A lamp is hanging along the axis of a circula | vedclass

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(D) Let the intensity of the lamp be $I_0$ and the height of the lamp above the center of the table be $h$.The illuminance at the center of the table

Vedclass · Accepted Answer

$\frac{r}{\sqrt{3}}$

Vedclass · Answer

(D) Let the intensity of the lamp be $I_0$ and the height of the lamp above the center of the table be $h$.The illuminance at the center of the table is given by $E_{center} = \frac{I_0}{h^2}$.The illuminance at the edge of the table (at distance $r$ from the center) is given by $E_{edge} = \frac{I_0 \cos 	heta}{d^2}$,where $d = \sqrt{r^2 + h^2}$ and $\cos 	heta = \frac{h}{d}$.Thus,$E_{edge} = \frac{I_0 h}{(r^2 + h^2)^{3/2}}$.Given that $E_{edge} = \frac{1}{8} E_{center}$,we have:$\frac{I_0 h}{(r^2 + h^2)^{3/2}} = \frac{1}{8} \frac{I_0}{h^2}$$\frac{h}{(r^2 + h^2)^{3/2}} = \frac{1}{8h^2}$$8h^3 = (r^2 + h^2)^{3/2}$Taking the cube root of both sides:$2h = (r^2 + h^2)^{1/2}$Squaring both sides:$4h^2 = r^2 + h^2$$3h^2 = r^2$$h = \frac{r}{\sqrt{3}}$.

$A$ lamp is hanging along the axis of a circular table of radius $r$. At what height should the lamp be placed above the table,so that the illuminance at the edge of the table is $\frac{1}{8}$ of that at its center?

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