Question

A photographic plate is placed directly, in front of a small diffused source in the sharp of a circular disc. It takes twelve seconds 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

Hint:We know the intensity of the light is needed for a good photograph to capture. Therefore, the angle at which the photographic plate is tilted is given, but most importantly, the time of the exposure of the light to the photographic plate determines the exposure.

Formula used: The formula of the radiation that pass through is given by,
$ \Rightarrow I = {I_0} \times \cos \theta \times t$
Where the incident radiation is ${I_o}$, the angle at which it strikes is equal to $\theta $, the resultant intensity is $I$ and time of exposure is t.

Complete step by step answer:
The source, providing the light must have some light intensity.
Let the intensity of incident radiation of the light be I not (${I_o}$).
We know that the time interval for which the light has been exposed is 12 sec.
Therefore, incident radiation is passing for 12 seconds.
Thus, the exposure of the light for twelve seconds can be written as $12{I_o}$
Now, if the source of light is tilted at an angle of sixty degrees $60^\circ $, the photographic plate is receiving light from the angle of $60^\circ $.
Hence, the radiation passing through is given by the expression:
$ \Rightarrow I = {I_0}\operatorname{Cos} \theta \times t$
Putting the value in the above expression
$ \Rightarrow 12{I_o} = {I_0}\operatorname{Cos} {60^0} \times t$
$ \Rightarrow {I_0} \times \dfrac{1}{2} \times t = 12{I_o}$
$ \Rightarrow t = \dfrac{{2 \times 12{I_o}}}{{{I_0}}}$
$ \Rightarrow t = 24s.$

Thus, time needed for the exposure will be 24 seconds.

Hence the correct answer is the option C.

Note:
-Photographic plates were earlier used to capture photos or we can say the light intensity gets imprinted on the film card or film reel fixed in the camera.
-If the photo is clicked in a dark place then the photo will have dark or black colour when printed.