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:

$A$ telescope has an objective of focal length $50 \text{ cm}$ and an eyepiece of focal length $5 \text{ cm}$. The least distance of distinct vision is $25 \text{ cm}$. The telescope is focused for distinct vision on a scale $200 \text{ cm}$ away. The separation between the objective and the eyepiece is.......$\text{cm}$.

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 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 $\,cm$)

$A$ Cassegrain telescope uses two mirrors as shown in the figure. Such a telescope is built with the mirrors $20 \; mm$ apart. If the radius of curvature of the large mirror is $220 \; mm$ and the small mirror is $140 \; mm$,where will the final image of an object at infinity be?

$A$ 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 $(25 \; cm)$?