To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.
Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is
[tex]I = \frac{P}{A}[/tex]
The area of a sphere is given by
[tex]A = 4\pi r^2[/tex]
So replacing we have to
[tex]I = \frac{P}{4\pi r^2}[/tex]
Since the question tells us to find the proportion when
[tex]r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}[/tex]
So considering the two intensities we have to
[tex]I_1 = \frac{P_1}{4\pi r_1^2}[/tex]
[tex]I_2 = \frac{P_2}{4\pi r_2^2}[/tex]
The ratio between the two intensities would be
[tex]\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}[/tex]
The power does not change therefore it remains constant, which allows summarizing the expression to
[tex]\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2[/tex]
Re-arrange to find [tex]I_2[/tex]
[tex]I_2 = I_1 (\frac{r_1}{r_2})^2[/tex]
[tex]I_2 = 100*(\frac{1}{5})^2[/tex]
[tex]I_2 = 4W/m^2[/tex]
Therefore the intensity at five times this distance from the source is [tex] 4W/m^2[/tex]
