Inverse Square Law concerning Light

[MangoOverlord]

[a]1. Homework Statement [/b]
How does the distance from a point source of light affect the visibility of that light from the said source?

Homework Equations

This is a bit tricky without sub points, but here goes:

I₁ x d₁²=I₂ x d₂²
where:
I=intensity d²=distance per sq. unit

[c]3. The Attempt at a Solution [/b]

By using the ISL formula above, I plan to find the intensity of light at predefined intervals (i.e. every 10 cm) by substituting variables in the equation. After which, I will use a voltmeter and CdS cell to verify my calculations. I'm pretty sure I'm doing everything right, but some suggestions would be nice.

 

[mgb_phys]

You might want to be careful defining "visibility"
What you are measuring is how the power received by a fixed area (the lightmeter) = the irradiance of the light varies with distance.

 

[MangoOverlord]

did i say visibility? I meant intensity. Ty.

 

[mgb_phys]

I meant intensity. Ty.

intensity is the power of the source
the correct term is irradiance = power/area