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The proper expsoure time for a photographic print is`20s`at a distance of `0.6m`froma `40`candles power lamb.How long will you expose the same print at a distance of `1.2m`form a `20`candles power lamb? |
Answer» In the 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_(2)^(1))xxt_(1)=(L_(2))/(r_(2)^(2))xxt_(2)` `So (40)/(0.6^(2))xx20=(20)/((1.2)^(2))t i.e,t=160sec` |
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