What is the wavelength of the satellite signal?

AI Thread Summary
The discussion revolves around calculating the wavelength of a satellite signal based on the interference of direct and reflected beams. A radio receiver, positioned 4 meters above a lake, detects variations in signal intensity as the satellite rises above the horizon at angles of 3 and 6 degrees. The key to solving the problem lies in determining the phase difference caused by the differing path lengths of the two beams. Participants emphasize the importance of drawing a diagram to visualize the geometry of the situation and suggest using small angle approximations for calculations. The wavelength is initially calculated as 2 meters, but further clarification and corrections are needed to confirm this result.
lektor
Messages
56
Reaction score
0
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 \theta = 3 degrees above the horizon and then again at \theta = 6 degrees above the horizon. What is the wavelength of the satelittle signal? The receiver is h = 4.0 m above the lake surface.

GL
 
Physics news on Phys.org
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?
 
I finished with a wavelength of 2m.

Can someone please go through as well and confirm or say otherwise :)
 
Last edited:
lektor said:
I finished with a wavelength of 2m.
That's not the answer I get. Show your work.
 
i started with the concept that.

sin\theta = tan\theta for small angles.

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

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

with H being L

x = 4 tan 3
x = 4 tan 6

\delta x = x1 - x2

x = 0.21037

In trig with x being the adjacent.

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

4 = dsin 3
4 = dsin 6

\delta d = d1 - d2

d = 38.26708893

then into the approximation formula

n\lambda = \frac {dx}{L}

\lambda = \frac {0.21037 X 38.26708893}{4}

= 2m 2sf
 
Last edited:
It's perfectly OK to use a small angle approximation for these angles. (You can even go further and use \sin\theta = \theta 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 \theta.
 
[edit] yup got the diagram sorted but i haven't be taugh how to calculate phase difference and I am not sure how it will help here :\[/edit]
 
Last edited:
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.
 
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 :)
 

Attachments

  • diagram.jpg
    diagram.jpg
    5.9 KB · Views: 568
  • #10
Try this: Draw two parallel rays coming from the satellite (at some angle \theta). 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.
 
Back
Top