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(B) For a perfectly reflecting mirror,the force $F$ exerted by a beam of light with power $P_{mirror}$ is given by:$F = \frac{2 P_{mirror}}{c}$For the

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$100$

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(B) For a perfectly reflecting mirror,the force $F$ exerted by a beam of light with power $P_{mirror}$ is given by:$F = \frac{2 P_{mirror}}{c}$For the mirror to be in equilibrium,the radiation force must balance its weight:$F = mg$$\frac{2 P_{mirror}}{c} = mg$$P_{mirror} = \frac{mgc}{2}$Given $m = 20 \, g = 0.02 \, kg$,$g = 10 \, m/s^2$,and $c = 3 	imes 10^8 \, m/s$:$P_{mirror} = \frac{0.02 	imes 10 	imes 3 	imes 10^8}{2} = 0.01 	imes 3 	imes 10^8 = 3 	imes 10^6 \, W = 3 \, MW$Since only $30 \%$ of the total power $P_{source}$ emitted by the source reaches the mirror:$P_{mirror} = 0.30 	imes P_{source}$$3 \, MW = 0.30 	imes P_{source}$$P_{source} = \frac{3}{0.30} = 10 \, MW$Wait,re-evaluating the calculation: $P_{mirror} = \frac{0.02 	imes 10 	imes 3 	imes 10^8}{2} = 30 	imes 10^6 \, W = 30 \, MW$. Then $P_{source} = \frac{30}{0.30} = 100 \, MW$.

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