In normal adujstment, 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 :

Solution

Physics sir · 3 steps

Two facts, two unknowns: the tube length is fₒ + fₑ and the magnification is fₒ/fₑ.

1. Normal adjustment. The final image is at infinity, so the focus of the objective sits exactly on the focus of the eyepiece. The distance between the lenses is then fₒ + fₑ = 30 cm.

   m = fₒ/fₑ = 2 ⇒ fₒ = 2fₑ

2. Solve together. Put fₒ = 2fₑ into the length equation: 2fₑ + fₑ = 30, so fₑ = 10 cm and fₒ = 20 cm.
3. So. The focal length of the objective is 20 cm, option (a).

The idea to keep. For a telescope in normal adjustment, remember the pair: length L = fₒ + fₑ and magnification m = fₒ/fₑ. Any two given quantities fix the other two.

Answer from the NTA final answer key. Solution by PrepWiser.