JEE Mains · Physics · STD 12 - 10. Wave optics
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
- A \(8.1 \times 10^{6}\)
- B \(10.0 \times 10^{7}\)
- C \(8.2 \times 10^{5}\)
- D \(1.0 \times 10^{-8}\)
Answer & Solution
Correct Answer
(C) \(8.2 \times 10^{5}\)
Step-by-step Solution
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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