Question

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

Answer 1

Let momentum of one photon is $p$ and after reflection velocity of the mirror is $v$ conservation of linear momentum

$Np\hat{i}=-Np\hat{i}+mv\hat{i}$
$mv\hat{i}=2pN\hat{i}$
$mv=2Np$
since $v$ is velocity of mirror (spring mass system) at mean position,
$v=A\Omega$
Where $A$ is maxiflection of mirror from mean position . ($A=1\mu m)$ and $\Omega$ is angular frequency of mirror spring system,

momentum of $1$ photon, $p=\frac{h}{\lambda }$
$mv=2Np$
$mA\Omega =2N\frac{h}{\lambda }$
$N=\frac{m\Omega }{h}\times \frac{\lambda A}{2}$

given, $\frac{m\Omega }{h}=\frac{{10}^{24}}{4\pi }{m}^{-2}$
$\lambda =8\pi \times {10}^{-6}m$
$N=\frac{{10}^{24}}{4\pi }\times \frac{8\pi \times {10}^{-6}\times {10}^{-6}}{2}$

$N={10}^{12}=X\times {10}^{12}$
Hence $X=1$