Radio Receiver problem (Path length difference)

[lemonlimesoap]

Homework Statement

A radio receiver is set up on a mast in the middle of a calm lake to track the radio signal from a satellite orbiting the Earth. As the satellite rises above the horizon, the intensity of the signal varies periodically. The intensity is at a maximum when the satellite is 1 = 3° above the horizon and then again at 2 = 6° above the horizon. What is the wavelength of the satellite signal? The receiver is h = 4.0 m above the lake surface.

Homework Equations

r=dsin(theta)
(theta)m = m*(lambda/d)
y=L*tan(theta)

ym = (m*lambda*L)/d

The Attempt at a Solution

i know that the radio signal acts like light as it is reflected off the lake and the signal path length difference of the signal directly and the signal off the lake will give me the path length difference but the lambda solution i got was off according to my teacher.

 

[Delphi51]

Interesting problem. I had no idea how to approach it until you mentioned the reflected waves constructively interfering with the direct waves. So, no diffraction involved, just a matter of finding the hypotenuse of a triangle. Did you get about 75 m for the wavelength? I don't follow your equations, don't know what n, L, theta, ym and d are.

 

[ehild]

Show your work, please, with a drawing attached.

ehild

 

[ehild]

Did you get about 75 m for the wavelength? I don't follow your equations, don't know what n, L, theta, ym and d are.

There are enough data given. The two angles belong to consecutive maxima. The incoming light beam can be considered parallel as the satellite is very far away. The path difference between the directly incident ray and the reflected ray has to be calculated at the given angles, see picture.

ehild

 

[Delphi51]

Thanks again, ehild. I had the diagram but forgot to subtract AB from AC.
The path difference works out to a wavelength of about 40 cm at 3 degrees and exactly twice as much at 6 degrees.

 

[ehild]

"The intensity is at a maximum when the satellite is 1 = 3° above the horizon and then again at 2 = 6° above the horizon"

The path difference is equal to m*lambda for the first angle and (m+1)*lambda for the next one. :)

ehild