How does distance affect light intensity according to the inverse square law?

[joemama69]

Homework Statement

I have a general question about light and Lumens and the inverse square law?

So Intensity = Initial Lumens/Lenth^2

So @ a Length of 1, there is no loss of light intensity. What unit is this 1.
1 meter? 1 foot?

Homework Equations

The Attempt at a Solution

 

[CompuChip]

This is why you need to specify the units when you are doing physics.

You can plug in the length in any unit that you like, but you will get the intensity in some "weird" (i.e. non-standard) unit. The SI-units - which people will usually assume are meant if you don't specify them - are:
Intensity [lux] = Initial lumens [lm] / (Length [m])²

Since a lumen is defined as the illuminance of a light source of 1 cd on a surface perpendicular to the source at a distance of 1 m, your conclusion that there is "no loss of intensity" is correct by definition of the units. Although I prefer to put it as "at a distance of 1 m, the numerical values for intensity and illuminance co-incide (in the units chosen)" - after all they are different quantities so you can't directly compare them - hence you can't say there is no "loss".

 

[joemama69]

Ok great, exactly the info I needed. Much appreciated.