Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1784 Upvotes
Verified Answer
The correct answer is:
$\left\{1+\frac{r^2}{h^2}\right\}^{3 / 2}$
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
$\begin{aligned}\frac{E_1}{E_2} & =\frac{I / h^2}{I h /\left(h^2+r^2\right)^{3 / 2}}=\frac{\left(h^2+r^2\right)^{3 / 2}}{h^3} \\
& =\left(\frac{h^2+r^2}{h^2}\right)^{3 / 2}=\left(1+\frac{r^2}{h^2}\right)^{3 / 2}\end{aligned}$
$E_1=\frac{I}{(L O)^2}=\frac{I}{h^2}...(i)$
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
$\begin{aligned}\frac{E_1}{E_2} & =\frac{I / h^2}{I h /\left(h^2+r^2\right)^{3 / 2}}=\frac{\left(h^2+r^2\right)^{3 / 2}}{h^3} \\
& =\left(\frac{h^2+r^2}{h^2}\right)^{3 / 2}=\left(1+\frac{r^2}{h^2}\right)^{3 / 2}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.