In a terrestrial telescope,the focal length of the objective is $90\, cm$,the inverting lens is $5\, cm$,and the eye lens is $6\, cm$. If the final image is formed at $30\, cm$ from the eye lens,calculate the magnification.

Explain the construction of a refracting telescope with a figure and derive the equation for its magnification.

An astronomical telescope has an objective and an eyepiece of focal lengths $40\, cm$ and $4\, cm$ respectively. To view an object $200\, cm$ away from the objective,the lenses must be separated by a distance of.....$cm$.

$A$ planet is observed by an astronomical telescope having an objective lens of focal length $16 \, m$ and an eye-piece of focal length $2 \, cm$. Find the correct statement.

Magnifying power of an astronomical telescope for normal adjustment is $10$ and the length of the telescope is $110 \ cm$. The magnifying power of the same telescope,when the image is formed at the near point,is:

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