JEE Main 2020 · 8 January · morning · Single correct · +4 / −1

The magnifying power of a telescope with tube 60 cm is 5. What is the focal length of its eye piece ?

  1. (a) 40 cm
  2. (b) 10 cm
  3. (c) 30 cm
  4. (d) 20 cm
Answer and solution

Answer (b) 10 cm

Solution

Physics sir · 3 steps

The tube length is fₒ + fₑ and the magnification is fₒ / fₑ. Two equations, two unknowns.

  1. Two relations. For an astronomical telescope in normal adjustment, the tube length is L = fₒ + fₑ. The magnifying power is M = fₒ / fₑ.
  2. Use the magnification. M = 5 gives fₒ = 5 fₑ. Put this into the tube length.

    fₒ + fₑ = 5fₑ + fₑ = 6fₑ = 60 cm

  3. So. fₑ = 10 cm and fₒ = 50 cm. Check: 50 + 10 = 60 and 50/10 = 5. The eyepiece focal length is 10 cm, option (b).

The idea to keep. In normal adjustment the telescope tube is fₒ + fₑ long and M = fₒ/fₑ. Write fₒ = M fₑ, so the tube becomes (M+1) fₑ.

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

Study Ray Optics

Asked about 200 times in JEE Main since 2020. This question is from 2020.