Question

the intensity, I, of light varies inversely as the square of the distance, D2 , from the light source. If I is 150 units when D is 6 meters, find a distance when the intensity is 24

Answer 1

`\bf \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{\underline{I} varies inversely with }\underline{D^2}}{I = \cfrac{k}{D^2}} \qquad \qquad \textit{we also know that } \begin{cases} I=150\\ D=6 \end{cases}`

`\bf 150=\cfrac{k}{6^2}\implies 150=\cfrac{k}{36}\implies 5400=k~\hfill \boxed{I=\cfrac{5400}{D^2}} \\\\\\ \textit{when I = 24, what is \underline{D}?}\qquad \qquad 24=\cfrac{5400}{D^2}\implies D^2=\cfrac{5400}{24} \\\\\\ D^2=225\implies D=√(225)\implies D=15`