The illuminance of a small area changes from 900 lumen m^ -2 to 400...

The illuminance $E$ produced by a point source at a distance $r$ is given by $E = \dfrac{I}{r^2}$, where $I$ is the luminous intensity of the source.

Let the initial distance be $r$. The initial illuminance is $E_1 = 900 \text{ lumen m}^{-2}$.

Since the illuminance decreases to $E_2 = 400 \text{ lumen m}^{-2}$, the distance must have increased. The new distance is $r + 10 \text{ cm}$.

Using the formula for illuminance:

$E_1 = \dfrac{I}{r^2} = 900$

$E_2 = \dfrac{I}{(r + 10)^2} = 400$

Dividing the two equations:

$\dfrac{E_1}{E_2} = \dfrac{(r + 10)^2}{r^2} = \dfrac{900}{400}$

\left(\dfrac{r + 10}{r}\right)^2 = \dfrac{9}{4}$<br><br>Taking the square root of both sides:<br><br>$\dfrac{r + 10}{r} = \dfrac{3}{2}$<br><br>$2(r + 10) = 3r$<br><br>$2r + 20 = 3r$<br><br>$r = 20 \text{ cm}