A) 0.02 s
B) 0.04 s
C) 0.20 s
D) 0.40 s
Correct Answer: A
Solution :
The exposure time of camera lens is given by Time of exposure\[\propto \frac{1}{{{(Aperture)}^{2}}}\] Also,\[f-\text{number}=\frac{\text{Focal}\,\text{length}\,\text{(f)}}{\text{Aperture}\,\text{(A)}}\] or \[\text{Aperture}\,\text{(A)}=\frac{\text{Focal}\,\text{length}\,(f)}{f-number}\] Therefore, \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}}\] Given, \[{{T}_{1}}=\frac{1}{200},{{A}_{1}}=\frac{f}{2.8},{{A}_{2}}=\frac{f}{5.6}\] \[\therefore \] \[\frac{1/200}{{{T}_{2}}}={{\left( \frac{f/5.6}{f/2.8} \right)}^{2}}\] or \[\frac{1}{200\,{{T}_{2}}}={{\left( \frac{2.8}{5.6} \right)}^{2}}\] or \[{{T}_{2}}={{\left( \frac{5.6}{2.8} \right)}^{2}}\times \frac{1}{200}\] Note: Smaller the f-number larger will be the aperture and lesser will be the time of exposure and faster will be the camera. This is why movie cameras have very low \[f.\]numbers such as \[f/1.5.\]
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