Question

light intensity (i) is proportional (∝) to the inverse square of distance (d) of a subject from a light source. the relationship in intensities for subjects at different distances from the same source can be likewise seen as a ratio or proportional relationship.
i ∝ 1/d²
one could, therefore, determine the distance a subject was from a light source based on an intensity presented in the above fashion. for example, in the followings
0.015625 = 1/d²
one could manipulate the formula to solve:
0.015625d² = 1
d² = 1/0.015625 = 64
d = √64 = 8
determine the distance of a subject as follows:
0.04 = 1/d²

Explanation:

Step1: Cross - multiply the equation

Multiply both sides of $0.04=\frac{1}{d^{2}}$ by $d^{2}$ to get $0.04d^{2}=1$.

Step2: Solve for $d^{2}$

Divide both sides of $0.04d^{2}=1$ by $0.04$. So, $d^{2}=\frac{1}{0.04}=25$.

Step3: Solve for $d$

Take the square - root of both sides. Since $d$ represents a distance (a non - negative quantity), $d = \sqrt{25}=5$.

Answer:

$5$