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PhysicsNumerical+3Question 12 of 54

JEE Advanced 2019 · Paper 2, Physics question 12

A perfectly reflecting mirror of mass MM mounted on a spring constitutes a spring-mass system of angular frequency Ω\Omega such that 4πMΩh=1024 m−2\dfrac{4\pi M\Omega}h = 10^{24}\ \mathrm{m^{-2}} with hh as Planck's constant. NN photons of wavelength λ=8π×10−6\lambda = 8\pi\times10^{-6} m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 μm. If the value of NN is x×1012x\times10^{12}, then the value of xx is ______.

[Consider the spring as massless]

Light falls normally from the left on a vertical plane mirror, which is attached on its right side to a horizontal spring fixed to a wall.
Answer and worked solution

Official answer: 1

Impulse. Each reflected photon delivers 2h/λ2h/\lambda, so the mirror starts moving at
v=2NhλM.v=\frac{2Nh}{\lambda M}.

Amplitude. The spring-mirror system oscillates with amplitude A=vΩ=2NhλMΩA=\dfrac v\Omega=\dfrac{2Nh}{\lambda M\Omega}. Setting A=10−6A=10^{-6} m:
N=10−6λMΩ2h=10−6×8π×10−62⋅10244π=1012.N=\frac{10^{-6}\lambda M\Omega}{2h}=\frac{10^{-6}\times8\pi\times10^{-6}}{2}\cdot\frac{10^{24}}{4\pi}=10^{12}.
Answer: 1

Topics:Photon momentum and radiation pressureSimple harmonic motion
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