Answer:
[tex]8.2\times 10^{15}\ m[/tex]
Explanation:
[tex]\lambda[/tex] = Wavelength = 600 nm
d = Diameter of mirror = 42 m
D = Distance of object
x = Diameter of Jupiter = [tex]1.43\times 10^8\ m[/tex]
Angular resoulution is given by
[tex]\Delta\theta=1.22\frac{\lambda}{d}\\\Rightarrow \Delta\theta=1.22\frac{600\times 10^{-9}}{42}\\\Rightarrow \Delta\theta=1.74286\times 10^{-8}\ rad[/tex]
We also have the relation
[tex]\Delta\theta\approx=\frac{x}{D}\\\Rightarrow D\approx\frac{x}{\Delta\theta}\\\Rightarrow D\approx\frac{1.43\times 10^8}{1.74286\times 10^{-8}}\\\Rightarrow D\approx 8.2049\times 10^{15}\ m[/tex]
The most distant Jupiter-sized planet the telescope could resolve is [tex]8.2\times 10^{15}\ m[/tex]
