Assuming that the human pupil has a radius of 0.25 cm and a comfortable viewing distance of 25 cm. The minimum separation between two point objects that the human eye can resolve for the light of wavelength 500 nm is

For the eye, the minimum angular separation that can be distinguished is governed by the Rayleigh criterion for a circular aperture. We first state the formula:

$$\theta_{\min}=1.22\,\frac{\lambda}{D}$$

Here $$\theta_{\min}$$ is the least resolvable angle (in radians), $$\lambda$$ is the wavelength of light and $$D$$ is the diameter of the aperture, which in this case is the pupil.

We are given the radius of the pupil as $$r=0.25\ \text{cm}$$. Converting this into metres, we write

$$r=0.25\ \text{cm}=0.25\times10^{-2}\ \text{m}=2.5\times10^{-3}\ \text{m}.$$

Hence the diameter of the pupil is twice the radius:

$$D = 2r = 2\left(2.5\times10^{-3}\ \text{m}\right)=5.0\times10^{-3}\ \text{m}.$$

The wavelength of the light used is $$\lambda = 500\ \text{nm}$$. Converting to metres,

$$\lambda = 500\ \text{nm}=500\times10^{-9}\ \text{m}=5.0\times10^{-7}\ \text{m}.$$

Now we substitute these values into the Rayleigh formula:

$$\theta_{\min}=1.22\,\frac{\lambda}{D}=1.22\,\frac{5.0\times10^{-7}\ \text{m}}{5.0\times10^{-3}\ \text{m}}.$$

Dividing the powers of ten first, we have

$$\frac{5.0\times10^{-7}}{5.0\times10^{-3}}=\frac{5.0}{5.0}\times10^{-7-(-3)}=1\times10^{-4}=10^{-4}.$$

Hence

$$\theta_{\min}=1.22\times10^{-4}\ \text{radian}.$$

The problem finally asks for the linear or spatial separation of two point objects that can just be resolved when they are at the comfortable viewing distance of the eye, given here as $$d=25\ \text{cm}$$. Converting the distance into metres, we get

$$d=25\ \text{cm}=25\times10^{-2}\ \text{m}=0.25\ \text{m}.$$

The small-angle approximation relates the minimum separation $$s_{\min}$$ on the object to the minimum angle by

$$s_{\min} = d\,\theta_{\min}.$$

Substituting $$d=0.25\ \text{m}$$ and $$\theta_{\min}=1.22\times10^{-4}\ \text{rad}$$, we find

$$s_{\min}=0.25\ \text{m}\times1.22\times10^{-4}=0.305\times10^{-4}\ \text{m}.$$

Multiplying,

$$0.305\times10^{-4}\ \text{m}=3.05\times10^{-5}\ \text{m}.$$

To express this in micrometres (1 μm = 10⁻⁶ m):

$$3.05\times10^{-5}\ \text{m}=3.05\times10^{-5}\ \text{m}\times\frac{10^{6}\ \mu\text{m}}{1\ \text{m}}=30.5\ \mu\text{m}.$$

This is approximately $$30\ \mu\text{m}.$$ Comparing with the given options, this corresponds to option C.

Hence, the correct answer is Option C.