An isotropoic point source of light is suspeneded h metre vericlaly above the centre of circular table of radius r metre.Then the ratio of illumenances at the centre to that at the edge of the table is

Options

A. `1+(r^(2)/(h^(2)))`
B. `1+(h^(2)/(r^(2)))`
C. `{1+r^(2)/(h^(2))}^(3//2)`
D. `{1+r^(2)/(h^(2))}^(3//2)`

Answer

c

Explanation

According to question the situation is shown in the figure below

`E_(1)=(I)/((LO)^(2))=(Icostheta)/(h^(2)+r^(2))`
From figure `cos theta = (h)/(sqrt(h^(2)+R^(2))^(3)/(2)=(h^(2)+r^(2))/(h^(3))^(3)/(2)`
Dividing Eq (i) by Eq(ii) we get
`(E_(1))/(e_(2))=(I//h^(2))/(Ih//H^(2)+R^(2))^(3)/(2)=(1+(r^(2)/(H^(2)))^(3)/(2)`