Interference In Waves
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Interference In Waves

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Interference In Waves Interference In Waves Presentation Transcript

  • Interference in waves Water waves.
  •  
    • We can use ripple tanks to explore the ways in which waves interact.
    • The water waves produced behave similarly to all other transverse waves, the behaviour of water waves will have parallels in the behaviour of light for example.
    View slide
  • Twin sources
    • If we use twin sources, which are in phase and have the same amplitude some very characteristic patterns appear.
    View slide
  • Draw concentric lines emanating from two point sources. Use alternate solid and dashed lines to represent peaks and troughs
  • Mark points of destructive interference.
  • Connect these marks with straight lines using a ruler
  •  
    • If we use double slits this also produces a similar interference pattern as it also gives 2 coherent sets of waves which will affect each other.
  •  
  • Young’ double slits.
    • Young’s double slits is a classic experiment examining the interference of light.
    • In this experiment a coherent light source is directed at a pair of slits, the light diffracts as it exits the slits and the overlapping light interferes producing a classic interference pattern.
    • In most explanations and diagrams a single slit is placed between the source and the double slit to produce a coherent light source.
  • We can predict the separation of the maxima produced on the screen using the formula:  = ws D Where: w is the width between the centres of the slits. s is the separation of the fringes. D is the distance between the slits and the screen  is the wavelength of the light source.
    • Interference pattern intensity for double slits.
    intensity distance The higher the intensity the brighter the fringes. Centre fringe is usually double the width of the others
    •  =ws/D D  =ws s= D  /w
    • s = (0.2 x 6E-7)/1E-3
    • s = 1.2 E-4
     = 6E-7 D = 20cm w = 1mm Calculate the separation of the fringes (s) screen
    • Questions
    • A light source of 420nm is shining through a set of double slits 2mm apart, calculate the distance of the screen if the separation of the bright fringes is 1.6 cm.
    • Calculate the wavelength of the light source if the minima are 1.08mm apart, the slit separation is 3mm and the screen is 4.5m away.
    • If the separation of the maxima is 3mm the screen is at a distance of 6m and the light source has a wavelength of 600nm, calculate the slit separation.
    • Answers
    • A light source of 420nm is shining through a set of double slits 2mm apart, calculate the distance of the screen if the separation of the bright fringes is 1.6 cm.
    •  = ws/D D = ws/ 
    • D = (2E-3 x 1.6E-2)/420E-9
    • D = 76 m
    • Calculate the wavelength of the light source if the minima are 1.08mm apart, the slit separation is 3mm and the screen is 4.5m away.
    •  = ws/D
    •  = (3E-3 x 1.08 E-3)/4.5
    •  = 7.2E-7 (720nm)
    • If the separation of the maxima is 3mm the screen is at a distance of 6m and the light source has a wavelength of 600nm, calculate the slit separation.
    •  = ws/D s =  D/w
    • s = (600E-9 x 6)/3E-3
    • s = 1.2mm
  • Interference of sound
    • We can apply exactly the same principles used for calculating the position of the maxima in Young’s double slits to sound waves.
    Sig gen D s w speakers detector As we move the detector (microphone and CRO/ear) we can detect the maxima and minima for any constant tone being played.
    • Questions.
    • A constant tone of 450Hz is being played by speakers 2m apart, what is the separation of the maxima at a distance of 2m? What is the separation at 4m?
    • At 180 KHz what would the separation of the speakers have to be to produce maxima at 2cm intervals at a distance of 4m?
    • Questions.
    • A constant tone of 450Hz is being played by speakers 2m apart, what is the separation of the maxima at a distance of 2m? What is the separation at 4m?
    •  = ws/D s=  D /w (  = c/f = 330/450 = 0.73)
    • s = (0.73 x 2)/2 = 0.73m
    • s= (0.73 x 4)/2 = 1.46m
    • At 180 KHz what would the separation of the speakers have to be to produce maxima at 2cm intervals at a distance of 4m?
    •  = ws/D w=  D /s (  = c/f = 330/180E3 = 1.83E-3)
    • w = (1.83E-3 x 4)/0.02
    • w = 0.37m