Electromagnetic waves > Propagation of electromagnetic waves > Problem 12.2.13

Statement

$12.2.13.$

Estimate the size of the light spot on the moon from the laser beam. The
laser is located on the Ground, its beam radius is 10 cm, the wavelength is
10−5cm. (The spot boundary is estimated from the condition that in the spot
area the rays coming from separate sections of the wave do not cancel each
other out.)
b. Estimate the size of the radar antenna emitting three-centimeter electro-
magnetic waves inside the angle of 0.01 rad.

Solution

First the Data

Beam radius: $a = 10\ \text{cm} = 0.1\ \text{m}$.

Wavelength: $\lambda = 10^{-5}\ \text{cm} = 10^{-7}\ \text{m}$

use visible light
$\lambda \approx 5\times10^{-5}\ \text{cm} = 5\times10^{-7}\ \text{m}$

We will adopt $ \lambda = 5\times10^{-7}\ \text{m}$.
· Earth–Moon distance:
$L \approx 3.84\times10^{8}\ \text{m}$.

The laser beam emerges from a circular aperture of radius a. Diffraction produces an Airy pattern on the Moon. The first intensity minimum (where the waves begin to cancel) defines the limit of the central spot. For a circular aperture, the angle of the first minimum is:

$\theta \approx \frac{0.61\lambda}{a}$

$R = L\, \theta = \frac{0.61\, \lambda L}{a}$.

Multiply $\lambda L$:

$\lambda L = (5\times10^{-7})\, (3.84\times10^{8}) = 192\ \text{m}$

Divide by a:

$\frac{\lambda L}{a} = \frac{192}{0.1} = 1920\ \text{m}$.

Apply the factor 0.61:

$R = 0.61 \times 1920 = 1171.2\ \text{m}$.

Rounding to order of magnitude:

$\boxed{R \approx 1\ \text{km}}$.

b) Radar antenna of 3 cm

Data

Wavelength: $\lambda = 3\ \text{cm} = 0.03\ \text{m}$.
· Total angular divergence of the beam: $\theta \approx 0.01\ \text{rad}$.

A circular parabolic antenna of diameter D produces a beam whose angular width is limited by diffraction. The angular radius of the first minimum is:

$\theta \approx \frac{\lambda}{D}$.

Solve for D:

$D \approx \frac{\lambda}{\theta} = \frac{0.03}{0.01} = 3\ \text{m}$.

The radius of the antenna is half the diameter:

$R = \frac{D}{2} = 1.5\ \text{m}$.

$\boxed{R \approx 1.5\ \text{m}}$.

Answer

$\boxed{R \approx 1\ \text{km}}$

$\boxed{R \approx 1.5\ \text{m}}$