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)`

Options

A. `(3 sqrt(3))/(8) I_(0)`
B. `(I_(o))/(8)`
C. `(3)/(8) I_(o)`
D. `(sqrt(3))/(8) I_(o)`

Answer

A

Explanation

`P_(1) P_(2) = 2a sin 60^(@)`
Now,
Intensity at `P_(2)`,
`I_(P_(2)) = (L)/((P_(1) P_(2))^(2)) `
= `(L)/((2a sin 60^(@))^(2) ) = (L)/(3a^(2))`
Intensity at `P_(3)` ,
`I_(P_(3)) = (L)/((P_(1) P_(3))^(2) + a^(2) ) cos 30^(@)`
= `(L)/((2a sin 60^(@))^(2) + a^(2) ) . (sqrt(3))/(2) = (sqrt(3L))/(8a^(2))`
`rArr I_(P_(3)) = (3 sqrt(3))/(8) I_(P_(2)) = (3sqrt(3))/(8) I_(0)`.