The proper exposure time for a photographic print is `20 s` at a distance of `0.6 m` from a `40` candles power lamb.How long will you expose the same print at a distance of `1.2 m` from a `20` candles power lamb?

Explanation

In case of camera, for proper exposure `I_(1)D_(1)^(2)t_(1) = I_(2)D_(2)^(2)t_(2)`
As here `D` is constant and `I = (L//r^(2)), (L_(1))/(r_(1)^(2)) xx t_(1)= (L_(2))/(r_(2)^(2))xx t_(2)` So `(40)/((0.6)^(2))xx20 = (20)/((1.2)^(2))t implies t=160"s"`