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 $\text{ cm}$)

In a terrestrial telescope,the focal length of the objective is $90\, cm$,the inverting lens is $5\, cm$,and the eye lens is $6\, cm$. If the final image is formed at $30\, cm$ from the eye lens,calculate the magnification.

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 $\,cm$)

There are four convex lenses $L_{1}, L_{2}, L_{3}$ and $L_{4}$ with focal lengths $2 \ cm, 4 \ cm, 6 \ cm$ and $8 \ cm$ respectively. Two of these lenses form a telescope of length $10 \ cm$ and magnifying power $4$. The objective and eye lenses are respectively:

$A$ small object is placed at a distance of $4 \,m$ from the objective of a telescope of focal length $2 \,m$. The focal length of the eyepiece is $0.2 \,m$. The final image of the object

$A$ small telescope has an objective of focal length $140 \ cm$ and an eyepiece of focal length $5.0 \ cm$. The magnifying power of the telescope for viewing a distant object is: