question_answer

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)            \[\frac{\sqrt{3}}{8}{{I}_{0}}\]

Correct Answer: A

Solution :

                   From the geometry of the figure  \[{{p}_{1}}{{p}_{2}}=2a\sin {{60}^{o}}\]                                                      so, \[{{I}_{{{P}_{2}}}}=\frac{L}{{{p}_{1}}p_{2}^{2}}\]                    \[=\frac{L}{{{(2a\sin {{60}^{o}})}^{2}}}\]\[=\frac{L}{3{{a}^{2}}}\]                    and \[{{I}_{{{P}_{3}}}}=\frac{L}{({{P}_{1}}P_{2}^{2}+{{a}^{2}})}\cos {{30}^{o}}\]                    =\[\frac{L}{[{{(2a\sin {{60}^{o}})}^{2}}+{{a}^{2}}]}\frac{\sqrt{3}}{2}\]\[=\frac{\sqrt{3}\,L}{8\,{{a}^{2}}}\]                    \[\Rightarrow {{I}_{{{P}_{3}}}}=\frac{3\sqrt{3}}{8}{{I}_{{{P}_{2}}}}=\frac{3\sqrt{3}}{8}{{I}_{0}}\]             All options are wrong.