The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 Å is used, the minimum separation between two points, to be seen as distinct, will be: (a) 0.12 μ m (b) 0.38 μ m (c) 0.24 μ m (d) 0.48 μ m

Answer: (c). Minimum separation d = (1.22λ)/(2 (NA)), where NA = nsin θ. d = 1.22×5000×10⁻¹⁰2×1.25 = 2.44×10⁻⁷ m So the answer is (c) 0.24 μ m.

The value of numerical aperature of the objective lens of a… | JEE Main 2019 (12 Apr, Shift 1)

Q19JEE Main 2019 · 12 APR · SHIFT 1Medium

Resolving Power

The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 $\text{Å}$ is used, the minimum separation between two points, to be seen as distinct, will be:

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Answer: (c)

Minimum separation $d = \dfrac{1.22\lambda}{2\,(NA)}$ , where $NA = n\sin\theta$ .
$d = \dfrac{1.22\times5000\times10^{-10}}{2\times1.25} = 2.44\times10^{-7}$ m
So the answer is (c) $0.24\ \mu$ m.

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