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(D) The intensity of light $I$ falling on the film is proportional to the area of the aperture,which is proportional to the square of the diameter $d^

Vedclass · Accepted Answer

$4$ fold

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(D) The intensity of light $I$ falling on the film is proportional to the area of the aperture,which is proportional to the square of the diameter $d^2$.Exposure time $t$ is inversely proportional to the intensity of light $I$ for a given amount of light energy required for exposure.Therefore,$t \propto \frac{1}{I} \propto \frac{1}{d^2}$.Let $t_1$ be the initial time with diameter $d$ and $t_2$ be the new time with diameter $d/2$.$\frac{t_2}{t_1} = \left( \frac{d}{d/2} \right)^2 = (2)^2 = 4$.Thus,the exposure time must be increased by $4$ fold.

$A$ camera objective has an aperture diameter $d.$ If the aperture is reduced to diameter $d/2,$ the exposure time under identical conditions of light should be made

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