If the tube length of an astronomical telescope is $105 \, cm$ and the magnifying power is $20$ for normal setting,calculate the focal length of the objective in $cm$.

$A$ 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 the telescope is in normal adjustment?

In an astronomical telescope,the focal length of the objective lens is $100 \, cm$ and of the eyepiece is $2 \, cm$. The magnifying power of the telescope for the normal eye is:

Magnification produced by an astronomical telescope for normal adjustment is $10$ and the length of the telescope is $1.1 \, m$. The magnification when the image is formed at the least distance of distinct vision $(D = 25 \, cm)$ is

An object is clearly seen through an astronomical telescope of length $50 \ cm$. The focal lengths of its objective and eyepiece,respectively,can be

The magnifying power of a telescope is $9.$ When it is adjusted for parallel rays,the distance between the objective and eyepiece is $20\; cm.$ The focal lengths of the lenses are: