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

When the diameter of the aperture of the objective of an astronomical telescope is increased,its

$A$ 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 . . . . . .

Diameter of the moon is $3.5 \times 10^3 \, km$ and its distance from earth is $3.8 \times 10^5 \, km$. It is seen by a telescope whose objective and eyepiece have focal lengths $4 \, m$ and $10 \, cm$ respectively. What will be the angular diameter of the image of the moon in degrees?

The magnifying power of an astronomical telescope is $8$. If the distance between the two lenses is $54 \, cm$,find the focal lengths of the eyepiece and the objective lens.

$A$ Cassegrain telescope uses two mirrors as shown in the figure. Such a telescope is built with the mirrors $20 \; mm$ apart. If the radius of curvature of the large mirror is $220 \; mm$ and the small mirror is $140 \; mm$,where will the final image of an object at infinity be?