Q. A light source is located at $P_{1}$ as shown in the figure. All sides of the polygon are equal. The intensity of illumination at $P_{2}$ is $I_{0}$. What will be the intensity of illumination at $P_{3}$
A) $\frac{3\sqrt{3}}{8}I_{0}$
B) $\frac{I_{0}}{8}$
C) $\frac{3}{8}I_{0}$
D) $\frac{\sqrt{3}}{8}I_{0}$
Explanation
(a) From the geometry of the figure
$P_{1}P_{2} = 2a\sin{60{^\circ}}$
So, $I_{P_{2}} = \frac{L}{\left( P_{2}P_{2} \right)^{2}}$
$= \frac{L}{\left( 2a\sin{60{^\circ}} \right)^{2}} = \frac{L}{3a^{2}}$
and $I_{P_{3}} = \frac{L}{(P_{1}P_{2}^{2} + a^{2})}\cos{30{^\circ}}$
$= \frac{L}{\lbrack\left( 2a\sin{60{^\circ}} \right)^{2} + a^{2}\rbrack}\frac{\sqrt{3}}{2} = \frac{\sqrt{3}L}{8\ a^{2}}$
$\Rightarrow I_{P_{3}} = \frac{3\sqrt{3}}{8}I_{P_{2}} = \frac{3\sqrt{3}}{8}I_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.