A perfectly reflecting mirror of mass $M$ mounted on a spring constitutes a spring-mass system of an

Question

A perfectly reflecting mirror of mass $M$ mounted on a spring constitutes a spring-mass system of angular frequency $\Omega$ such that $\frac{4 \pi M \Omega}{h}=10^{24} \mathrm{~m}^{-2}$ with $h$ as Planck's constant. $N$ photons of wavelength $\lambda=8 \pi 	imes 10^{-6} \mathrm{~m}$ strike the mirror simultaneously at normal incidence such that the mirror gets displaced by $1 \mu \mathrm{~m}$. If the value of $N$ is $x 	imes 10^{12}$, then the value of $x$ is [Consider the spring as massless]

Professor PC Thomas Classes · Answer

(b) : Let momentum of one photon be \ $p$ and after reflection, velocity of the mirror be $v$.\ So, from conservation of linear momentum we get,
$
N p \hat{i}=-N p \hat{i}+M v \hat{i} 	ext { or } M v \hat{i}=2 N p \hat{i}
$
$M v=2 N p$ (Considering only the magnitude)...(i)
The velocity of the mirror (spring mass system) \ at mean position, $v=A \Omega$, where $A=1 \mu \mathrm{~m}$, \is the maximum deflection of mirror from mean position \and $\Omega$ is the angular frequency of mirror spring system.
For $N$ number of photons, momentum, \ $N p=\frac{N h}{\lambda}$
? From equation (i), $M A \Omega=\frac{2 N h}{\lambda} \quad$ \ or $\quad N=\frac{M \Omega}{h} 	imes \frac{\lambda A}{2}$
Given, $\frac{M \Omega}{h}=\frac{10^{24}}{4 \pi} \mathrm{~m}^{-2}$,
$
\begin{aligned}
& \lambda=8 \pi 	imes 10^{-6} \mathrm{~m} \
& 	herefore \quad N=\frac{10^{24}}{4 \pi} 	imes \frac{8 \pi 	imes 10^{-6} 	imes 10^{-6}}{2}=1 	imes 10^{12} \
& \because \quad N=x 	imes 10^{12}=1 	imes 10^{12},
\end{aligned}
$
we get $x=1$.