A light source is located at P_1 as shown in | vedclass

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Question

(A) From the geometry of the regular hexagon with side length $a$,the distance $P_1P_2$ is the length of the diagonal connecting two vertices separate

Vedclass · Accepted Answer

$\frac{3\sqrt{3}}{8}I_0$

Vedclass · Answer

(A) From the geometry of the regular hexagon with side length $a$,the distance $P_1P_2$ is the length of the diagonal connecting two vertices separated by one vertex. Using the law of cosines or geometry,$P_1P_2 = 2a \sin(60^\circ) = a\sqrt{3}$.The intensity of illumination $I$ at a point is given by $I = \frac{L \cos 	heta}{r^2}$,where $L$ is the luminous intensity of the source,$r$ is the distance,and $	heta$ is the angle between the light ray and the normal to the surface.At $P_2$,the light ray is incident normally,so $	heta = 0^\circ$ and $\cos 0^\circ = 1$. Thus,$I_0 = \frac{L}{(a\sqrt{3})^2} = \frac{L}{3a^2}$.For point $P_3$,the distance $r = P_1P_3$. In a regular hexagon,the distance between $P_1$ and $P_3$ is $r = \sqrt{(a\sqrt{3})^2 + a^2} = \sqrt{3a^2 + a^2} = 2a$.The angle $	heta$ between the light ray $P_1P_3$ and the normal at $P_3$ is $30^\circ$. Thus,$I_{P_3} = \frac{L \cos 30^\circ}{r^2} = \frac{L (\sqrt{3}/2)}{(2a)^2} = \frac{L \sqrt{3}}{8a^2}$.Substituting $L = 3a^2 I_0$,we get $I_{P_3} = \frac{(3a^2 I_0) \sqrt{3}}{8a^2} = \frac{3\sqrt{3}}{8} I_0$.

$A$ 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 $P_3$?

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