In an astronomical telescope in normal adjustment,a straight black line of length $L$ is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is $l$. The magnification of the telescope 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 $cm$.

The magnifying power of a refracting type of astronomical telescope is $m$. If the focal length of the eyepiece is doubled,then the magnifying power will become:

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 $\,cm$)

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 diameter of the moon is $3.5 \times 10^3 \text{ km}$ and its distance from the earth is $3.8 \times 10^5 \text{ km}$. If it is seen through a telescope whose focal lengths for the objective and eye lens are $4 \text{ m}$ and $10 \text{ cm}$ respectively,then the angle subtended by the moon on the eye will be approximately.......$^o$