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Show that the radiation exerted by an EM wave of intensity I on a surface kept in vacuum is (I)/(c ). |
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Answer» Solution :Pressure `=("Force")/("Area")=(F)/(A) therefore P=(F)/(A)` Rate of change of momentum is force, `therefore F=(dp)/(dt)` Now `E=mc^(2)` `therefore U=(mc)C "" [because E=U]` `therefore U=Pc "" [because mc = " P momentum"]` By TAKING differentiation both side w.r.t TIME, `(dU)/(dt)=c(dP)/(dt)` `therefore (dU)/(dt)xx(1)/(c )=F "" [because (dP)/(dt)=F]` Now `P=(F)/(A)=(dV)/(dt)xx(1)/(Ac)` `therefore P=(I)/(c )"" [because (dV)/(Adt)=" Intensity I"]` |
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