Is there an easy way to understand inverse square law?

11 years 8 months ago #246343 by Tsoto
I've looked at links I found on Google searches and been reading en.wikipedia.org/wiki/Inverse-square_law

I'm completely lost and need some help understanding this. So if I double the distance between my flash and a surface that is reflecting that the light from my flash the intensity drops 1/4?

Is that how it works?


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11 years 8 months ago #246349 by KCook
With direct flash (aimed straight at the subject), yes, you have this right. With bounce flash things get more complicated. Thank goodness for TTL and chimping!

blinky

Canon 50D, Olympus PL2
kellycook.zenfolio.com/

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11 years 8 months ago #246369 by John Landolfi
Let me try:

Let's call

I=actual intensity of light
O=Flash output
d=distance to subject.

The law says that I=O/(d squared)

If the distance doubles to 2d, then you have to divide by (2d)squared. So

I=O/(2d)squared=O/4(d squared), So the intensity falls by a factor of 4.

If the distance trebles to 3d, then you divide by (3d)squared=9(d squared), so the intensity falls by a factor of 9.
And so on.

:cheers:


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11 years 8 months ago #246449 by John Landolfi
Anybody care to explain diffraction? Maybe this could be a thread for contributing explanations for the many physical principles we use and obey in photography...


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11 years 8 months ago #246482 by KCook
Why bother? Cambridgeincolour is the bomb for theory -

www.cambridgeincolour.com/tutorials/diff...ion-photography.htm/

Kelly

Canon 50D, Olympus PL2
kellycook.zenfolio.com/

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11 years 8 months ago #246511 by John Landolfi
OK Kelly. You're out! Anyone else care to contribute?


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11 years 8 months ago - 11 years 8 months ago #247525 by Henry Peach

Tsoto wrote: So if I double the distance between my flash and a surface that is reflecting that the light from my flash the intensity drops 1/4?


Yep. 2 stops ( 1 stop is double or half the light). 1/2 x 1/2 = 1/4th.
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11 years 8 months ago #247612 by Stealthy Ninja



What I look like right now:
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11 years 8 months ago #247614 by KCook
You look like you are tracking a wounded gazelle :evil:

Canon 50D, Olympus PL2
kellycook.zenfolio.com/

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11 years 8 months ago #247620 by Stealthy Ninja

KCook wrote: You look like you are tracking a wounded gazelle :evil:


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11 years 8 months ago #247719 by Brad Eckel Photography
This start of this article is pretty straight forward regarding this question.

check this site out from NASA. I have not posted on here before so it would let me do the entire link... so put the 3w's and the dot in front of it.

imagine.gsfc.nasa.gov/YBA/M31-velocity/1overR2-more.html


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11 years 8 months ago #247729 by Brad Eckel Photography
Think of it this way, as you move further away from and object (increase the distance), the SAME object looks smaller (looks like it decreases in size) hence the word inverse in the relatonship.

From basic math a circle uses the radius sqaured and a sqaure has two equal sides. When calculating the area of these, the area becomes larger non linearily due to the squaring of the radius or multiply of the length times the width (since L=W in a sqaure L x W = L x L = L^2 (or L sqaured). Similarly in ALL units of measure for an area for instance a room is measure in feet squared.

Light spreads out from... more or less... a single "flash" point, in straight, but not in parallel lines. Light, is a form of energy, which once the flash is activated is sent out on various angles from the flash point. This engery value does not increase or decrease, but merely gets dispersed over a larger area as it travels further away from the flash point (until it hits an object and is either absorded, reflected or refracted... another story)

In the NASA diagram, you will see each progressively larger area shades lighter as the distance increase from the from the flash point.

Note: "flash" point, can be replaced for any light source (lamp, reflector, sun etc) base princple is still the same.


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11 years 8 months ago #247798 by John Landolfi
If this is meant to "clarify" the concept, I'd hate to see you try to seriously obfuscate it....Just kidding....:rofl:


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11 years 8 months ago #249406 by Crammer

John Landolfi wrote: Let me try:

Let's call

I=actual intensity of light
O=Flash output
d=distance to subject.

The law says that I=O/(d squared)

If the distance doubles to 2d, then you have to divide by (2d)squared. So

I=O/(2d)squared=O/4(d squared), So the intensity falls by a factor of 4.

If the distance trebles to 3d, then you divide by (3d)squared=9(d squared), so the intensity falls by a factor of 9.
And so on.

:cheers:



:goodpost: FYI to the original poster there is a video in the tips section here with Mark Wallace that covers this too.


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11 years 8 months ago #249527 by John Landolfi
BTW, this law also models the propagation of any form of energy, not just light


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