Inverse sqaure law (1 Viewer)

hscishard

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Inverse square law

Intensity prop. to 1/d^2

So if 0<distance < 1, the intensity increases?
Then it'll decrease after distance >1 ?
 
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Lets assume the wave is visible light, makes it easier to explain. Intensity (lux) is the amount of energy (lumens) received per square metre. In your example,

At 1 metre from the light source, the intensity of light is 2 lx (2 lm/m2). At 3 metres from the light source, the intensity will be 1/32 of the intensity from 1 metre (1/9 * 2 lx = 0.22 lx). At 0.5 metres from the light source, the intensity will be 1/0.52 of the intensity from 1 metre (4 * 2 lx = 8 lx).
 

hscishard

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Oh..I never knew it was Intensity at 1 metre x 1/d^2. Thought it was from the actual source.

You got that from jacaranda physics 1, right?

Wait.. so the intensity when d--> 0 means Intensity-->infinity? This doesn't need to be answered.
 
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Oh..I never knew it was Intensity at 1 metre x 1/d^2.
I just assumed it was from 1 metre, an actual question would say from where.

Like, the intensity of light 4 m from the source is measured to be 2 lx, find the intensity from 5 m.

So from 5 m the intensity is 1/(5/4)2 * 2 lx = 16/25 * 2 lx = 1.28 lx.
 

hscishard

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I just assumed it was from 1 metre, an actual question would say from where.

Like, the intensity of light 4 m from the source is measured to be 2 lx, find the intensity from 5 m.

So from 5 m the intensity is 1/(5/4)2 * 2 lx = 16/25 * 2 lx = 1.28 lx.
No like the general forumla thing.
Intensity from 1 metre x 1/d^2 = Intensity at d

From your example.
a x 1/16 = 2
a = 32 This is I from a metre
Then 32 x 1/25 = 32/25 = 1.28
 

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