Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light = 500 nm ] (a) 1 m (b) 5 m (c) 3 m (d) 6 m

Answer: (b). θ = (1.22λ)/(D) = 1.22×500×10⁻⁹3×10⁻³ = 2.03×10⁻⁴ rad Dots resolved while x/L ≥ θ, so Lmax = (x)/(θ). Lmax = 10⁻³2.03×10⁻⁴ ≈ 4.9 m So the answer is (b) 5 m.

Two point white dots are 1 mm apart on a black paper. They are viewed… | JEE Main 2005 (AIEEE)

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Resolving Power

Two point white dots are $1\text{ mm}$ apart on a black paper. They are viewed by eye of pupil diameter $3\text{ mm}.$ Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light $= 500\text{ nm}$ ]

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Answer: (b)

$\theta = \dfrac{1.22\lambda}{D} = \dfrac{1.22\times500\times10^{-9}}{3\times10^{-3}} = 2.03\times10^{-4}$ rad
Dots resolved while $\dfrac{x}{L} \ge \theta$ , so $L_{max} = \dfrac{x}{\theta}$ .
$L_{max} = \dfrac{10^{-3}}{2.03\times10^{-4}} \approx 4.9$ m
So the answer is (b) 5 m.

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