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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2020
  • Medium
  • Numerical

The aperture diameter of a telescope is \(5\; \mathrm{m} .\) The separation between the moon and the earth is \(4 \times 10^{5} \;\mathrm{km} .\) With light of wavelength of \(5500\;\mathring A\), the minimum separation between objects on the surface of moon, so that they are just resolved, is close to......\(m\)

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Answer & Solution

Correct Answer

(C) \(60\)

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Detailed explanation

Let distance is \(x\) then

\(\mathrm{d} \theta=\frac{1.22 \lambda}{\mathrm{D}}(\mathrm{D}=\text { diameter })\)

\(\frac{\mathrm{x}}{\mathrm{d}}=\frac{1.22 \lambda}{\mathrm{D}}(\mathrm{d}=\text { distance between earth }\;and\; \text { moon })\)

\(\mathrm{x}=\frac{1.22 \times\left(5500 \times 10^{-10}\right) \times\left(4 \times 10^{8}\right)}{5}=53.68 \mathrm{m}\)

most appropriate is \(60 \mathrm{m}\)

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