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(D) The intensity of the wave is $I = 6 \ W/m^2$ and the area of the mirror is $A = 40 \ cm^2 = 40 	imes 10^{-4} \ m^2$.For a perfectly reflecting su

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$1.6 	imes 10^{-10} \ kg \cdot m/s$

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(D) The intensity of the wave is $I = 6 \ W/m^2$ and the area of the mirror is $A = 40 \ cm^2 = 40 	imes 10^{-4} \ m^2$.For a perfectly reflecting surface (mirror) held perpendicular to the wave,the radiation pressure exerts a force,and the momentum transferred per second is equal to the force exerted.The momentum transferred per second is given by $p = \frac{2U}{c}$,where $U$ is the energy incident per second $(U = I 	imes A)$ and $c$ is the speed of light $(3 	imes 10^8 \ m/s)$.Substituting the values:$p = \frac{2 	imes I 	imes A}{c} = \frac{2 	imes 6 	imes 40 	imes 10^{-4}}{3 	imes 10^8}$$p = \frac{480 	imes 10^{-4}}{3 	imes 10^8} = 160 	imes 10^{-12} = 1.6 	imes 10^{-10} \ kg \cdot m/s$.

$A$ plane electromagnetic wave of wave intensity $6 \ W/m^2$ strikes a small mirror of area $40 \ cm^2$,held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be:

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