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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?

## Homework Equations

M=f(objective)/f(eyepiece)

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

or θ=(1.22*λ)/D

1/f=1/d(object)+1/d(image)

m=h(image)/h(object)=-d(image)/d(object)

## 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 d

_{o}=3.84e8 m and f=2.2 m to figure out that d

_{i}≈2.2 m and then using the magnification equation of h

_{i}/h

_{o}=-d

_{i}/d

_{o}to get h

_{i}=-4.0104e-5 m.

This image can then be used as the object for the eyepiece and since we want h

_{i}=0.0001m (the distance apart for the final image is supposed to be 0.10 mm) and d

_{i}=-.25m (25 cm from the person and negative because the image will be on the same side as the object if d

_{o}is positive), we can again use h

_{i}/h

_{o}=-d

_{i}/d

_{o}to get d

_{o}=-0.10026m. With 1/f=1/d

_{o}+1/d

_{i}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.