An object is photographed when placed at a distance of 2m from a camera. If the same object be placed at a distance of 4m from the camera, how will the time of exposure be affected?

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Solution

$H E R E D I S T A N C E I S I N C R E A S E D T W O T I M E S F O R O B T A I N I N G S A M E Q U A L I T Y O F O B J E C T H E N C E T O T A L A M O U N T O F L I G H T i n t e n s i t y F A L L I N G P E R U N I T t i m e M U S T B E S A M E \frac{I}{{R_{1}}^{2}} \times T_{1} = \frac{I}{{R_{2}}^{2}} \times T_{2} (H E R E I = I N T E N S I T Y O F L I G H T, R_{1} = 2 C M A N D R_{2} = 4 C M) \frac{T_{1}}{2^{2}} = \frac{T_{2}}{4^{2}} \frac{T_{1}}{4} = \frac{T_{2}}{16} 4 T_{1} = T_{2} H E N C E T I M E O F E X P O S U R E I N C R E A S E S 4 T I M E S A S O F P R E V I O U S .$

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