A photographic plate placed a distance of 5 cm from a weak point source is exposed for 3 s. If the plate is kept at a distance of 10 cm from the source, the time needed for the same exposure is _____________ .

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KollegeApply · Accepted Answer

12s Here, d_1 = 5 cm = 0 . 05 m d_2 = 10 cm = 0 . 1 m t_1 = 3 s t_2 = ? Let the actual incident illuminance be E_o Let the iluminance at 3cm distance be E_{d1} Let the iluminance at 10cm distance be E_{d2} \cos$\theta$ = 1 E_{d1} = \frac{E_o}{{d_1}^2} Now, t_1 \alpha\frac{1}{E_{d1}} \Rightarrow t_1 = \frac{k}{E_{d1}} \Rightarrow t_1 = k 5^2/E_o \Rightarrow k/E_o = $\frac{3}{25}$ Similarly, \Rightarrow t_2 = \frac{k {10}^2}{E_o} \Rightarrow t_2 = $\frac{3}{25}$ $\times$ {10}^2 = 12 s

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