A perfectly reflecting mirror of mass M mount | vedclass

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Question

(B) The momentum of one photon is $p = \frac{h}{\lambda}$.When $N$ photons strike the mirror,the total momentum transferred to the mirror is $\Delta P

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$1$

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(B) The momentum of one photon is $p = \frac{h}{\lambda}$.When $N$ photons strike the mirror,the total momentum transferred to the mirror is $\Delta P = 2Np = \frac{2Nh}{\lambda}$.By the conservation of linear momentum,this impulse imparts an initial velocity $v$ to the mirror of mass $M$ at the equilibrium position:$Mv = \frac{2Nh}{\lambda} \implies v = \frac{2Nh}{M\lambda}$.For a spring-mass system,the maximum displacement (amplitude $A$) is related to the velocity at the mean position by $v = A\Omega$.Given $A = 1 \mu	ext{m} = 10^{-6} 	ext{ m}$,we have:$A\Omega = \frac{2Nh}{M\lambda} \implies N = \frac{M\Omega A \lambda}{2h}$.Rearranging the given expression $\frac{4 \pi M \Omega}{h} = 10^{24} 	ext{ m}^{-2}$,we get $\frac{M\Omega}{h} = \frac{10^{24}}{4\pi} 	ext{ m}^{-2}$.Substituting the values:$N = \left( \frac{10^{24}}{4\pi} ight) 	imes \frac{A \lambda}{2} = \frac{10^{24}}{4\pi} 	imes \frac{10^{-6} 	imes 8\pi 	imes 10^{-6}}{2}$.$N = \frac{10^{24} 	imes 8\pi 	imes 10^{-12}}{8\pi} = 10^{12}$.Since $N = x 	imes 10^{12}$,we find $x = 1$.

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