JEE Challenger
More from Dual Nature of Radiation and Matter
PhyHardAdv 2019 • Paper 2

Displacement of Spring Mounted Mirror by Photon Radiation Pressure

A spring-mass system with a perfectly reflecting mirror of mass MM has an angular frequency Ω\Omega satisfying 4πMΩh=1024 m−2\frac{4\pi M\Omega}{h} = 10^{24}\text{ m}^{-2}, where hh represents Planck's constant. NN photons, each with wavelength λ=8π×10−6 m\lambda = 8\pi \times 10^{-6}\text{ m}, strike the mirror simultaneously at normal incidence, displacing it by 1 μm1\text{ }\mu\text{m}. If N=x×1012N = x \times 10^{12}, find the value of xx. [Assume the spring is massless]

Question Diagram 1
Official Numerical Answer1

Topics & Concepts

Dual Nature of Radiation and MatterPhotoelectric EffectDe Broglie WavelengthSHM

Step-by-Step Solution

When NN photons each with wavelength λ\lambda strike the perfectly reflecting mirror normally and are reflected back, each photon undergoes a change in momentum given by: Δpphoton=2p=2hλ\Delta p_{\text{photon}} = 2p = \frac{2h}{\lambda}

The total momentum imparted (impulse JJ) to the mirror of mass MM by the NN photons is: J=NΔpphoton=2NhλJ = N \Delta p_{\text{photon}} = \frac{2Nh}{\lambda}

Since the collision happens simultaneously (instantly), this impulse imparts an initial velocity v0v_0 to the mirror from its equilibrium position: J=Mv0  ⟹  v0=2NhMλJ = M v_0 \implies v_0 = \frac{2Nh}{M\lambda}

After receiving this initial velocity, the spring-mass system undergoes Simple Harmonic Motion (SHM) with angular frequency Ω\Omega. The relationship between the amplitude AA (maximum displacement) and the maximum velocity v0v_0 in SHM is: v0=AΩv_0 = A\Omega

Equating the two expressions for v0v_0: 2NhMλ=AΩ\frac{2Nh}{M\lambda} = A\Omega

Rearranging to solve for the number of photons NN: N=Aλ2(MΩh)N = \frac{A\lambda}{2} \left(\frac{M\Omega}{h}\right)

We are given the relation: 4πMΩh=1024 m−2  ⟹  MΩh=10244π m−2\frac{4\pi M\Omega}{h} = 10^{24} \text{ m}^{-2} \implies \frac{M\Omega}{h} = \frac{10^{24}}{4\pi} \text{ m}^{-2}

Substitute this into the equation for NN: N=Aλ2(10244π)N = \frac{A\lambda}{2} \left(\frac{10^{24}}{4\pi}\right)

Now, substitute the given numerical values:

  • Wavelength of photon, λ=8π×10−6 m\lambda = 8\pi \times 10^{-6} \text{ m}
  • Maximum displacement (amplitude), A=1 μm=10−6 mA = 1\text{ }\mu\text{m} = 10^{-6} \text{ m}

N=(10−6 m)(8π×10−6 m)2×10244πN = \frac{(10^{-6} \text{ m})(8\pi \times 10^{-6} \text{ m})}{2} \times \frac{10^{24}}{4\pi}

N=(4π×10−12)×10244π=1012N = \left(4\pi \times 10^{-12}\right) \times \frac{10^{24}}{4\pi} = 10^{12}

Comparing this with N=x×1012N = x \times 10^{12}: x=1x = 1

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