By how much should the distance from the source be increased to reduce the surface illuminance to 30 lumens?

Question

This is my problem and I have no idea how to solve it:

The illuminance of a surface varies inversely with the square of its distance from the light source.

If the illuminance of a surface is 120 lumens per square meter when its distance from a certain light source is 6 meters, by how many meters should the distance of the surface from the source be increased to reduce its illuminance to 30 lumens per square meter?

Any help would be greatly appreciated.

Answer 1

For a given distance $d$, you can get the luminance with $$d^2=\frac{k}{l}$$ where $k$ is a constant. $l$ is your luminance, which varies inversely (hence the reciprocal of $l$) and with the square of the distance (hence $d^2$).

As such you have $$(d_1)^2=\frac{k}{l_1}$$ and $$(d_2)^2=\frac{k}{l_2}$$

Since these are proportional, you can combine these to make

$$\frac{(d_1)^2}{(d_2)^2}=\frac{kl_2}{kl_1}$$

$k$ can be eliminated to give

$$\frac{(d_1)^2}{(d_2)^2}=\frac{l_2}{l_1}$$

Plugging in your numbers, $$\frac{6^2}{(d_2)^2}=\frac{30}{120}$$ $$\frac{36}{(d_2)^2}=0.25$$ $$\frac{36}{0.25}=(d_2)^2$$ $$12=d_2$$

And you're looking for the difference, so $$d_2-d_1=12-6=6$$

Answer 2

Let I represent illuminance, d represent distance from light source. Then, "The illuminance of a surface varies inversely with the square of its distance from the light source" implies the following equation: $$I=c/d^2$$ where $c$ is a constant. So, if I=120 when d=6, then $$120=c/(6^2)\implies c=120*36=4320$$ Now we know the constant $c$ so we can plug in 30 for $I$ and see what $d$ is: $$I=c/d^2\implies 30=4320/d^2\implies d^2=144\implies d=\pm12$$ But since distance cannot be negative, we use the positive root, namely $d=12$.

Answer 3

We want the illumination to decrease by a factor $\frac{30}{120}=\frac{1}{4}$.

So we want the distance to increase by a factor $\frac{1}{\sqrt{1/4}}=2$.

The distance should be doubled, so it should be increased by $6$ metres.