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Interference of radia waves

  1. Nov 29, 2009 #1

    a.a

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    1. The problem statement, all variables and given/known data

    Radio waves of wavelength 250 meters from a star reach a radio telescope by two seperate paths. One is a direct path to the receiver which is situated on the edge of a cliff by the ocean. The second is by reflection off the water. The first minimum of destructive interference occurs when the star is 25 degrees above the horizon. Find the height of the cliff.

    We know the path difference is 125 meters since first minimum occurs at 25 degrees, so the difference is half a wavelength.

    I just found phase difference (=half wavelength) and found path difference (from: phase difference = 2pi*phase difference/wavelength)
    Then I used height = phase difference*sin theta
    but that didnt work
    any advice
     
  2. jcsd
  3. Nov 29, 2009 #2

    Doc Al

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    Staff: Mentor

    Re: Inteferance

    Draw yourself a diagram showing the two rays (one direct; one reflected) and use it to find the difference in path length between them.
     
  4. Nov 29, 2009 #3

    a.a

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    Re: Inteferance

    I did.. I got 5504.29 as my path diff,
    PD= (lambda)^2/4pi

    thats from phase difference = half lambda = 2pi PD/lambda
    Where am i going wrong?
    After I found PD.. from my diagram i found that H= PD sin 22/sin46
     

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  5. Nov 29, 2009 #4

    Doc Al

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    Re: Inteferance

    I don't quite understand what you're doing here. As you said in your first post, the path difference must be λ/2.
    Some problems with your diagram:
    (1) The light rays from the star are parallel.
    (2) What you've labeled as PD Δx is not the path difference.
    (3) Your diagram shows the angle as 22 degrees, not 25.
     
  6. Nov 29, 2009 #5

    a.a

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    Re: Inteferance

    Sorry, my numbers are different, the angle from horizontal is 22 and my wavelength is 263 m.. so the question would read:
    Radio waves of wavelength 263 meters from a star reach a radio telescope by two seperate paths. One is a direct path to the receiver which is situated on the edge of a cliff by the ocean. The second is by reflection off the water. The first minimum of destructive interference occurs when the star is 23 degrees above the horizon. Find the height of the cliff.

    Okay.. does this diagram look better?
    Phase difference is half lambda.. path difference is something else, isn't it?
     

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  7. Nov 29, 2009 #6

    a.a

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    Re: Inteferance

    phase difference=(2pi* path difference)/labmda
     
  8. Nov 29, 2009 #7

    Doc Al

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    Re: Inteferance

    OK.

    Not really. I want to see two parallel rays coming from that star. (The star is very far away.) Those rays make an angle of 22 degrees with the horizontal.
    Imagine that the two rays start out with the same phase as they leave the star. The difference in their path lengths must equal λ/2.
     
  9. Nov 29, 2009 #8

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    Re: Inteferance

    is this better?
     

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  10. Nov 29, 2009 #9

    a.a

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    Re: Inteferance

    where would i go from here?
     
  11. Nov 29, 2009 #10

    Doc Al

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    Re: Inteferance

    Much better! But what you've labeled as λ/2 is not right. You need to compare that distance with the distance that the other ray travels (the hypotenuse of the triangle). The difference will be the path difference.
     
  12. Nov 29, 2009 #11

    a.a

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    Re: Inteferance

    So, in this diagram half lamda = H-X

    so how do we go about finding H or X?
    nothing is given...:S
    Probably a lot of trigs, do the angles in my second diagram make sence?
    If so then I get:
    sin 22= hyp (h)/(half lambda +X)
    and I dont really know where to go from there..
     
  13. Nov 29, 2009 #12

    Doc Al

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    Re: Inteferance

    OK.
    H and X are part of a right triangle, so they are related. (Find the angle of that triangle.)
     
  14. Nov 29, 2009 #13

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    Re: Inteferance

    related by pythagorean? How does that help us?
     

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  15. Nov 29, 2009 #14

    Doc Al

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    Re: Inteferance

    Even easier is to use some trig to relate H and X. Add to that the equation H - X = λ/2. Then you can solve for H. Once you have that, use some more trig to find the height of the cliff.
     
  16. Nov 29, 2009 #15

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    Re: Inteferance

    sorry about the late reply,
    I only got as far as :
    sin 46 = X/H = (H-half lambda)/H
    Solved for H and got 468.54
    Then: sin 22 = h/H
    and got h= 175.5
    Is that right?
     
  17. Nov 30, 2009 #16

    a.a

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    Re: Inteferance

    nope...I got it wrong...?
     
  18. Nov 30, 2009 #17

    Doc Al

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    Re: Inteferance

    Looks right to me.
     
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