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# 4.4

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### 4.4

1. 1. Topic 4.4 Wave behaviour
2. 2. Wave Behaviour Reflection in one dimension When a wave moves through a medium, the velocity and shape of that wave remains constant. This is so no matter what the medium.
3. 3. This diagram shows a pulse travelling along a string
4. 4. This diagram shows the pulse after it has been reflected
5. 5. Notice The pulse keeps its shape It is inverted It has undergone a 180o phase change Or π change in phase
6. 6. This is because the instant the pulse hits the fixed end, the rope attempts to move the fixed end upwards It exerts an upwards force on the fixed end By Newton’s third law, the wall will exert an equal but opposite force on the rope This means that a disturbance will be created in the rope which, however is downwards and will start moving to the left
7. 7. If the end of the rope is not fixed but free to move the situation is different Most of the pulse would carry on in the same direction, some would be reflected but the reflected pulse is in the same phase as the original pulse There is a change of direction, but no inversion here
8. 8. Similar situations occur in springs and columns of air It is also why metals are shiny. The light incident on the surface is reflected back.
9. 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 slinky Fill in notes
10. 10. Wave Behaviour Reflection in two dimensions
11. 11. 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. Image from aplusphysics
12. 12. Light is shone from above a ripple tank onto a piece of white card beneath The bright areas represents the crests The dark areas represent the troughs
13. 13. These wavefronts can be used to show reflection (and refraction and diffraction and interference) of water waves Image from agilegeoscience
14. 14. Normal Angle of incidence Angle of reflection=
15. 15. Image from agilegeoscience
16. 16. Candle in front of mirror The rays diverge(spread apart) from the tip of the flame and continue to diverge upon reflection. These rays appear to originate from a point located behind the mirror.
17. 17. Candle in front of mirror This is called a virtual image because the light does not actually pass through the image but behaves as though it virtually did. The image appears as far behind the mirror as the object is in front of it and the object and the image is the same. It is different with a curved mirror.
18. 18. Light and sound reflections Image from P Hewitt, conceptual physics
19. 19. 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
20. 20. 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
21. 21. O I
22. 22. 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
23. 23. 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.
24. 24. Wave Behaviour Refraction
25. 25. 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?
26. 26. 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.
27. 27. Refraction for light Partial reflection Incident ray Incident ray Refracted ray Refracted ray Partial reflection
28. 28. Interpreting diagrams involving refraction It may help to imagine the ranks of a marching band. Obviously, the cadence does not change. Thus the period and the frequency do not change. But the speed and the wavelength do change. CONCRETE DEEP MUD
29. 29. 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
30. 30. 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 .
31. 31. 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 = v1 t while, –wavelet from A has moved, –smaller distance s2 = v2 t, –in the denser medium 2, –where the velocity is less.
32. 32. 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 .
33. 33. 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 .
34. 34. Refraction & Huygen’s Principal The new wavefront at the end of this time; envelope of tangents to, the wavelet wavefronts at A1 B2 C2 D1 . The direction of movement of the wavefront has changed; refracted towards the normal at point A.
35. 35. Deriving Snell’s Law Since the wave; travels in a direction, perpendicular to its wavefront, ∠IAD = 90o and ∠AA1 D1 = 90o .
36. 36. 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 ,
37. 37. Deriving Snell’s Law so α = 90o - θ, so α = i. In triangle D1 AA1 , β + γ + 90o = 180o . ⇒β = 90o - γ, so β = R
38. 38. Deriving Snell’s Law In triangle ADD1 , sin i = v1 t/AD1 In triangle AA1 D1 , sin R = v2 t/AD1 Dividing
39. 39. Deriving Snell’s Law • 1n2 = a constantsin sin i R v t AD v t AD v t v t v v n= ÷ = = =1 1 2 1 1 2 1 2 1 2
40. 40. 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.
41. 41. 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 sin i R v v n= =1 2 1 2
42. 42. Snell’s Law
43. 43. 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
44. 44. Combining them! Rearranging n2/n1 = sin θ1 / sin θ2 But n1 = 1 n2 = sin θ1 / sin θ2 Therefore anb = nb / na
45. 45. 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
46. 46. Eg 5 Light strikes a glass block at an angle of 60o to the surface. If air nglass = 1.5, calculate: (a) the angle of refraction (R) and (b) the angle of deviation (D).
47. 47. Part (a) i = 300 air nglass = 1.5 air nglass sin R = sin R = R = 19.5o R = 20o (2 sig digits) Part (b) i = 300 R = 19.5o D = i - R D = 30 – 19.5 D = 10.5o D = 11o (2 sig digits)
48. 48. Eg 6 Light of wavelength 500 nm is incident on a block of glass (air nglass ) at 60o to the glass surface. Calculate (a) the velocity, and (b) the wavelength of the light waves in the glass.
49. 49. Part a vair = 3.0 x 108 m s-1 air nglass = 1.5 vair/ vglass = air nglass vglass = vair/ air nglass vglass = 3.0 x 108 /1.5 vglass = 2.0 x 108 m s-1
50. 50. Part b air nglass = 1.5 i = 30o λair = 500 nm = 5.0 x 10-7 m vair = 3.0 x 108 m s-1 λair/ λglass = air nglass λglass = 3.3 x 10-7 m
51. 51. Critical angle The critical angle is the angle of incidence that results in the refracted ray travelling along the boundary between the two media. nt/ni = sinθc Image from Physics classroom
52. 52. Total internal reflection Total internal reflection occurs when the angle of incidence is greater than the critical angle. Image from Hyper Physics
53. 53. Questions 1.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 1.5 = 1.8/x Therefore, x = 1.2 m 2. Ang = real/apparent = 4/2.72 Therefore, = 1.47 3. Sin i / sin r = sin 60 / sin r = 1.33 sin r = sin 60/ 1.33 = 0.651 Angle is 41 degrees
54. 54. 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
55. 55. If the waves enter the shallow area at an angle then a change in direction occurs. Shallow water
56. 56. 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!
57. 57. Deep water If the waves enter the deep area at an angle then a change in direction occurs
58. 58. 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!
59. 59. 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.
60. 60. Diffraction Diffraction is the spreading out of a wave as it passes by 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.
61. 61. When the wavelength is comparable to the opening then diffraction takes place. There is considerable sideways spreading, i.e. considerable diffraction
62. 62. 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
63. 63. 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.
64. 64. 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.
65. 65. The new wavefront would then be the surface that is tangential 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.
66. 66. 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
67. 67. 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 + =
68. 68. 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. + =
69. 69. Path difference Constructive interference occurs when the path difference between the rays is equal to a whole number of wavelengths. C.I. = nλ Destructive interference occurs when the path difference between the rays is equal to a multiple of half a wavelength. D.I. = (n + ½) λ
70. 70. Double slit interference s D λ d monochromatic light Young’s double-slit experiments = λD / d
71. 71. Coherent light having a wavelength of 675 nm is incident on an opaque card having two vertical slits separated by 1.25 mm. A screen is located 4.50 m away from the card. What is the distance between the central maximum and the first maximum? SOLUTION: Use s = λD / d. λ = 675×10-9 m, D = 4.50 m, and d = 1.25×10-3 m. Thus s = λD / d = 675×10-9 ×4.50 / 1.25×10-3 = 0.00243 m. Example