In normal adjustment,for a refracting telescope,the distance between the objective and the eyepiece is $30\,cm$. The focal length of the objective,when the angular magnification of the telescope is $2$,will be $.....\,cm$.

The magnifying power of an astronomical telescope for normal adjustment is given by:

If the tube length of an astronomical telescope is $105 \, cm$ and the magnifying power is $20$ for normal setting,calculate the focal length of the objective in $cm$.

The magnifying power of an astronomical telescope for relaxed vision is $16$. On adjusting,the distance between the objective and eye lens is $34\, cm$. Then the focal length of the objective and eye lens will be respectively:

There are four convex lenses $L_{1}, L_{2}, L_{3}$ and $L_{4}$ with focal lengths $2 \ cm, 4 \ cm, 6 \ cm$ and $8 \ cm$ respectively. Two of these lenses form a telescope of length $10 \ cm$ and magnifying power $4$. The objective and eye lenses are respectively:

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