Chapters 11, 12, 24. Refraction and Interference of Waves

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1 Chapters 11, 12, 24 Refraction and Interference of Waves

2 Beats Two overlapping waves with slightly different frequencies gives rise to the phenomena of beats.

3 Beats The beat frequency is the difference between the two sound frequencies. f beat = f 1 f 2 In this case, f beat = = 2 Hz

4 Refraction Wave transmitted from one medium to another à Refraction In this figure, v 1 > v 2 f is same in each medium but v and λ are different λ = v f sinθ 1 = λ 1 h = v 1 fh λ 1 = v 1 f and λ 2 = v 2 f sinθ 2 = λ 2 h = v 2 fh sinθ 1 v 1 = 1 fh = sinθ 2 v 2 sinθ 1 v 1 = sinθ 2 v 2 Snell s Law

5 Example: A sound wave in air is incident on a pool of water at an angle of 10 o with respect to the normal to the surface. Find the angle of the wave with respect to the normal to the surface after it is transmitted into the water. sinθ 1 v 1 = sinθ 2 v 2 sinθ water v water = sinθ air v air sinθ water = v water v air sinθ air = sin10o = θ water = sin = 49 o

6 Chapters 11 and 24 Diffraction of Waves

7 17.3 Diffraction The bending of a wave around an obstacle or the edges of an opening is called diffraction.

8 Diffraction Diffraction is the bending of waves around obstacles or the edges of an opening. Huygens principle Every point on a wave front acts as a source of tiny wavelets that move forward with the same speed as the wave; the wave front at a latter instant is the surface that is tangent to the wavelets.

9 Diffraction Approximate expression for angular bending of waves of wavelength λ around an object or opening of size D θ (radians) λ D

10 Diffraction Example: Assume the doorway is D = 1.0 m wide and the sound in the room has a frequency of 500 Hz. Find the diffraction angle of the sound through the doorway. D = 1.0 m f = 500 Hz θ? λ = v f = 343 m/s 500 Hz = m θ (radians) λ D = m 1.0 m = rad 39o

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12 Diffraction around barriers Shadow region infringed upon by diffracting waves Diffraction around a hill allows radio reception in the valley

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14 Example of diffraction of electromagnetic waves around a hill. An AM radio station broadcasts at a frequency is 1230x10 3 Hz, and a microwave transmitter broadcasts waves at 2.50 GHz. Approximately how much will each of the waves be bent when they encounter a hill that is 1000 ft high? How far behind the hill must a receiver antenna be located to pick up the AM radio broadcast? Will it pick up the microwave signal?

15 ! }! h λ AM = λ µw = v = m s f AM Hz = 244 m v f µw = m s Hz = 0.12 m }! d θ AM λ AM h = 244 m 1000 ft m ft = 0.80 rad 46o A lot of diffraction! θ µw λ µw h = 0.12 m 1000 ft m ft = rad o hardly any diffraction! tanθ AM = h d d = For the microwave: d = h tanθ AM = h tanθ µw 1000 ft = 970 ft o tan 46 = 1000 ft = 470 miles Won t pick up o tan0.023 the µw signal!

16 Diffraction The extent of the diffraction increases as the ratio of the wavelength to the width of the opening increases. θ (radians) λ W

17 Diffraction Light waves also exhibit diffraction effects.

18 Diffraction This top view shows five sources of Huygens wavelets.

19 Diffraction These drawings show how destructive interference leads to the first dark fringe on either side of the central bright fringe. Dark fringes for single-slit diffraction sinθ = m λ W m =1, 2,3...

20 Example for single-slit diffraction. Light passes through a slit and shines on a flat screen that is located L = 0.40 m away. The wavelength of the light in a vacuum is λ = 410 nm. The distance between the midpoint of the central bright fringe and the first dark fringe is y. Determine the width 2y of the central bright fringe when the width of the slit is (a) W = 5.0 x 10-6 m and (b) W = 2.5 x 10-6 m. first dark fringe W { θ θ y y midpoint of central bright fringe L first dark fringe " ( a) θ = sin 1 λ % " $ ' = sin % $ ' = 4.7 o # W & # & 2y = 2L tanθ = 2( 0.40)tan 4.7 o = m ( b) 2y = 0.13 m à The width of the central bright fringe is greater for smaller W

21 Diffraction through a circular aperture of diameter D by a wave of wavelength λ Diffraction pattern of light incident on a circular aperture sin θ =1. 22 λ D where θ is the angle from the center of the central maximum to the first minimum

22 Diffraction Example. A 1500 Hz sound and a 8500 Hz sound each emerges from a loudspeaker through a circular opening that has a diameter of 0.30 m. Find the diffraction angle θ for each sound (assume v sound = 343 m/s). λ = v f sinθ =1.22 λ D =1.22 v fd " 1500 Hz: θ = sin v % ( $ ' = sin 1 * 1.22 # fd & )* " 8500 Hz: θ = sin v % ( $ ' = sin 1 * 1.22 # fd & )* ( )( 0.30) ( )( 0.30) + -,- = 68o + -,- = 9.4o

23 Diffraction Poisson spot Diffraction pattern formed by an opaque disk or sphere.

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