JEE Main 2022 · 28 June · morning · Single correct · +4 / −1

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 Aᵒ is used to see th object will be :

  1. (a) 8.1 × 10⁶
  2. (b) 10.0 × 10⁷
  3. (c) 8.2 × 10⁵
  4. (d) 1.0 × 10⁻⁸
Answer and solution

Answer (c) 8.2 × 10⁵

Solution

Physics sir · 4 steps

Resolving power is just aperture over 1.22λ, so put both in metres and divide.

  1. Use the formula. The smallest angle a telescope can resolve is Δθ = 1.22λ/D. Resolving power is the reciprocal of this angle.

    RP = 1/Δθ = D/(1.22 λ)

  2. Same units. D = 24.4 cm = 0.244 m and λ = 2440 Å = 2.44 × 10⁻⁷ m. So 1.22λ = 2.9768 × 10⁻⁷ m.
  3. Divide. RP = 0.244/(2.9768 × 10⁻⁷) = 8.197 × 10⁵. It is a pure number because the metres cancel.
  4. So. The resolving power is about 8.2 × 10⁵, which is option (c).

The idea to keep. For a telescope, RP = D/1.22λ. A bigger aperture or a shorter wavelength resolves better. Always convert D and λ to the same unit before dividing.

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

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Asked about 140 times in JEE Main since 2020. This question is from 2022.