# Intensity and PowerWatch

#1
If you double the distance between a source of radiation and a detector, how will the intensity of the radiation detected change?

I get the answer as 0.5 but this is wrong.

Can anyone help?

I=p/a
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5 years ago
#2
(Original post by zed963)
If you double the distance between a source of radiation and a detector, how will the intensity of the radiation detected change?

I get the answer as 0.5 but this is wrong.

Can anyone help?

I=p/a
The intensity (power) depends on the inverse square of the distance.
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#3
(Original post by Stonebridge)
The intensity (power) depends on the inverse square of the distance.
And how does that work?
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5 years ago
#4
(Original post by zed963)
And how does that work?
http://en.wikipedia.org/wiki/Inverse-square_law
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5 years ago
#5
(Original post by zed963)
And how does that work?
Intensity is power per unit surface area. For a point source, the power output is assumed equally distributed over a sphere with the point source at the centre. Since the surface area of a sphere is 4pi*radius^2 then the intensity decreases proportional to the square of the radius (the inverse square law)

i.e. Intensity is proportional to 1/(r^2)
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#6
(Original post by F1 fanatic)
Intensity is power per unit surface area. For a point source, the power output is assumed equally distributed over a sphere with the point source at the centre. Since the surface area of a sphere is 4pi*radius^2 then the intensity decreases proportional to the square of the radius (the inverse square law)

i.e. Intensity is proportional to 1/(r^2)
So if we were to take a source of radiation that emits 200w and the detector detects it 2m away then the intensity would be 200/4 = 50w/m^2

But if the distance increases to 4m will the intensity then be 12.5w/m^2

Is the above correct so far?
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#7
So if I :

Double Distance = 1/4
Triple "" = 1/9

Etc.

Is that all there is to this law?
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5 years ago
#8
(Original post by zed963)
So if we were to take a source of radiation that emits 200w and the detector detects it 2m away then the intensity would be 200/4 = 50w/m^2

But if the distance increases to 4m will the intensity then be 12.5w/m^2

Is the above correct so far?
Not quite because you've changed from talking about power to talking about intensity. If you're doing that you need to worry about the conversion factors (ie the actual rather than relative surface area). If instead you had an intensity of 200w/m^2 at a distance of 1m then at 2m it would be 1/4 of what it was at 1 m, and at 4m it would be 1/16 of what it was at 1m.

Make sense?
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#9
(Original post by F1 fanatic)
Not quite because you've changed from talking about power to talking about intensity. If you're doing that you need to worry about the conversion factors (ie the actual rather than relative surface area). If instead you had an intensity of 200w/m^2 at a distance of 1m then at 2m it would be 1/4 of what it was at 1 m, and at 4m it would be 1/16 of what it was at 1m.

Make sense?
Yes I've understood it. Is this law like the easiest law because it didn't take long for me to understand it?
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