Question

The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and the eye-piece is 101 cm. Calculate the focal lengths of the objective and of the eyepiece.

Answer 1

it is given that, distance between the objective and the eye-piece , $f_0+f_e$= 101 cm.....(1)

magnification, m = $\frac{f_0}{f_e}$ = 100

or, $f_0=100f_e$ ......(2)


from equations (1) and (2),

$100f_e+f_e=101$

or, $101f_e=101\implies f_e=1$

and $f_0$ = 100$f_e$ = 100

hence, focal length of objective = 100cm

focal length of eyepiece = 1cm

Answer 2

Answer:

100 cm and 1 cm respectively

Explanation: