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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2022
  • Easy
  • MCQ

The aperture of the objective is \(24.4\,cm\). The resolving power of this telescope. If a light of wavelength \(2440 \mathring A\) is used to see the object will be

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Answer & Solution

Correct Answer

(C) \(8.2 \times 10^{5}\)

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Detailed explanation

\(R.P =\frac{ d }{1.22 \lambda}=\frac{24.4 \times 10^{-2}}{1.22 \times 2440 \times 10^{-10}}=8.2 \times 10^{5}\)
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