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Light PHYSICS – UNIT TWO Light What is light? Light exhibits properties of both waves and particles. Light as a wave and a particle https://www.youtube.com/watch?v=J1yIApZtLos Properties of light https://www.youtube.com/watch?v=bP4i1KUfB3o Waves Light as a wave Light behaves in a way that is consistent with our conceptual and mathematical understanding of waves, as it…… • Reflects like a wave • Refracts like a wave • Diffracts like a wave • Undergoes interference like a wave Now we will take a closer look at waves to understand what all this means Waves What are waves? • A disturbance that travels through a medium from one place to another. • They originate from a vibration that initially disturbs the medium. • They transport energy as they travel through the medium Lets view an example of energy transfer via waves. Guitar video The string is plucked, a wave transfers this energy to sound. You can see that different sided waves produce different sounds https://mrsamsingyr11physics.wordpress.com/ Waves Transverse vs Longitudinal Waves Longitudinal Waves are waves in which the particles of the medium are In Transverse Waves, the particles of the medium are displaced in a direction perpendicular (right angled) to the direction of the energy transport https://www.youtube.com/watch?v=7cDAYFTXq3E displaced in a direction parallel to the energy transport. Waves Transverse vs Longitudinal Waves Longitudinal Waves are waves in which the particles of the medium are In Transverse Waves, the particles of the medium are displaced in a direction perpendicular (right angled) to the direction of the energy transport https://www.youtube.com/watch?v=7cDAYFTXq3E displaced in a direction parallel to the energy transport. Waves Longitudinal Waves Longitudinal Waves are waves in which the particles of the medium are displaced in a direction parallel to the direction of the energy transport. Direction of the vibration through the medium The wave moves in this direction Waves Transverse Waves Transverse Waves are waves in which the particles of the medium are displaced in a direction perpendicular (right angled) to the direction of the energy transport The wave moves in this direction Direction of the vibration through the medium Transverse Waves Pulsed vs Continuous Wave Pulsed Wave • When a single disturbance is passed through a medium • The wave carries the energy which gradually oscillates away once the energy is used eg. An Explosion or a Sudden Impact Continuous Wave • When a continuous disturbance is passed through a medium generated by repetitive motion • Energy is carried away in the form of a continuous wave eg. A vibration of sound waves from a speaker NOW DO Problems from the alternative text Ex1.1A Q 1, 2, 3, 4, 5, 6, 7 Transverse Waves Amplitude • Is a measure of how much energy the wave has. • Is measured from the midpoint to the peak of the crest (or bottom of the trough). • Can be defined as the maximum displacement from the average position. Amplitude Crest Trough Transverse Waves Wavelength (π) • Is the distance between two peaks (or the distance between two troughs) • Can be defined as ‘the distance the wave has travelled during one complete cycle’. • Given by the symbol π ( pronounced ‘lam-da’ ) • Measured in metres (as it’s a length) Wavelength ( λ ) Crest Trough Transverse Waves Frequency (π) • The Frequency of a wave is defined as ‘the number of complete cycles in one second’. • Units: Hertz (Hz) • 1 Hz = 1 cycle per second One complete cycle Transverse Waves Period (π) • The Period of a wave is defined as ‘the time taken for one complete wave cycle’. • Measured in seconds ππππππ = 1 πππππ’ππππ¦ Which can be rearranged to give: πΉππππ’ππππ¦ = Period - One complete cycle 1 ππππππ Transverse Waves Velocity of a wave In a constant medium, the velocity of the wave is given by πππππππ‘π¦ = πΉππππ’ππππ¦ × π€ππ£ππππππ‘β → π£ = πλ Which can be rearranged to give: π= π£ λ or λ= π£ π Or as usual, if we know a distance travelled and a time taken to travel that distance, we can use πππ π‘ππππ π£= π‘πππ Transverse Waves Examples eg1. A wave undergoes 42 vibrations in 60 seconds. a) Find the frequency of the wave. b) Find the period of a wave. πΉππππ’ππππ¦ = πΆπ¦ππππ πππ π πππππ 42 = 60 = 0.7 π»π§ ππππππ = ππππ π‘ππππ π‘π πππππππ‘π πππ ππ¦πππ 1 = πΉππππ’ππππ¦ 1 0.7 = = 1.43 π ππππππ Transverse Waves Examples eg2. On a guitar string, the trough of the wave travels 10cm in 2 seconds. Find the speed of the wave in metres per second.. πππππ = = πππ π‘ππππ π‘πππ 0.10 π 2π = 0.05 π/π Transverse Waves Examples eg3. A bird lands on a lake, causing vibrations resulting in a ripple of 30 waves in 5 seconds. a) Find the frequency of the waves πΉππππ’ππππ¦ = πΆπ¦ππππ πππ π πππππ 30 = 5 = 6 π»π§ b) Find the period of each wave ππππππ = ππππ π‘ππππ π‘π πππππππ‘π πππ ππ¦πππ 1 = πΉππππ’ππππ¦ 1 6 = = 0.167 π ππππππ Transverse Waves Examples eg4. If waves on a lake vibrate at a frequency of 30 Hz and the wave crests are 1cm apart, what is the speed of the waves? π£ = πλ = 30 × 0.01 = 0.30 π/π NOW DO Problems from the alternative text Q1.2A Q1 – 7 Q1.3A Q1 – 4 NOW DO Problems from the alternative text Worksheet – Waves equations Q1.2A Q9 – 10 Behaviour of light Light can be absorbed, reflected or transmitted We will look closer at these behaviours, but firstly we need to look at another model for describing light. The Ray model We can understand many phenomena involving light without having to use sophisticated models such as the wave model or the particle model. Instead, we describe light and it’s interactions with objects more simply according to the path it takes. To model and draw light, we can think of light travelling in a straight line path, called light rays. This is called the ray model. The Ray model Light rays transmitted from a light source Transmitted light rays from the sun hit a tree – the shadow formed shows than light does not pass through the tree – where does it go? Some light is reflected – enabling us to see the tree Some light is absorbed – providing energy to the tree. Straight rays of light transmit through a curved medium – the light refracts and converges Rays of light reflect off the fish in the medium (water) and transmit through to a different medium (air) causing the rays to diverge due to the refraction. Some Light Terminology Luminous – Objects that we see as they give off their own light. eg. sun, lamp. Non-Luminous – Objects that we see as light reflects off them. eg. people, book. Incandescent – A luminous object that produces light due to being heated. Brightness and colour changes depending on how hot the object is. eg. Gas heater/cooker – blue at source = hot! Yellow flame = cooler Some Light Terminology Opaque – Opaque objects reflects some light and absorbs the rest. No light passes through. eg. A brick wall, tiled floor. Transparent – Objects that allow a significant amount of light to pass through. Some light may also be absorbed and/or reflected. eg. Clear windows, cling wrap. Translucent – Objects that allow a some light to pass through, but no clear image can be seen through the material. eg. Frosted bathroom windows, tissue paper. Behaviour of Light Rays Ex 1.1B Q1 – 6 Have a think about Question 2, attempt it and then we can do it together Reflection off perfectly flat surfaces Light rays travel off a surface into the direction of the mirror and reflect off into the direction of the eye. Plane Mirror Reflection A normal household mirror is constructed with 3 separate layers: • Transparent Glass • Thin coating of Aluminium or Silver deposited on the glass to reflect light • A backing layer of protective paint When a beam of light strikes the surface of the mirror; ~4% of the light is reflected off the glass, ~96% of the light is reflected of the metal. Because this is near perfect reflection, we use the plane mirror as our model to display reflection Plane Mirror Reflection The light ray that approaches the mirror is called the incident ray. The ray leaving the mirror is called the reflected ray. The normal is perpendicular (right angle) to the mirror. The angle of incidence (π) is the angle between the incident ray of the normal. The angle of reflection π is the angle between the reflected ray and the normal. πππππ ππ πππππππππ π = πππππ ππ πππππππ‘πππ (π) Behaviour of Light Rays – Reflection off irregular surfaces When a surface is not perfectly flat like a plane mirror, the reflection is known as diffuse reflection. Each light ray still reflects off the surface with πππππ ππ πππππππππ π = πππππ ππ πππππππ‘πππ (π), however because the surface is irregular, the reflected rays emerge in different directions. Behaviour of Light Rays – Forming images with a plane mirror • Images are a likeness of an object. • When viewing an object with a mirror, we see a virtual image. • Rays of light from the cats ear reflect off the mirror and enter the viewers eye and appear to come from a region behind the mirror. • The real image and the virtual image are located at equal distances from the mirror. “Real Image” “Virtual Image” Behaviour of Light Rays – Drawing ray diagrams Summary, Drawing ray diagrams for plane mirror reflection https://www.youtube.com/watch?v=OHqYB42WfZk Drawing ray diagrams – Plane Mirrors “Virtual Image” “Real Image” Drawing ray diagrams – Plane Mirrors Plane Mirror Reflection – The Image you see is: UPRIGHT, LATERALLY INVERTED Behaviour of Light Rays - Reflection Ex 1.2B Q1, 2, 4, 6, 7 Refraction • Light travels in straight line paths while travelling in a uniform medium. • When it passes through another medium, the light ray changes direction, or its path bends. • Refraction occurs as the light changes speed in different media, due to the change in density of different substances. Behaviour of Light Rays - Refraction • In 1621, Dutch physicist, Willebrand Snell investigated refraction of light • He found that the ratio of the sines of the angles of incidence and angles of refraction was constant for all angles of incidence for a given pair of media. πππππ ′ π πΏππ€: π ππ ππ π ππ ππ = ππππ π‘πππ‘ This ‘constant’ value (n) changes depending on the pair of media Behaviour of Light Rays - Refraction • Snell repeated the experiment & found the ratio changed with different substances. • This showed that different substances bend light by different amounts. • The value of the ratio (of any medium compared to a vacuum) is called the refractive index, given by the symbol n. • We can rewrite πππππ ′ π πΏππ€ to solve angles and n values for any refracted ray: π1 π ππ π1 = π2 π ππ π2 Behaviour of Light Rays - Refraction eg1. A ray of light travelling through air (nβ1), strikes a glass block with a refractive index of 1.8 at an angle of incidence of 40°. What is the angle of refraction? π1 = 1 π2 = 1.8 π1 = 40° π2 = ? π1 π ππ π1 = π2 π ππ π2 1 sin 40 = 1.8 sin π2 π2 = sin−1 ( sin(40) ) 1.8 π2 = 20. 92° Behaviour of Light Rays - Refraction eg2. A ray of light travelling through air (nβ1), strikes a plastic block at an angle of incidence of 35°, refracting at an angle of 15°. What is the refractive index of the plastic block? π1 = 1 π2 =? π1 = 35° π2 = 15° π1 π ππ π1 = π2 π ππ π2 1 sin 35 = π sin(15) sin( 35 ) π= sin( 15 ) π = 2. 22 Behaviour of Light Rays - Refraction Table: Some common refractive index values These were derived by: πππππ ππ πππβπ‘ ππ π π£πππ’π’π π= πππππ ππ πππβπ‘ ππ π‘βπ πππππ’π Behaviour of Light Rays - Refraction ππ < ππ ππππ 1 > ππππ π΄ π£πππ 1 < π£πππ π΄ ππ > ππ ππππ 1 < ππππ π΅ π£πππ 1 > π£πππ π΄ ππ = ππ ππππ 1 = ππππ πΆ π£πππ 1 = π£πππ π΄ Behaviour of Light Rays - Refraction As the speed of light in a vacuum (c) = π × πππ π/π, we can use this to calculate the speed of light in different media: ππππππ’π πππππ ππ πππβπ‘ ππ π£πππ’π’π π 3 × 108 = = = πππππ ππ πππβπ‘ ππ π‘βπ πππππ’π π£πππππ’π π£πππππ’π ∴ π£πππππ’π = π ππππππ’π sin ππ sin π1 π£1 π2 ∴ = = = sin ππ sin π2 π£2 π1 eg1. A ray of light travels through a medium at a speed of 1.2 × 108 π/π . a) What is the refractive index of the medium? π =? b) The light enters a different medium with a refractive index of 1.7. What speed is the light travelling at in this medium? π£ = 1.2 × 108 π/π ππππππ’π = ππππππ’π ππππππ’π π π£πππππ’π π£1 π2 = π£2 π1 3 × 108 = π£πππππ’π π2 π£1 = π£2 × π1 8 108 × 3 × 10 = 1.2 × 108 = 2.5 π£1 = 1.2 × 2.5 1.7 π£ ≅ 1.7647 × 108 π/π eg2. A ray of light travels through a medium that has a refractive index of π = 1.96 a) At what speed does the light travel through the medium? π = 1.96 π£ =? π£πππππ’π = π ππππππ’π π£πππππ’π 3 × 108 = ππππππ’π π£πππππ’π 3 × 108 = 1.96 = 1.53 × 108 π/π b) The ray leaves the medium at an angle of incidence of 25°, into air. What angle is the light at once entering air? sin π1 π2 = sin π2 π1 sin ππππ ππππππ’π = sin ππππππ’π ππππ sin ππππ 1.96 = sin 25 1 ππππ = sin−1 (1.96 sin 25) ππππ = 55.9° Angle of Deviation The term angle of deviation refers to the difference between the Angle of incidence and the Angle of refraction eg. If the ππ = 40° and ππ = 25°, find the angle of deviation. Angle of deviation = 40° − 25° = 15° Apparent Depth • Water is more dense than air so it has a larger refractive index than air. • This causes light to travel slower in water than it does in air. • When viewed from above, objects appear to be shallower than they actually are. • Where the object is called ‘Real Depth’ • Where the object appears to be is ‘Apparent Depth’ We can calculate these using: π πππ π·πππ‘β π΄πππππππ‘ π·πππ‘β = π ππππππ‘ππ£π πΌππππ₯ Apparent Depth π πππ π·πππ‘β π΄πππππππ‘ π·πππ‘β = π ππππππ‘ππ£π πΌππππ₯ eg1. Water has a refractive index ≅ 1.33. A fish appears to be 1.2 metres beneath the surface, but how far is it in reality? π πππ π·πππ‘β 1.2 = 1.33 π πππ π·πππ‘β = 1.33 × 1.2 π πππ π·πππ‘β = 1.596 π Apparent Depth π πππ π·πππ‘β π΄πππππππ‘ π·πππ‘β = π ππππππ‘ππ£π πΌππππ₯ eg2. Water has a refractive index ≅ 1.33. If the fish is 4 metres beneath the surface, how deep does it appear to be? 4 π΄πππππππ‘ π·πππ‘β = 1.33 π΄πππππππ‘ π·πππ‘β = 4 1.33 π΄πππππππ‘ π·πππ‘β = 3.01 π Behaviour of Light Rays - Refraction Ex2.2B Q1-10 Internal Reflection Depending on: • The angle of incidence, and • The medium that light is travelling from/to; a single light ray may be: * Refracted only * Both Reflected and refracted * Reflected only Internal Reflection • Internal reflection involves light moving to a less dense medium (ie. Lower refractive index ‘n’ ) • As the angle of incidence of a light ray increases, more of the light ray is reflected and less refracted, as seen below. • As the angle of incidence increases, so does the angle of refraction Internal Reflection – Critical Angle • As we increase the angle of incidence, eventually this results in a reflected angle of almost 90° - the transmitted ray runs along the surface of the water • When this occurs, the angle of incidence is called the critical angle ππ Total Internal Reflection • If the angle of incidence increases beyond the ‘critical angle’, no light is refracted – it is all reflected • This is called Total Internal Reflection Critical Angle The critical angle can be calculated using Snell’s law. π1 π ππ π1 = π2 π ππ π2 For these problems always use angle of refraction (π2 ) = 90° eg. If a light ray travels from water to air, calculate the critical angle. π1 = 1.3 π1 =? 1.3 π ππ π1 = 1 π ππ 90° sin 90° −1 sin ( 1.3 π2 = 1 π1 = π2 = 90° π1 = 50.28° ) Behaviour of Light Rays Ex 2.3B Q1 – Q3 & Q5 – 6 Research Questions Ex 2.3B Q7 – 8 (all answers can be found within the text I sent to you)
