Inverse Square Law

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Ka

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Who would like to explain the Inverse Square Law, to someone (obviously somewhat thick) who still doesn't grasp it?

Thanks,

Ka
 

Flotsam

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If you move your light twice as far away from the subject as it was before, the light falling on the subject will be one quarter (not one half) of the intensity.

It truly only applies to a point source but it is a good thing to keep in mind whenever you are using artificial lights.

ie: 3 feet at f5.6 becomes f2.8 at 6 feet
 

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Imagine a sqare spotlight shining on a sheet of paper. Say that at one distance, the illuminated spot is a one meter square.

Now move the paper twice as far away. Since the light spreads from the spotlight at the same angle as before, the illuminated spot is now a two meter square.

So the light that previously fell on one square meter now falls on 4 (2*2) square meter. Each area of the paper now gets only one quarter of the light!

Move the paper the same distance again, and you have a three meter square of light. The square of 3 is 9, so the light intensity is one ninth!

It works the same way with film and magnification: A lens which barely covers 4x5" at infinity will cover 8x10" at a 1:1 scale, since it has to be moved out to twice the distance from the film. Twice the distance gives four times the coverage, and one quarter of the light at any point on the film.
 
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Ka

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Thank you all. Now I only have to actually do what you said in order to fully grasp it. I'm so kinetic.

Ka
 

Flotsam

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A simple way to see it demonstrated is to look at the manual exposure dial on the back of a portable flash unit. You will see that it indicates that as the light to subject distance doubles, you must open the lens two stops to maintain proper exposure.
Of course, if you are using a continuous light source you can also use your shutter speed to make the two stop adjustment.
 

127

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It's really all about geometry - nothing clever that you can't figure out from first principles on a sheet of paper:

Mark a point on a piece of paper, and draw lines radially away from it (saw 12 to start with - like on a clock. You may need to add more lines latter to get more acurate results). Thats your light source. Draw a line representing your object, so that a few rays of light land upon it. Each ray carries a fixed amount of energy. The amount of energy landing on your object is simply the number of rays which it it.

As you move the object further away, less rays of the rays it it. In fact in this case if you double the distance the number would approximatly half because we're working on a flat sheet of paper (it's onnly approximmate because we have only a few rays - in fact there would be billions of rays - aka photons, so in real life the result is almost exact). With a real ilght source the light wouldn't just be spreading up and down the page, but into and out of the page too, so a doubleing of distances half's the rays twice. A tripling of distance would divide the rays by three twice.

Another though experiment is to imagine a ballon, with a light source in the middle. All the energy from the light is illuminating the surface of the balloon. If you inflate the ballon the surface are increases, so the light is shared more thinly. In fact for each doubling or the radius, the surface area goes up by 4. If you marked a set of points on the balloon representing light rays, and an area on the ballon the number of points in the area represent the amount of light hitting it. Then inflate the balloon to twice its diameter, the area would now be four times the size, but has exactly the same rays hitting it, so the illumination (candles per square meter, or candles per square foot) has divided by four.

Ian
 
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Ka

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Hi Ian,

I only just saw this bit you wrote about the balloon and paper.... now that does make sense to me. I can see it, touch it. Thanks.
 
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127 said:
[snip]

Another though experiment is to imagine a ballon, with a light source in the middle. All the energy from the light is illuminating the surface of the balloon. If you inflate the ballon the surface are increases, so the light is shared more thinly. In fact for each doubling or the radius, the surface area goes up by 4. If you marked a set of points on the balloon representing light rays, and an area on the ballon the number of points in the area represent the amount of light hitting it. Then inflate the balloon to twice its diameter, the area would now be four times the size, but has exactly the same rays hitting it, so the illumination (candles per square meter, or candles per square foot) has divided by four.

Ian

Ian:
Clever explanation the baloon one. Easy to grasp.

Jorge O
 
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