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Question

(C) The energy flux $I$ is given as $25\,W/cm^2.$ The total power $P$ incident on the surface is $P = I 	imes A = 25\,W/cm^2 	imes 25\,cm^2 = 625\,W

Vedclass · Accepted Answer

$5.0	imes 10^{-3}\,Ns$

Vedclass · Answer

(C) The energy flux $I$ is given as $25\,W/cm^2.$ The total power $P$ incident on the surface is $P = I 	imes A = 25\,W/cm^2 	imes 25\,cm^2 = 625\,W.$For a completely absorbing surface,the radiation pressure exerts a force $F = \frac{P}{c},$ where $c$ is the speed of light $(3 	imes 10^8\,m/s)$.$F = \frac{625}{3 	imes 10^8}\,N.$The momentum $p$ transferred to the surface in time $t$ is given by $p = F 	imes t.$Given time $t = 40\,min = 40 	imes 60\,s = 2400\,s.$$p = \left( \frac{625}{3 	imes 10^8} ight) 	imes 2400 = \frac{625 	imes 2400}{3 	imes 10^8} = \frac{1500000}{3 	imes 10^8} = 500000 	imes 10^{-8} = 5.0 	imes 10^{-3}\,Ns.$

Light is incident normally on a completely absorbing surface with an energy flux of $25\,W\,cm^{-2}.$ If the surface has an area of $25\,cm^2,$ the maximum momentum transferred to the surface in $40\,min$ time duration will be

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