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

Q: Why is the change in momentum of the photons $2Np$ and not just $Np$?

Because the mirror is perfectly reflecting. The photons hit the mirror with momentum $Np$ and bounce back with momentum $-Np$. The change is final minus initial: $-Np - (Np) = -2Np$. By conservation of momentum, the mirror gains $+2Np$.

Q: Why do we equate the mirror's velocity to $A\Omega$?

The photons strike the mirror at its equilibrium (mean) position. In Simple Harmonic Motion, the velocity of the oscillating body is maximum at the mean position, and this maximum velocity is given by the formula $v_{\text{max}} = A\Omega$.

Q: Does the mass of the photons matter in this collision?

Photons are massless particles. However, they still carry momentum according to de Broglie's relation $p = h/\lambda$. This momentum is what exerts the force on the mirror.

Q: Why is the spring considered massless?

If the spring had mass, some of the kinetic energy imparted by the photons would go into moving the spring itself, complicating the SHM equations. Assuming a massless spring ensures all kinetic energy is initially with the mirror.

Q: What if the photons struck the mirror at an angle instead of normal incidence?

If they struck at an angle $\theta$ to the normal, only the normal component of their momentum ($p \cos\theta$) would be reversed and transferred to the spring. The change in momentum would be $2Np \cos\theta$.

Question

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

Sigma Percentile · Accepted Answer

### The Setup: A Collision of Two Worlds

Imagine a perfectly reflecting mirror attached to a spring. This isn't just any ordinary mechanics problem; it's a beautiful intersection of classical mechanics and quantum physics. Suddenly, a burst of $N$ photons strikes the mirror simultaneously. What happens? The photons, despite having no mass, carry momentum. When they hit the mirror and bounce back, they transfer this momentum to the mirror, causing it to compress the spring and start oscillating.

This is the essence of radiation pressure acting as an impulsive force. The mirror gets a sudden "kick" from the light, and we need to figure out exactly how many photons were in that burst to cause a specific displacement.

### The Quantum Kick: Momentum of Photons

First, let's delve into the quantum side of things. According to de Broglie's relation, the momentum $p$ of a single photon is given by Planck's constant $h$ divided by its wavelength $\lambda$:

$$p = \frac{h}{\lambda}$$

But we don't just have one photon; we have a burst of $N$ photons. And here is the crucial detail: the mirror is **perfectly reflecting**. When a photon hits a perfectly reflecting surface, it bounces back with the exact same speed but in the opposite direction.

The initial momentum of the photons is $Np$ towards the mirror. The final momentum is $-Np$ away from the mirror. Therefore, the change in momentum of the photons is:

$$\Delta p_{	ext{photons}} = -Np - Np = -2Np$$

By the law of conservation of linear momentum, the momentum lost by the photons must be gained by the mirror. So, the mirror acquires a momentum of $2Np$. If the mirror has a mass $M$ and acquires a velocity $v$, we can write:

$$Mv = 2Np$$

### The Classical Response: Simple Harmonic Motion

Now, let's shift our focus to the classical mechanics of the spring-mass system. The mirror receives this momentum almost instantaneously. This means it acquires its velocity $v$ right at its mean position (equilibrium position) before the spring even has a chance to compress significantly.

In Simple Harmonic Motion (SHM), the velocity is maximum at the mean position. The formula for the maximum velocity $v_{	ext{max}}$ in terms of the amplitude $A$ and the angular frequency $\Omega$ is:

$$v_{	ext{max}} = A\Omega$$

Since the velocity $v$ acquired from the photons is the maximum velocity, we can substitute $v = A\Omega$ into our momentum equation:

$$M(A\Omega) = 2N\left(\frac{h}{\lambda}ight)$$

### The Master Equation

We now have a beautiful equation linking the macroscopic world (mass, amplitude, frequency) with the microscopic world (number of photons, Planck's constant, wavelength). Our goal is to find $N$, so let's rearrange the equation to isolate $N$:

$$N = \frac{M\Omega A\lambda}{2h}$$

This is our master equation. It tells us exactly how many photons are needed to produce a given amplitude of oscillation.

### Plugging in the Numbers

The problem provides us with a very specific and somewhat unusual given value:

$$\frac{4\pi M\Omega}{h} = 10^{24} 	ext{ m}^{-2}$$

Let's isolate the term $\frac{M\Omega}{h}$ from this given information so we can substitute it directly into our master equation:

$$\frac{M\Omega}{h} = \frac{10^{24}}{4\pi}$$

Now, let's gather the rest of our given values:
- The amplitude $A$ is $1 \mu	ext{m}$, which is $10^{-6} 	ext{ m}$.
- The wavelength $\lambda$ is $8\pi 	imes 10^{-6} 	ext{ m}$.

Let's substitute everything into the master equation:

$$N = \left(\frac{10^{24}}{4\pi}ight) 	imes \frac{(8\pi 	imes 10^{-6}) 	imes (10^{-6})}{2}$$

### The Final Calculation

Now comes the satisfying part: the cancellation.

First, let's handle the constants. The $\pi$ in the numerator cancels with the $\pi$ in the denominator. We have an $8$ in the numerator and a $4 	imes 2 = 8$ in the denominator. They cancel out perfectly!

$$N = 10^{24} 	imes 10^{-6} 	imes 10^{-6}$$

Using the laws of exponents, we add the powers:

$$N = 10^{24 - 6 - 6} = 10^{12}$$

The problem states that $N = x 	imes 10^{12}$. Comparing our result with this format, it is crystal clear that:

$$x = 1$$

And there we have it! A seemingly complex problem involving quantum mechanics and classical oscillations elegantly unravels into a simple integer. Always remember to break down such problems into their fundamental physical principles: the momentum of light, conservation laws, and the kinematics of SHM.

Enter Numerical Value:

Visualized Solution

$Visualizing the System$

$Momentum of a Photon$

$Conservation of Linear Momentum$

$Maximum Velocity in SHM$

$Substituting Velocity$

$Using Given Values$

$Final Substitution$

$Calculating N$

$The Way Forward$

The Sigma Insight: Photon Theory of Light

The Setup

The Quantum Kick

The Classical Response

The Master Equation

Plugging in the Numbers

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Momentum of a Photon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Linear Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximum Velocity in SHM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using Given Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating N

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward