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Jul 12th 2009, 12:17 AM

#1  Member
Join Date: Jun 2009
Posts: 79
 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

 
Jul 12th 2009, 08:37 PM

#2  Senior Member
Join Date: Feb 2009 Location: Singapore
Posts: 206

Originally Posted by hana1 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

 
Jul 12th 2009, 09:13 PM

#3  Senior Member
Join Date: Apr 2008 Location: HK
Posts: 886

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.

 
Jul 12th 2009, 10:40 PM

#4  Member
Join Date: Jun 2009
Posts: 79

Originally Posted by arze 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? 
 
Jul 14th 2009, 11:53 PM

#5  Senior Member
Join Date: Feb 2009 Location: Singapore
Posts: 206

Originally Posted by hana1 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
$\displaystyle D=\frac{d}{m}$ 
 
Jul 15th 2009, 02:28 AM

#6  Member
Join Date: Jun 2009
Posts: 79

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?

 
Jul 15th 2009, 03:09 AM

#7  Senior Member
Join Date: Feb 2009 Location: Singapore
Posts: 206

Originally Posted by hana1 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.

 
Jul 15th 2009, 03:52 AM

#8  Physics Team
Join Date: Feb 2009
Posts: 1,425

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.

 
Jul 15th 2009, 04:53 AM

#9  Senior Member
Join Date: Feb 2009 Location: Singapore
Posts: 206

Oops! ok, no wonder i thought something was wrong. thank you for correcting.

 
Jul 27th 2009, 08:16 PM

#10  Member
Join Date: Jun 2009
Posts: 79

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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