The objective lens of a telescope has a focal length of $100 \, cm$. When the final image is formed at the least distance of distinct vision,the distance between the lenses is $105 \, cm$. Calculate the focal length of the eyepiece in $cm$.

$A$ Cassegrain telescope uses two mirrors as shown in the figure. Such a telescope is built with the mirrors $20 \; mm$ apart. If the radius of curvature of the large mirror is $220 \; mm$ and the small mirror is $140 \; mm$,where will the final image of an object at infinity be?

The focal length of the objective and the eyepiece of a telescope are $100\, cm$ and $5\, cm$ respectively. If the final image is formed at the least distance of distinct vision,the magnification of the telescope is:

The focal length of the objective and the eyepiece of a telescope are $50\,cm$ and $5\,cm$ respectively. If the telescope is focused for distinct vision on a scale distant $2\,m$ from its objective,then its magnifying power will be

$A$ telescope has an objective of focal length $100 \ cm$ and an eye-piece of focal length $5 \ cm$. The magnifying power of the telescope is

An astronomical refracting telescope is being used by an observer to observe planets in normal adjustment. The focal lengths of the objective and eyepiece used in the construction of the telescope are $20\,m$ and $2\,cm$ respectively. Consider the following statements about the telescope:
$(a)$ The distance between the objective and eyepiece is $20.02\,m$.
$(b)$ The magnification of the telescope is $1000$.
$(c)$ The image of the planet is erect and diminished.
$(d)$ The aperture of the eyepiece is smaller than that of the objective.
The correct statements are: