Question

A point light source are located at $P_1$ as shown in the figure in which all sides of the polygon are equal. The intensity of illumination at $P_2$ is $8\frac{W}{m^2}$ , what will be the intensity of illumination at $P_3.$

• A. $3\sqrt{3}\frac{W}{m^2}$
• B. $6\frac{W}{m^2}$
• C. $8\frac{W}{m^2}$
• D. $\sqrt{3}\frac{w}{m^2}$

Answer 1

Answer: (B)

$6\frac{W}{m^2}$

We know that

sum of all the internal angle of a hexagon$={720}^o$

So, $\angle P_{1}A{P}_2=\frac{720}{2}={120}^o$

Since it is regular hexagon, so all the angles will be equal.

Let the side of the helygon be $a$

$△A{P}_1{P}_2$ is an isosceles triangle

$\angle P_{1}A{P}_2+\angle A{P}_1{P}_2+\angle A{P}_2{P}_1=180$

${120}^o+\theta +\theta =180$

$2\theta =180-120$

$\theta =\frac{60}{2}={30}^o$

Now

$P_1{P}_2=2A{P}_1\cos \theta$

$=2a\cos {30}^o$

$=2a\left(\frac{\sqrt{3}}{2}\right)$

$\Rightarrow$ $P_1{P}_2=\sqrt{3}a$

Let intensity of source located at $P_1$

In $△P_1{P}_2{P}_3$ (right angled triangle)

${\left(P_1{P}_3\right)}^2={\left(P_1{P}_2\right)}^2+{\left(P_2{P}_3\right)}^2$

$\Rightarrow {\left(P_1{P}_3\right)}^2={\left(\sqrt{3}a\right)}^2+a^2=3a^2+a^2=4a^2\Rightarrow P_1{P}_3=\sqrt{4a^2}=2a$

Now intensity at $P_3=\frac{I_0}{4\pi {\left(P_1{P}_3\right)}^2}=\frac{96\pi a^2}{4\pi {\left(2a\right)}^2}=\frac{24a^2}{4a^2}=6\frac{W}{m^2}$