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 relative difference between the focal lengths of the objective and the eye lens in a microscope and a telescope is given as:

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

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

Consider a refracting telescope whose objective has a focal length of $1 \ m$ and the eyepiece has a focal length of $1 \ cm$. The magnifying power of this telescope will be . . . . . . .

For a telescope,the focal length of the objective lens is $15\, cm$ and the focal length of the eyepiece is $10\, mm$. If the tube length is $16\, cm$,find the magnification.