Light with an energy flux of 18 ; W/cm^2 fall | vedclass

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Question

(D) The total energy $U$ falling on the surface is given by the product of energy flux,area,and time.$U = (18 \; W/cm^2) 	imes (20 \; cm^2) 	imes (3

Vedclass · Accepted Answer

$1.2 	imes 10^{-6} \; N$

Vedclass · Answer

(D) The total energy $U$ falling on the surface is given by the product of energy flux,area,and time.$U = (18 \; W/cm^2) 	imes (20 \; cm^2) 	imes (30 	imes 60 \; s)$$U = 18 	imes 20 	imes 1800 \; J = 6.48 	imes 10^5 \; J$For a non-reflecting (completely absorbing) surface,the momentum $p$ delivered by the light is given by $p = U/c$,where $c = 3 	imes 10^8 \; m/s$ is the speed of light.$p = \frac{6.48 	imes 10^5 \; J}{3 	imes 10^8 \; m/s} = 2.16 	imes 10^{-3} \; kg \cdot m/s$The average force $F$ exerted on the surface is the rate of change of momentum:$F = \frac{p}{t} = \frac{2.16 	imes 10^{-3} \; kg \cdot m/s}{1800 \; s}$$F = 1.2 	imes 10^{-6} \; N$

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 $20 \; cm^2$,find the average force exerted on the surface during a $30$ minute time span.

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