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 . . . . . . .

The focal lengths of the objective and eye lenses of a telescope are respectively $200 \, cm$ and $5 \, cm$. The maximum magnifying power of the telescope will be:

$A$ Galileo telescope has an objective of focal length $100 \, cm$ and magnifying power $50$. The distance between the two lenses in normal adjustment will be....$cm$

$A$ small telescope has an objective lens of focal length $140 \; cm$ and an eyepiece of focal length $5.0 \; cm$. What is the magnifying power of the telescope for viewing distant objects when
$(a)$ the telescope is in normal adjustment (i.e.,when the final image is at infinity)?
$(b)$ the final image is formed at the least distance of distinct vision $(25 \; cm)$?

For a telescope in normal adjustment,the length of the telescope is found to be $27 \, cm$. If the magnifying power of the telescope at normal adjustment is $8$,the focal lengths of the objective and the eyepiece are respectively:

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: