A lamp is hanging at a height of 4,m above a | vedclass

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(C) The illuminance $E$ at a point directly below a source of luminous intensity $L$ at a distance $r$ is given by $E = \frac{L}{r^2}$.Initially,the h

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$78$

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(C) The illuminance $E$ at a point directly below a source of luminous intensity $L$ at a distance $r$ is given by $E = \frac{L}{r^2}$.Initially,the height $r_1 = 4\,m$,so $E_1 = \frac{L}{4^2} = \frac{L}{16}$.After lowering the lamp by $1\,m$,the new height $r_2 = 4\,m - 1\,m = 3\,m$,so $E_2 = \frac{L}{3^2} = \frac{L}{9}$.The percentage increase in illuminance is given by $\frac{E_2 - E_1}{E_1} 	imes 100$.Substituting the values: $\left( \frac{L/9 - L/16}{L/16} ight) 	imes 100 = \left( \frac{16}{9} - 1 ight) 	imes 100$.$= \left( \frac{16 - 9}{9} ight) 	imes 100 = \frac{7}{9} 	imes 100 \approx 77.77\% \approx 78\%$.

$A$ lamp is hanging at a height of $4\,m$ above a table. The lamp is lowered by $1\,m$. The percentage increase in illuminance will be.....$\%$

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