Question

The illuminance of a small area changes from 900 lumen m⁻² to 400 lumen m⁻² when it is shifted along its normal by 10 cm. Assuming that it is illuminated by a point source placed on the normal, find the distance between the source and the area in the original position.

Answer 1

Let the luminous intensity of the source be l and the distance between the source and the area, in the initial position, be x.
Given,
Initial illuminance (E_A)​ = 900 lumen/m²
Final illuminance (E_B)​​ = 400 lumen/m²
Illuminance on the initial position is given by,
$E_{\mathrm{A}}=l\frac{\cos \theta }{x^2}$ ......(1)
Illuminance at final position is given by
$E_{\mathrm{B}}=\frac{l\cos \theta }{(x+10{)}^2}$.......(2)
Equating luminous intensity from $\left(1\right)$ and $\left(2\right)$, we get
$l=\frac{E_{\mathrm{A}}{x}^2}{\cos \theta }=\frac{E_B(x+10{)}^2}{\cos \theta }$

$$\begin{gathered} \Rightarrow 900x^2=400(x+10{)}^2 \\ \Rightarrow \frac{x}{x+10}=\frac{2}{3} \end{gathered}$$
⇒ 3x = 2x + 20
$\Rightarrow$ x = 20 cm
The distance between the source and the area at the initial phase was 20 cm.