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:

The focal lengths of the objective lens and the eye lens of a Galilean telescope are $30\, cm$ and $3.0\, cm$ respectively. The telescope produces a virtual,erect image of an object situated far away from it at the least distance of distinct vision from the eye lens. In this condition,the magnifying power of the Galilean telescope should be:

The focal length of the objective and eyepiece of a telescope are respectively $200 \, cm$ and $5 \, cm$. The maximum magnifying power of the telescope will be

$A$ distant hot air balloon subtends $0.25^o$ at the objective lens of an astronomical telescope. The image subtends $1.5^o$ at the eyepiece when viewed with a relaxed eye. The objective and eyepiece are in a $35 \ cm$ long tube. Find the focal length of the eyepiece lens in $cm$.

$A$ 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 the telescope is in normal adjustment?

$A$ 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.......$\text{cm}$.