The focal length of the objective lens of a telescope is $30 \text{ cm}$ and that of its eye lens is $3 \text{ cm}$. It is focused on a scale at a distance $2 \text{ m}$ from it. The distance of the objective lens from the eye lens to see the clear image is: (in $\text{ cm}$)

The objective and eyepiece of an astronomical telescope are double convex lenses with refractive index $1.5$. When the telescope is adjusted to infinity,the separation between the two lenses is $16 \,cm$. If the space between the lenses is now filled with water and the telescope is again adjusted for infinity,then the present separation between the lenses is: (in $\,cm$)

What is the primary purpose of the eyepiece in a telescope?

$A$ telescope has an objective lens of focal length $200 \, cm$ and an eyepiece with focal length $2 \, cm$. If this telescope is used to see a $50 \, m$ tall building at a distance of $2 \, km$,what is the height of the image of the building formed by the objective lens in $cm$?

To increase the magnifying power of a telescope ($f_o$ = focal length of the objective and $f_e$ = focal length of the eye lens):

$A$ telescope has an objective of focal length $100 \ cm$ and an eyepiece of focal length $5 \ cm$. The least distance of distinct vision is $25 \ cm$. The telescope is focused for distinct vision on a scale $3 \ m$ away from the objective. The magnification produced is . . . . . .