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Light with an energy flux of 18 W//cm^(2) falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm^(2), the total momentum delivered (for complete absorption) during 30 minutes is: (a) 36xx10^(-5) kg m/s (b) 36xx10^(-4) kg m/s (c) 108xx10^(4) kg m/s (d) 1.08xx10^(7) kg m/s |
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Answer» <P>`36xx10^(-5)` kg m/s `therefore E=IAt` `=(20W)/(cm^(2))xx 30 cm^(2)xx30xx60 sec` `=108xx10^(4)` J Final momentum `P_(f)=(U)/(C )=(108xx10^(+4))/(3xx10^(8))` `therefore P_(f)=36xx10^(-4)NS` INITIAL momentum `P_(i)=0` `therefore` Momentum imparted to wall, `Delta P =P_(f)-P_(i)` `=36xx10^(-4)-0` `=36xx10^(-4)` kg m/s. |
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