An isotropic point source of light is suspended $h$ metre vertically 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

• A

  $1+\left(\frac{r^2}{h^2}\right)$
• B

  $1+\left(\frac{h^2}{r^2}\right)$
• Correct answer
  C

  $\left\{1+\frac{r^2}{h^2}\right\}^{3 / 2}$
• D

  $\left\{1+\frac{h^2}{r^2}\right\}^{3 / 2}$

Solution

According to question, the situation is shown in the figure below $E_1=\frac{I}{(L O)^2}=\frac{I}{h^2}...(i)$ The illumination at the edge $A$ is given by $E_2=\frac{I \cos \theta}{(L A)^2}=\frac{I \cos \theta}{\left(h^2+r^2\right)}...(ii)$ From figure, $\cos \theta=\frac{h}{\sqrt{\left(h^2+r^2\right)}}$ $\therefore E_2=\frac{I h}{\left(h^2+r^2\right)^{3 / 2}}$ Dividing Eq. (i) by Eq. (ii), we get $\begin{aligned}\frac{E_1}{E_2} & =\frac{I / h^2}{I h /\left(h^2+r^2\right)^{3 / 2}}=\frac{\left(h^2+r^2\right)^{3 / 2}}{h^3} \\ & =\left(\frac{h^2+r^2}{h^2}\right)^{3 / 2}=\left(1+\frac{r^2}{h^2}\right)^{3 / 2}\end{aligned}$

Hence, the correct answer is option: 'C'.