Answer :
Answer:
The diameter of the aperture is 0.321 m
Explanation:
The expression for the angle with wavelength is equal to:
[tex]\theta _{R} =sin^{-1} (\frac{1.22\lambda }{d} )[/tex] (eq. 1)
Where
λ = wavelength
d = diameter of the lens
The expression for the angle of the satellite is:
[tex]\theta _{m} =tan^{-1} (\frac{delta-x}{h} )[/tex] (eq. 2)
Where
h = distance between the Earth and the Moon
Like small angles:
[tex]sin\theta _{R} =\theta _{R} \\tan\theta _{m} =\theta _{m}[/tex]
Matching both equations and clearing Δx:
[tex]delta-x=h(\frac{1.22\lambda }{d} )[/tex] (eq. 3)
Where
λ = 633 nm = 633x10⁻⁹m
For the width of the central maxima is equal:
[tex]w=2*delta-x\\delta-x=\frac{w}{2}[/tex]
Where
w = 1.85 km
Replacing:
[tex]delta-x=\frac{1.85}{2} = 0.925km[/tex]
From equation 3, the diameter of the aperture is:
[tex]d=h(\frac{1.22\lambda }{delta-x} )[/tex]
h = 3.84x10⁵km = distance between the Earth and the Moon
Replacing:
[tex]d=3.84x10^{5}km *(\frac{(1.22)*(633x10^{-9}m) }{0.925km} )=0.321m[/tex]