Question

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

Answer 1

Assuming human pupil to have a radius r = 0.25 cm or diameter d = 2r = 0.5 cm
The wavelength of light is λ = 500 nm = 5 $\times$ ${10}^{-7}$ m.
sin $\theta =\frac{1.22\lambda }{d}$
$\sin$ $\theta$ = $\frac{1.22\times 5\times {10}^{-7}}{0.5\times {10}^{-2}}$ = 1.22 $\times$ ${10}^{-4}$
The distance of comfortable viewing is D = 25 cm = 0.25 m.
Let x be the minimum separation between two objects that the human eye can resolve.
$\sin$ $\theta$ = $\frac{x}{D}$
D $\sin$ $\theta$ = 0.25 $\times$ 1.22 $\times$ ${10}^{-4}$ = 3 $\times$ ${10}^{-5}$ m = 30 $\mu$m.