So we havent been taught this yet but have been set questions on it.
Anyway, here it is:
http://postimg.org/image/z2rqi75uv/
I got 2.8 * 107 for 6)a
4x larger for 6)b
however I am stuck on 6c
I dont actually know what the "cross sectional area* of a wave is? Is its amplitude * wavelength or what?
I have no idea how to work it out as I'm not sure which variables change?
intensity = power/ cross sectional area
is the power constant? Does it change?
I'm sure this is real basic, but my revision guide is absolutely useless and has a small paragraph on it not giving enough information to help anyone in anyway, and I havent been taught it either. So please dont just say "think about it" or go right into it, an explanation would be great. Thanks

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 28012015 20:56

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 28012015 21:42
(Original post by Danny.L)
So we havent been taught this yet but have been set questions on it.
Anyway, here it is:
http://postimg.org/image/z2rqi75uv/
I got 2.8 * 107 for 6)a
4x larger for 6)b
however I am stuck on 6c
I dont actually know what the "cross sectional area* of a wave is? Is its amplitude * wavelength or what?
I have no idea how to work it out as I'm not sure which variables change?
intensity = power/ cross sectional area
is the power constant? Does it change?
I'm sure this is real basic, but my revision guide is absolutely useless and has a small paragraph on it not giving enough information to help anyone in anyway, and I havent been taught it either. So please dont just say "think about it" or go right into it, an explanation would be great. Thanks
the surface area, A, of a sphere is related to the distance from the centre, r, by
A=4πr^2 e.g. if the radius doubles the area increases by factor 2^2 or 4 and if the radius triples the area increases by factor 2^3 or 8
in each case the 60W radiated at the centre must equal 60W crossing the surface of the sphere however far away it is from the source.

part C is giving you 5.6*10^2 W m^2
which is 5.6*10^2 Watts per meter squared... power/cross sectional area... it's already an intensity, you don't need to do any more working out.
the power of a wave is proportional to Amplitude^2 if you hold frequency, speed and mass constant.
hope that helps. 
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 28012015 21:58
(Original post by Joinedup)
The reason for the 1/r^2 law is that if you have a point source of sound (or light etc.) giving out a known power (say 60W) which is radiating equally in every direction and you have it located at the centre of a transparent sphere with a known surface area (say 1m^2) you can say that the intensity at the sphere surface must be 60/1 60W/m^2 all the power radiated from the source must cross the surface (by conservation of energy)
the surface area, A, of a sphere is related to the distance from the centre, r, by
A=4πr^2 e.g. if the radius doubles the area increases by factor 2^2 or 4 and if the radius triples the area increases by factor 2^3 or 8
in each case the 60W radiated at the centre must equal 60W crossing the surface of the sphere however far away it is from the source.

part C is giving you 5.6*10^2 W m^2
which is 5.6*10^2 Watts per meter squared... power/cross sectional area... it's already an intensity, you don't need to do any more working out.
the power of a wave is proportional to Amplitude^2 if you hold frequency, speed and mass constant.
hope that helps.
it is 16 times larger than our original value so its amplitude bust be 4x larger, then did 4 * 0.45?
Now need help on 7 though. For b
I used intensity is proportional to 1/r^2
the distance is 1.5*10^11 m
so our 1.5kw of energy is a (1/(1.5*10^11))"th" of the actual intensity, so did
1.5kw * distance squared from sun to get 3.15 * 10 ^22
however this seems enormous.
as when i then go onto C and re arrange to do intensity * cross sectional area = power
I get 1.9396 * 10^41
So my mistake seems to be calculating the intensity of the sun, however cannot see my mistake.
The intensity of the sun is 1/distance^2 so 1/(1.5*10^11)^2
therefore if we get 1.5kw, surely if we times it by distance squared that is equal to the full intensity of the sun? It makes logical sense to me 
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 28012015 22:08
(Original post by Danny.L)
for part c I did 5.6*10^2 / 3.5 * 10^3
it is 16 times larger than our original value so its amplitude bust be 4x larger, then did 4 * 0.45?
Now need help on 7 though. For b
I used intensity is proportional to 1/r^2
the distance is 1.5*10^11 m
so our 1.5kw of energy is a (1/(1.5*10^11))"th" of the actual intensity, so did
1.5kw * distance squared from sun to get 3.15 * 10 ^22
however this seems enormous.
as when i then go onto C and re arrange to do intensity * cross sectional area = power
I get 1.9396 * 10^41
So my mistake seems to be calculating the intensity of the sun, however cannot see my mistake.
The intensity of the sun is 1/distance^2 so 1/(1.5*10^11)^2
therefore if we get 1.5kw, surely if we times it by distance squared that is equal to the full intensity of the sun? It makes logical sense to me 
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 28012015 22:10
(Original post by Texx)
Well you would expect the sun to have a very large power output. It's the sun.
However, when I have looked in my book which uses similar values for distance and for intensity of light when it reaches earth, its power of the sun is 3.7*10^26
to the power of 41 is considerably larger and thus making me think i'm incorrect. 
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 28012015 22:21
(Original post by Danny.L)
yes, very smart of you.
However, when I have looked in my book which uses similar values for distance and for intensity of light when it reaches earth, its power of the sun is 3.7*10^26
to the power of 41 is considerably larger and thus making me think i'm incorrect.
Also, calm down, it's amazing you get touchy because I offered the suggestion that you weren't in fact incorrect.Last edited by Texx; 28012015 at 22:28. 
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 28012015 22:30
(Original post by Texx)
Because you need to root the distance. And covert kW to SI units.
Also, calm down, it's amazing you get touchy because I offered the suggestion that you weren't in fact incorrect.
Edit: it's the surface area of the sun you'll want to include in the total power calculation, also.
Anyway,
I understand the SI units, that's stupid of me.
However, the root of distance I'm trying to get my head around. care to explain more why? Not sure I fully understand 
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 28012015 22:35
(Original post by Texx)
Because you need to root the distance. And covert kW to SI units.
Also, calm down, it's amazing you get touchy because I offered the suggestion that you weren't in fact incorrect.
(1.5*10^11)^2 less bright than on the surface of it, correct?
So why would we root it?
if we move 10 times further away the light is 100 times dimmer.
So if we are told the intensity is 10w at 10 m away the intensity is 1/10^2 of the actual magnitude.
so 10w is 1/100 th of the actual magnitude. So 10w * 100 (distance squared) surely equals original intensity. Which would be 1000w
Do you see where I'm coming from. I'm not sure I understand why a root is coming into place 
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 28012015 22:51
(Original post by Texx)
Because you need to root the distance. And covert kW to SI units.
Also, calm down, it's amazing you get touchy because I offered the suggestion that you weren't in fact incorrect.
I took the energy earth recieves, 1400w * 2.25 * 10^22 (distance squared) and got 3.15*10^25, which looks suspiciously close to what I was expecting the power of the sun to be.
It's almost like I have unintentionally done intensity * cross sectional area opposed to scaling up the intensity.
although if I were to do intensity * cross sectional area, wouldn't I actually do 4pi(1.5*10^11)^2
the 4pi gives the circle created at the distance squared.
I am so confused, I have no idea why my original idea isn't working but is now seem to have accidently worked out my power of the sun with it.
I need someone to stepbystep walk though this w/ me 
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 28012015 22:52
(Original post by Danny.L)
Not touchy, just looked sarcastic in your original response, so i responded accordingly.
Anyway,
I understand the SI units, that's stupid of me.
However, the root of distance I'm trying to get my head around. care to explain more why? Not sure I fully understand 
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 28012015 22:55
(Original post by Texx)
Yes, you're right, sorry I had a dumb moment. Hang on, I'll try to explain it properly. 
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 28012015 22:58
(Original post by Danny.L)
After working it out.
I took the energy earth recieves, 1400w * 2.25 * 10^22 (distance squared) and got 3.15*10^25, which looks suspiciously close to what I was expecting the power of the sun to be.
It's almost like I have unintentionally done intensity * cross sectional area opposed to scaling up the intensity.
although if I were to do intensity * cross sectional area, wouldn't I actually do 4pi(1.5*10^11)^2
the 4pi gives the circle created at the distance squared.
I am so confused, I have no idea why my original idea isn't working but is now seem to have accidently worked out my power of the sun with it.
I need someone to stepbystep walk though this w/ me
Imagine the earth is a point on a sphere with radius 1.5x10^11m, every point of this sphere is receiving light from the sun with the same intensity of 1.4x10^3 W per m. The surface area of this sphere is 4 x pi x r^2. The total power of the sun is the surface area multiplied by the intensity of the light at this radius. Does that make sense?
Sorry for the stupid moment with mentioning rooting it... I think I'll go have a nap. 
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 28012015 23:11
(Original post by Texx)
Alright:
Imagine the earth is a point on a sphere with radius 1.5x10^11m, every point of this sphere is receiving light from the sun with the same intensity of 1.4x10^3 W per m. The surface area of this sphere is 4 x pi x r^2. The total power of the sun is the surface area multiplied by the intensity of the light at this radius. Does that make sense?
Sorry for the stupid moment with mentioning rooting it... I think I'll go have a nap.
Seemed logical to me though.
I understand the principle, I guess. What I am unsure about is why the distance is the radius, I am now imagining looking sideon with the sun at the centre and earth to the far right, you're saying we treat the distance from the 2 as the radius, I guess I can understand that.
Okay so the light at this radius is 1400w.
the surface area is 4 Pi r^2 although I cant see how that works now, as surely as you get towards the point the intensity of light increases.
I am imagining it as a 2d circle looking at it with the earth on the right, sun middle and the surface area is the whole circle, which I do not understand as all of those bits arent being supplied with 1400w, but will be receiving more energy as they are closer. But I guess I'll just accept it.
My question is.
Is there any possible way to work out intensity directly?
It wants intensity first, but the only way I see is to calculate power and then we can get intensity from that?
Thanks so far, most helpful person of tonight. This has to be in tomorrow as well 
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 28012015 23:25
(Original post by Danny.L)
So whatever I was doing originally was complete nonsense?
Seemed logical to me though.
I understand the principle, I guess. What I am unsure about is why the distance is the radius, I am now imagining looking sideon with the sun at the centre and earth to the far right, you're saying we treat the distance from the 2 as the radius, I guess I can understand that.
Okay so the light at this radius is 1400w.
the surface area is 4 Pi r^2 although I cant see how that works now, as surely as you get towards the point the intensity of light increases.
I am imagining it as a 2d circle looking at it with the earth on the right, sun middle and the surface area is the whole circle, which I do not understand as all of those bits arent being supplied with 1400w, but will be receiving more energy as they are closer. But I guess I'll just accept it.
My question is.
Is there any possible way to work out intensity directly?
It wants intensity first, but the only way I see is to calculate power and then we can get intensity from that?
Thanks so far, most helpful person of tonight. This has to be in tomorrow as well
(edit: that wasn't very clear, I mean your initial method for working out the intensity at the surface should yield the correct answer. but the total power output is found by picturing the sphere, and can't be done the way you originally tried.)
I actually only realised it because I had a vague memory of doing a similar question at Alevel. It's one of those things that makes sense when you are explained it but wouldn't necessarily occur to you otherwise. Yeah it's quite hard to visualise, the distance to the earth is the radius and not the diameter because the sun is emitting light in every direction, so it's emitting light of the same intensity in a direction away from the earth too. The tough thing is you're working out the surface area of a kind of imaginary sphere, made up of points that are all the same distance from the sun as the earth is. Try thinking about it in 3D.
You're right, a point closer to the sun receives light with greater intensity, but it's not the surface area of a circle you're working out. It's hard to explain without a drawing, I've tried to attach one. Imagine the yellow patch is a square metre on the surface of the sphere. It's receiving 1.4 kW of power. Every square metre on that sphere is receiving the same.Last edited by Texx; 28012015 at 23:30. 
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 28012015 23:42
(Original post by Texx)
Well, the intensity of the light at the surface should be correct, just not the total power output.
(edit: that wasn't very clear, I mean your initial method for working out the intensity at the surface should yield the correct answer. but the total power output is found by picturing the sphere, and can't be done the way you originally tried.)
I actually only realised it because I had a vague memory of doing a similar question at Alevel. It's one of those things that makes sense when you are explained it but wouldn't necessarily occur to you otherwise. Yeah it's quite hard to visualise, the distance to the earth is the radius and not the diameter because the sun is emitting light in every direction, so it's emitting light of the same intensity in a direction away from the earth too. The tough thing is you're working out the surface area of a kind of imaginary sphere, made up of points that are all the same distance from the sun as the earth is. Try thinking about it in 3D.
You're right, a point closer to the sun receives light with greater intensity, but it's not the surface area of a circle you're working out. It's hard to explain without a drawing, I've tried to attach one. Imagine the yellow patch is a square metre on the surface of the sphere. It's receiving 1.4 kW of power. Every square metre on that sphere is receiving the same.
I have looked at in another way,
using:
Power = intensity * surface area
Power = 1400 * 4Pi (1.5 *10^11)^2
power = 3.96 * 10^26
so intensity = (3.26 * 10^26)/(4Pi(7*10^8)^2)
Intensity = 64.3 Gwatts per m^2
However doing it my original way, the intensity of light on the sun must be 2.25 *10^22 larger than that on earth so
1400 * (2.25 * 10^22) = 3.15 *10^22
Which one is correct?
It seems to me the 64.3 Gw is more reasonable as if I use the 3.15 *10^22 that means the power becomes something * 10^41, which I am certain is wrong!
What do you think?Last edited by Danny.L; 28012015 at 23:42. Reason: Missed bracket 
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 28012015 23:49
(Original post by Texx)
Well, the intensity of the light at the surface should be correct, just not the total power output.
(edit: that wasn't very clear, I mean your initial method for working out the intensity at the surface should yield the correct answer. but the total power output is found by picturing the sphere, and can't be done the way you originally tried.)
I actually only realised it because I had a vague memory of doing a similar question at Alevel. It's one of those things that makes sense when you are explained it but wouldn't necessarily occur to you otherwise. Yeah it's quite hard to visualise, the distance to the earth is the radius and not the diameter because the sun is emitting light in every direction, so it's emitting light of the same intensity in a direction away from the earth too. The tough thing is you're working out the surface area of a kind of imaginary sphere, made up of points that are all the same distance from the sun as the earth is. Try thinking about it in 3D.
You're right, a point closer to the sun receives light with greater intensity, but it's not the surface area of a circle you're working out. It's hard to explain without a drawing, I've tried to attach one. Imagine the yellow patch is a square metre on the surface of the sphere. It's receiving 1.4 kW of power. Every square metre on that sphere is receiving the same.
When i used it to calculate intensity on Neptune I got 1.556 W/m^2 which is exactly the same as the value if you were to divide neptune distance from sun by earths distance = 30
30^2 =900
1400 (earths light) / 900 = 1.555(rec)
So I believe the formula is correct, just curious as to why the inverse square law doesnt work!
For some reason I feel very agitated when I don't understand something, this being one of those times! 
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 29012015 00:22
(Original post by Danny.L)
So whatever I was doing originally was complete nonsense?
Seemed logical to me though.
I understand the principle, I guess. What I am unsure about is why the distance is the radius, I am now imagining looking sideon with the sun at the centre and earth to the far right, you're saying we treat the distance from the 2 as the radius, I guess I can understand that.
Okay so the light at this radius is 1400w.
the surface area is 4 Pi r^2 although I cant see how that works now, as surely as you get towards the point the intensity of light increases.
I am imagining it as a 2d circle looking at it with the earth on the right, sun middle and the surface area is the whole circle, which I do not understand as all of those bits arent being supplied with 1400w, but will be receiving more energy as they are closer. But I guess I'll just accept it.
My question is.
Is there any possible way to work out intensity directly?
It wants intensity first, but the only way I see is to calculate power and then we can get intensity from that?
Thanks so far, most helpful person of tonight. This has to be in tomorrow as well
or you could also do it by ratios which would look like
I_{1} R_{1}^2 = I_{2} R_{2}^2
I_{1} is intensity at earth
R_{1} is earth's orbital radius
I_{2} is intensity at sun's surface
R_{2} is the radius of the sun
it's *as if* you're treating the sun as if it was a sphere with a point source inside it 
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 29012015 21:18
(Original post by Joinedup)
Power first would be a good way of getting to the intensity at the surface of the sun
or you could also do it by ratios which would look like
I_{1} R_{1}^2 = I_{2} R_{2}^2
I_{1} is intensity at earth
R_{1} is earth's orbital radius
I_{2} is intensity at sun's surface
R_{2} is the radius of the sun
it's *as if* you're treating the sun as if it was a sphere with a point source inside it
therefore they equal each other.
He also spotted what I had done wrong for my working by multiplying 1400 Wm2 by (1.5 *10^11)^2. That was finding the intensity 1m away from the centre of the sun, but we wanted it at the surface, so needed to then divide by (7 * 10*)^2 (which is radius of sun squared). I think that's correct! How does that sound to you? 
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 29012015 21:20
(Original post by Texx)
Well, the intensity of the light at the surface should be correct, just not the total power output.
(edit: that wasn't very clear, I mean your initial method for working out the intensity at the surface should yield the correct answer. but the total power output is found by picturing the sphere, and can't be done the way you originally tried.)
I actually only realised it because I had a vague memory of doing a similar question at Alevel. It's one of those things that makes sense when you are explained it but wouldn't necessarily occur to you otherwise. Yeah it's quite hard to visualise, the distance to the earth is the radius and not the diameter because the sun is emitting light in every direction, so it's emitting light of the same intensity in a direction away from the earth too. The tough thing is you're working out the surface area of a kind of imaginary sphere, made up of points that are all the same distance from the sun as the earth is. Try thinking about it in 3D.
You're right, a point closer to the sun receives light with greater intensity, but it's not the surface area of a circle you're working out. It's hard to explain without a drawing, I've tried to attach one. Imagine the yellow patch is a square metre on the surface of the sphere. It's receiving 1.4 kW of power. Every square metre on that sphere is receiving the same.
It was because doing that finds the intensity of light at the centre of it, 1m away, but we wanted the surface which is 7*10^8 m away, so the intensity at that point is (1400 * (1.5*10^11)^2) all / by (7*10^8)^2
Like we work out the intensity at the core, then work our way back out. Hope that's correct, I figured that'd be right. 
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 29012015 22:10
(Original post by Danny.L)
Thanks for your help btw, my teacher spotted why 1400 * (1.5*10^11)^2 was wrong straight away, was quite impressed, actually!
It was because doing that finds the intensity of light at the centre of it, 1m away, but we wanted the surface which is 7*10^8 m away, so the intensity at that point is (1400 * (1.5*10^11)^2) all / by (7*10^8)^2
Like we work out the intensity at the core, then work our way back out. Hope that's correct, I figured that'd be right.
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