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Waves question

  1. Apr 15, 2005 #1
    Recently found this question, was hoping someone could help me out or get me started on it >_<

    A radio reciever is set up on a mast in the middle of a calm lake to track the radio signal from a satellite orbiting Earth. As the satelllite rises above the horizon, the intensity of the signal varies periodically. The intensity is at a maximum when the satellite is [tex]\theta = 3 degrees[/tex] above the horizon and then again at [tex]\theta = 6 degrees[/tex] above the horizon. What is the wavelength of the satelittle signal? The receiver is h = 4.0 m above the lake surface.

  2. jcsd
  3. Apr 15, 2005 #2

    Doc Al

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    Consider the interference between the direct beam and the beam that reflects off the water. Find the phase difference between the two beams (due to path length difference). How must that phase difference change when the angle of the satellite changes from 3 to 6 degrees?
  4. Apr 15, 2005 #3
    I finished with a wavelength of 2m.

    Can someone please go through aswell and confirm or say otherwise :)
    Last edited: Apr 16, 2005
  5. Apr 16, 2005 #4

    Doc Al

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    That's not the answer I get. Show your work.
  6. Apr 16, 2005 #5
    i started with the concept that.

    [tex]sin\theta = tan\theta[/tex] for small angles.

    first real attempt at using it as we were told it may be usefull in the scholarship questions we were assigned.

    [tex]dsin\theta = \frac {dx}{L}[/tex]
    [tex]sin\theta = \frac {x}{L}[/tex]
    [tex]sin\theta = tan\theta[/tex] small angles ( 3 and 6 in this case )
    [tex]tan\theta = sin\theta[/tex]

    with H being L

    [tex]x = 4 tan 3[/tex]
    [tex]x = 4 tan 6[/tex]

    [tex] \delta x = x1 - x2[/tex]

    [tex] x = 0.21037 [/tex]

    In trig with x being the adjacent.

    This was a point where i was a bit lost.. but i think it was on track..
    [tex]adjacent = dsin\theta [/tex]

    [tex] 4 = dsin 3 [/tex]
    [tex] 4 = dsin 6 [/tex]

    [tex] \delta d = d1 - d2 [/tex]

    [tex] d = 38.26708893 [/tex]

    then into the approximation formula

    [tex] n\lambda = \frac {dx}{L} [/tex]

    [tex] \lambda = \frac {0.21037 X 38.26708893}{4}[/tex]

    = 2m 2sf
    Last edited: Apr 16, 2005
  7. Apr 16, 2005 #6

    Doc Al

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    It's perfectly OK to use a small angle approximation for these angles. (You can even go further and use [itex]\sin\theta = \theta[/itex] if you measure the angle in radians.) But I can't follow what you are doing since you don't define the triangles you are working with. A picture would really help!

    Go back to my first post and reread my hints. Start by drawing a picture of the reflected beam (use the law of reflection) and the direct beam. Then figure out the phase difference between those two beams in terms of h and [itex]\theta[/itex].
  8. Apr 16, 2005 #7
    [edit] yup got the diagram sorted but i haven't be taugh how to calculate phase difference and im not sure how it will help here :\[/edit]
    Last edited: Apr 16, 2005
  9. Apr 17, 2005 #8

    Doc Al

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    The reason for the variation in intensity as the source changes angle is due to the interference between the direct and reflected beams. As the angle changes, the beams go in and out of phase, because they travel different distances. That phase difference is key to this problem.

    If you want to continue working this problem, post your diagram. To find the phase difference between the two beams, find the difference in path lengths. Hint: the reflected beam travels a greater distance.
  10. Apr 18, 2005 #9
    Here is my diagram. I've tried to make it as clear as possible, the n ='s refer to my interpretation of the reflected rays forming an interference pattern with 3 degrees been n = 1 and 6 being n = 2.

    Hopefully this can help :)

    Attached Files:

  11. Apr 18, 2005 #10

    Doc Al

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    Try this: Draw two parallel rays coming from the satellite (at some angle [itex]\theta[/itex]). Have one go directly to the receiver; have the other reflect off the water and then go to the receiver. What you want to find is how much farther does the reflected ray have to travel to get to the target.
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