QJEEM20Ph92e0: JEE Main 2020 Physics - Class 12 JEEM20Ph92e0 Solution: The aperture diameter of a telescope is 5m. The separation…

Question

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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

\overset{o}{A}

, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :

Accepted Answer

θ

= 1.22

\frac{λ}{a}

Distance O₁O₂ = (

θ

)d

= (1.22

\frac{λ}{a}

)d

=

\frac{(1.22) (5500 \times 10^{- 10}) (4 \times 10^{5}) \times 10^{3}}{5}

= 5368 × 10^–2 m

= 53.68 m

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