Answer
The distance between the two small objects is $3.05*10^{-5}m$
Work Step by Step
Rayleigh's criterion states that the minimum angular separation of two objects $\theta_{R}$ for them to be just resolved is given by:
$$sin\theta_{R}=1.22\frac{\lambda}{d}$$
where $\lambda$ is the incident wavelength on the circular aperture of diameter $d$.
In this question $d=5.0\ mm$, $\lambda=500\ nm$. Therefore to resolve two objects using this circular aperture, the minimum angular separation $\theta_{R}$ must be
$$\theta_{R}=arcsin(1.22\frac{500*10^{-9}}{5*10^{-3}})$$
Thus,
$$\theta_{R}=6.99*10^{-3}\ degrees$$
The distance of the objects is $250mm$ from the eye (our circular aperture). Hence we now have a triangular setup where the two objects have a vertical distance of $250mm$ from the eye, and have a minimum angular separation of $\theta_{R}=6.99*10^{-3}\ degrees$ from the eye. Hence the minimum distance between the objects can be calculated.
If we can bisect the triangle vertically, we get two halves of the triangle with the angle $\theta_{R}$ being divided into two. Hence now we have a right angle triangle with the height being $250mm$ and the angle opposite to the base is $0.5*\theta_{R}=3.495*10^{-3}$ degrees. Hence using trigonometry we get
$$tan(0.5\theta_{R})=\frac{base}{height}=\frac{base}{250*10^{-3}m}$$
Therefore plugging in our numbers we get:
$$base=1.52*10^{-5}m$$
Hence the distance between the two objects is $$2*base=3.05*10^{-5}m$$
