The illumination, $I$ units, of a bulb varies inversely as the square of the distance, $d$ metres. It is given that the illumination is 8 units when the distance is 3 m.

1. Express the inverse-square variation using a constant of proportionality.

Since $I$ varies inversely as the square of $d$,
$I \propto \frac{1}{d^2}$
so
$I=\frac{k}{d^2}$
where $k$ is a constant.

2. Substitute the given illumination and distance to find the constant, then state the equation.

Given $I=8$ when $d=3$,
$8=\frac{k}{3^2}$
$8=\frac{k}{9}$
$k=72$
Hence,
$I=\frac{72}{d^2}.$

1. Express the new distance as a fraction of the original distance and find the corresponding change factor for illumination.

Let the original distance be $d$.

New distance $= \dfrac14 d$.

Since $I \propto \dfrac{1}{d^2}$,
$\frac{I_{\text{new}}}{I_{\text{old}}}=\frac{d^2}{\left(\frac14 d\right)^2}=\frac{d^2}{\frac{1}{16}d^2}=16.$

2. Convert the change in illumination into a percentage change.

New illumination is $16$ times the original illumination.

Percentage change
$=\frac{16I-I}{I}\times 100\%$
$=15\times 100\%=1500\%.$