Understanding Laser Beam Diffraction

This problem involves understanding how a laser beam spreads out as it travels over a long distance due to diffraction. We are given the initial size of the laser beam, its wavelength, and the distance to the moon. We need to estimate the size of the beam when it returns to Earth after reflecting off the moon.

Laser Beam Parameters and Assumptions

The key information provided is:

• Initial beam diameter, $d_0$ = 1 cm = $0.01$ m
• Laser wavelength, $\lambda$ = 500 nm = $5 \times 10^{-7}$ m
• Distance to the Moon, $L$ = 384,000 km = $3.84 \times 10^8$ m

We assume the beam propagates through a vacuum and experiences ideal diffraction-limited spreading. This means the spreading is solely due to the wave nature of light and is not worsened by atmospheric effects or imperfections.

Physics of Beam Spreading

Laser beams, especially over long distances, tend to spread out. For lasers, this spreading is often modeled using the principles of Gaussian beams. The beam radius (distance from the center to the point where intensity drops to $1/e^2$ of the maximum) at a distance $z$ from the beam waist is given by:

$w(z) = w_0 \sqrt{1 + \left(\frac{z}{z_R}\right)^2}$

where:

• $w_0$ is the initial beam radius (waist size) at $z=0$.
• $z_R = \frac{\pi w_0^2}{\lambda}$ is the Rayleigh range, a characteristic distance over which the beam radius grows significantly.

In this problem, the initial beam diameter $d_0 = 1$ cm, so the initial radius is $w_0 = d_0/2 = 0.005$ m.

Calculating the Rayleigh Range

First, let's calculate the Rayleigh range ($z_R$) for this laser beam:

$z_R = \frac{\pi w_0^2}{\lambda} = \frac{\pi (0.005 \text{ m})^2}{5 \times 10^{-7} \text{ m}}$

$z_R = \frac{\pi \times 2.5 \times 10^{-5} \text{ m}^2}{5 \times 10^{-7} \text{ m}} = \frac{\pi \times 2.5}{0.05} \text{ m} = 50\pi \text{ m} \approx 157 \text{ m}$

Beam Spreading in the Far-Field

The distance to the moon ($L \approx 3.84 \times 10^8$ m) is vastly larger than the Rayleigh range ($z_R \approx 157$ m). This means the beam is propagating deep into the far-field region ($z \gg z_R$). In this far-field region, the Gaussian beam formula simplifies significantly. The term $(z/z_R)^2$ becomes much larger than 1, so:

$w(L) \approx w_0 \sqrt{\left(\frac{L}{z_R}\right)^2} = w_0 \frac{L}{z_R}$

Substituting the expression for $z_R$ back into this equation:

$w(L) \approx w_0 \frac{L}{(\pi w_0^2 / \lambda)} = \frac{\lambda L}{\pi w_0}$

This formula shows that in the far-field, the beam radius grows linearly with distance $L$. The diameter $D(L)$ is twice the radius, $D(L) = 2w(L)$.

$D(L) \approx \frac{2 \lambda L}{\pi w_0}$

Since $w_0 = d_0/2$, we can also write this in terms of the initial diameter $d_0$:

$D(L) \approx \frac{2 \lambda L}{\pi (d_0/2)} = \frac{4 \lambda L}{\pi d_0}$

Estimating the Final Beam Diameter

Now, let's plug in the values to calculate the estimated diameter $D(L)$ upon returning to Earth:

• $\lambda = 5 \times 10^{-7}$ m
• $L = 3.84 \times 10^8$ m
• $d_0 = 0.01$ m

Using the formula $D(L) \approx \frac{4 \lambda L}{\pi d_0}$:

$D(L) \approx \frac{4 \times (5 \times 10^{-7} \text{ m}) \times (3.84 \times 10^8 \text{ m})}{\pi \times (0.01 \text{ m})}$

$D(L) \approx \frac{(20 \times 10^{-7}) \times (3.84 \times 10^8)}{\pi \times 10^{-2}} \text{ m}$

$D(L) \approx \frac{7.68 \times 10^2 \text{ m}^2}{\pi \times 10^{-2} \text{ m}} = \frac{768}{\pi \times 0.01} \text{ m}$

$D(L) \approx \frac{768}{0.0314159} \text{ m} \approx 24446 \text{ m}$

Final Diameter in Kilometers

Converting the result to kilometers:

$D(L) \approx 24446 \text{ m} \times \frac{1 \text{ km}}{1000 \text{ m}} \approx 24.45 \text{ km}$

Conclusion on Beam Diameter Estimate

The calculated diameter of the laser beam upon returning to Earth is approximately $24.45$ km. Comparing this value to the options provided:

• 200 m
• 20 m
• 20 km
• 200 km

The value $24.45$ km is closest to the option 20 km. Therefore, 20 km is the best estimate for the beam diameter.