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(D) The intensity $I$ of the wave is $6 \, W/m^2$. The area $A$ of the mirror is $40 \, cm^2 = 40 	imes 10^{-4} \, m^2$.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 $I$ of the wave is $6 \, W/m^2$. The area $A$ of the mirror is $40 \, cm^2 = 40 	imes 10^{-4} \, m^2$.For a perfectly reflecting surface,the momentum $p$ transferred by an electromagnetic wave in time $t$ is given by $p = \frac{2U}{c}$,where $U$ is the energy incident on the surface.The energy incident per second is $U = I 	imes A 	imes t = 6 	imes 40 	imes 10^{-4} 	imes 1 = 240 	imes 10^{-4} \, J = 0.024 \, J$.The momentum transferred per second is $p = \frac{2 	imes 0.024}{3 	imes 10^8} = \frac{0.048}{3 	imes 10^8} = 0.016 	imes 10^{-8} = 1.6 	imes 10^{-10} \, kg \cdot m/s$.

$A$ 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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