stdevlin03
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radio waves of wavelength 320m travel directly to receiver R 120km from transmitter t they are also reflected from the base of the ionosphere which is at an effective height of 80km above the surface.A)calculate path differenceB)sky waves are reflected 180 degrees between the incident and reflected waves uses answer from A to suggest whether they are constructive or destructiveC)over a few seconds, the height of the ionosphere varies state and explain any effects this would have on radio signal
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phys981
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I would start by sketching a diagram to show the path of the waves which travel 120km directly and those which reflect off the ionosphere. Assume that the point of reflection is exactly half way along the 120km. You should then be able to use Pythagoras to work out exactly how far these reflected waves travel.

Now you can find the path difference between the reflected waves and those which travel directly along the surface. And then, of course, you need to know whether that path difference is a whole number multiple of the wavelength or not to work out the phase difference.

Normally of course, a path diff of a whole nº of wavelengths means waves arriving in phase and a diff of (n + 1/2) means antiphase. BUT there's a twist here, as we are told that the reflection produces a phase change of 180º.

So, do the waves meet in phase, or not...?

If you need more details help let me know but have a go first using the above.
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stdevlin03
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(Original post by phys981)
I would start by sketching a diagram to show the path of the waves which travel 120km directly and those which reflect off the ionosphere. Assume that the point of reflection is exactly half way along the 120km. You should then be able to use Pythagoras to work out exactly how far these reflected waves travel.

Now you can find the path difference between the reflected waves and those which travel directly along the surface. And then, of course, you need to know whether that path difference is a whole number multiple of the wavelength or not to work out the phase difference.

Normally of course, a path diff of a whole nº of wavelengths means waves arriving in phase and a diff of (n + 1/2) means antiphase. BUT there's a twist here, as we are told that the reflection produces a phase change of 180º.

So, do the waves meet in phase, or not...?

If you need more details help let me know but have a go first using the above.
I had a go and drew a diagram and used Pythagoras but after seeing this I feel i may be wrong could you elaborate or finish the question please
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phys981
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Ok, how far have you got so far?

How far do the reflected waves travel?
What does this give for the path differece? How many wavelenghts is this?
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stdevlin03
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(Original post by phys981)
Ok, how far have you got so far?

How far do the reflected waves travel?
What does this give for the path differece? How many wavelenghts is this?
I finished the question, the reflected waves travel 200km this minus the 120km is 80km which is the path difference then do 80000m / 320 gives 250 wavelengths which shows a constructive wave then I took the reflected part of the wave which is 100km so the pd for this is 20 which is 62.8 wavelengths which lines up perfectly with the 180⁰ out of phase reflected wave that also shows a destructive wave. But after reading ur comment i feel this may not have been the correct way to the answer
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phys981
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That sounds right to me! 200m travelled by the reflected wave (100m to the ionosphere and 100m back) - so path diff is 80km. You are correct, this means a whole number of wavelengths which would mean in phase. BUT, when the wave is reflected from the ionosphere, it undergoes a phase change of 180º so what was in phase is now in antiphase.

I'm not sure why you took the 100m and of course, 62.8 waves is not exactly in antiphase, it's just not in phase. Ignore that bit and go with the path diff of 80.
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stdevlin03
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(Original post by phys981)
That sounds right to me! 200m travelled by the reflected wave (100m to the ionosphere and 100m back) - so path diff is 80km. You are correct, this means a whole number of wavelengths which would mean in phase. BUT, when the wave is reflected from the ionosphere, it undergoes a phase change of 180º so what was in phase is now in antiphase.

I'm not sure why you took the 100m and of course, 62.8 waves is not exactly in antiphase, it's just not in phase. Ignore that bit and go with the path diff of 80.
what would be the answer to part c and it was 62.5 sorry so it is in antiphase
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phys981
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Ah, ok 62.5 makes sense.

As for part c, thinking back to your diagram of the waves that showed the ionosphere at 80km - of the height of that ionosphere keeps changing, what does that do to the path difference? And hence to the phase difference and interference?
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stdevlin03
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I wrote that because it is at its minimum the increased distance will increase the pd which will mean that the reception will get a lot stronger
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phys981
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Actually that's right, it's at its minimum so any change will be a stronger signal, but I suspect that what they're getting at here (I may be wrong) is that as the height of the ionosphere varies, the strength of the signal varies as well, so I suppose you'd get that effect of the radio signal coming and going - constantly changing.
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stdevlin03
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(Original post by phys981)
Actually that's right, it's at its minimum so any change will be a stronger signal, but I suspect that what they're getting at here (I may be wrong) is that as the height of the ionosphere varies, the strength of the signal varies as well, so I suppose you'd get that effect of the radio signal coming and going - constantly changing.
thanks for the help
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You're welcome
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