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IBD Physics Topic 4 Section 5

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- 1. Waves Topic 4.5 Wave Properties
- 2. Wave Behaviour Reflection in one dimension
- 3. This diagram shows a pulse travelling along a string
- 4. This diagram shows the pulse after it has been reflected
- 5. Notice The pulse keeps its shape It is inverted It has undergone a 180o phase change Or π change in phase
- 6. This is because the instant the pulse hits thefixed end, the rope attempts to move thefixed end upwardsIt exerts an upwards force on the fixed endBy Newton’s third law, the wall will exert anequal but opposite force on the ropeThis means that a disturbance will becreated in the rope which, however isdownwards and will start moving to the left
- 7. If the end of the rope is not fixed but freeto move the situation is differentMost of the pulse would carry on in thesame direction, some would be reflectedbut the reflected pulse is in the samephase as the original pulseThere is a change of direction, but noinversion here
- 8. Similar situations occur insprings and columns of airIt is also why metals areshiny. The light incident onthe surface is reflected back.
- 9. Slinky Investigation• Use a slinky or combination to investigate:• 1. Reflection of wave at fixed boundary (ie hold the slinky firm at one end)• 2. Reflection of wave at free boundary (do not hold other end of slinky)• 3. Transfer of energy between a heavy and a light slinkyFill in notes
- 10. Huygen’s Principle• Huygen’s principle states that every point on a wavefront may be regarded as a point source of secondary circular wavelets. The new wavefront is formed along the common tangent to these secondary wavelets.
- 11. Wave Behaviour Reflection in two dimensions
- 12. The direction of incident and reflected waves is best described by straight lines called rays. The incident ray and the reflected ray make equal angles with the normal. The angle between the incident ray and the normal is called the angle of incidence and the angle between the reflected ray and the normal is called the angle of reflection.
- 13. Light is shone from above a rippletank onto a piece of white cardbeneathThe bright areas represents thecrestsThe dark areas represent thetroughs
- 14. These wavefronts can be used toshow reflection (and refraction anddiffraction and interference) of waterwaves
- 15. NormalAngle of Angle ofincidence = reflection
- 16. The Law for Reflection• The angle of incidence is equal to the angle of reflection• Also - The incident ray, the reflected ray and the normal lie on the same plane• Use this rule for any ray or wave diagram involving reflection from any surface
- 17. • For circular waves hitting a flat reflector, the reflected waves appear to come from a source, which is the same distance behind the reflector as the real source is in front of it• Also a line joining these 2 sources is perpendicular to the reflecting surface
- 18. O I
- 19. • If a plane wave is incident on a circular reflector then the waves are reflected so that they – Converge on a focus if the surface is concave – Appear to come from a focus if the surface is convex
- 20. Echos• In the case of sound, a source of sound can be directed at a plane, solid surface and the reflected sound can be picked up by a microphone connected to an oscilloscope.• The microphone is moved until a position of maximum reading on the oscilloscope is achieved.• When the position is recorded it is found that again the angle of incidence equals the angle of reflection.
- 21. Wave Behaviour Refraction
- 22. Examples• Place a pencil in a beaker of water – what happens?• Place a coin in a mug, move your head so that you cant see the coin, pour in water – what happens?• Why is it hard to catch fish with your hands?
- 23. • The speed of a wave depends only on the nature and properties of the medium through which it travels.• This gives rise to the phenomenon of refraction• Refraction is the change of direction of travel of a wave resulting from a change in speed of the wave when it enters the other medium at an angle other than a right angle.
- 24. Refraction for light Refracted ray Partial reflection Partial reflection Incident ray Refracted ray Incident ray
- 25. • This law enable us to define a property of a given optical medium by measuring θ1 and θ2 when medium 1 is a vacuum• The constant is then the property of medium 2 alone and it is called the refractive index (n).• We usually write • n = (Sin i) / (Sin r)• n is also a ratio of the speeds in the 2 mediums i.e. n = cvacuum / vmedium
- 26. Combining them!• Rearranging n2/n1 = sin θ1 / sin θ2• But n1 = 1• n2 = sin θ1 / sin θ2• Therefore anb = nb / na
- 27. Using Refractive Index• Refractive index is written for materials in the form of light entering from a vacuum or air into the material.• The refractive index of a vacuum or air is 1• It can also be shown that, for two mediums (1 and 2)• n1 sin θ1 = n2 sin θ2• Care needs to be taken when dealing with light leaving a material
- 28. Eg 5• Light strikes a glass block at an angle of 60 o to the surface. If airnglass = 1.5, calculate:• (a) the angle of refraction (R) and• (b) the angle of deviation (D).
- 29. Part (b)• Part (a) i = 300• i = 300 R = 19.5o• airnglass = 1.5 D=i-R• n air glass D = 30 – 19.5• sin R = D = 10.5o• sin R = D = 11o (2 sig digits)• R = 19.5o• R = 20o (2 sig digits)
- 30. Eg 6• Light of wavelength 500 nm is incident on a block of glass (airnglass) at 60o to the glass surface. Calculate (a) the velocity, and (b) the wavelength of the light waves in the glass.
- 31. Part a• vair = 3.0 x 108 m s-1• airnglass = 1.5• vair/ vglass = airnglass• vglass = vair/ airnglass• vglass = 3.0 x 108 /1.5• vglass = 2.0 x 108 m s-1
- 32. Part b• airnglass = 1.5• i = 30o∀ λair = 500 nm = 5.0 x 10-7 m• vair = 3.0 x 108 m s-1∀ λair/ λglass = airnglass∀ λglass = 3.3 x 10-7 m
- 33. Refraction of Waves in 1 & 2 Dimensions• Light bends towards the normal when; – it enters a more optically dense medium.• Light bends away from the normal; – when it enters a less optically dense medium.• The amount the incident ray is deviated; – depends on the nature of the transparent material
- 34. Refraction & Huygen’s Principal• Consider a wavefront advancing through medium 1; – travelling at velocity v1 – falling at an angle on to a medium 2, – travelling at velocity is v2.
- 35. Refraction & Huygen’s Principal• The wavefront with the points; – A, B, C, D are a source of secondary wavelets.• After a time t, the secondary source wavelet from D; – has moved a distance s1 = v1t while, – wavelet from A has moved, – smaller distance s2 = v2t, – in the denser medium 2, – where the velocity is less.
- 36. Refraction & Huygen’s Principal• The time for the wavelet to travel; – from B to B2 is identical, – for the wavelet to travel from C to C2.• The wavelet from C; – spends a longer time in, – less dense medium 1 in travelling, – from C to C1 than for, – B to travel from B to B1.
- 37. Refraction & Huygen’s Principal• Thus there is less time for the wave; – to travel C1 to C2 in the denser medium 2, – than for the wavelet to travel from B1 to B2.• The distance CC1 is thus less than BB1.
- 38. Refraction & Huygen’s Principal• The new wavefront at the end of this time; – envelope of tangents to, – the wavelet wavefronts at A1B2C2D1.• The direction of movement of the wavefront has changed; – refracted towards the normal at point A.
- 39. Deriving Snell’s Law • Since the wave; – travels in a direction, – perpendicular to its wavefront, ∀ ∠IAD = 90o and ∀ ∠AA1D1 = 90o.
- 40. Deriving Snell’s Law • From ∠IAD, – i + θ = 90o, – so i = 90o - θ. • From ∠N’AD1, – R + γ = 90o, – so R = 90o - γ. • But from ∠D1AN, α + θ = 90o,
- 41. Deriving Snell’s Law • so α = 90o - θ, – so α = i. • In triangle D1AA1, β + γ + 90o = 180o. ∀ ⇒β = 90o - γ, – so β = R
- 42. Deriving Snell’s Law • In triangle ADD1, – sin i = v1t/AD1 • In triangle AA1D1, – sin R = v2t/AD1 • Dividing
- 43. Deriving Snell’s Law sin i = v1t ÷ v2 t = v1t v1 = =1 n2 • 1n2 = a constant sin R AD1 AD1 v2 t v2
- 44. Snell’s Law• Snell’s law states that the ratio of the sine of the angle of incidence to the sine of the refraction is constant and equals the ratio of the velocity of the wave in the incident medium to the velocity of the refracting medium. sin i v1 = =1n2 sin R v2
- 45. Snell’s Law
- 46. Snell’s Law• When a wave is incident from a vacuum; – on to a medium M, – the refractive index is written nM – is called the absolute refractive index – of medium M.
- 47. Questions1.The refractive index of water is 1.5. A swimming pool is filled to a depth of 1.8m with water. How deep would it appear to someone standing on the side of the pool?2. A microscope is focussed on a scratch at eh bottom of a beaker. Turpentine is poured into the beaker to a depth of 4cm, it is found that the lens must be raised by 12.8mm in order to bring back into focus. What is the refractive index of turpentine?3. A ray of light enters a pond at an angle of 30 deg to the horizontal. What is its direction as it travels through the water? The refractive index of water is 1.33 1. Ang = real/apparent 3. Sin i / sin r = sin 60 / sin r = 1.33 1.5 = 1.8/x sin r = sin 60/ 1.33 = 0.651 Therefore, Angle is 41 degrees x = 1.2 m 2. Ang = real/apparent = 4/2.72 Therefore, = 1.47
- 48. • In a ripple tank this is achieved by using a flat piece of plastic, giving two regions of different depth• As the wave passes over the plastic it enters shallow water and slows down.• As v = f λ,• if v decreases• And f is constant (the source hasn’t changed)∀ λ must also decrease• So the waves get closer together
- 49. • If the waves enter the shallow area at an angle then a change in direction occurs. Shallow water
- 50. • This is because the bottom of the wavefront as drawn, hits the shallow water first so it slows, and hence travels less distance in the same time as the rest of the wavefront at the faster speed travel a larger distance!
- 51. • If the waves enter the deep area at an angle then a change in direction occurs Deep water
- 52. • This is because the top of the wavefront hits the deep water first so it speeds up, and hence travels more distance in the same time as the rest of the wavefront at the slower speed travel a smaller distance!
- 53. Refraction of Sound• A sound wave is also able to be refracted.• This is due to the fact that the speed of sound is affected by temperature and the medium through which it travels.
- 54. Diffraction• Diffraction is the spreading out of a wave as it goes passed an obstacle or through an aperture• When the wavelength is small compared to the aperture the amount of diffraction is minimal• Most of the energy associated with the waves is propagated in the same direction as the incident waves.
- 55. • When the wavelength is comparable to the opening then diffraction takes place.• There is considerable sideways spreading, i.e. considerable diffraction
- 56. examples• Light through net curtains• Fog Horns – two frequencies to ‘fill in’ gaps• AM radio – can bend round small buildings• Dolphins use two different frequency waves to ‘see’ and ‘fine tune’ their surroundings
- 57. • Diffraction also takes place when a wave moves passed an obstacle• If the wavelength is much smaller than the obstacle, little diffraction takes place• If the wavelength is comparable to the obstacle size, then diffraction takes place
- 58. Using Huygens’ Principle• Remember that Huygens idea was to consider every single point on the wavefront of the wave as itself a source of waves.• In other words a point on the wavefront would emit a spherical wavelet or secondary wave,of same velocity and wavelength as the original wave.
- 59. • Therefore as a wave goes through a gap or passed an obstacle the wavelets at the edges spread out.• Huygens’ construction can be used to predict the shapes of the wave fronts.
- 60. • The new wavefront would then be the surface that is tangent to all the forward wavelets from each point on the old wavefront.• We can easily see that a plane wavefront moving undisturbed forward easily obeys this construction.
- 61. The Principle of Linear Superposition• Pulses and waves (unlike particles) pass through each other unaffected and when they cross the total displacement is the vector sum of the individual displacements due to each pulse at that point.• Try this graphically with two different waves
- 62. Interference• Most of the time in Physics we are dealing with pulses or waves with the same amplitude.• If these cross in a certain way we will get full constructive interference, here the resultant wave is twice the amplitude of each of the other 2 + =
- 63. • If the pulses are 180o (π) out of phase then the net resultant of the string will be zero. This is called complete destructive interference. + =

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