For an astronomical telescope,the focal length of the objective lens is $f_{0}$ and the eyepiece lens is $f_{e}$. Then the tube length of the telescope is . . . . . . .

When a telescope is adjusted for parallel light,the distance of the objective from the eyepiece is found to be $80 \,cm$. The magnifying power of the telescope is $19$. The focal lengths of the lenses are

Explain the construction of a refracting telescope with a figure and derive the equation for its magnification.

$(a)$ $A$ giant refracting telescope at an observatory has an objective lens of focal length $15 \; m$. If an eyepiece of focal length $1.0 \; cm$ is used,what is the angular magnification of the telescope?
$(b)$ If this telescope is used to view the moon,what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^{6} \; m$,and the radius of the lunar orbit is $3.8 \times 10^{8} \; m$.

Four convergent lenses have focal lengths $100 \, cm, 10 \, cm, 4 \, cm$ and $0.3 \, cm$. For a telescope with maximum possible magnification,we choose the lenses of focal length

The focal lengths of the objective and eyepiece of an astronomical telescope are $60\, cm$ and $5\, cm$ respectively. When the final image is formed at the least distance of distinct vision $(25\, cm)$, the magnifying power and the length of the telescope will be, respectively: