# Wave Transmittion ProblemWatch

This discussion is closed.
Thread starter 8 years ago
#1
I'm trying to solve the following question: A ship's transmitter (operating at wavelengths greater than 3m) loses contact with a land based receiving station, of height 150m above sea level at a distance of 2km due to destructive interference between the directed and reflected waves. If the transmitter on the ships mast is 25m above the sea, what wavelength is the signal?

I'm struggling to imagine how destructive interference results from reflection off the water.
0
8 years ago
#2
(Original post by Russels_Teapot)
I'm trying to solve the following question: A ship's transmitter (operating at wavelengths greater than 3m) loses contact with a land based receiving station, of height 150m above sea level at a distance of 2km due to destructive interference between the directed and reflected waves. If the transmitter on the ships mast is 25m above the sea, what wavelength is the signal?

I'm struggling to imagine how destructive interference results from reflection off the water.
The direct wave travels a distance TR and the reflected wave travels a distance TP + PR. For destructive interference, TP + PR = TR + half a wavelength.
In other words, the reflected wave is half a wavelength behind the direct wave when it reaches R. Or to put it another way, the reflected wave has a path difference of half a wavelength compared to the direct wave.
It's then just a matter of geometry in the diagram. There are plenty of right angle triangles, and the angles theta are the same because it's reflection.

Thread starter 8 years ago
#3
(Original post by Stonebridge)
The direct wave travels a distance TR and the reflected wave travels a distance TP + PR. For destructive interference, TP + PR = TR + half a wavelength.
In other words, the reflected wave is half a wavelength behind the direct wave when it reaches R. Or to put it another way, the reflected wave has a path difference of half a wavelength compared to the direct wave.
It's then just a matter of geometry in the diagram. There are plenty of right angle triangles, and the angles theta are the same because it's reflection.

The problem with this is that I can't figure out how to calculate the lengths RP and TP.
0
8 years ago
#4
(Original post by Russels_Teapot)
The problem with this is that I can't figure out how to calculate the lengths RP and TP.
Here's a clue:

The triagles RBP and TAP are similar triangles, which means that the ratio BP/PA = 25/150

You also know that BP + PA = 2000m (given in question)
so

PA = 2000 - BP

For the ratio

BP/(2000 - BP) = 25/150

Find BP (and thus PA) from this. Then use these values in the triangles to find the 3rd sides, the ones you want.
5 years ago
#5
Actually, there is a phase change of pi at the air-sea boundary. This means that for destructive interference to occur, the difference between (TP + PR) and TR must be an integer number of wavelengths. Use this fact, and that the wavelength is greater than 3 to deduce the actual wavelength.
0
5 years ago
#6
This thread is 3 years old and is now closed.
5 years ago
#7
(Original post by sAuhsoj)
Actually, there is a phase change of pi at the air-sea boundary. This means that for destructive interference to occur, the difference between (TP + PR) and TR must be an integer number of wavelengths. Use this fact, and that the wavelength is greater than 3 to deduce the actual wavelength.
This thread is 3 years old and is now closed.
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