If the focal length of the eyepiece of a tele | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnifying power $(m)$ of an astronomical telescope in normal adjustment is given by the formula: $m = -\frac{f_o}{f_e}$,where $f_o$ is the fo

Vedclass · Accepted Answer

$\frac{m}{2}$

Vedclass · Answer

(C) The magnifying power $(m)$ of an astronomical telescope in normal adjustment is given by the formula: $m = -\frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.If the focal length of the eyepiece is doubled,the new focal length becomes $f_e' = 2f_e$.The new magnifying power $(m')$ will be: $m' = -\frac{f_o}{f_e'} = -\frac{f_o}{2f_e} = \frac{1}{2} \left( -\frac{f_o}{f_e} \right) = \frac{m}{2}$.Therefore,the magnifying power becomes half of its original value.

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

Similar Questions

The length of a telescope is $36 \, cm$. The focal lengths of its lenses can be

What is the diameter of the objective used in astronomical telescopes in present use (in $cm$)?

If a telescope is reversed,i.e.,viewed from the objective side,what happens to the appearance of the object?

An astronomical telescope has objective and eye-piece lenses of powers $0.5\, D$ and $20\, D$ respectively. Its magnifying power will be:

What is a telescope? Discuss the types of telescopes that are used in general.

Vedclass Test Series

Exam Paper Generator

Online Exam Module