The human eye has an approximate angular resolution of $\phi = 5.8 \times 10^{-4} \, rad$ and a typical photoprinter prints a minimum of $300 \, dpi$ (dots per inch, $1 \, inch = 2.54 \, cm$). At what minimal distance $z$ should a printed page be held so that one does not see the individual dots?

(N/A) Let $S_{1}$ and $S_{2}$ be two consecutive dots on the paper.
As shown in the figure, suppose the images of two consecutive dots $S_{1}$ and $S_{2}$ (on the paper at distance $Z$ from the eye) can be seen clearly and just separated by the eye. Here, the distance $d_{m}$ between $S_{1}$ and $S_{2}$ is called the 'linear limit of resolution' of the eye, and the angle $\alpha_{\min} = \phi$ is called the 'angular limit of resolution' of the eye.
According to the definition of the measurement of an angle in radians:
$\text{Angle} = \frac{\text{arc}}{\text{radius}}$
Therefore, $\alpha_{\min} = \frac{d_{m}}{Z}$
Given that the printer prints $300 \, dots$ per $2.54 \, cm$, the distance between two consecutive dots is:
$d_{m} = \frac{2.54 \, cm}{300} \approx 8.467 \times 10^{-3} \, cm$
Substituting the values into the formula:
$Z = \frac{d_{m}}{\phi} = \frac{2.54 \, cm / 300}{5.8 \times 10^{-4} \, rad}$
$Z = \frac{2.54}{300 \times 5.8 \times 10^{-4}} \, cm$
$Z = \frac{2.54}{0.174} \, cm \approx 14.6 \, cm$
If the paper is kept at a distance greater than $14.6 \, cm$, the images of $S_{1}$ and $S_{2}$ cannot be seen as separated. Hence, the required minimum distance is $Z = 14.6 \, cm$.