Leaving Cert Physics Notes

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weekly grinds brochure.indd 5 24/07/2020 16:46 Light & Sound Notes Contents: Light and Sound

Chapter 1. Waves 1.1 Introduction to Waves ...... 2 1.2 Periodic Time and Frequency ...... 2 1.3 The Wave Equation ...... 4 1.4 Types of Wave ...... 6 Numerical Problems ...... 8

Chapter 2. Reflection 2.1 Laws of Refection ...... 10 2.2 Refection of light at Plane Surfaces ...... 11 2.3 Refection of light at Curved Surfaces ...... 12 Numerical Problems ...... 19

Chapter 3. Refraction 3.1 Laws of Refraction ...... 23 3.2 Real and Apparent Depth ...... 27 3.3 Total Internal Refection ...... 28 3.4 Lenses ...... 31 3.5 Power of a Lens ...... 36 3.6 Eye ...... 37 Numerical Problems ...... 40

Chapter 4. Wave Nature of Light 4.1 Interference ...... 44 4.2 Difraction ...... 46 4.3 Difraction and Interference of Light ...... 47 4.4 Polarisation ...... 51 4.5 Dispersion ...... 53 4.6 Electromagnetic Spectrum ...... 56 4.7 Spectrometer ...... 58 Numerical Problems ...... 59

Chapter 5. Wave Nature of Sound 5.1 Wave Nature of Sound ...... 62 5.2 Properties of Sound ...... 63 5.3 Characteristics of Sound ...... 64 5.4 Resonance ...... 65 5.5 Doppler Efect ...... 66 5.6 Standing Waves in Strings and Pipes ...... 69 5.7 Sound Intensity Level ...... 76 Numerical Problems ...... 78

Leaving Cert Questions (2002 - 2019) ...... 81

Chapter 4 • Wave Nature of Light 4.1 Interference Interference is the physical efect which results from two or more waves overlapping and occurs for both types of wave (transverse and longitudinal).

Only waves of the same type will interfere. Obviously, a sound wave will not interfere with a light wave. However, even waves of the same type need to be fairly similar in terms of their frequency and amplitude in order to interfere. Waves like this are said to be coherent and they are produced from coherent sources.

Definition: Interference is what results when waves from two or more coherent sources overlap and then combine to form a new wave.

Consider two examples of interference when two waves of identical amplitude meet. 1. Constructive interference When two waves meet in step (in phase) they reinforce each other to produce a wave of greater amplitude. The new wave is the sum of the amplitudes of the other waves. +

Definition: When waves from two sources meet and the amplitude of the resulting wave is greater than the amplitudes of each of the individual waves, the waves are said to undergo constructive interference.

2. Destructive interference When two waves meet that are half a wavelength out of step (out of phase) the two waves cancel each other. +

Definition: When waves from two sources meet and the amplitude of the resulting wave is less than the amplitudes of each of the individual waves, the waves are said to undergo destructive interference.

Condition for constructive interference: Path diference = 01λλ,,23λλ,,...... nnλ,.∈ N The path diference between the waves must be a whole number of wavelengths.

1 3 5 1 Condition for destructive interference: Path diference = 2 λλλ,,,2 2 ...... (nn!+2),λ N.

Definition: Coherent Sources give rise to waves of the same frequency and amplitude which are emitted in phase (step).

Example 1: Radio waves of frequency 30 kHz are received at a location 1500 km from a transmitter. The radio reception temporarily fades due to destructive interference between the waves travelling parallel to the ground and the waves refected from a layer (ionosphere) of the earth’s atmosphere, as indicated in the diagram. (a) Calculate the wavelength of the radio waves. (b) What is the minimum distance that the refected wave should travel for destructive interference to occur at the receiver? (c) The layer at which the waves are refected is at a height h above the ground. Calculate the minimum height of this layer for destructive interference to occur at the receiver.

Ionosphere

Transmitter Receiver Ground

Solution

4.2 Diffraction

Definition: Difraction is the spreading of waves around corners.

Explaining Diffraction If the width of the gap is approximately Consider a water wave that meets an the same size as the wavelength of the obstacle. Take the view of looking down wave, λ ≈ d, the wave will spread out from above the waves. The waves will into circular waves. extend around the corner. λ Water wave

If a wave passes through a gap it will spread out. Note: If light is passed through a slit whose size is of the order of 10–7 m this spreading out efect is achieved. This indicates that light is a wave motion with a very small wavelength. For sound, it is very easy to hear around corners.

What are the efects when difraction and interference are put together? A wave is sent towards an obstacle with two gaps in it. The size of the gaps is about the same as the wavlength of the wave. Circular waves spreads out from each gap. The gaps, S1 and S2, are two coherent sources.

S 1

S 2

These two set of coherent waves will undergo interference. There will be lines of maximum amplitude where constructive interference (C.I.) occurs and lines of zero amplitude where destructive interference (D.I.) S S occurs. 1 2

An interference pattern is formed as a result of interference of waves that have already difracted.

C.I. D.I. C.I. D.I. C.I.

Light: oung carried out this experiment with light and obtained bright (constructive interference) and dark fringes (destructive interference) alternately on the screen. This interference pattern was excellent proof that light behaves as a wave motion.

Sound: A similar experiment was also carried out with sound producing loud (constructive interference) and low sounds (destructive interference). This interference pattern was excellent proof that sound behaves as a wave motion.

Interference of waves following difraction setting up an interference pattern is excellent proof of a wave motion.

4.3 Diffraction and Interference of Light oung carried out the following experiment and obtained bright and dark images on a screen.

Demonstration to show that light is a wave

S Monochromatic light S S 2 1 Young's Slits

Interference Pattern Young’s Experiment is proof that light is a wave motion. A source S of monochromatic

light (one colour only and so only one wavelength) is passed through two slits, S1 and S2, which act as coherent sources. Waves spread out from each slit (difraction) and merge with each other (interference). Bright (constructive interference) and dark (destructive interference) images are formed on a screen.

The images obtained in oung’s experiment were very faint and close together. A much more distinct pattern was obtained by using many slits very close to each other. Such a set up is called a diraction grating.

A difraction grating consists of a large number of parallel lines ruled on a sheet of transparent material. Light cannot pass through the lines but can pass through the spaces between them. These spaces then act like a series of parallel slits.

Example 2: A difraction grating consists of 500 lines per millimetre (mm). What is the distance, d, between successive lines? Solution

A difraction grating can be used to fnd the wavelength of light.

A laser passes a beam of monochromatic light through a difraction grating such that bright and dark images are formed at various angles on a screen.

A bright image is formed at an angle of 0o. This is called the zero order image (n = 0). Bright images called frst and second order images are formed in a symmetric pattern at certain angles either side of this central image.

Screen n = 2

Diffraction n = 1 Grating θ Laser n = 0

n = 1

n = 2

The difraction grating formula is: nλ = d sin θ Formulae and Tables Book: Page 59 (Light and sound) n: Order of the image λ: Wavelength of light d: Slit separation of the difraction grating θ: Angle at the which the order of image is formed.

Example 3: A difraction grating, having 80 lines per mm, is arranged between a monochromatic light source (e.g. a laser) and a screen placed 2∙5 apart. Bright spots are seen on the screen at the points A, B, C, D and E. Given that the wavelength of the light from the source is 620 nm calculate the distance AE. Solution Screen A

Diffraction B Grating

Laser C 2.5 m

−7 Example 4: A difraction grating has 500 lines per mm. If light of wavelength 6 × 10 m is incident of the grating how many images will be formed? Solution

Proof of formula: nλ = d sin θ Consider the rays which emerge in a direction making an angle θ with the normal C to the grating. If the path diferences between θ these waves are whole number of wavelengths d B then they will arrive in phase. θ A This occurs if CB = nλ. It can be seen that CB = sinsθθ⇒ CB= CA in = d sinθ CA where d is the distance between the slits.

Therefore the necessary condition for a bright image is: nλ = d sin θ

STS: Soap ubbles and Petrol Films - Colours produced by Interference The colours seen in soap bubbles or on a flm of petrol on water are due to interference of light waves. When light falls on these surfaces, some of the light is refected from the flm and some from the surface of the water.

Interference occurs when light from each surface meets. Diferent wavelengths are refracted at diferent angles in the petrol. Depending on the thickness of the flm or the angle at which you Petrol view it, diferent wavelengths interfere Water constructively and hence that particular colour is seen.

4.4 Polarisation If a taut rope is shook up and down, a transverse wave travels along the rope in a plane. Such a wave is plane polarised. If the rope is shaken up and down vertically, the wave is vertically plane polarised. If the rope is shaken from side to side, the wave is horizontally plane polarised.

A vibrating electron emits an electromagnetic wave that is plane polarised. The plane of polarisation will be along the direction of the vibrating charge. A vertically accelerating electron emits vertically polarised light.

A X Y

A common light source emits non-polarised light. This is because the vibrating electrons producing the light vibrate in many directions and so the waves travel in many planes. If an unpolarised transverse wave is passed through a narrow slot X only the vibrations in the direction AB will pass through the slot. The other vibrations will be blocked. The transmitted wave will be plane polarised.

A second slot Y is now placed behind the slot X. If the two slots are parallel the plane polarised wave passes through. If the slots are perpendicular to each other the wave will be blocked. The frst slot X is called a polariser. The second slot Y is called an analyser. Light waves can be polarised but sound waves cannot. This is because light waves are transverse whereas sound waves are longitudinal. A back and forth movement of particles as in a sound wave does not have planes of polarisation and so cannot be plane polarised or blocked as above.

Only transverse waves can be polarised.

Definition: A wave is polarised when it vibrates in a single plane after passing through a polariser. Unpolarised waves vibrate in all planes.

Light as a Transverse Wave Motion Light can be polarised by a substance called polaroid. Polaroid is a crystal structure with the molecules aligned parallel to one another.

Demonstration to show that light is a transverse wave If two pieces of polaroid are rotated with respect to each other the light transmitted through the overlapping section goes from maximum brightness to darkness.

Light gets through Light gets blocked

Only transverse waves can be polarised. The fact that light can be polarised shows it is a transverse wave.

STS: Polarisation Polarisation y Reection: Light refected from a glass or water surface is found to be partially plane polarised. This refected light is a nuisance causing a glare. By viewing through a piece of polaroid this glare or shine can be considerably reduced. Polaroid sunglasses achieve this. Polaroid flters can also be used on cameras.

Stress Polarisation: A piece of perspex is put between two polaroid materials which are at right angles to each other. If the perspex is bent putting it under strain, colours can be seen indicating where the stress occurs. This phenomenon is called photoelasticity and is used by engineers to analyse stresses in components.

4.5 Dispersion Newtons Experiment Newton observed that a ray of white sunlight was split into a spectrum of colours as it passed through a glass prism. The white light has been deviated and dispersed.

Deviation: This is the change in the direction of a ray of light as it goes from one medium to another. Each colour of light has a slightly diferent deviation which is caused by the refractive index of glass being slightly diferent for each colour. Violet light has the greatest deviation.

Dispersion: The prism disperses all the colours which already exist in white light. White light is not a single colour it is a combination of all the colours in the spectrum.

White Light λ Red Orange Prism Yellow Green Blue Indigo λ Violet

Definitions: Deviation is the change in the direction of a ray of light as it goes from one medium to another. Dispersion is the breaking up of light into its colours.

Recombining the Spectrum White light can be separated into a spectrum of diferent colours of light. It can also be reformed from the separate colours. This can be done in two ways: 1. Using two prisms: The second prism reverses the deviation and dispersion of the frst prism so that the colours of the spectrum recombine where the colours overlap.

White

Red White Light Violet

Red Orange 2. Newton’s disc: This is a card coloured with all the colours of the spectrum in equal areas. If the disc is spun rapidly Yellow Violet the persistence of vision of the human eye remembers all the colours it sees as the colour changes places. They are Green Indigo combined in the eye to produce white. Blue

Dispersion using a diffraction grating According to the difraction formula the position of a particular image formed by a difraction grating depends on the wavelength of light used. White light is made up of diferent colours with diferent wavelengths. When white light is passed through a grating, the diferent coloured images are formed at diferent angles.

A visible spectrum is formed for each value of n, except at n = 0. At this position, regardless of the wavelength, all colours hit the screen at 0o. Therefore, white light is seen as the colours are in the same position. Dispersion today is usually carried out with gratings rather than prisms. R n =2 V Diffraction R Grating n =1 V White n =0

V n =1 R V n =2 R

Example 5: Monochromatic red light is passed through a difraction grating and onto a screen. What would be the efect, if any, on the separation of the images on the screen if: (a) the screen was moved nearer the grating, (b) the separation of the lines on the grating were reduced, (c) blue light was used instead of red light? Note: Blue light has a shorter wavelength than red light. Solution The separation of the fringes depends on θ which in turn depends on sin θ. A reduction in sin θ means the images get closer together. Write the difraction nλ formula in this form: sinθ = d

Screen Diffraction Grating

Red

STS Rainbows: When white light enters a raindrop it is both refracted and internally refected. The refraction causes the diferent wavelengths to be dispersed and the rainbow is seen. Diamonds: The sparkle of a diamond is due to its high refractive index. Most of the light entering the diamond is refected due to total internal refection at the cut faces. Compact Disc (CD): The colours seen on a CD when white light falls on it are due to the disc behaving as a refection difraction grating.

Colours Red (R), Green (G) and Blue (B) are called the primary colours of light. When the three primary colours of light are added together they produce white (W) light.

R G B → W

The secondary colours are ellow (), Magenta (M) and Cyan (C). They are formed by mixing two primary colours together. R G → Y R B → M B G → C White light can be obtained by mixing a secondary colour (made up of two primary colours) with the other primary colour. These pairs of colours are called complementary colours.

R R + C → W G M → W R Y G B → W Yellow Magenta W W M C

G B B Cyan

Definitions The primary colours of light are those which cannot be made by adding (or mixing) any other colours of light together. The secondary colours of light are made by adding two primary colours together. Complementary colours of light are a primary colour and a secondary colour which when mixed together make white light.

STS: Colours Creating light of a given colour by mixing the primary colours together is done with stage lighting and colour television.

Absorption and Reflection of colours The colour of an object depends on the colours it refects. A white object refects all colours and absorbs none. A black object refects no colours and absorbs all colours. An object that appears red will refect red light and absorb all other colours.

R B G G B R G R B

Green Red Blue If red light falls onto a red object it will appear red as it refects the red light. If red light falls onto a green object it will appear black as it absorbs the red light refecting no light. Red Filters: A red flter allows red light to pass through and blocks all other colours. R G

4. The Electromagnetic Spectrum Visible light (the colours) is a small part of a much bigger spectrum called the electromagnetic spectrum. All the radiations in this spectrum are produced by vibrating charges.

If electrons vibrate up and down a few thousand times per second a radio wave is produced. If the electrons could be made to vibrate a few million-billion (≈ 1015) times per second then a visible light wave is produced.

All electromagnetic waves travel at the same speed in a vacuum. They difer from one another in their frequency and wavelength. The frequency of the wave is the same as the frequency of the vibrating charge producing it.

1011 10–9 108 10–7 105 10–2 103

Gamma Rays X-Rays UV Visible IR Microwaves Waves Radio Waves Emitted in Used in medicine All the colours Used in Used in nuclear reactions communications communications

An electric charge is surrounded by an electric feld. When the charge vibrates the electric feld vibrates with it. A changing electric feld gives rise to a magnetic feld perpendicular to it.

Magnetic Field Electric Field

STS Infra-red (IR) radiation This is radiation beyond the red end of the visible spectrum (radiation having longer wavelengths than visible light). Like light, it is emitted by hot bodies. However, those bodies emitting mainly IR radiation are generally at lower temperatures than those emitting light. They produce a more noticeable heating efect than light and can be detected using a sensitive thermometer.

Infra-Red Camera: IR radiation efects photographic plates and can be used to take photographs in the dark. Most bodies give of IR radiation. This radiation can pass through fog and mist. Thermal Imaging: In medicine, body heat emitted by the skin can be photographed and thermal images of the body produced. This can be used to diagnose abnormalities in the body. reenhouse Eect: The sun heats up the earth. The earth re-radiates this heat at a slightly longer wavelength - in the IR range. Gases in the atmosphere, particularly carbon dioxide, trap this radiation and keep the earth warm. Over the last number of years, carbon dioxide emissions have greatly increased due to the burning of fossil fuels. This has lead to an increased warming of the earth called the greenhouse efect. This may cause the polar caps to melt raising the level of seawater around the globe leading to increased fooding in years to come.

ltra-violet () radiation This is radiation beyond the violet end of the visible spectrum (radiation having shorter wavelengths than visible light). It is emitted by hot bodies. The temperatures are generally higher than those which emit mainly visible light. UV radiation causes certain substances like vaseline to fuoresce. The substance absorbs the UV radiation and re-emits it as visible light. This property can be used to detect UV radiation.

Oone Layer: UV radiation causes sun-tanning. Too much exposure to UV light can cause skin cancer. The ozone layer in the atmosphere absorbs much of the UV light reaching the earth’s surface. Many years ago holes started to appear in the ozone layer, probably caused by the overuse of chemicals used in aerosols called CFC’s (chlorinated fuorocarbons). The banning of these substances has led to a recovery in the ozone layer.

4. The Spectrometer The spectrometer is an instrument used for the production and study of optical spectra. It can be used to measure the angles at which bright images are formed from a difraction grating. It consists of a collimator which produces a narrow beam of parallel light. A diraction grating is clamped on a rotating table which is set perpendicular to the collimator. A telescope which views the fringes can rotate about in a circle over a scale which measures angle. If it is rotated to the left or right other fringes will be viewed at certain angles.

Levelling Screw n = 2

Diffraction n = 1 grating Monochromatic Light Source θ n = 0 Telescope Collimator θ

n = 1

Turntable n = 2

Parts of a Spectrometer A Collimator: A tube with an convex lens at one end and an adjustable aperture at the other end. It produces a parallel beam of light. B Turntable: This can revolve about a vertical axis. It has three levelling screws and a vernier scale. C Telescope: This has two convex lenses - the objective and the eyepiece. The eyepiece is ftted with cross-wires and is adjustable. It focuses light and measures the angle of the displaced rays. A difraction grating is placed on the turntable.

Phsics Experiment Boo Experiment L5: To measure the wavelength of light using a difraction grating and spectrometer (Page 112) Results: 2, page 116

Answers to Examples Example 1: (a) 10 km, (b) 5 km, (c) 61∙3 km Example 2: 2 × 106 m Example 3: 0∙5 m Example 4: 7 Example 5: (a) Closer, (b) Further, (c) Closer

Numerical Problems Interference of Waves

1. A signal generator is connected to two loudpeakers which act as coherent sources of sound waves. The waves produced spread out and interfere producing 0.2 m an interference pattern 4 m away. If the frst position 2 m of destructive interference is 0∙2 m from the centre, fnd the path diference between the waves. What is the frequency and wavelength of the waves if the velocity of sound in air is 340 m s1? 4 m

S 2. Two sources of coherent sound waves, S1 and S2, 1 produce an interference pattern 1.5 m away. Find . O the path diference between the waves at O. If 1 4 m . . this is the frst position of destructive interference 1 5 m 0 6 m fnd the wavelength. If the velocity of sound in air 1.4 m is 340 m s1, fnd the frequency. S 2

3. Sound from a loudspeaker L can travel two separate paths, A and B, to an opening O. The length of the apparatus can be varied by moving the sliding tube in and out. As the tube is moved, the loudness at O increases and decreases. Explain this.

If the frequency of the note is 2000 Hz calculate the distance moved by the sliding tube between successive positions of maximum and minimum loudness given that the speed of sound in air is 340 m s1. L

A B

Sliding Tube

4. Radio waves of frequency 25 kHz are received at a location C from a transmitter A, 1000 km apart. The radio reception temporarily fades due to destructive interference between the waves travelling parallel to the ground and the waves refected from a layer (ionosphere) of the earth’s atmosphere. Ionosphere

A Ground C 1000 km (i) Calculate the wavelength of the radio waves. (ii) What is the minimum path diference between both waves? (iii) The layer at which the waves are refected is at a height h above the ground. Calculate the minimum height of this layer for destructive interference to occur at the receiver.

Diffraction of Light 5. A difraction grating has 5000 lines per cm. The frst order image is at an angle of 22o. Find the wavelength of the light.

6. Monochromatic light is incident on a difraction grating with a grating spacing d of o 3 × 10−6 m. The second order image is formed on a screen at an angle of 35 . What is the wavelength of the light?

7. Monochromatic light of wavelength 6∙5 × 10−7 m is incident normally on a difraction grating having 6000 lines per cm. At what angles to the normal are the frst and second order images obtained?

8. Light of wavelength 6 × 10−7 m is falls on a difraction grating. The frst order image is formed at an angle of 24o to the normal. What is the grating element (i.e. number of lines per m or cm)?

9. A beam of monochromatic light of wavelength 5 × 10−7 m is falls normally on a grating of 5000 lines per cm. A screen is placed 0.3 m from the grating. Find the distance of the second order image from the central image.

10. Light of wavelength 6 × 10−7 m is is passed through a difraction grating having 500 lines per mm. At what angles to the normal will fringes be formed?

11. A difraction grating, having 100 lines per mm, is arranged between a monochromatic light source (e.g. a laser) and a screen producing an interference pattern of bright and dark fringes. The screen and difraction grating are 50 cm apart. Given that the wavelength of the light from the source is 700 nm (nano n = 10–9) calculate the distance between the second order bright fringes either side of the normal.

12. Using the same difraction grating, the second order spectrum of light of wavelength 6∙25 × 10−7 m coincides with the fourth order spectrum of light of a diferent wavelength. What is the wavelength of the second light?

Answers Interference of Waves 1. 0.097 m, 0.194 m, 1752.6 Hz

2. 0.8 m, 1.6 m, 212.5 Hz

3. 0.17 m, 0.0425 m

4. (i) 12 km (ii) 6 km (iii) 54.85 km

Diffraction of Light 5. 7∙5 × 10−7 m

6. 8∙6 × 10−7 m

7. 22.9o, 51.1o

8. 6779 lines per cm

9. 17∙3 cm

10. On the normal and at angles 17.46o, 36.87o and 64.16o on either side of the normal.

11. 14.1 cm

12. 3∙125 × 10−7 m

© The Academy Page 61 Kieran Mills & Tony Kelly Light & Sound Notes Leaving Cert Questions Chapter 2: Reflection Experimental Section A Exp. L1: Focal Length of a Mirror (Graph) 2007 uestion 3 ...... 84 Exp. L1: Focal Length of a Mirror (Graph not necessary) [2013 Question 3] ...... 86 Theoretical Section B Refection, convex mirrors 2014 uestion 12 (b) ...... 88 Concave mirrors 2017 uestion 12 (c) ...... 89

Chapter 3: Refraction Experimental Section A Exp. L1: Focal Length of a Mirror (Graph not necessary) [2019 Question 2] ...... 90 Exp. L2: Snell’s Law (Graph) [2018 Question 2] ...... 91 Exp. L2: Snell’s Law (Graph) [2014 Question 2] ...... 93 Exp. L2: Snell’s Law (Graph) [2005 Question 3] ...... 95 Exp. L2: Snell’s Law (Graph) [2010 Question 3] ...... 97 Exp. L4: Focal Length of a Lens (Graph not necessary) [2003 Question 3] ...... 99 Exp. L4: Focal Length of a Lens (Graph not necessary) [2009 Question 2] ...... 100 Exp. L4: Focal Length of a Lens (Graph not necessary) [2012 Question 2] ...... 101 Theoretical Section B Lenses, power of a lens, eye defects 2006 uestion 7 ...... 102 Lenses, power of a lens, eye [2008 Question 9] ...... 104 Power of a lens, eye [2002 Question 12 (b)] ...... 106 Optical Fibres [2004 Question 12 (b)] ...... 107 Optical Fibres [2009 Question 12 (c)] ...... 108 Real and Apparent depth, total internal refection 2011 uestion 12 (b) ...... 109 Refraction of light problems [2012 Question 12 (b)] ...... 110 Refraction of light, converging lens [2015 Question 12 (b)] ...... 111 Critical angle, total internal refection 2019 uestion 12 (c) ...... 112

Chapter 4: Wave Nature of Light Experimental Section A Exp. L5: Wavelength of light [2004 Question 2] ...... 113 Exp. L5: Wavelength of light [2006 Question 2] ...... 115 Exp. L5: Wavelength of light [2008 Question 3] ...... 116 Exp. L5: Wavelength of light [2011 Question 3] ...... 117 Exp. L5: Wavelength of light [2015 Question 3] ...... 118 Exp. L5: Wavelength of light [2016 Question 3] ...... 120 Exp. L5: Wavelength of light [2018 Question 3] ...... 122 Theoretical Section B Interference of Waves (Ionosphere Problem) 2002 uestion 7 ...... 124 Interference and Difraction, types of wave, polarisation 2005 uestion 7 ...... 126 Interference and Difraction, energy in a photon 2014 uestion 7 ...... 128 Difraction and Dispersion 2009 uestion 7 ...... 130 Electromagnetic Spectrum 2012 uestion 7 ...... 132 Difraction and Dispersion 2013 uestion 12 (b) ...... 133

Chapter 5: Wave Nature of Sound Experimental Section A Exp. S1: Speed of Sound [2014 Question 3] ...... 134 Exp. S1: Speed of Sound [2006 Question 3] ...... 136 Exp. S2: Sonometer (Frequency vs. length) [2004 Question 3] ...... 137 Exp. S2: Sonometer (Frequency vs. length) [2012 Question 3] ...... 138 Exp. S2: Sonometer (Frequency vs. length) [2016 Question 2] ...... 140 Exp. S3: Sonometer (Frequency vs. tension) [2002 Question 3] ...... 142 Exp. S3: Sonometer (Frequency vs. tension) [2009 Question 3] ...... 144 Exp. S3: Sonometer (Frequency vs. tension) 2017 uestion 3 ...... 146 Theoretical Section B Doppler Efect (Bats Problem) 2003 uestion 7 ...... 147 Doppler Efect (Moving Star Problem) 2007 uestion 7 ...... 149 Doppler Efect 2010 uestion 7 ...... 150

Doppler Efect, polarisation, colours 2017 uestion 7 ...... 152 Interference, standing waves in pipes, sound intensity 2011 uestion 8 ...... 154 Resonance, standing waves in pipes 2013 uestion 7 ...... 156 Resonance, standing waves in pipes [2015 Question 9] ...... 157 Resonance, standing waves, sound intensity 2018 uestion 7 ...... 159 Difraction, power of a lens, resonance in closed pipe 2019 uestion 10 ...... 161 STS: (Radio Waves in mobile phones) [2010 Question 11] ...... 163 Vibrating strings [2005 Question 12 (c)] ...... 165 Sound intensity 2007 uestion 12 (b) ...... 166 Musical notes and Doppler Efect 2008 uestion 12 (b) ...... 167 Resonance, characteristics of sound [2010 Question 12 (c)] ...... 168

LC 200: Experiment L1 (Focal Length of a Mirror - Graph necessary)

3. In an experiment to measure the focal length of a concave mirror, an approximate value for the focal length was found. The image distance v was then found for a range of values of the object distance u. The following data was recorded.

u/cm 15.0 20.0 25.0 30.0 35.0 40.0 v/cm 60.5 30.0 23.0 20.5 18.0 16.5

How was an approximate value for the focal length found? What was the advantage of finding the approximate value for the focal length? (10)

Describe, with the aid of a labelled diagram, how the position of the image was found. (12)

Calculate the focal length of the concave mirror by drawing a suitable graph based on (18) the recorded data.

u/cm 15.0 20.0 25.0 30.0 35.0 40.0 v/cm 60.5 30.0 23.0 20.5 18.0 16.5

LC 2013: Experiment L1 (Focal Length of a Mirror - Graph not necessary)

3. The following is part of a student’s report on an experiment to measure the focal length of a concave mirror.

“I started with the object 6 cm from the mirror but couldn’t get an image to form on the screen. I moved the object back a few centimetres and tried again, but I couldn’t get an image to form on the screen until the object was 24 cm from the mirror. From then on I moved the object back 8 cm each time and measured the corresponding image distances. I wrote my results in the table.”

u/cm 24.0 32.0 40.0 48.0 v/cm 72.7 40.3 33.0 27.8

Draw a labelled diagram of the apparatus used. (9)

Give two precautions that should be taken when measuring the image distance. (6)

Explain why the student was unable to form an image on the screen when the object was close (6) to the mirror.

Use all of the data in the table to calculate a value for the focal length of the mirror. (15)

Describe how the student could have found an approximate value for the focal length of the (4) mirror before starting the experiment.

LC 201: Reflection, convex mirrors

12. (b) What is reflection? (3)

Spherical mirrors can be either convex or concave.

Draw a ray diagram to show the formation of an image in a convex mirror. (9)

A person looks at her image in a shiny spherical decoration when her face is 30 cm from the surface of the decoration. The diameter of the decoration is 20 cm. Find the position of the image. (12)

Concave mirrors, rather than convex mirrors, are used by dentists to examine teeth. Explain why. (4)

LC 2017: Concave mirrors

12. (c) It has been recently suggested that the 17th century Dutch artist Rembrandt used a concave mirror to help him etch self-portraits by projecting an inverted image of himself onto a copper sheet. Draw a ray diagram to illustrate how Rembrandt used a concave mirror in this way. (10)

Rembrandt used a concave mirror of focal length 60 cm so that the image on the copper sheet was only half the size of the object. Calculate (i) the distance from the sheet to the mirror and (ii) the distance from the object to the mirror. A concave mirror can also be used to produce an upright image. Explain why this image was not of use to Rembrandt. (18)

LC 2019: Experiment L1 (Focal Length of a Concave - Graph not necessary)

2. In an experiment to determine the focal length of a concave mirror a student first found the approximate focal length of the mirror. He then placed an object in front of the mirror and measured the object distance u and the corresponding image distance v. He repeated this procedure for different values of u. The following data were recorded.

u (cm) 20.0 30.0 40.0 50.0 v (cm) 61.0 29.5 24.0 20.5

How did the student find the approximate focal length? Why did the student find the approximate focal length at the start of the experiment?

Describe, with the aid of a labelled diagram, how the position of the image was found.

State two precautions that should be taken when measuring v. (27)

Use all of the data to calculate the focal length of the mirror. (13)

LC 2018: Experiment L2 (Snell's Law)

2.A student measured the angle of incidence i and the angle of refraction r for a ray of light passing through a transparent block. She repeated this experiment for different values of i and used her data to investigate the relationship between the angle of incidence and the angle of refraction. The following data were recorded.

i (degrees) 20 30 40 50 60 70 r (degrees) 14 20 26 31 35 38

Describe, with the aid of a labelled diagram, how the student determined the angle of refraction. (12)

Draw a suitable graph to show the relationship between the angle of incidence and the angle of refraction. State this relationship and explain how your graph verifies it. Use your graph to determine the refractive index of the material used. (24)

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What would be observed if the incident ray was perpendicular to the block? (4)

i (degrees) 20 30 40 50 60 70 r (degrees) 14 20 26 31 35 38 sin i sin r

LC 201: Experiment L2 (Snell's Law)

2. In an experiment to measure the refractive index of a substance, a student used a rectangular block of the substance to measure the angle of incidence i and the corresponding angle of refraction r for a ray of light as it passed from air into the substance. The student repeated the procedure for a series of different values of the angle of incidence and recorded the following data.

i (degrees) 20 30 40 50 60 70 80 r (degrees) 13 20 27 23 36 40 43

One of the recorded angles of refraction is inconsistent with the others. Which one? (4)

Describe, with the aid of a labelled diagram, how the student found the angle of refraction. (12)

Calculate a value for the refractive index of the substance by drawing a suitable graph based on the recorded data. (18)

i (degrees) 20 30 40 50 60 70 80 r (degrees) 13 20 27 23 36 40 43

Using a graph to calculate a value for the refractive index is a more accurate method than calculating the refractive index for each pair of angles and then finding the mean. Give two reasons for this. (6)

LC 200: Experiment L2 (Snell's Law)

3. In an experiment to verify Snell’s law, a student measured the angle of incidence i and the angle of refraction r for a ray of light entering a substance. This was repeated for different values of the angle of incidence. The following data was recorded.

i/degrees 20 30 40 50 60 70 r/degrees 14 19 26 30 36 40

Describe, with the aid of a diagram, how the student obtained the angle of refraction. (9)

Draw a suitable graph on graph paper and explain how your graph verifies Snell’s law. (18)

rom your graph, calculate the refractive index of the substance. (9)

The smallest angle of incidence chosen was 20o. Why would smaller values lead to a less (4) accurate result?

i/degrees 20 30 40 50 60 70 r/degrees 14 19 26 30 36 40

LC 2010: Experiment L2 (Snell's Law)

3. In an experiment to verify Snell’s law, a student recorded the following data.

i/° 30 40 50 55 60 65 70 r /° 19 26 30 33 36 38 40

Draw a labelled diagram of the apparatus used. On your diagram, indicate an angle i and its corresponding angle r. 12)

Using the recorded data, draw a suitable graph and explain how your graph verifies Snell’s law.

Using your graph, calculate the refractive index of the substance used in the experiment. (22)

The student did not record any values of the angle i below 30°. Give two reasons why. (6)

i/degrees 30 40 50 55 60 65 70 r/degrees 19 26 30 33 36 38 40

LC 2003: Experiment L4 (Focal Length of a Lens - Graph not necessary)

3. The following is part of a student’s report of an experiment to measure the focal length of a converging lens.

“I found the approximate focal length of the lens to be 15 cm. I then placed an object at different positions in front of the lens so that a real image was formed in each case.”

The table shows the measurements recorded by the student for the object distance u and the image distance v.

u/cm 20.0 25.0 35.0 45.0

v/cm 66.4 40.6 27.6 23.2

How did the student find an approximate value for the focal length of the lens? (6)

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LC 200: Experiment L4 (Focal Length of a Lens - Graph not necessary)

2. A student was asked to measure the focal length of a converging lens. The student measured the image distance v for each of three different object distances u. The student recorded the following data.

u/cm 20.0 30.0 40.0 v/cm 65.2 33.3 25.1

Describe how the image distance was measured. (12)

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8IBUEJGGJDVMUZXPVMEBSJTFJGUIFTUVEFOUQMBDFEUIFPCKFDUDNGSPNUIFMFOT (7)