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(B) The intensity of light $I$ follows the inverse square law with respect to distance $d$, given by $I \propto \frac{1}{d^2}$.To obtain the same phot

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(B) The intensity of light $I$ follows the inverse square law with respect to distance $d$, given by $I \propto \frac{1}{d^2}$.To obtain the same photographic effect (exposure), the total energy received by the film must be constant, where $	ext{Energy} = 	ext{Intensity} 	imes 	ext{Time}$.Thus, $I_1 	imes t_1 = I_2 	imes t_2$.Substituting the relation $I \propto \frac{1}{d^2}$, we get $\frac{t_1}{d_1^2} = \frac{t_2}{d_2^2}$.Given $d_1 = 20 \,cm$, $t_1 = 2 \,s$, and $d_2 = 40 \,cm$.Substituting these values: $\frac{2}{20^2} = \frac{t_2}{40^2}$.$t_2 = 2 	imes \left(\frac{40}{20}ight)^2 = 2 	imes (2)^2 = 2 	imes 4 = 8 \,s$.

$A$ good photographic print is obtained at a distance of $20 \,cm$ with an exposure time of $2 \,s$. What will be the exposure time required to get an equally good result at a distance of $40 \,cm$?

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