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Question
Draw a labelled ray diagram of an astronomical telescope to show the image formation of a distant object. Write the main considerations required in selecting the objective and eyepiece lenses in order to have large magnifying power and high resolution of the telescope.
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Solution
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.
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| 1 | Charge of a proton | e | 1.6 × 10-19 C |
| 2 | Speed of light in vacuum | c | 3 × 108 ms-1 |
| 1 u = 931 MeV | |||
In a telescope, the lens facing the object is called the:
In the working idea of a telescope's magnifying power, the beam of incident rays is inclined at an angle \[\alpha\] with the principal axis, while the emergent beam is inclined at a greater angle \[\beta\]. This produces:
According to Rayleigh's criterion for the angular separation between two resolvable objects, the correct relation is:
Which type of telescope uses a converging lens as eyepiece and gives an inverted image?
