Astronomical telescope An astronomical telescope is in normal adjustment.The diameter of the incident beam falling on the objective is d.If the angular magnification of the telescope is m,the diameter of the emerged beam is ? Ans given is d/m In normal adjustment ,would mean the image is at infinity,& ang.magnification= angle subtented at the eye by the image at infinity/ angle subtented at the eye by the object at infinity In astro.telescope,fo(of objective)> fe(of eyepiece), & the image formed by the objective will fall on fe,but I don't really know how to calculate the height of the final image formed(which is what we're asked to find,right?). Hope someone can help. Thank you 
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correct me if i'm wrong :P 
I guess you cannot find the height of the final image as you 've mentioned, in normal adjustment, the image formed at infinity. So there is no definite position for final image. 
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but the answer given in the paper is actually d/m.We're supposed to arrive at this answer,but I don't know how? As werehk said,i think we can't find the height of the final image,but I wonder if we're actually asked to find the height of the intermediate image formed by the objective? :( 
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$\displaystyle D=\frac{d}{m}$ (Worried) 
Hmm...we do get the answer that way... but isn't m the angular magnification of the whole optical instrument? I'm not sure if we can use 'm' as the linear magnification of the objective lens only? 
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if that's the case, then i suppose that from http://www.physicshelpforum.com/phys...40a7d1091.gif we can find D where it is the diameter of the image. then, rearranging http://www.physicshelpforum.com/phys...1c5225731.gif (Worried) . How do you get this? Shouldn't magnification be image size / obj size ? Or m = D/d , in which case D = dm which is why i thought this approach is wrong. 
Oops! ok, no wonder i thought something was wrong. thank you for correcting. 
I'm not sure if this question is related to the eyering,but assuming the incident rays fill the objective lens completely whose diameter is d,then I guess , m = d/D , where D is the diameter of the eyering or the diameter of the image formed of the objective by the eyelens. Therefore D=d/m 
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