Take the pupil of an eye to have a diameter of 6.0 mm…

Step 1. The angle. $\theta_{\min} = \dfrac{1.22\lambda}{D} = \dfrac{1.22 \times 550 \times 10^{-9}\ \text{m}}{6.0 \times 10^{-3}\ \text{m}} = 1.12 \times 10^{-4}\ \text{rad}$.

Step 2. The smallest distance at $300$ m. $s = \theta_{\min} L = 1.12 \times 10^{-4} \times 300\ \text{m} = 0.0336\ \text{m} = 3.36\ \text{cm}$.

Step 3. The greatest distance for points $5.0$ cm apart. $L = \dfrac{s}{\theta_{\min}} = \dfrac{5.0 \times 10^{-2}\ \text{m}}{1.12 \times 10^{-4}} = 447\ \text{m}$.