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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2019
  • Medium
  • MCQ

The value of numerical aperature of the objective lens of a microscope is \(1.25\). If light of wavelength \(5000\,\mathop A\limits^o \) is used, the minimum separation between two points, to be seen as distinct, will be....\(\mu m\)

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Answer & Solution

Correct Answer

(C) \(0.24\)

Step-by-step Solution

Detailed explanation

Numerical aperature of the microscope is given as

\({\text{NA}} = \frac{{0.61\,\lambda }}{{\text{d}}}\)

Where \(\mathrm{d}=\) minimum sparaton between two points to be seen as distinct

\({\text{d}} = \frac{{0.61\,\,\lambda }}{{{\text{NA}}}} = \) \(\frac{{(0.61) \times \left( {5000 \times 10\,{{\text{m}}^{ - 10}}} \right)}}{{1.25}}\) \( = 2.4 \times {10^{ - 7}}{\text{m}}\)

\( = 0.24\,\mu m\)

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