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If the focal length of a lens of a camera is 5f and that of another is \[2.5\]f, what is the time of exposure for the second if for the first one is \[\dfrac{1}{{200}}s\]? (where f is the focal length/unit aperture)
A) \[\dfrac{1}{{200}}s\]
B) \[\dfrac{1}{{800}}s\]
C) \[\dfrac{1}{{6400}}s\]
D) \[\dfrac{1}{{3200}}s\]

Answer
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Hint:Aperture is defined as the opening of the lens in the camera through which the light ray is passing. The focal length of the camera also called as F-number provides information about how much amount of light will pass through the lens of the camera if aperture is opened to its maximum.
It means that the aperture of the camera is inversely proportional to its F-number.

Complete step by step answer:
Given
Step I: Time of exposure for first camera, \[{t_1} = \dfrac{1}{{200}}s\]
F-number for first camera, \[{f_1}\] = \[5f\]
F-number for second camera, \[{f_2}\] = \[2.5f\]


Step II: The time of exposure of a camera is directly proportional to its focal length. Using the relation of ratio of time of exposures
\[\dfrac{{{t_1}}}{{{t_2}}} = {(\dfrac{{{f_1}}}{{{f_2}}})^2}\]

Step III: Substituting values in the above relation,
\[\dfrac{1}{{200}}.\dfrac{1}{{{t_2}}} = {(\dfrac{{5f}}{{2.5f}})^2}\]
\[\dfrac{1}{{200}}.\dfrac{1}{{{t_2}}} = {(2)^2}\]
\[{t_2} = \dfrac{1}{{200 \times 4}}\]
\[{t_2} = \dfrac{1}{{800}}s\]

The time period for the exposure of the second camera is \[\dfrac{1}{{800}}s\].
Option B, is the correct answer.

Note: The aperture of a camera depends inversely on its f-number. If the aperture is high, then the amount of light entering the camera will be less. It will give a better shot for a group of people or in a landscape. The shot will be more focused. On the other hand, if the aperture of the camera has a low value, then the amount of light entering the lens is more. It will give a clear image with a blurry background. In this case the shot will be less focused. It can be concluded that aperture has an effect on the focus of the image but it does not affect the sharpness of the shot.