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

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Modern Physics

A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass system of angular frequency

Ω

such that

\frac{4 π M Ω}{h} = 1 0^{24} m^{- 2}

with h as Planck's constant. N photons of wavelength

λ = 8 π \times 1 0^{- 6} m

strike the mirror simultaneously at normal incidence such that the mirror gets displaced by

1 μ m

. If the value of N is

x \times 1 0^{12}

, then the value of x is ____.
[Consider the spring as massless]

Answer: 1

From momentum conservation, after N photons reflect:

m v = 2 N p

where

p = \frac{h}{λ}

is the momentum of one photon.

The mirror oscillates with

v = A Ω

where

A = 1 μ

m is the amplitude.

Substituting:

m A Ω = 2 N \frac{h}{λ}

N = \frac{m Ω}{h} \times \frac{λ A}{2}

Given

\frac{m Ω}{h} = \frac{1 0 ^{24}}{4 π}

^{- 2}

λ = 8 π \times 1 0^{- 6}

m, and

A = 1 0^{- 6}

N = \frac{1 0 ^{24}}{4 π} \times \frac{8 π \times 1 0 ^{- 6} \times 1 0 ^{- 6}}{2} = 1 0^{12}

Therefore,

x = 1.00

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Numerical answer