Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# 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. - Physics

Sum

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.

Concept: Light Process and Photometry
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 22 Photometry
Exercise | Q 10 | Page 455
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