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.
Solution
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 (EA) = 900 lumen/m2
Final illuminance (EB) = 400 lumen/m2
Illuminance on the initial position is given by,
`E_A=l costheta/x^2............(1)`
Illuminance at final position is given by
`E_B=(lcostheta)/(x+10)^2............(2)`
Equating luminous intensity from (1) and (2), we get
`l=(E_Ax^2)/costheta=(E_B(x+10)^2)/costheta`
`rArr900x^2=400(x+10)^2`
`rArrx/(x+10)=2/3`
⇒ 3x = 2x + 20
⇒ x = 20 cm
The distance between the source and the area at the initial phase was 20 cm.