# The Inverse Square Law of Illuminance (Laws of Illumination)

Power SystemsUtilisation of Electrical PowerUtilization of Electrical EnergyElectrical Engineering

The light received by any surface depends upon the distance of that surface from the source of light. The inverse square law gives the relation between the distance of the surface and the illumination of the surface.

## Statement of Inverse Square Law

The inverse square law states that "the illumination of a surface is inversely proportional to the square of distance of the surface from a point source of light."

### Explanation

Let us consider a point source of light 'S' having luminous intensity 'I' lumens per steradian. If two surfaces having areas A1 and A2 are placed at distances 'r' and 'R' meters away respectively from the point source. Let these surfaces be enclosed in the same solid angle 'ω'. Then, the total luminous flux received by the surface is given by,

$$\mathrm{Luminous\: flux,\phi \: =\:\mathit{I}\times \omega }$$

Hence, the luminous flux received by the area A1 is,

$$\mathrm{\phi_{1} \: =\:\mathit{I}\times \omega \: \: \: \cdot \cdot \cdot \left ( 1 \right )}$$

$$\mathrm{\because Solid \: angle,\omega \: =\:\frac{Area}{Distance^{2}} }$$

Therefore, solid angle for area A1 is given by,

$$\mathrm{\omega \: =\:\frac{\mathit{A}_{1}}{\mathit{r}^{2}}\: \: \cdot \cdot \cdot \left ( 2 \right ) }$$

Thus, the total luminous flux on surface area A1 is,

$$\mathrm{\phi _{1}\: =\:\frac{\mathit{I\times A_{\mathrm{1}}}}{\mathit{r^{\mathrm{2}}}}\: \: \: \cdot \cdot \cdot \left ( 3 \right )}$$

$$\mathrm{\because Illumination,\mathit{E}\: =\:\frac{Luminous \: flux\left ( \phi \right )}{Area\, \left ( \mathit{A} \right )} }$$

Thus, the illumination for surface A1 is

$$\mathrm{\mathit{E} _{1}\: =\:\frac{\phi _{1}}{\mathit{A}_{1}}\: =\:\frac{\mathit{I\times A_{\mathrm{1}}}}{\mathit{r^{\mathrm{2}}}}\times \frac{1}{\mathit{A}_{1}}\: =\:\mathit{\frac{I}{r^{\mathrm{2}}}}\: \: \: \cdot \cdot \cdot \left ( 4 \right )}$$

Similarly, the total luminous flux on the surface A2, that is placed 'R' meters away from the source of light and enclosed in the same solid angle 'ω' is given by,

$$\mathrm{\phi _{2}\: =\:\mathit{I}\times \omega \: =\:\mathit{I\times \frac{A_{\mathrm{2}}}{\left ( R \right )^{\mathrm{2}}}} }$$

$$\mathrm{\therefore \phi _{2}\: =\:\mathit{ \frac{I\times A_{\mathrm{2}}}{R^{\mathrm{2}}}}}$$

Thus, the illumination of the surface A2 is given by,

$$\mathrm{\mathit{E} _{2}\: =\:\frac{\phi _{2}}{\mathit{A}_{2}}\: =\:\frac{\mathit{I\times A_{\mathrm{2}}}}{\mathit{R^{\mathrm{2}}}}\times \frac{1}{\mathit{A}_{2}}}$$

$$\mathrm{\therefore \mathit{E} _{2}\: =\:\frac{\mathit{I}}{\mathit{R}^{\mathrm{2}}}\: \cdot \cdot \cdot \left ( 5 \right )}$$

Hence, from equations (4) & (5), we have,

$$\mathrm{\mathit{E}_{1}:\mathit{E}_{2}::\frac{\mathit{I}}{\mathit{r}^{2}}:\frac{\mathit{I}}{\mathit{R}^{2}}}$$

This expression is known as the inverse square law of illumination. From this expression, it can be seen that the illumination varies inversely as square of the distance between the source and the surface. This relation can be applied to all light sources.