An astronomical telescope is set for normal adjustment and the distance between its objective and eye piece is 1.05 cm. The magnifying power of the telescope is 20. What is the focal length of the objective ?

Question

An astronomical telescope is set for normal adjustment and the distance between its objective and eye piece is 1.05 cm. The magnifying power of the telescope is 20. What is the focal length of the objective ? (A) 2 m (B) 1 m (C) 0.5 m (D) 0.25 m.

Tardigrade Admin · Accepted Answer

Here M = f0 / fe and L= f 0 + fe= 1 .05. Hence fe = 105 - f o. And 20 =fo /(105- f 0). This gives f0 -- 1m and fe = 0 .05 m.

Q. An astronomical telescope is set for normal adjustment and the distance between its objective and eye piece is 1.05 cm. The magnifying power of the telescope is 20. What is the focal length of the objective ?

2 m

1 m

0.5 m

0.25 m.

Here M = $f_{0} / f_{e} an d L = f_{0} + f_{e} = 1.05.$
$He n ce f_{e} = 105 - f_{o} . A n d 20 = f_{o} / (105 - f_{0}) .$
$T hi s g i v es f_{0} - - 1 m an d f_{e} = 0.05 m .$

Q. An astronomical telescope is set for normal adjustment and the distance between its objective and eye piece is 1.05 cm. The magnifying power of the telescope is 20. What is the focal length of the objective ?

2 m

1 m

0.5 m

0.25 m.

Here M = f0​/fe​andL=f0​+fe​=1.05. Hencefe​=105−fo​.And20=fo​/(105−f0​). Thisgivesf0​−−1mandfe​=0.05m.

Here M = $f_{0} / f_{e} an d L = f_{0} + f_{e} = 1.05.$
$He n ce f_{e} = 105 - f_{o} . A n d 20 = f_{o} / (105 - f_{0}) .$
$T hi s g i v es f_{0} - - 1 m an d f_{e} = 0.05 m .$