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Radio waves received by radar

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=58645&stc=1&d=1368288914.jpg


    2. Relevant equations



    3. The attempt at a solution
    The path difference between the two rays being received by antenna is ##h/cos(\alpha)## but how to relate the path difference with the intensity?
     

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  3. May 11, 2013 #2

    rude man

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    What is the phase difference requirement between two signals for totally constructive interference?
     
  4. May 11, 2013 #3
    Phase difference is equal to ##2n\pi## or path difference is ##n\lambda##.
     
  5. May 11, 2013 #4

    rude man

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    Correct. So you need n*lambda path difference. Solve for h?
     
  6. May 11, 2013 #5
    What is the value of n? :confused:

    I end up with ##h=n\lambda \cos \alpha##.
     
  7. May 11, 2013 #6

    TSny

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    n can be any positive integer. Are you sure that you got the correct expression for the path difference? Did you take into account the law of reflection? I wonder if you are meant to include a phase shift due to reflection?
     
  8. May 11, 2013 #7

    rude man

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    Do you want to make the height higher than it has to be? You can make it ##h=\lambda \cos \alpha## or ##h=2\lambda \cos \alpha## or ....

    I would make it as low as possible while satisfying your equation! :smile:
     
  9. May 11, 2013 #8
    I don't know about considering the phase shift. There is nothing mentioned in the question about it. I think I got the right expression for path difference considering no phase shift. The given answer is ##\lambda/(4\cos \alpha)##.
     
  10. May 11, 2013 #9

    rude man

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    Good point. I assumed the OP gave the correct path difference and I didn't consider any phase change due to reflection.

    Judging from the given answer the path length was not what the OP came up with.
     
  11. May 11, 2013 #10
    The light reflected from the surface of water which reaches the antenna has to travel an extra distance of ##h/\cos \alpha## (which can be easily calculated from simple trigonometry). This is the path difference. What have I done wrong? :confused:
     
  12. May 11, 2013 #11

    TSny

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    The two waves travel the same distance to the red dots shown in the figure.
     

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  13. May 13, 2013 #12
    Sorry for the late reply.

    The length AC is ##h/\cos \alpha##.

    ##\angle ACB=2\alpha-\pi/2##
    ##\sin \angle ACB=AB/AC=AB\cos \alpha/h \Rightarrow -\cos 2\alpha=AB\cos \alpha/h##
    ##\Rightarrow AB=-h\cos 2\alpha/\cos \alpha##

    Am I doing this right?
     

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  14. May 13, 2013 #13

    TSny

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    Yes, that looks ok.
     
  15. May 13, 2013 #14
    Path difference is ##AC-AB=\frac{h}{\cos \alpha}+\frac{h\cos 2\alpha}{\cos \alpha}##
    [tex]= \frac{h(1+\cos 2\alpha)}{\cos \alpha}=2h\cos \alpha[/tex]

    The path difference should be equal to ##n\lambda## and for min height, ##n=1##
    [tex]\Rightarrow h=\frac{\lambda}{2\cos \alpha}[/tex]

    But this is wrong. :confused:
     
  16. May 13, 2013 #15

    TSny

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    That all looks good. You might be expected to include a phase shift of the wave that reflects off the water.
     
  17. May 13, 2013 #16
    If I include the phase shift, I do get the right answer. Thank you TSny! :smile:
     
  18. May 13, 2013 #17

    TSny

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    Good!

    There's a trick that can make finding the path difference easier. In the figure, the tower is extended a distance h below the ground. You can use the law of reflection to show that the triangles abd and cbd are congruent. Then ba = bc and ec is the path difference.
     

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  19. May 13, 2013 #18
    That makes it a lot easier, thanks! :cool:
     
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