The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 Å is used to see the object will be: (a) 8.1 × 106 (b) 10.0 × 107 (c) 8.2 × 105 (d) 1.0 × 10 - 8

Answer: (c). Resolving power of telescope = (D)/(1.22λ). RP = 0.2441.22×2440×10⁻¹⁰ = 0.2442.977×10⁻⁷ RP ≈ 8.2×10⁵ So the answer is (c) 8.2×10⁵.

The aperture of the objective is 24.4 cm. The resolving power of this… | JEE Main 2022 (28 Jun, Shift 1)

Q13JEE Main 2022 · 28 JUN · SHIFT 1Medium

Resolving Power

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 $\text{Å}$ is used to see the object will be:

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Answer: (c)

Resolving power of telescope $= \dfrac{D}{1.22\lambda}$ .
$RP = \dfrac{0.244}{1.22\times2440\times10^{-10}} = \dfrac{0.244}{2.977\times10^{-7}}$
$RP \approx 8.2\times10^{5}$
So the answer is (c) $8.2\times10^{5}$ .

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