A lamp rated at 100 , cd hangs over the middl | vedclass

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Question

(A) Let $I_0$ be the luminous intensity of the lamp and $h$ be the height above the table. The illumination at the center is $E_c = \frac{I_0}{h^2}$.I

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(A) Let $I_0$ be the luminous intensity of the lamp and $h$ be the height above the table. The illumination at the center is $E_c = \frac{I_0}{h^2}$.In the first case,$I_1 = 100 \, cd$ and $h_1 = 2 \, m$. So,$E_{c1} = \frac{100}{2^2} = 25 \, lx$.In the second case,$I_2 = 25 \, cd$ and $h_2$ is the new height. Since $E_{c2} = E_{c1}$,we have $\frac{25}{h_2^2} = 25$,which gives $h_2 = 1 \, m$.The radius of the table is $r = 1.5 \, m$.The illumination at the edge is given by $E_e = \frac{I_0 \cos 	heta}{d^2} = \frac{I_0 h}{d^3} = \frac{I_0 h}{(h^2 + r^2)^{3/2}}$.For the first case,$E_{e1} = \frac{100 	imes 2}{(2^2 + 1.5^2)^{3/2}} = \frac{200}{(4 + 2.25)^{3/2}} = \frac{200}{(6.25)^{3/2}} = \frac{200}{15.625} = 12.8 \, lx$.For the second case,$E_{e2} = \frac{25 	imes 1}{(1^2 + 1.5^2)^{3/2}} = \frac{25}{(1 + 2.25)^{3/2}} = \frac{25}{(3.25)^{3/2}} \approx \frac{25}{5.858} \approx 4.267 \, lx$.Given $E_{e2} = X \cdot E_{e1}$,we have $X = \frac{E_{e2}}{E_{e1}} = \frac{25}{(3.25)^{3/2}} 	imes \frac{(6.25)^{3/2}}{200} = \frac{25}{200} 	imes \left(\frac{6.25}{3.25}ight)^{3/2} = \frac{1}{8} 	imes \left(\frac{25}{13}ight)^{3/2} = \frac{1}{8} 	imes \frac{125}{13\sqrt{13}} \approx 0.333 \approx \frac{1}{3}$.

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