To increase the magnifying power of a telescope ($f_o$ = focal length of the objective and $f_e$ = focal length of the eye lens):

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 $\,cm$)

In a Galilean telescope,if the powers of an objective and eye lens are respectively $+1.25 \, D$ and $-20 \, D$,then for relaxed vision,the length and magnification will be

The diameter of the moon is $3.5 \times 10^3 \text{ km}$ and its distance from the earth is $3.8 \times 10^5 \text{ km}$. If it is seen through a telescope whose focal lengths for the objective and eye lens are $4 \text{ m}$ and $10 \text{ cm}$ respectively,then the angle subtended by the moon on the eye will be approximately.......$^o$

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.

If the focal length of the eyepiece of a telescope is doubled, its magnifying power $(m)$ will be