A laser beam is aimed at the Moon from a distance 3.84 ×10^8 m away. The angular spread of the

A laser beam is aimed at the Moon from a distance 3.84 ×10^8...

A laser beam is aimed at the Moon from a distance $3.84 \times 10^{8} \mathrm{~m}$ away. The angular spread of the beam is given by the diffraction formula (Rayleigh's criterion), $\sin \theta=1.22 \lambda / D,$ where $D$ is the diameter of the laser tube or rod. $(a)$ Calculate the size of the beam on the Moon for $D=10 \mathrm{~cm}$ and $\lambda=600 \mathrm{nm} .$ ( $b$ ) Repeat the calculation if the laser beam is projected toward the Moon through a $1.0-\mathrm{m}$ -diameter telescope.

Key Concepts

Diffraction

Diffraction is a fundamental wave phenomenon that occurs when a wave encounters an obstacle or passes through an aperture. It leads to the spreading of the wave beyond the geometric shadow of the aperture, and in optics it is responsible for the limits of resolution and the behavior of beams propagating over long distances.

Rayleigh's Criterion

Rayleigh's criterion provides a quantitative measure of the angular resolution limit due to diffraction. It specifies the minimum angular separation at which two point sources of light can be distinguished, and it is commonly used to estimate the angular spread of a beam emerging from an aperture. The criterion shows that the angular spread is proportional to the wavelength and inversely proportional to the aperture diameter.

Aperture Diameter and Wavelength Relationship

The relationship between the aperture diameter and the wavelength is critical in determining diffraction effects. A larger aperture results in a smaller diffraction angle, while a longer wavelength increases the diffraction angle. This relationship is encapsulated in the formula related to Rayleigh's criterion, which helps calculate the beam divergence based on the optical system's parameters.

Angular Beam Spread and Beam Size at a Distance

The angular spread of a beam due to diffraction determines how much the beam expands over a long distance. By using the diffraction angle, one can estimate the size of the beam at a target location by multiplying this small angle (in radians) by the distance to the target. This concept is essential in applications where beam divergence directly influences the illumination or resolution over large distances.

Small Angle Approximation

The small angle approximation simplifies the relationship between the angular spread and the linear beam size at a given distance. When the diffraction angle is small, the tangent of the angle approximates the angle itself (in radians), making it straightforward to calculate the beam's enlargement over long distances by a linear multiplication of the diffraction angle and the distance.

Transcript

0:00 Hi there.

00:01 So for this problem, we have a laser beam that is aimed at the moon from a distance that we are going to call that distance d.

00:17 Well, better if we call the distance r.

00:20 So calling the distance r, we have 3 .84 times 10 to the 8 meters.

00:33 And the angular spread of the beam is given by the diffraction formula or ryle's criterion, that is sine of theta, is equal to 1 .22 times the wavelength divided by d, which is the diameter of the laser tube or rod.

01:03 So for part a of this problem, what we need to find or calculate is the size of the beam on the moon from distance d, which is equal to 10 centimeters, that is the diameter of the laser, and a wavelength, we use a wavelength of 600 nanometers.

01:29 And we need to, well, so to calculate that, we use a wavelength of 600 nanometers.

01:34 The equation that is given.

01:39 So we know that sign of theta is equal to solving with the equation that we are given.

01:56 So we substitute those values in here.

01:59 We will have 1 .22 times the wavelength.

02:04 That is 600 nanometers.

02:09 Nano means 10 to the minus 9 meters.

02:12 And this divided by the diameter.

02:19 So the diameter is 10 centimeters, so in meters is 10 to the minus 2 meters.

02:29 And this will give us a value of 7 .32 times 10 to the minus 6.

02:37 Now, we know that when the angle is very tiny, we can approximate the sign of theta to, be just theta.

02:48 So if we do that in here, we will have that theta is approximate to the value that we obtain.

02:57 Now, this is given in radiance.

03:00 Now, we can use this because we know that theta is related to s, which is the diameter of the being on the moon, divided by the distance to the moon.

03:13 So we can solve to obtain x...

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