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

The magnifying power of an astronomical telescope in the normal adjustment position is $100$. The distance between the objective and the eye piece is $101 \,cm$. Find the focal length of the objective lens in $cm$.

Two convex lenses of focal lengths $0.3 \, m$ and $0.05 \, m$ are used to make a telescope. The distance kept between the two is .... $m$

$A$ small telescope has an objective lens of focal length $144\;cm$ and an eyepiece of focal length $6.0\;cm$. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Write a note on Cassegrain telescope.

In an astronomical telescope,the focal lengths of the objective lens and the eyepiece are $150 \, cm$ and $6 \, cm$ respectively. In the case where the final image is formed at the least distance of distinct vision,the magnifying power is: