Question

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
(a) 3 s
(b) 12 s
(c) 24 s
(d) 48 s

Answer 1

Correct option (b)

Here,

$$\begin{gathered} d_1=5cm=0.05\mathrm{m} \\ {d}_2=10cm=0.1\mathrm{m} \\ {t}_1=3s \\ {t}_2=? \\ \mathrm{Let}the\mathrm{actual}\mathrm{incident}\mathrm{illuminance}be{E}_o \\ \mathrm{Let}the\mathrm{iluminance}at3\mathrm{cm}\mathrm{distance}be{E}_{d1} \\ \mathrm{Let}the\mathrm{iluminance}at10\mathrm{cm}\mathrm{distance}be{E}_{d2} \\ \cos \theta =1 \\ {E}_{d1}=\frac{E_o}{{d_1}^2} \\ \mathrm{Now}, \\ {\mathrm{t}}_1\mathrm{\alpha }\frac{1}{{\mathrm{E}}_{\mathrm{d}1}} \\ \Rightarrow {\mathrm{t}}_1=\frac{\mathrm{k}}{{\mathrm{E}}_{\mathrm{d}1}} \\ \Rightarrow {\mathrm{t}}_1=\frac{\mathrm{k}{5}^2}{{\mathrm{E}}_{\mathrm{o}}} \\ \Rightarrow \frac{\mathrm{k}}{{\mathrm{E}}_{\mathrm{o}}}=\frac{3}{25} \\ S\mathrm{imilarly}, \\ \Rightarrow {\mathrm{t}}_2=\frac{\mathrm{k}{10}^2}{{\mathrm{E}}_{\mathrm{o}}} \\ \Rightarrow {\mathrm{t}}_2=\frac{3}{25}\times {10}^2=12\mathrm{s} \end{gathered}$$