A perfectly reflecting mirror of mass mounted on a spring constitutes a spring-mass system of angular frequency such that with as Planck's constant. photons of wavelength strike the mirror simultaneously at normal incidence such that the mirror gets

Question

A perfectly reflecting mirror of mass $M$ mounted on a spring constitutes a spring-mass system of angular frequency such that $= 10^{24} m^{- 2},$ with $h$ as Planck's constant. $N$ photons of wavelength $λ = 8 π \times 10^{- 6} m$ strike the mirror simultaneously at normal incidence such that the mirror gets displaced by $1 μ m$ . If the value of $N$ is $x \times 10^{12}$ , then the value of $x$ is ________. [Consider the spring as massless] [2019]

Smart Achievers · Accepted Answer

(1)

From conservation of momentum principle, change in momentum of photon = change in momentum of mirror

$2 (N P) = M V_{\max}$

$\Rightarrow 2 [N (\frac{h}{λ})] = M V_{m a x}$

$∴$ $2 \frac{N h}{λ} = M (A Ω)$ $[∵ V_{\max} = A Ω]$

$N = (\frac{M Ω}{h}) \frac{A λ}{2} = \frac{10^{24}}{4 π} \times \frac{10^{- 6} \times 8 π \times 10^{- 6}}{2}$ $[∵ \frac{M Ω}{h} = \frac{10^{24}}{4 π}, A = 1 hm, λ = 8 π \times 10^{- 6}]$

$∴$ $N = 1 \times 10^{12} = x \times 10^{12}$

$∴$ $x = 1$

Solution

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Solution

Ans. (1) From conservation of momentum principle, change in momentum of photon = change in momentum of mirror 2(NP)=MVmax ⇒2[N(hλ)]=MVmax ∴ 2Nhλ=M(AΩ) [∵ Vmax=AΩ] N=(MΩh)Aλ2=10244π×10-6×8π×10-62 [∵ MΩh=10244π, A=1 hm, λ=8π×10-6] ∴ N=1×1012=x×1012 ∴ x=1

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