A small lamp is hung at a height of 8 text{ f | vedclass

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(C) Let $L$ be the luminous intensity of the lamp. The height of the lamp above the table is $h = 8 	ext{ feet}$. The radius of the table is $r = \fr

Vedclass · Accepted Answer

$2\sqrt{2}:1$

Vedclass · Answer

(C) Let $L$ be the luminous intensity of the lamp. The height of the lamp above the table is $h = 8 	ext{ feet}$. The radius of the table is $r = \frac{16}{2} = 8 	ext{ feet}$.The illuminance at the centre $A$ is given by $I_A = \frac{L}{h^2}$.The illuminance at a point $B$ on the circumference is given by $I_B = \frac{L}{d^2} \cos 	heta$,where $d = \sqrt{h^2 + r^2}$ is the distance from the lamp to point $B$,and $\cos 	heta = \frac{h}{d}$.Substituting these values,$I_B = \frac{L}{h^2 + r^2} \cdot \frac{h}{\sqrt{h^2 + r^2}} = \frac{Lh}{(h^2 + r^2)^{3/2}}$.The ratio of intensities is $\frac{I_A}{I_B} = \frac{L}{h^2} \cdot \frac{(h^2 + r^2)^{3/2}}{Lh} = \frac{(h^2 + r^2)^{3/2}}{h^3} = \left(1 + \frac{r^2}{h^2}ight)^{3/2}$.Substituting $h = 8$ and $r = 8$,we get $\frac{I_A}{I_B} = \left(1 + \frac{8^2}{8^2}ight)^{3/2} = (1 + 1)^{3/2} = 2^{3/2} = 2\sqrt{2}$.Thus,the ratio is $2\sqrt{2}:1$.

$A$ small lamp is hung at a height of $8 \text{ feet}$ above the centre of a round table of diameter $16 \text{ feet}$. The ratio of intensities of illumination at the centre and at points on the circumference of the table will be

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