Question

A photographic plate is placed directly in front of a small diffused source in the shape of a circular disc. It takes 12 s to get a good exposure. If the source is rotated by 60° about one of its diameter, the time needed to get the same exposure will be
(a) 6 s
(b) 12 s
(c) 24 s
(d) 48 s

Answer 1

Correct option (c)

$$\begin{gathered} \mathrm{Here}, \\ {t}_1=12\mathrm{s} \\ {\mathrm{\theta }}_1=0^0 \\ {\mathrm{\theta }}_2={60}^0 \\ {\mathrm{t}}_2=? \\ L\mathrm{et}the\mathrm{distance}ber. \\ L\mathrm{et}the\mathrm{incident}\mathrm{luminosity}be{E}_o. \\ \mathrm{We}\mathrm{have}, \\ {E}_{\theta 1}=\frac{E_o{\mathrm{cos\theta }}_1}{r^2} \\ {t}_1\alpha \frac{1}{E_{\theta 1}} \\ \Rightarrow t_1=\frac{r^{2}k}{E_o{\mathrm{cos\theta }}_1} \\ \Rightarrow 12=\frac{r^{2}k}{E_o\cos 0} \\ \Rightarrow \frac{r^{2}k}{E_o}=12 \\ S\mathrm{imilarly}, \\ {\mathrm{t}}_2=\frac{{\mathrm{r}}^2\mathrm{k}}{{\mathrm{E}}_{\mathrm{o}}{\mathrm{cos\theta }}_2}=\frac{12}{\cos \left({60}^0\right)} \\ \Rightarrow {\mathrm{t}}_2=12\times 2=24\mathrm{s} \end{gathered}$$