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 2mm. Find the least value of d (in cm), so that the marks can be seen as separate? The mean wavelength of visible light may be taken as 5000 $\stackrel{0}{A}$

$$\begin{gathered} Angular limit of resolution of eye, \theta =\frac{\lambda }{d} , \\ where, d is diameter of eye lens. \\ Also, if y is the minimum separation between two objects at dis\tan ce D from eye then \\ \theta =\frac{y}{D} \\ \Rightarrow \frac{y}{D}=\frac{\lambda }{d} \\ \Rightarrow y=\frac{\lambda D}{d}\mathrm{..........................}\left(1\right) \\ Here, wavelength \lambda =5000\stackrel{0}{A}=5\times {10}^{-7}m \\ D=50m \\ Diameter of eye lens = 2 mm = 2\times {10}^{-3}m \\ From eq.\left(1\right) minimum separation is \\ y=\frac{5\times {10}^{-7}\times 50}{2\times {10}^{-3}}=12.5\times {10}^{-3}m=1.25cm \end{gathered}$$