In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:

Question

In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be: (A) 20 cm (B) 30 cm (C) 10 cm (D) 15 cm

Tardigrade Admin · Accepted Answer

f 0+ f e =30 m =( f 0/ f e ) 2=( f 0/ f e ) ⇒ f 0=2 f e So f 0+( f 0/2)=30 f 0=20 cm

Q. In normal adjustment, for a refracting telescope, the distance between objective and eye piece is $30 c m$ . The focal length of the objective, when the angular magnification of the telescope is 2, will be:

$20 c m$

$30 c m$

$10 c m$

$15 c m$

$f_{0} + f_{e} = 30$
$m = \frac{f _{0}}{f _{e}}$
$2 = \frac{f _{0}}{f _{e}} \Rightarrow f_{0} = 2 f_{e}$
So $f_{0} + \frac{f _{0}}{2} = 30$
$f_{0} = 20 c m$

Q. In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:

20cm

30cm

10cm

15cm