Wave questions - how do I solve these?

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Graphix
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Hello,

so...my physics teacher gave us homework on waves but the problem is she hasn't taught us any of it! So, I've been using my textbooks to help me but I'm stuck on two particular questions - I have the answers but I'm just not sure how to get there:

1) A ray of light is incident on a rectangular block of glass of index of refraction 1.450 at an angle of 40 degrees. If the thickness of the block is 4.00cm, find the amount d by which the ray is deviated.
(d is shown to be the distance between the actual ray of light after it has left the block of glass and where the ray of light would have been if it had travelled in a straight direction.)

I thought this would be simple. I worked out the angle of refraction to be 26.3 degrees and then I tried to use trigonometry to solve it and I get 1.38cm. The actual answer is 1.06cm...

2) A radio station emits radio waves of wavelength 1600m which reach a house directly and after reflecting froma mountain behind the house. If reception at the house is very poor, what is the shortest possible distance between the house and the mountain?

The answer is 400m, so 4 times less than the wavelength of the original wave...

Is this a really trivial question...?


Thanks everybody!
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F1 fanatic
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(Original post by Graphix)
Hello,

so...my physics teacher gave us homework on waves but the problem is she hasn't taught us any of it! So, I've been using my textbooks to help me but I'm stuck on two particular questions - I have the answers but I'm just not sure how to get there:

1) A ray of light is incident on a rectangular block of glass of index of refraction 1.450 at an angle of 40 degrees. If the thickness of the block is 4.00cm, find the amount d by which the ray is deviated.
(d is shown to be the distance between the actual ray of light after it has left the block of glass and where the ray of light would have been if it had travelled in a straight direction.)

I thought this would be simple. I worked out the angle of refraction to be 26.3 degrees and then I tried to use trigonometry to solve it and I get 1.38cm. The actual answer is 1.06cm...

2) A radio station emits radio waves of wavelength 1600m which reach a house directly and after reflecting froma mountain behind the house. If reception at the house is very poor, what is the shortest possible distance between the house and the mountain?

The answer is 400m, so 4 times less than the wavelength of the original wave...

Is this a really trivial question...?


Thanks everybody!
I think your method with 1) is nearly correct but I think you might have misinterpreted what the distance d is. You have calculated the distance along the face of the block, but the beams emerge parallel and you want the perpendicular distance to their direction (so you need to use trigonometry again on exit. It works out as d=1.38*sin50 = 1.06cm

for 2) it's as straightforward as it looks. The waves must be in anti-phase (wavelength/2) but must travel there and back so the distance is half of this (lambda/4)
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Graphix
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(Original post by F1 fanatic)
I think your method with 1) is nearly correct but I think you might have misinterpreted what the distance d is. You have calculated the distance along the face of the block, but the beams emerge parallel and you want the perpendicular distance to their direction (so you need to use trigonometry again on exit. It works out as d=1.38*sin50 = 1.06cm

for 2) it's as straightforward as it looks. The waves must be in anti-phase (wavelength/2) but must travel there and back so the distance is half of this (lambda/4)
Thanks so much

Where did you get 50 degrees from?

And...how do you know the waves must be in anti-phase?
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(Original post by Graphix)
Thanks so much

Where did you get 50 degrees from?

And...how do you know the waves must be in anti-phase?
well since the system is the same into the block as out then the angle to the normal on exit in air will be 40 degrees again. If you draw it you'll see the triangle you want is 90-40=50 degrees. You could equally well use cos40 instead of sin50 of course.

And the waves must be in antiphase because this is the requirement that their amplitude is zero, the waves cancel out and so there is a very poor signal. You're assuming that the signal is as bad as it can be (which is when the waves cancel out to give zero signal, ie antiphase)
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Graphix
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(Original post by F1 fanatic)
well since the system is the same into the block as out then the angle to the normal on exit in air will be 40 degrees again. If you draw it you'll see the triangle you want is 90-40=50 degrees. You could equally well use cos40 instead of sin50 of course.

And the waves must be in antiphase because this is the requirement that their amplitude is zero, the waves cancel out and so there is a very poor signal. You're assuming that the signal is as bad as it can be (which is when the waves cancel out to give zero signal, ie antiphase)
Thank you so much F1 fanatic.

I don't think this is the first time you've helped me in the physics forum so I'm really grateful for your time and I wish I could give you some rep!
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(Original post by Graphix)
Thank you so much F1 fanatic.

I don't think this is the first time you've helped me in the physics forum so I'm really grateful for your time and I wish I could give you some rep!
No problem. I'm not particularly short of rep so your appreciation will be sufficient . Did that make sense? Not really sure I explained the antiphase thing very well. Basically a good signal means high amplitude so if the signal is poor then the amplitude is weak. It is really a case of insight - reading between the lines - as to why you assume the house is at the worst possible signal. You have nothing else to go on, there is no cut-off defined as to what signal amplitudes are "poor" and so we have to assume the worst case. It also makes the problem significantly easier to do that, and would mean that it was poor however strong the initial amplitude. The weak signal results from the wavelength and will be weak whether the original transmitted signal was a powerful high amplitude signal or a very weak one.
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(Original post by F1 fanatic)
No problem. I'm not particularly short of rep so your appreciation will be sufficient . Did that make sense? Not really sure I explained the antiphase thing very well. Basically a good signal means high amplitude so if the signal is poor then the amplitude is weak. It is really a case of insight - reading between the lines - as to why you assume the house is at the worst possible signal. You have nothing else to go on, there is no cut-off defined as to what signal amplitudes are "poor" and so we have to assume the worst case. It also makes the problem significantly easier to do that, and would mean that it was poor however strong the initial amplitude. The weak signal results from the wavelength and will be weak whether the original transmitted signal was a powerful high amplitude signal or a very weak one.
Yeah - I noticed the high rep after I posted my message...

I already knew about antiphase - it was just a matter of me linking that into the idea of poor quality - and you explained it very well, and very simply. Much better than my teacher...
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(Original post by Graphix)
Yeah - I noticed the high rep after I posted my message...

I already knew about antiphase - it was just a matter of me linking that into the idea of poor quality - and you explained it very well, and very simply. Much better than my teacher...
Perhaps I should consider a career change.
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(Original post by F1 fanatic)
Perhaps I should consider a career change.
Yes - the world of physics teaching would be a much better place with you involved!
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Day6
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Cna you explain how did you get 1,38 cm. Please.
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Can you explain how did you get 1,38 cm. Please.
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