Question

A student is studying a book placed near the edge of a circular table of radius R. A point source of light is suspended directly above the centre of the table. What should be the height of the source above the table so as to produce maximum illuminance at the position of the book?

Answer 1

The height of the source above the table so as to produce maximum illuminance at the position of the book is given by $\frac{R}{\sqrt{2}}$

Explanation:

Step 1:

Let the source height be h, and let the usual direction of the luminous intensity be $I_0$.

Therefore, illuminance is granted on book (E) by,

$E=\frac{l_{0} \cos \theta}{r^{2}}$

From this figure we get

$\cos \theta=\frac{h}{r}$

Step 2:

When we replace the cosθ value, we get

$E=\frac{l_{0} h}{r^{3}}$

$\text { But } \mathrm{r}=\sqrt{R^{2}+h^{2}}$

$E=\frac{l_{0} h}{\left(r^{2}+h^{2}\right)^{3 / 2}}$

Step 3:

For maximum illuminance,     $\frac{d E}{d H}=0$  

$\frac{d E}{d H}=\frac{l_{0}\left[\left(r^{2}+h^{2}\right)^{3} / 2-\frac{3}{2} h \times\left(r^{2}+h^{2}\right)^{1 / 2} \times 2 h\right]}{\left(R^{2}+h^{2}\right)^{3}}=0$

$\left(r^{2}+h^{2}\right)^{1 / 2}\left[R^{2}+h^{2}-3 h^{2}\right]=0$

$R^{2}-2 h^{2}=0$

$h=\frac{R}{\sqrt{2}}$

Therefore the height of the source is $\frac{R}{\sqrt{2}}$