How Does Distance Affect Light Intensity According to the Inverse Square Law?

[Nul]

Homework Statement

A source of light is used to illuminate a screen. 1 m from the source, the intensity of the light falling on the screen is 12 000 lux. Calculate the intensity on the screen at distances of:
(a) 2 m
(b) 3 m
(c) 4 m

Homework Equations

I 1/I 2 = d^2 2/d^2 1
or something

The Attempt at a Solution

I don't understand at all. I'm hoping you guys would be able to explain what I am meant to do here because the teacher fails at teaching.
Just give me the equation for (a) so that I have a better understanding of what the is going on which would hopefuly enable me to do (b) and (c).
Thank you.

 

[Fightfish]

I'm going to assume that the source is either a point source or a spherical, isotropic one.
The inverse square law applies for intensity, in that the intensity of the light at a point away from the source is inversely proportional to the square of the distance of that point from the source.
Mathematically,

I \varpropto \frac{1}{r^2}​

Thus, problems can be solved using this proportionality:

I_{1} = \frac{k}{{r_{1}}^2}

I_{2} = \frac{k}{{r_{2}}^2}

\frac{I_{1}}{I{_2}} = \frac{{r_{2}}^{2}}{{r_{1}}^{2}}
​

 

[Nul]

yes... i know the formula
i have it in my book
but i don't know HOW to use them
can you just show me where the numbers should go for (a) then I would probably understand
please

 

[Fightfish]

Using the final equation I posted, substitute the values:

I_{1} = 12 000, r_{1} = 1, r_{2} = 2​

Then it follows that

I_{2} = I_{1}\,\frac{{r_1}^2}{{r_2}^2} = 12 000\,\frac{1^2}{2^2} = 3 000 \,lux​

From this, the inverse square law becomes apparent: at double the distance, intensity decreases to a quarter, and so on.

 

[Nul]

ok thanks man
^^

 

[Nul]

so if i were doing (b) would the answer be 1333.333 lux

 

[Fightfish]

so if i were doing (b) would the answer be 1333.333 lux

Yes, but you need to round off appropriately.