The magnifying power of a telescope in normal adjustment is 20 , and the focal length of the eyepiece is `5xx10^(-2)` m. what is the magnifying power obtained when the system is adjusted so that the final image of a distant object is formed `25xx10^(-2)` m away from the eyepiece?

The magnifying power of an astronomical telescope for normal vision is 10 . The focal length of the objective is 20 cm . What will be the magnifying power for focussing the final image formed at a distance of 25 cm from the eyepiece ?

The length of the tube of an astronomical telescope is 80 cm and the magnifying power of the instrument is 19 . Calculate the focal lengths of the objective and the eyepiece.

The focal lengths of the lenses of the objective and eyepiece of a microscope are 2 cm and 5 cm respectively. The distance between them is 20 cm . At what distance from the objective should the object be placed so that the final image is formed 25 cm from the eyepiece ? what will be the magnification ?

A Galileo's telescope has an objective of focal length 100 cm and magnifying power 50. The distance between the two lenses in normal adjustment is

The focal lenghts of the objective and the eyepiece of a terrestrial telescope are 180 cm and 5 cm respectively. The focal length of the erecting lens is 3.5 cm. what is the distance between the objective and the eyepiece ? what is its magnifying power ? Can it be used for the purpose of astronomy ?

If in compound microscope m_1 and m_2 be the linear magnification of the objective lens and eye lens respectively, then what is the magnifying power of the microscope ?