to the photoelectrons that are emitted?

10 If the intensity of the incident radiation that falls on a metal surface is

increased what happens to the photoelectrons that are emitted? 11 (a) Light of wavelength 480 nm just gives photoelectric emission from the

certain metal surface. What is the work function of that surface in Joules?

(b) If the wavelength is reduced to 400 nm what is the energy of the

electrons emitted?

Page 77 of 92

E

K

f

12 The diagram shows monochromatic light falling on a photocell.

The photocell is connected so that there is a reverse potential difference across the

cathode and the anode.

(a) Explain the following observations.

(i) Initially there is a current which is measured by the microammeter. As the reverse potential difference is increased

the current reading on the microammeter decreases. (ii) When the potential difference reaches a certain value V, the

stopping potential, the current is zero.

(b) What would be the effect on the value of the stopping potential VS of (i) increasing the intensity of the incident radiation whilst keeping

its frequency constant. (ii) increasing the frequency of the incident radiation whilst

keeping its intensity constant?

13 Photoelectrons are emitted from the surface of a metal when radiation above a certain frequency, fo , is incident upon it. The maximum kinetic energy of the emitted electrons is EK.

(a) On the axes below sketch a graph to show how EK varies with frequency f.

(b) State how the work function, , of the metal can be obtained from the graph.

(c) Explain why this graph always has the same gradient irrespective of the metal

used.

Page 78 of 92

14 The diagram shows a coulombmeter (an instrument for measuring charge) set up to demonstrate the photoelectric effect.

The clean zinc plate is negatively charged. Ultraviolet light is shone onto the zinc plate and the plate discharges. The coulombmeter reading gradually falls

to zero. When the experiment is repeated with red light the plate does not discharge.

(a) Explain these effects in terms of the particle theory of light. (b) What would happen to the charged plate if

(i) the intensity of the red light were increased (ii) the intensity of the ultraviolet light were increased?

(c) Zinc has a work function of 3.6 eV. Calculate the maximum kinetic

energy of the photoelectrons when the zinc is illuminated with ultraviolet light of wavelength 250 nm.

15 The photoelectric effect supports a particle theory of light but not a wave

theory of light.

Below are two features of the photoelectric effect. For each feature explain

why it supports the particle theory and not the wave theory.

(a) Feature 1: The emission of photoelectrons from a metal surface can take place instantaneously.

Explanation (b) Feature 2: Incident light with a frequency below a certain threshold

frequency cannot release electrons from a metal surface.

Explanation

Page 79 of 92

16 (a) Define the intensity of an electromagnetic wave.

(b) Two beams of monochromatic electromagnetic radiation, A and B, have equal intensities. Their wavelengths are:

Beam A 300 nm Beam B 450 nm

In the table below, E denotes the energy of a photon and N denotes the number of photons passing per second through unit area normal to the beam. The

subscripts A and B refer to the two beams. In the second column of the table, state the value of each ratio, and in the third column explain your answer.

Ratio Value Explanation

B

AE

E

B

AN

N

(c) The table below gives the work functions of four metals.

Metal Work function/eV

Potassium 2.26

Magnesium 3.68

Tungsten 4.49

Iron 4.63

Define the term work function.

(d) A metal plate made from one of these metals is exposed to beams A

and B in turn. Beam A causes electrons to be emitted from the plate, but beam B does not. Calculate the photon energies in each beam and

hence deduce from which metal the plate is made.

Page 80 of 92

17 A monochromatic light source is placed 120 mm above the cathode of a photocell.

(a) The light source consumes 6 W of power and is 15% efficient. Calculate the light intensity at the cathode. State an assumption that

you made.

A potential difference is applied between the cathode and the anode of the

photocell and the sensitive ammeter detects the current.

The table below shows the currents that are obtained with this apparatus for

two different intensities and two different wavelengths of light, using two different cathode materials. Work function energies are given.

Wavelength of incident

radiation/nm

Cathode

material

Work

function/eV

Photocurrent/A

when intensity of incident radiation is

1 W m-2 5 W m

-2

320 Aluminium 4.1 0 0

640 Aluminium 4.1 0 0

320 Lithium 2.3 0.2 x 10-12 1.0 x 10-12

640 Lithium 2.3 0 0

(b) Show that the incident photons of = 320 nm and) = 640 nm have

energies of approximately 4 eV and 2 eV respectively.

(c) Account for the photocurrent readings shown in the table.

(d) Calculate the stopping potential for the photoelectrons released by lithium when irradiated by light of wavelength 320 nm.

Page 81 of 92

18 The table below shows the results of an experiment like Millikan‟s using sodium as the metal plate.

Stopping voltage Vs /V Frequency of light f / 1014 Hz

0.43 5.49

1.00 6.91

1.18 7.41

1.56 8.23

2.19 9.61

3.00 11.83

(a) Plot a graph of Vs against f.

The following equation applies to the photoelectric effect:

SeVhf

where is the work function of the metal and e is the charge on an electron.

(b) What information about the electrons emitted does the value of the

term SeV give?

Page 82 of 92

Spectra and energy levels The atoms in the tube are excited and as a result, they emit radiation.

When the radiation from excited atoms is

viewed through a diffraction grating lines are seen at different angles.

These correspond to images formed by light of different wavelengths (and

frequencies). The emission line spectrum pattern is

characteristic of the gas in the tube. http://jersey.uoregon.edu/elements/Elements.html

Line spectra & electron transitions.SWF

Electron energy levels

Electrons are held in atoms at only certain energy levels.

Normally an electron will lie in its lowest energy state - the ground state.

When an electron is given sufficient energy to rise to one of the higher energy states the

atom is said to be „excited‟.

The electron may remain above the ground state temporarily, but it will usually drop back to the ground state, either directly or

via another energy level, giving out energy as it does so.

Animations\BohrModel.swf

The energy levels for atoms are fixed.

Only certain changes in energy are possible.

To move between any two of these levels an electron needs to give out or

receive a definite amount of energy.

Small jumps mean small energy changes, which correspond to low

frequencies of radiation and large .

Big jumps mean large energy changes, which correspond to high frequencies of

radiation and small .

Gas discharge tube containing e.g. sodium vapour

Narrow slit

Diffraction

Grating

(0) Ground state

-13.6 eV

1st excited state

-3.4 eV (10.2)

2nd

excited state

-1.5 eV

-0.85

Ionised state

-0 eV

(12.1)

(12.75)

(13.6)

E1

E2

E3

E = E3 – E2 = h f

E = E2 – E1 = h f

Page 83 of 92

The atoms of each element have a characteristic set of energy levels, and so emit a characteristic set of frequencies when they are excited.

You can identify an element from the frequencies of radiation it emits when excited.

Emission spectra of different atoms.SWF

http://www.dartmouth.edu/~chemlab/info/resources/mashel/MASHEL.html

Example problem

Calculate the frequency and wavelength of a quantum of radiation emitted when an electron in level 4 falls to level 2. Energy of level 4 = - 1.36 x 10

-19 J

Energy of level 2 = - 5.42 x 10-19

J

Energy difference (E) = + 4.06 x 10-19

J Therefore

greenbluenmm

h

Ef

/490109.41012.6

1000.3

1012.61063.6

1006.4

7

14

8

14

34

19

A typical absorption spectrum

Emission & absorption spectra.SWF

http://www.dartmouth.edu/~chemlab/info/resources/mashel/MASHEL.html http://phys.educ.ksu.edu/vqm/html/emission.html http://www.launc.tased.edu.au/online/sciences/physics/linespec.html http://www.launc.tased.edu.au/online/sciences/physics/photo-elec.html http://lectureonline.cl.msu.edu/~mmp/kap28/PhotoEffect/photo.htm

R G V White V G R

Page 84 of 92

(1 eV = 1.60 x 10-19

J)

19 The diagram shows some of the energy

levels for an atom of hydrogen. Photons are emitted when an electron moves down from one level to another.

(a) When an electron moves from level 2 to level 1, what is

(i) its loss of energy in eV (ii) its loss of energy in J

(iii) the frequency of the emitted photon (iv) the wavelength of the emitted photon (v) the part of the electromagnetic spectrum

in which this radiation occurs?

(b) Repeat part (a) for an electron moving from leve1 3 to level 2.

(c) Repeat part (a) for an electron moving from leve1 4 to level 3.

20 The two lowest excited states of a hydrogen atom are 10.2 eV and 12.1 eV

above the ground state. (a) Calculate three wavelengths of radiation that could be produced by

transitions between these states and the ground state. (b) In which parts of the spectrum would you expect to find these

wavelengths?

21 The figure shows an energy level diagram.

Sketch a possible line spectrum for the light emitted when electrons make the transitions shown.

Label the lines, using the letters shown in the diagram, and indicate on your spectrum diagram

which end corresponds to the higher frequency.

22 The four lowest energy levels for an atom consist of the ground state and three

levels above that. How many transitions are possible between these four levels?

Page 85 of 92

23 The figure shows three energy levels for a particular atom.

When an electron moves from level1 to the ground state the light emitted is blue.

In what part of the spectrum would you expect to find the radiation emitted when an electron moves

from level 2 to the ground state?

24 Refer to the diagram in Q19 .

(a) If the atom is in the ground state, how much energy must be given to it

to ionise it?

(b) Suppose an electron of energy 2.2 eV collides with the atom. Explain the possible results if (i) the atom is in the ground state

(ii) its electron is at the -3.41 eV level

(c) What is the wavelength of the photon that could raise an electron from the -0.849 eV level to the -0.545 eV level?

(d) If an electron returns from the -0.849 eV level to the ground state, what is the wavelength of the photon emitted?

25 Some of the energy levels for the sodium atom are -1.51 eV, -1.94 eV, -3.03

eV (two levels very close together) and -5.14 eV; which is the ground state. Draw a labelled diagram for these levels, and describe and explain what might

happen if cool sodium vapour (i.e. sodium whose atoms are in the ground state) is bombarded with

(a) electrons whose k.e. is 2.00 eV

(b) electrons whose k.e. is 2.50 eV (c) light of wavelength 590 nm.

Page 86 of 92

26 Four of the energy levels of a lithium atom are shown below.

(a) Draw on the diagram all the possible transitions which the atom could make when going from the -3.84 eV level to the -5.02 eV level.

(b) Photons of energy 3.17 eV are shone onto atoms in lithium vapour. Mark on the diagram, and label with a T, the transition which could occur.

(c) One way to study the energy levels of an atom is to scatter electrons from it

and measure their kinetic energies before and after the collision. If an electron

of kinetic energy 0.92 eV is scattered from a lithium atom which is initially in the -5.02 eV level, the scattered electron can have only two possible kinetic

energies.

State these two kinetic energy values, and explain what has happened to the

lithium atom in each case. (You should assume that the lithium atom was at rest both before and after the collision.)

Kinetic energy 1

Explanation

Kinetic energy 2

Explanation

Page 87 of 92

27 The diagram shows the lowest four energy levels of atomic hydrogen.

(a) Calculate the ionisation energy in joules for atomic hydrogen.

(b) On the diagram above draw

(i) a transition marked with an R which shows a photon released with

the longest wavelength, (ii) a transition marked with an A which shows a photon absorbed

with the shortest wavelength.

(c) Describe how you would produce and observe the emission spectrum of

hydrogen in the laboratory.

(d) What would such a spectrum look like?

Page 88 of 92

28 The diagram shows some of the energy levels of a mercury atom.

a. Calculate the ionisation energy in joules for an electron in the -10.4 eV level.

b. A proton of kinetic energy 9.2 eV collides with a mercury atom. As a

result, an electron in the atom moves from the -10.4 eV level to the -1.6 eV level. What is the kinetic energy in eV of the proton after the

collision?

c. A transition between which two energy levels in the mercury atom

will give rise to an emission line of wavelength 320 nm?

Page 89 of 92

Conservation of energy for waves

Inverse square law

A point source of waves emits energy equally in all directions

If energy is conserved then as the waves spread out the same energy is spread over a larger area.

Inverse square law.SWF

http://mort.isvr.soton.ac.uk/SPCG/Tutorial/Tutorial/Tutorial_files/Web-basics-pointsources.htm

The energy flux or intensity is defined as

24 r

P

area

power

This relationship is called the inverse square law. When the distance is doubled the energy flux is reduced by a factor of four.

29 (a) Explain what is meant by the inverse square law of electromagnetic waves such as visible light.

(b) Explain how this inverse square law is consistent with the law of conservation of energy.

30 A light with an output in the visible range of 50 W is switched on at night at

the top of a high tower. Calculate the intensity of the light from the tower at

distances of (a) l00 m (b) 200 m (c) 300 m.

31 A 60 W light bulb converts electrical energy to visible light with an efficiency

of 8%. Calculate the visible light intensity 2 m away from the light bulb. 32 The minimum intensity that can be detected by a given radio receiver is

2.2 10-5 W m-2.

Calculate the maximum distance that the receiver can be from a 10 kW

transmitter so that it is just able to detect the signal.

33 In listening to a person talking to you who is standing 4.0 m away the intensity

of the sound at your ear is 1.2 W m-2. What is the power output power of the

speaker's voice?

34 A radio-operated garage door opener responds to signals with intensity greater

than 20 W m-2. For a 250 mW transmitter unit that broadcasts equally in all

directions, what is the maximum distance from the garage at which the transmitter will open the door?

Page 90 of 92

35 A communication satellite is in orbit above the Earth‟s surface.

(a) The satellite‟s-electrical system is powered by 20 000 photovoltaic cells, each of area 10 cm2. The intensity of the sunlight falling on the cells is 1.4 kWm-2. The cells produce 5.0 kW of electrical power.

Calculate the efficiency of the cells in transferring solar energy to electrical energy.

(b) (i) The satellite generates a signal of power 5.0 kW and orbits at a height of 3.6 x 104 km above the Earth‟s surface. Calculate the

intensity which is detected at the Earth‟s surface if the satellite transmits uniformly in all directions. Assume there is no

absorption of the signal along its path.

(ii) In practice, reflectors on the satellite focus all the 5.0 kW of

transmitted power onto a small area of the Earth‟s surface. If this area is a circle of diameter 1000 km, calculate the intensity

that would be detected there. Assume there is no absorption of the signal along its path.

36 A leaf of a plant tilts towards the Sun to receive solar radiation of intensity

1.1 kW m-2, which is incident at 50 to the surface of the leaf.

(a) The leaf is almost circular with an average radius of 29 mm. Show that the power of the radiation perpendicular to the leaf is approximately 2 W.

(b) Calculate an approximate value for the amount of solar energy received by the

leaf during 2.5 hours of sunlight.

Page 91 of 92

37 The graph shows how the intensity of light from a light-emitting diode (LED)

varies with distance from the LED.

(a) Use data from the graph to show that the intensity obeys an inverse square law.

(b) What does this suggest about the amount of light absorbed by the air?

Page 92 of 92

38 Radio waves and sound waves are sometimes confused by the general public.

(a) Complete the table to give three ways in which they differ.

Radio waves Sound waves

(b) It is proposed to place a solar power station in orbit around the Earth. The

solar power station will convert sunlight to microwave energy. Microwave

collectors on Earth will convert the microwaves into electricity.

(c) The solar power station orbits the Earth at a constant distance from the surface of 36000 km. The total area of the collectors is equivalent to a rectangle with dimensions of 120m by 250m. The collectors are used to generate 600 kW of power. Calculate the intensity of the microwaves at the collectors. State any assumption that you make.

(d) Calculate the total power which the orbiting station would have to emit if it

transmitted microwaves equally in all directions. State any assumption that

you make.

(e) Suggest a more efficient method of transmitting the microwave energy to the collectors on Earth.