MCQ
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 diameters, the time needed to getthe same exposure will be:
  • A
    $6s$
  • B
    $12s$
  • $24s$
  • D
    $48s.$

Answer

Correct option: C.
$24s$
Here,
$\text{t}_1=12\text{s}$
$\theta_1=0^0$
$\theta_2=60^0$
$\text{t}_2=?$
Let the distance be $r.$
Let the incident luminosity be $E_o$.
We have,
$\text{E}_{\theta1}=\frac{\text{E}_\text{o}\cos\theta_1}{\text{r}^2}$
$\text{t}_1 \alpha\frac{1}{\text{E}\theta_1}$
$\Rightarrow\text{t}_1=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos\theta_1}$
$\Rightarrow12=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos0}$
$\Rightarrow \frac{\text{r}^2\text{k}}{\text{E}_\text{o}}=12$
Simiolarly,
$\text{t}_2=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos\theta_2}=\frac{12}{\cos(60^0)}$
$\Rightarrow\text{t}_2=12\times2=24\text{s}$

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