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(D) The illumination $E$ on the photographic paper is proportional to the intensity of the source $L$ and inversely proportional to the square of the

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$20$

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(D) The illumination $E$ on the photographic paper is proportional to the intensity of the source $L$ and inversely proportional to the square of the distance $r$ from the source,i.e.,$E \propto \frac{L}{r^2}$.For the same quality print,the total exposure (illumination $	imes$ time) must be constant,so $E_1 t_1 = E_2 t_2$.This implies $\frac{L_1}{r_1^2} t_1 = \frac{L_2}{r_2^2} t_2$.Given: $L_1 = 60 \, cd$,$r_1 = 2 \, m$,$t_1 = 10 \, s$ and $L_2 = 120 \, cd$,$r_2 = 4 \, m$.Substituting the values: $\frac{60}{2^2} 	imes 10 = \frac{120}{4^2} 	imes t_2$.$\frac{60}{4} 	imes 10 = \frac{120}{16} 	imes t_2$.$150 = 7.5 	imes t_2$.$t_2 = \frac{150}{7.5} = 20 \, s$.

$A$ satisfactory photographic print is obtained when the exposure time is $10 \, s$ at a distance of $2 \, m$ from a $60 \, cd$ lamp. The time of exposure required for the same quality print at a distance of $4 \, m$ from a $120 \, cd$ lamp is $... \, s$.

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