GR9277 #22



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Comments 
kstephe6 20121105 09:33:02  The faster the answer the better.
the distance is the sum of the two focal lengths, so it must be between 1 and 2 meters. Choice D. DONE
djh101 20140829 16:06:32 
You win.

hipparcos 20180405 01:57:20 
Nice one!

  sky 20091106 13:14:32  This can also be solved by similar triangles. Draw 2 rays passing through the common focal point of the two lenses and ending in the center of each lens. This will give two similar triangles that share equal vertical angles at the focus. the magnification factor tells us that the base of the triangle touching the objective lens should have a base length of 10 units. They give its height as the longer focal length of 1.0m. The smaller eyepiece should have a base length of 1 unit, again due to the magnification. let x be the focal length of the eyepiece. The similar triangle proportion yields,
=
or,
x=0.1m
addition of the focal lengths gives 1.1m (D)
now that I wrote it out I think this is probably the derivation of the equation used in the solution already given. lol   emailzac 20071102 18:39:17  Since the distance between the two eye pieces is the sum of the focal lengths, you know it can't be A B or C, and and the focal length of the eyepiece is usually smaller than that of the objective (at least a can't think of a case when its not) so you know the answer has to be less than 2, hence D.  

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