Suppose that you wish to construct a telescope that can resolve features 7.0 km across on the Moon, 384,000 km away. You have a 2.2 m-focal-length objective lens whose diameter is 10.5 cm. What focal-length eyepiece is needed if your eye can resolve objects 0.10 mm apart at a distance of 25 cm?
RP=resolving power=s=fθ=(1.22*λ*f)/D where s=distance between objects, f=focal length, D=diameter, and θ=angle between objects
The Attempt at a Solution
I'm sure this problem uses the resolution equations somehow but I'm not sure how to go through two lenses with that.
I tried to figure it out another way using the lens and magnification equations above. I used do=3.84e8 m and f=2.2 m to figure out that di≈2.2 m and then using the magnification equation of hi/ho=-di/do to get hi=-4.0104e-5 m.
This image can then be used as the object for the eyepiece and since we want hi=0.0001m (the distance apart for the final image is supposed to be 0.10 mm) and di=-.25m (25 cm from the person and negative because the image will be on the same side as the object if do is positive), we can again use hi/ho=-di/do to get do=-0.10026m. With 1/f=1/do+1/di then, I got f=-0.07156 m.
Not only did I go about this in a roundabout way that I'm sure the problem writer did not intend, my method also yielded the wrong answer. (I tried a positive value for f as well but that's not correct either.) I'm not concerned with correcting what I did wrong above because I'm sure there's a simpler and more direct way to get the correct answer but I'm not seeing it. Any help would be appreciated, thanks.