Sun Interference astronomy question

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A radio wave reflecting off the ocean surface undergoes a 180-degree phase shift, leading to destructive interference when combined with a direct wave from the sun. Australian astronomers studied this phenomenon using an antenna positioned 25 meters above sea level and radio waves at a frequency of 60 MHz, resulting in a wavelength of 5 meters. The condition for destructive interference is met when the path difference between the direct and reflected waves equals (n - 1/2)λ. Participants in the discussion emphasized the importance of accurately calculating the total path lengths of both wave paths and accounting for the phase shift. The conversation highlights the complexities of reflection interference in radio astronomy.
lektor
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A radio wave undergoes a phase shift of 180 degrees when it reflects from the calm surface of the ocean. In early days of radioastronomy, Australian Astronomers observed the interference between a radio wave arriving at the their antena on a direct path from the sun and on a path involving one reflecting on the sea surface. If their antenna was located on the coastline 25m above the sea level and the radio waves had a frequency of 60 MHZ, find the least angle above the horizon for the sun that results in destructive interference of the waves at the receivng antenna.

The identity cos 2θ = 1-2sin^2θ


Ok, so that is the problem right now i have a diagram and understand destructive interference will occur when (n - 1/2 )λ

Wavelength in this situation is

V = Fλ
therefore λ = 60x10^6 / c
λ = 5m

the attachment is of my current diagram ( sorry for the crude drawings, only had paint avalible).

Cheers.
 

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  • diagram.JPG
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lektor said:
A radio wave undergoes a phase shift of 180 degrees when it reflects from the calm surface of the ocean. In early days of radioastronomy, Australian Astronomers observed the interference between a radio wave arriving at the their antena on a direct path from the sun and on a path involving one reflecting on the sea surface. If their antenna was located on the coastline 25m above the sea level and the radio waves had a frequency of 60 MHZ, find the least angle above the horizon for the sun that results in destructive interference of the waves at the receivng antenna.
The identity cos 2θ = 1-2sin^2θ
Ok, so that is the problem right now i have a diagram and understand destructive interference will occur when (n - 1/2 )λ
Wavelength in this situation is
V = Fλ
therefore λ = 60x10^6 / c
λ = 5m
the attachment is of my current diagram ( sorry for the crude drawings, only had paint avalible).
Cheers.
Some hints: your drawing is misleading: you should consider rays coming from the sun *parallel*, and you should calculate, from a common wavefront onwards (perpendicular to the incoming rays from the sun), what is the total path a) directly, and b) through reflection, and translate that in a number of wavelengths. You should then add half a wavelength to the reflected path (the 180 degree phase flip) and compare.
 
Hi vanesch,

I revised my diagram as to how you asked and attempted to redo the question with unfortunately no avail.

These questions to do with reflection interference have been plagueing me for most of this year and if it is not too much would you possibly be able to do the question and show each step of working.. possibly with notes on the working.

Cheers if this is possible :)
 
Hi,
I attached a diagram that should be more suggestive. Sorry, I have drawn it quite quickly...
The idea is to calculate the path of beam 1 and of beam 2 up to R, the position of the receiver, starting from a (perpendicular) wavefront: points x and y.
In fact, you can even move the wavefront (points x and y) to the point of reflexion on the sea surface...
 

Attachments

  • pbreflexion.GIF
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