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(A) The intensity $I$ of the electromagnetic wave is $6 \, W/m^2$. The area $A$ of the mirror is $30 \, cm^2 = 30 	imes 10^{-4} \, m^2$.For a perfect

Vedclass · Accepted Answer

$1.2 	imes 10^{-10}$

Vedclass · Answer

(A) The intensity $I$ of the electromagnetic wave is $6 \, W/m^2$. The area $A$ of the mirror is $30 \, cm^2 = 30 	imes 10^{-4} \, m^2$.For a perfectly reflecting surface,the radiation pressure $P$ is given by $P = \frac{2I}{c}$,where $c = 3 	imes 10^8 \, m/s$ is the speed of light.The force $F$ exerted on the mirror is $F = P 	imes A = \frac{2I}{c} 	imes A$.The momentum transmitted per second is equal to the force exerted on the mirror.Substituting the values: $F = \frac{2 	imes 6 	imes 30 	imes 10^{-4}}{3 	imes 10^8}$.$F = \frac{360 	imes 10^{-4}}{3 	imes 10^8} = 120 	imes 10^{-12} = 1.2 	imes 10^{-10} \, kg \cdot m/s$.

$A$ plane electromagnetic wave of intensity $6 \, W/m^2$ strikes a small mirror of area $30 \, cm^2$,held perpendicular to the approaching wave. The momentum transmitted in $kg \cdot m/s$ by the wave to the mirror each second will be:

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