To solve the problem, it is necessary to apply the concepts related to the diffraction given in circular spaces. By definition it is expressed as
[tex]\Delta \theta = 1.22\frac{\lambda}{d}[/tex]
Where,
\lambda = Wavelength
d = Optical Diameter
[tex]\theta =[/tex] Angular resolution
In turn you can calculate the angle through the diameter and the arc length, that is,
[tex]\Delta \theta = \frac{x}{D}[/tex]
Where,
x = The length of the arc
D = Distance
From known data we know that Jupiter's diameter is,
[tex]x_J = 1.43*10^8m[/tex]
[tex]D = 20*9.4608*10^{15}[/tex]
[tex]\lambda = 600*10^{-9}m[/tex]
Replacing we have that,
[tex]\frac{x}{D} = 1.22\frac{\lambda}{d}[/tex]
[tex]\frac{1.43*10^8}{20*9.4608*10^{15} } = 1.22\frac{600*10^{-9}}{d}[/tex]
Re-arrange to find d,
[tex]d = 968.5m = 0.968Km[/tex]
Therefore the minimum diameter of an optical telescope to resolve a Jupiter-sized planet is 0.968Km.
