Q. A bulb of 100 $watt$ is hanging at a height of one meter above the centre of a circular table of diameter $4\ m$. If the intensity at a point on its rim is $I_{0}$, then the intensity at the centre of the table will be
A) $I_{0}$
B) $2\sqrt{5}I_{0}$
C) $2I_{0}$
D) $5\sqrt{5}I_{0}$
Explanation
(d) The illuminance at $B$
$I_{B} = \frac{L}{1^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
And illuminance at point $C$
$I_{C} = \frac{L\cos\theta}{\left( \sqrt{5} \right)^{2}} = \frac{L}{\left( \sqrt{5} \right)^{2}} \times \frac{1}{\sqrt{5}}$
$\Rightarrow I_{C} = \frac{L}{5\sqrt{5}}\ \ \ \ \ \ \ \ldots(iii)$
From equation (i) and (ii) $I_{B} = 5\sqrt{5}\ I_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.