Maximizing Telescope Resolution: Finding the Optimal Eyepiece Focal Length

AI Thread Summary
To resolve features 7.0 km across on the Moon with a 2.2 m focal-length objective lens and a diameter of 10.5 cm, the resolution limit due to diffraction is calculated using the formula θ=(1.22*λ)/D, resulting in θ=6.39e-6 radians. The relationship between the eyepiece and the objective lens is not clearly understood, particularly in determining the appropriate focal length for the eyepiece. The attempt to calculate the eyepiece focal length using θ=s/f yielded an incorrect result. Further clarification on how to connect the eyepiece focal length to the resolving power is needed. Assistance is requested to solve this problem before the upcoming deadline.
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Homework Statement



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? What is the resolution limit (radians) set by the size of the objective lens (that is, by diffraction)? Use λ=550 nm.

Homework Equations



Resolving power=RP=s=f*θ=(1.22*λ*f)/D where s=distance between two resolvable points, f=focal length of objective lens, D=diameter, and θ=angle between objects

The Attempt at a Solution



Using θ=(1.22*λ)/D, I got θ=6.39e-6 rad, which is correct for the second question. I don't really understand the relationship to the eyepiece in the first question though. θ=8e-3 rad for the first situation, but I'm not sure this is needed. I tried θ=s/f (even though this f refers to the objective lens) just to see what would happen and it gave f=0.125 m, which is incorrect. Any help is appreciated, thanks.
 
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