The aperture diameter of a telescope is 5m. The separation between the moon and the earth is 4 × 105 km. With light of wavelength of 5500 Å, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to: (a) 20 m (b) 200 m (c) 600 m (d) 60 m

Answer: (d). θ = (1.22λ)/(D) = 1.22×5500×10⁻¹⁰5 = 1.34×10⁻⁷ rad Separation on the moon x = Lθ = 4×10⁸ × 1.34×10⁻⁷ x ≈ 54 m So the answer is (d) 60 m (closest).

The aperture diameter of a telescope is 5m. The separation between… | JEE Main 2020 (9 Jan, Shift 1)

Q16JEE Main 2020 · 9 JAN · SHIFT 1Medium

Resolving Power

The aperture diameter of a telescope is 5m. The separation between the moon and the earth is 4 × 105 km. With light of wavelength of 5500 $\text{Å}$ , the minimum separation between objects on the surface of moon, so that they are just resolved, is close to:

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Answer: (d)

$\theta = \dfrac{1.22\lambda}{D} = \dfrac{1.22\times5500\times10^{-10}}{5} = 1.34\times10^{-7}$ rad
Separation on the moon $x = L\theta = 4\times10^{8} \times 1.34\times10^{-7}$
$x \approx 54$ m
So the answer is (d) 60 m (closest).

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