NEET-XII-Physics
22: Photometry
- #10The 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.Ans : 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}_{\,\mathrm{\,A\,}}=l\frac{\,\mathrm{\,cos\,}\theta }{{x}^{2}}`` ......(1)
Illuminance at final position is given by
`` {E}_{\,\mathrm{\,B\,}}=\frac{l\,\mathrm{\,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}}{\,\mathrm{\,cos\,}\theta }=\frac{{E}_{B}(x+10{)}^{2}}{\,\mathrm{\,cos\,}\theta }``
`` ``
`` \Rightarrow 900{x}^{2}=400(x+10{)}^{2}``
`` \Rightarrow \frac{x}{x+10}=\frac{2}{3}``
⇒ 3x = 2x + 20
`` \Rightarrow `` x = 20 cm
The distance between the source and the area at the initial phase was 20 cm.
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