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Old Jul 12th 2009, 12:17 AM   #1
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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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Old Jul 12th 2009, 08:37 PM   #2
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Originally Posted by hana1 View Post
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
i assume that you would know the values of d and m. substitute these values into the equation you gave $\displaystyle d/m$ and you get the diameter of the emerging beam. isn't that the answer?
correct me if i'm wrong :P
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Old Jul 12th 2009, 09:13 PM   #3
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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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Old Jul 12th 2009, 10:40 PM   #4
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Originally Posted by arze View Post
i assume that you would know the values of d and m. substitute these values into the equation you gave $\displaystyle d/m$ and you get the diameter of the emerging beam. isn't that the answer?
correct me if i'm wrong :P
I wish it were that easy,arze.
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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Old Jul 14th 2009, 11:53 PM   #5
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Originally Posted by hana1 View Post
I wish it were that easy,arze.
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?
if that's the case, then i suppose that from $\displaystyle m=\frac{d}{D}$ we can find D where it is the diameter of the image. then, rearranging
$\displaystyle D=\frac{d}{m}$
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Old Jul 15th 2009, 02:28 AM   #6
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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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Old Jul 15th 2009, 03:09 AM   #7
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Originally Posted by hana1 View Post
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?
then i'm probably wrong here. but if we suppose that the eye say, is at the eyepiece, then, the eyepiece would not magnify the image. if this is the case then i suppose that my attempt was correct.
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Old Jul 15th 2009, 03:52 AM   #8
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if that's the case, then i suppose that from we can find D where it is the diameter of the image. then, rearranging
.

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.
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Old Jul 15th 2009, 04:53 AM   #9
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Oops! ok, no wonder i thought something was wrong. thank you for correcting.
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Old Jul 27th 2009, 08:16 PM   #10
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I'm not sure if this question is related to the eye-ring,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 eye-ring or the diameter of the image formed of the objective by the eyelens.
Therefore D=d/m
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