Resolution power of telescope

  • Thread starter grouper
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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.
 

Answers and Replies

  • #2
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Still haven't figured this one out if anybody has any suggestions at all; it's due this weekend.
 

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