Question

laser produces a coherent beam of light that does not spread (diffract) as much in comparison to light from other sources, like an incandescent bulb. Lasers therefore have been used for very accurate measurements of large distances, such as the distance between the Moon and the Earth. In one such experiment, a laser pulse (wavelength 633 nm) is fired at the Moon. What should be the size of the circular aperture of the laser source in order to produce a central maximum of 1.85-km diameter on the surface of the Moon

Answer 1

The diameter of the aperture is 0.321 m

Step-by-step explanation:

The expression for the angle with wavelength is equal to:

`\theta _(R) =sin^(-1) ((1.22\lambda )/(d) )` (eq. 1)

Where

λ = wavelength

d = diameter of the lens

The expression for the angle of the satellite is:

`\theta _(m) =tan^(-1) ((delta-x)/(h) )` (eq. 2)

Where

h = distance between the Earth and the Moon

Like small angles:

`sin\theta _(R) =\theta _(R) \\tan\theta _(m) =\theta _(m)`

Matching both equations and clearing Δx:

`delta-x=h((1.22\lambda )/(d) )` (eq. 3)

Where

λ = 633 nm = 633x10⁻⁹m

For the width of the central maxima is equal:

`w=2*delta-x\\delta-x=(w)/(2)`

Where

w = 1.85 km

Replacing:

`delta-x=(1.85)/(2) = 0.925km`

From equation 3, the diameter of the aperture is:

`d=h((1.22\lambda )/(delta-x) )`

h = 3.84x10⁵km = distance between the Earth and the Moon

Replacing:

`d=3.84x10^(5)km *(((1.22)*(633x10^(-9)m) )/(0.925km) )=0.321m`