Question

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 illuminance's at the centre to that at the edge of the table is?

• A. $1+\left(\frac{r^2}{h^2}\right)$
• B. $1+\left(\frac{h^2}{r^2}\right)$
• C. ${\{1+\frac{r^2}{h^2}\}}^{3/2}$
• D. ${\{1+\frac{h^2}{r^2}\}}^{3/2}$

Answer 1

Answer: (C)

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

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