A source is at 4m height above the centre of a circular table of a circular table of radius 3m. The ratio of illuminance at $O$ and $P$ will be [image:9_image27.emf]

Question

Dr. Abubacker · Accepted Answer

(b) $E_{0} = \frac{I}{r^{2}} = \frac{I}{(4)^{2}} = \frac{I}{16}$
$E_{p} = \frac{I\cos	heta}{r^{'2}} = \frac{I 	imes (415)}{(5)^{2}}$
$= \frac{4I}{125}$
$	herefore\frac{E_{0}}{E_{p}} = \frac{I}{16} 	imes \frac{125}{4I} = \frac{125}{64}$

Q. A source is at 4m height above the centre of a circular table of a circular table of radius 3m. The ratio of illuminance at $O$ and $P$ will be

Explanation

Reviewed by Dr. Abubacker

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