What is the wavelength of the satellite signal?

In summary: You should find that the reflected ray has to travel a length of x - h where x is the distance from the satellite to the receiver. So, with 3 degrees being n = 1 and 6 degrees being n = 2, the reflected ray has to travel 3 + h or 6 - h to get to the receiver. In summary, the wavelength of the signal from a satellite is 2 meters when it is at 3 degrees above the horizon and 4 meters when it is at 6 degrees above the horizon.
  • #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.

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  • #2
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?
  • #3
I finished with a wavelength of 2m.

Can someone please go through as well and confirm or say otherwise :)
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  • #4
lektor said:
I finished with a wavelength of 2m.
That's not the answer I get. Show your work.
  • #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
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  • #6
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].
  • #7
[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]
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  • #8
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.
  • #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 :)


  • diagram.jpg
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  • #10
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.

1. What is the definition of wavelength in relation to satellite signals?

Wavelength is the distance between two consecutive peaks or troughs of a wave. In satellite communication, it refers to the distance between the crests of electromagnetic waves that are used to transmit signals from the satellite to the receiver on Earth.

2. How is the wavelength of a satellite signal measured?

The wavelength of a satellite signal is typically measured in meters or nanometers. It can be calculated by dividing the speed of light (299,792,458 meters per second) by the frequency of the signal in hertz (cycles per second).

3. Why is the wavelength of a satellite signal important?

The wavelength of a satellite signal is important because it determines the amount of information that can be transmitted at a given time. A shorter wavelength allows for more data to be transmitted, while a longer wavelength may be more prone to interference and signal degradation.

4. What factors can affect the wavelength of a satellite signal?

The wavelength of a satellite signal can be affected by factors such as atmospheric conditions, distance between the satellite and receiver, and the frequency of the signal. Interference from other electronic devices can also impact the wavelength.

5. How does the wavelength of a satellite signal change as it travels through space?

The wavelength of a satellite signal remains constant as it travels through space at the speed of light. However, it may be affected by the medium it is traveling through, such as the Earth's atmosphere, which can cause some distortion and attenuation of the signal.

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