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Twin Source Interference - Radio Towers

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data

    http://www.mediafire.com/view/?09c89b8986u5zls

    The above is a question from my problem sheet. I've got no idea how to attack this since its been very poorly explained in my lecture handouts.

    Can someone please help me ?
     
  2. jcsd
  3. Apr 9, 2013 #2

    Simon Bridge

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    The problem is basically Young's interference - you can look it up online.
     
  4. Apr 9, 2013 #3
    Could you please just give me a quick run down of the theory or how to approach the question ?
     
  5. Apr 9, 2013 #4

    Simon Bridge

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    Two waves will add together
    sometimes the addition will make the wave bigger, sometimes smaller, sometimes it will make the wave disappear completely ... the first is "constructive interference" and the last is "destructive interference".

    in 2D, if the wave sources are in-phase, then the maxima will occur at locations of constructive interference - i.e. when the path-difference from each source is an integer number of whole wavelengths.

    Have a look at:
    http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html

    The approach to the question is by analogy with 2-slit interference.
    The radio towers are the slits.
     
    Last edited: Apr 9, 2013
  6. Apr 9, 2013 #5
    dsinθ=mλ

    2λsinθ=mλ

    sinθ ≈ θ ≈ 0, 0.5, 1, 1.5, 2 etc.

    Theta is in radians correct ?

    For destructive :

    θ≈m/2 where m = 1,3,5,7 etc.

    a)Moving the sources closer together means a wider gap between each fringe ?

    b)Higher λ means wider fringe spacing ?

    c)The central fringe which is normally a maxima will be a minima... What else would happen ?

    Also, how could I calculate the positions of the maxima and minima if the sources have a phase difference of pi ?
     
  7. Apr 9, 2013 #6

    Simon Bridge

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    Yep - unless otherwise specified, theta is [ialways[/i] in radians.
    You should show your reasoning - why those numbers?
    Didn't you include some of those numbers in your previous answer for the maxima angles?

    Again - show your reasoning.

    hat about the angles for the maxima and minima?
    From the physics ... the angle for a maxima is normally where the path difference is an integer number of whole wavelengths - what does the phase difference do to this condition?
     
  8. Apr 9, 2013 #7
    If m is allowed to take those values then we have a non-integer number of wavelengths and hence this meets the primary condition for destructive interference.

    I've applied the formula x=λD/a where D is the distance from the screen and a is separation between the slits i.e. towers

    I'm not really sure :s Please help me !

    I really dont know. I'm a FIrst year chemistry student who's been plunged back into all the interference stuff from school... stuff I never really enjoyed. Could you please break it down for me ?
     
  9. Apr 9, 2013 #8

    Simon Bridge

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    but aren't 1,3,5... etc integers too?

    That works ... you could also have used the angle formula since you don't have a screen in this case.

    You need to go back to the derivation of the formulae and rework them for the phase difference.
    That is the break-down for you: it's geometry.
    The point of the exercise is to get you to do this so you will understand the phenomena: physics is not about memorizing and applying equations.
     
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