Question

A photographic plate is placed directly in front of a small diffused source in the sharp 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 get the same exposure will be

• A. $6s$
• B. $12s$
• C. $24s$
• D. $48s$

Answer 1

Answer: (C)

$24s$

Let intensity of incident radiation $=I_0$
This incident radiation passes for 12s
This implies the exposure is $12I_0$
If the photographic plate is at an angle $\theta$
Radiation passing through is given by:
$I=I_{0}cos\theta$ Given that $\theta ={60}^{\circ }C$
$I=\frac{I_0}{2}$
For the same exposure:
$12I_0=\frac{I_0}{2}\times t$
i.e. we get $t=24s$
Option C is correct.