Draw a labelled diagram of telescope when the image is formed at least distance of distinct vision

Astronomical telescope
When the final image is formed at the least distance of distinct vision:

Magnifying power, `M =β/α`

Since α and β are small, we have:

∴ `M= tanβ/tanα    ...... (1)`

In `ΔA'B'C_2, tanβ = (A'B')/(C_2B') `

In `ΔA'B'C_1, tanβ = (A'B')/(C_2B') `

From equation (i), we get:

`M = (A'B')/(C_2B') xx (C_1B')/(A'B')`

\[\Rightarrow\] `M = (C_1B')/(C_2B')`

Here, `C_1B' = +f_0`

\[\Rightarrow\] `C_2B' = -u_e`

\[\Rightarrow\] `M = f_0/ -u_e  .......... (2)`

Using the lens equation `(1/v-1/u=1/f)`for the eyepieces `(1/-D-1/-u_e=1/f_e,)`we get:

`(-1/D+1/u_e=1/f_e)`

\[\Rightarrow\] `(1/u_e=1/(f_e)+1/D)`

\[\Rightarrow\] `(f_0)/u_e =(f_0)/(f_e )(1+f_e/D)`

\[\Rightarrow\] `(-f_0)/u_e =(-f_0)/(f_e )(1+f_e/D) or M = -f_0/(f_e) (1+f_e/D) `

In order to have a large magnifying power and high resolution of the telescope, its objective lens should have a large focal length and the eyepiece lens should have a short focal length.

Derive the formula for angular magnification of a compound microscope, when the final image is formed at least distance of distinct vision. Draw the required ray diagram. 

The angular magnification of a compound microscope is the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye, when both are placed at the least distance of distinct vision.

This is the required expression for angular magnification. 

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

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