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Question
Draw a labelled ray diagram of an image formed by a refracting telescope with the final image formed at infinity. Derive an expression for its magnifying power with the final image at infinity
Solution
As the object lies at the very huge distance, therefore, the angle subtended by the object at C2 (where the eye is held) is almost the same as the angle subtended by the object at C1 (because C1 is close to C2). Let it be α, i.e `angleA ""^'C_1B^' = alpha`. Rays coming from the final image at infinity make `angleA ""^'C_2B^' = beta` on the eye. Therefore, by definition
Magnifying power, `m = beta/alpha ` .....1
As angle `alpha` and `beta` are small, therefore `alpha = tan alpha ` and `beta = tan beta`
From 1 `m = tan beta/tan alpha` ... (2)
In `triangle A'B'C_2`
`tan beta = (A'B")/(C_2B')`
In `triangle A'B'C_1`
`tan alpha = (A'B')/(C_1B')`
putting 2
`m = (A'B')/(C_2B') xx (C_1B')/(A'B') = (C_1B)/(C_2B')` or `m = f_0/(-f_e)` ...(3)
where `C_1B' = f_0` = focal length of objective lens,
`C_2B' = -f_e` = focal length of eye lens.
The negative sign of m indicates the final image is inverted w.r.t. the object
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