Question

an astronomical telescope has magnifying power 24 when used in normal vision.The length of the telescope is 1.00 m.The focal length of the objective lens is
(A)0.04m
(B)0.96m
(C)1.00m
(D)0.24m​

Answer 1

Explanation:

(B)0.96m

hope it will help you

Answer 2

Answer: The focal length of the objective lens is 0.96 m

Explanation:

Given :

magnifying power, m = 24

length of the telescope, L = 1.00 m

the focal length of the objective lens $f_o$ =?

The formula for magnifying power

$m = \frac{f_o}{f_e}\\f_e = \frac{f_o}{m}\\f_e = \frac{f_o}{24}$--------------------------------1

Where, $f_o$ = focal length of an objective lens

$f_e$ = focal length of the eyepiece

$L = f_o + f_e= 1 m\\\\f_e = 1-f_o$-----------------------2

Using Eq. 1 and Eq.2

$\frac{f_o}{24}=1-f_o\\25f_o= 24\\f_o =0.96 m$