Step-by-step Derivation

1. Momentum of a single photon: The momentum p of a single photon with wavelength λ is given by the de Broglie relation: $p = \frac{h}{\lambda}$ where h is Planck's constant.

2. Momentum transferred to the mirror: The mirror is perfectly reflecting. The photons strike at normal incidence and bounce back along the same path. The initial momentum of a photon is p, and its final momentum is -p. The change in momentum for one photon is $Δp_{photon} = p_{final} - p_{initial} = (-p) - (p) = -2p$. By the law of conservation of momentum, the momentum transferred to the mirror by one photon is $Δp_{mirror} = -Δp_{photon} = 2p$. $Δp_{mirror} = 2 \frac{h}{\lambda}$

3. Total momentum transferred by N photons: N photons strike the mirror simultaneously. The total momentum P transferred to the mirror is the sum of the momentum transferred by each photon: $P = N \times Δp_{mirror} = N \left( \frac{2h}{\lambda} \right)$

4. Initial velocity of the mirror: This momentum P is transferred to the mirror of mass M as an impulse, causing it to move with an initial velocity v. The mirror is initially at rest in its equilibrium position. $P = Mv$ Equating the two expressions for P: $Mv = \frac{2Nh}{\lambda} \quad (*)$

5. Simple Harmonic Motion (SHM) of the mirror: The mirror is part of a spring-mass system. After receiving the impulse at its equilibrium position, it undergoes Simple Harmonic Motion. The velocity v it acquires at the equilibrium position is the maximum velocity ($v_{max}$) of the SHM. The maximum velocity in an SHM is related to the amplitude A and the angular frequency Ω by: $v = v_{max} = A\Omega$ The problem states that the mirror gets displaced by 1 μm. This maximum displacement is the amplitude A of the oscillation. So, $A = 1 \mu m = 10^{-6} m$.

6. Solving for N: Substitute v = AΩ into equation (*): $M(A\Omega) = \frac{2Nh}{\lambda}$ Now, we rearrange this equation to solve for the number of photons, N: $N = \frac{MA\Omega\lambda}{2h}$

7. Using the given values: We are given the following values:

   • $A = 1 \mu m = 10^{-6} m$
   • $\lambda = 8\pi \times 10^{-6} m$
   • $\frac{4\pi M\Omega}{h} = 10^{24} m^{-2}$

   To use the given expression, let's rearrange our formula for N: $N = \frac{MA\Omega\lambda}{2h} = \left( \frac{A\lambda}{8\pi} \right) \left( \frac{4\pi M\Omega}{h} \right)$

8. Calculation: Substitute the numerical values into the rearranged formula: $N = \left( \frac{(10^{-6} m)(8\pi \times 10^{-6} m)}{8\pi} \right) \times (10^{24} m^{-2})$ The 8π terms in the numerator and denominator cancel out: $N = (10^{-6} \times 10^{-6} m^2) \times (10^{24} m^{-2})$ $N = 10^{-12} \times 10^{24}$ $N = 10^{12}$

9. Finding the value of x: The problem states that $N = x \times 10^{12}$. Comparing this with our calculated value $N = 1 \times 10^{12}$: $x \times 10^{12} = 1 \times 10^{12}$ Therefore, x = 1.

Final Answer

The value of x is 1.