A photographic plate is placed directly in front of a small diffused source in the shape of a circular disc. It takes 12s to get a good exposure. If the source is rotated by $60^\circ$ about one of its diameter, the time needed to get the same exposure will be ___________ .

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KollegeApply · Accepted Answer

24s Here, t_1 = 12 s $\theta$_1 = 0^0 $\theta$_2 = {60}^0 t_2 = ? Let the distance be r. Let the incident luminosity be E_o. We have, E_$\theta$1 = \frac{E_o {\cos$\theta$}_1}{r^2} t_1 \alpha 1/E_$\theta$1 \Rightarrow t_1 = \frac{r^2 k}{E_o {\cos$\theta$}_1} \Rightarrow 12 = r^2 k/E_o \cos0 \Rightarrow r^2 k/E_o = 12 Similarly, t_2 = \frac{r^2 k}{E_o {\cos$\theta$}_2} = \frac{12}{\cos( {60}^0 )} \Rightarrow t_2 = 12 $\times$ 2 = 24 s

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