Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is: (a) 100 (b) 300 (c) 1 (d) 30

Answer: (d). Pupil diameter D = 2 × 0.25 = 0.5 cm = 5×10⁻³ m. θ = (1.22λ)/(D) = 1.22×500×10⁻⁹5×10⁻³ = 1.22×10⁻⁴ rad x = 0.25 × 1.22×10⁻⁴ ≈ 3×10⁻⁵ m So the answer is (d) 30 μ m.

Assuming human pupil to have a radius of 0.25 cm and a comfortable… | JEE Main 2015

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Resolving Power

Assuming human pupil to have a radius of $0.25\text{ cm}$ and a comfortable viewing distance of $25\text{ cm}$ , the minimum separation between two objects that human eye can resolve at $500\text{ nm}$ wavelength is:

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Answer: (d)

Pupil diameter $D = 2 \times 0.25 = 0.5$ cm $= 5\times10^{-3}$ m.
$\theta = \dfrac{1.22\lambda}{D} = \dfrac{1.22\times500\times10^{-9}}{5\times10^{-3}} = 1.22\times10^{-4}$ rad
$x = 0.25 \times 1.22\times10^{-4} \approx 3\times10^{-5}$ m
So the answer is (d) $30\ \mu$ m.

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