Assuming that the human pupil has a radius of 0 . 25cm and a...

By Fraunhofer diffraction through a circular aperture,& $\theta = \dfrac{1 .22 \lambda}{d}$

$d =$ diameter of pupil~$= 2 \times \text{0.25} = \text{0.5}cm$
$\lambda = 500nm$. The first dark ring is formed by the light diffracted from the hole at an angle $\theta$ with the axis. Viewing distance,~$D = 25cm∴$~The minimum separation between $2$ objects $= D\theta= \dfrac{25 \times {10}^{- 2} \times 1.22 \times 500 \times {10}^{- 9}}{5 \times {10}^{- 3}}m$
$D\theta = 30 \times {10}^{- 6}m⇒ D\theta = 30\mu m$.