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CBSE Class 12-science Answered

Draw a ray diagram of astronomical telescope and derive the formula of its magnifying power when final image is formed at the least distance of distinct vision
Asked by Harshsharma245101 | 09 Oct, 2019, 08:05: PM
answered-by-expert Expert Answer
Figure drawn above shows the ray diagram of astronomical telescope.

Parallel rays from distant object is focussed at focal point of objective lens and first image AB is formed.
Image AB is within the focal length of eye piece, hence it is magnified to form second image A'B'
at the distance of distinct vision D from eye.
 
Magnification of telescope is defined as the ratio of angle β subtended by first image with eye piece
to the angle α subtended by same image with objective lens.
 
If the angles are small, then we have, magnification m = ( β / α ) ≈ tanβ / tanα  ..............(1)
 
In ΔABC, we have tanβ =  (AB/BC)  ,  In ΔABC' , we have tanα =  (AB/BC')
 
Using the above tangents of angles, eqn.(1) is written as  m = BC' / BC = fO / ue .................(2)
 
where fO is focal length of objective lens and ue is lens-to-object distance of eye piece.
 
Using lens equation, we have for eyepiece,  (1/v) - (1/u) = 1/f
 
For the above lens equation,  v = - D , u = - ue  and f = fE , where fE is focal length of eyepiece.
 
hence, we get,  -(1/D) + (1/ue ) = 1/fE   ..................(3)
 
from (3), we get,  ( fO / ue ) =  ( fO / fE ) + ( fO / D ) = ( fO / fE ) [ 1 + ( fE / D ) ]  ..................(4)
 
Hence using eqn.(2) and eqn.(4),  we get magnification as,  m = ( fO / fE ) [ 1 + ( fE / D ) ] 
Answered by Thiyagarajan K | 09 Oct, 2019, 10:48: PM
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