The magnifying power of a telescope with tube length $60 \; cm$ is $5$. What is the focal length of its eyepiece in $cm$?

An opera glass (Galilean telescope) measures $9 \, cm$ from the objective to the eyepiece. The focal length of the objective is $15 \, cm$. Its magnifying power is

$A$ planet is observed by an astronomical telescope having an objective lens of focal length $16 \, m$ and an eye-piece of focal length $2 \, cm$. Find the correct statement.

The focal lengths of the objective lens and the eye lens of a Galilean telescope are $30\, cm$ and $3.0\, cm$ respectively. The telescope produces a virtual,erect image of an object situated far away from it at the least distance of distinct vision from the eye lens. In this condition,the magnifying power of the Galilean telescope should be:

The length of an astronomical telescope for normal vision (relaxed eye) ($f_o$ = focal length of objective lens and $f_e$ = focal length of eye lens) is

$A$ telescope has an objective lens of focal length $150\,cm$ and an eyepiece of focal length $5\,cm$. If a $50\,m$ tall tower at a distance of $1\,km$ is observed through this telescope in normal setting,the angle formed by the image of the tower is $\theta$,then $\theta$ is close to.....$^o$