Question

The time of exposure is 4 s to print a photograph from a negative when the source is 1 m away. If the source is placed 0.5 m away the time of exposure will be
A. 4s
B. 2s
C. 1s
D. 0.5s

Answer 1

Hint: The time of exposure depends on how much radiation from the source hits the negative. And the amount of radiation given out per second over the complete surface area of the sphere will be the same. So, we will find out how the radiation amount will increase if the radius is decreased.

Complete step-by-step solution:
First, let us assume that the rate at which the radiation is emitted from the source remains the same. And that the radiation from the source is emitted uniformly over the whole sphere. The radiation gets collected at the surface of the sphere. The negative will be kept at the surface of the sphere and we will assume that all the area of the negative is at an equal distance from the source. So, when the radius of the sphere is halved. The surface area of the sphere will become one-fourth. But the amount of radiation emitted will be the same. Let A be the area of the photograph then the ratio of the surface covered in two different cases by the photograph will be
Ratio of area covered: $\dfrac{A}{4\pi {{r}^{2}}}$
First case: $\dfrac{A}{4\pi {{\left( 1 \right)}^{2}}}=\dfrac{A}{4\pi }$
Second case: $\dfrac{A}{4\pi {{\left( 0.5 \right)}^{2}}}=\dfrac{A}{\pi }$
Hence, the radiation collected by the photograph will become 4-times. So, the time taken to collect the same amount of radiation will also become one-fourth. Initially the time of exposure was 4 seconds, now it will become 1 seconds. Hence, the correct option is C, i.e. 1s.

Note: There must be some assumptions that must be taken to solve the question correctly. First is that we must assume that the radiation is emitted uniformly over the entire area. Then we must assume that the negative is small enough that all points on the negative are approximately the same distance from the source.