# The intensity of light received at a source varies inversely as the square of the distance from the source. A particular light has an intensity of 20 foot-candles at 15 feet. What is the lights intensity at 10 feet?

Jul 12, 2016

45 foot-candles.

#### Explanation:

$I \propto \frac{1}{d} ^ 2 \implies I = \frac{k}{d} ^ 2$ where k is a proportionality constant.

We can solve this problem in two ways, either solving for k and subbing back in or by using ratios to eliminate k. In many common inverse square dependences k can be quite a lot of constants and ratios often save on calculation time. We will use both here though.

$\textcolor{b l u e}{\text{Method 1}}$

${I}_{1} = \frac{k}{d} _ {1}^{2} \implies k = I {d}^{2}$

$k = 20 \cdot {15}^{2} = 4500 \text{ foot-candles} f {t}^{2}$

$\therefore {I}_{2} = \frac{k}{d} _ {2}^{2}$

${I}_{2} = \frac{4500}{{10}^{2}}$ = 45 foot-candles.

$\textcolor{b l u e}{\text{Method 2}}$

${I}_{1} = \frac{k}{d} _ {1}^{2}$

${I}_{2} = \frac{k}{d} _ {2}^{2}$

$\frac{{I}_{2}}{{I}_{1}} = \frac{k}{d} _ {2}^{2} \cdot {d}_{1}^{2} / k$

$\implies {I}_{2} = {I}_{1} \cdot {\left({d}_{1} / {d}_{2}\right)}^{2}$

${I}_{2} = 20 \cdot {\left(\frac{15}{10}\right)}^{2} = 45 \text{ foot-candles}$