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um the inverse square law isn't one single thing

its just saying that something is inversely proportional to the square of the distance. (dont even think its always distance, but usually is)

there are loads of examples of this
GazingAtStars
I have been searching on google for explanations of the inverse square law. But all I can find is complicated physicsy genious explanations (or maybe I'm just being stupid).

Does anyone know of any websites or books where I can get this information IN SIMPLE FORM?


are you studying radiography by chance?
Peace'n'loveman
are you studying radiography by chance?


Ha yes.

And I never did physics at A Level. And just a weeny tiny bit at GCSE (and that was a billion years ago).

So I want to try and get a good understanding of it before my exam :smile:
Reply 4
I found this explanation quite helpful. Well, the picture anyway.
cardine92
I found this explanation quite helpful. Well, the picture anyway.


Ooh thank you :smile:
Who'd have thought.. wikipedia :rolleyes: haha. But that is actually quite helpful :biggrin:
GazingAtStars
Ha yes.

And I never did physics at A Level. And just a weeny tiny bit at GCSE (and that was a billion years ago).

So I want to try and get a good understanding of it before my exam :smile:


Ok yeh its one of those horrible things you have to learn ,but soon forget the minutiae of. However ,the principle behind it is very important ,particularly regarding radiation protection. The important principle is that getting distance between you and the radiation source is important,because with the inverse square law the intensity of the original source is inversely proportional to the distance from the source.
Think about a point source which is producing radiation in all directions. The radiation spreads out as it moves away from the source so if you measure the intensity at a given distance away (say 10 cm), the value you get is total emission rate spread over the surface of a sphere with radius 10 cm.

If you now measure the intensity 20 cm away, the value you measure will be the same total emission rate, but now spread more thinly over the surface of a sphere with 20 cm radius

The surface area of a sphere is related to the square of the radius (4 pi x radius^2), so if you double the radius, you increase the surface area by four. So, by doubling the radius, you have spread your total emission rate over 4x the surface area; in other words, the measured intensity will have decreased by 4x.

Hope this helps

C




from the point, say 10 cm, the
Reply 8
Thank you everyone :smile:

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