JEE Main 2022 Physics question

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:

Answer: (3)

Solution

The resolving power is R.P. = $1/(1.22λ /a) = a/(1.22λ ) = 24.4

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Question ID VK-034908