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

Answer 1

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