The focal length of an objective of a telescope is $3 \ m$ and its diameter is $15 \ cm$. Assuming that for a normal eye,the diameter of the pupil is $3 \ mm$ for its complete use,the focal length of the eyepiece must be: (in $cm$)

Diameter of the moon is $3.5 \times 10^3 \, km$ and its distance from earth is $3.8 \times 10^5 \, km$. It is seen by a telescope whose objective and eyepiece have focal lengths $4 \, m$ and $10 \, cm$ respectively. What will be the angular diameter of the image of the moon in degrees?

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 $\,cm$)

$A$ telescope has an objective lens of focal length $150\,cm$ and an eyepiece of focal length $5\,cm$. If a $50\,m$ tall tower at a distance of $1\,km$ is observed through this telescope in normal setting,the angle formed by the image of the tower is $\theta$,then $\theta$ is close to.....$^o$

In an astronomical telescope in normal adjustment,a straight black line of length $L$ is drawn on the inside part of the objective lens. The eyepiece forms a real image of this line. The length of this image is $I$. The magnification of the telescope is:

The magnifying power of a telescope is $9$. When it is adjusted for parallel rays,the distance between the objective and eyepiece is $20 \ cm$. The focal lengths of the objective and eyepiece are respectively: