If the focal lengths of the objective and eyepiece of an astronomical telescope are $200 \, cm$ and $4 \, cm$ respectively,what will be the magnifying power for distant vision (normal adjustment)?

$A$ small telescope has an objective lens of focal length $140 \; cm$ and an eyepiece of focal length $5.0 \; cm$. What is the magnifying power of the telescope for viewing distant objects when
$(a)$ the telescope is in normal adjustment (i.e.,when the final image is at infinity)?
$(b)$ the final image is formed at the least distance of distinct vision $(25 \; cm)$?

An astronomical telescope has an objective and an eyepiece of focal lengths $40\, cm$ and $4\, cm$ respectively. To view an object $200\, cm$ away from the objective,the lenses must be separated by a distance of.....$cm$.

The focal lengths of the objective and eye lenses of a telescope are respectively $200 \, cm$ and $5 \, cm$. The maximum magnifying power of the telescope will be:

The relative difference between the focal lengths of the objective and the eye lens in a microscope and a telescope is given as:

The focal length of the objective of a terrestrial telescope is $80 \, cm$ and it is adjusted for parallel rays. If its magnifying power is $20$ and the focal length of the erecting lens is $20 \, cm$,then the full length of the telescope will be......$cm$.