A paper, with two marks having separation d, is held normal to the line of sight of an observer at a distance of 50m. The diameter of the eye-lens of the observer is 2 mm. Which of the following is the least value of d, so that the marks can be seen as separate ? The mean wavelength of visible light may be taken as  5000 $\text{Å}$

Angular limit of resolution of eye, $\theta =\frac{\lambda }{d}$, where, d is diameter of eye lens.

Also, if y is the minimum separation between two objects at distance D from eye then

$\theta =\frac{y}{D}$

$\Rightarrow \frac{y}{D}=\frac{\lambda }{d}\Rightarrow y=\frac{\lambda D}{d}\dots \left(1\right)$

Here, wavelength $\lambda =5000 \text{Å}=5\times {10}^{-7}\mathrm{m}$

D = 50 m 

Diameter of eye lens= 2 mm = 2 x ${10}^{-3}$m 

From eq. (1), minimum separation is 

$\mathrm{y}=\frac{5\times {10}^{-7}\times 50}{2\times {10}^{-3}}=12.5\times {10}^{-3}\mathrm{m}=1.25\mathrm{cm}$