Physics Help Forum (http://physicshelpforum.com/physics-help-forum.php)
-   Light and Optics (http://physicshelpforum.com/light-optics/)
-   -   Astronomical telescope (http://physicshelpforum.com/light-optics/2628-astronomical-telescope.html)

 hana1 Jul 12th 2009 12:17 AM

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

 arze Jul 12th 2009 08:37 PM

Quote:
 Originally Posted by hana1 (Post 7363) 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

 werehk Jul 12th 2009 09:13 PM

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.

 hana1 Jul 12th 2009 10:40 PM

Quote:
 Originally Posted by arze (Post 7375) 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? :(

 arze Jul 14th 2009 11:53 PM

Quote:
 Originally Posted by hana1 (Post 7379) 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
$\displaystyle D=\frac{d}{m}$ (Worried)

 hana1 Jul 15th 2009 02:28 AM

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?

 arze Jul 15th 2009 03:09 AM

Quote:
 Originally Posted by hana1 (Post 7401) 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.

 physicsquest Jul 15th 2009 03:52 AM

if that's the case, then i suppose that from http://www.physicshelpforum.com/phys...40a7d109-1.gif we can find D where it is the diameter of the image. then, rearranging
http://www.physicshelpforum.com/phys...1c522573-1.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.

 arze Jul 15th 2009 04:53 AM

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

 hana1 Jul 27th 2009 08:16 PM

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

All times are GMT -7. The time now is 09:03 AM.