The aperture diameter of a telescope is $$5$$ m. The separation between the moon and the earth is $$4 \times 10^5$$ km. With light of wavelength $$5500$$ Å, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to:

We have an astronomical telescope whose objective (the front mirror) has a circular aperture of diameter $$D = 5\ \text{m}$$. When two point objects on the surface of the moon are imaged, the ability of the telescope to distinguish them separately is governed by the Rayleigh criterion for a circular aperture.

According to the Rayleigh criterion, the minimum angular separation $$\theta_{\min}$$ that can just be resolved is given by the formula

$$ \theta_{\min} \;=\; 1.22\,\dfrac{\lambda}{D}. $$

Here $$\lambda$$ is the wavelength of the light used and $$D$$ is the diameter of the aperture. First, let us put every quantity into SI units:

$$ \lambda = 5500\ \text{\AA} = 5500 \times 10^{-10}\ \text{m} = 5.5 \times 10^{-7}\ \text{m}. $$

The distance between the earth and the moon, which we denote by $$L$$, is given as

$$ L = 4 \times 10^{5}\ \text{km} = 4 \times 10^{5} \times 10^{3}\ \text{m} = 4 \times 10^{8}\ \text{m}. $$

Now we substitute $$\lambda$$ and $$D$$ into the Rayleigh formula:

$$ \theta_{\min} = 1.22 \,\dfrac{5.5 \times 10^{-7}\ \text{m}}{5\ \text{m}}. $$

Simplify the numerator first:

$$ 1.22 \times 5.5 = 6.71, $$

so

$$ 1.22 \times 5.5 \times 10^{-7}\ \text{m} = 6.71 \times 10^{-7}\ \text{m}. $$

Dividing by the diameter $$D = 5 \ \text{m}$$, we get

$$ \theta_{\min} = \dfrac{6.71 \times 10^{-7}}{5} = 1.342 \times 10^{-7}\ \text{radian}. $$

The linear separation $$s$$ on the moon’s surface that corresponds to this angular separation is obtained by simple geometry:

$$ s = L \,\theta_{\min}. $$

Substituting $$L = 4 \times 10^{8}\ \text{m}$$ and $$\theta_{\min} = 1.342 \times 10^{-7}\ \text{rad}$$, we have

$$ s = \left(4 \times 10^{8}\right)\,\left(1.342 \times 10^{-7}\right)\ \text{m}. $$

Multiply the numbers explicitly:

$$ 4 \times 1.342 = 5.368, $$

and combine the powers of ten:

$$ 10^{8} \times 10^{-7} = 10^{1}. $$

Therefore

$$ s = 5.368 \times 10^{1}\ \text{m} = 53.68\ \text{m}. $$

Rounding to the nearest convenient value, the minimum resolvable separation is approximately $$54\ \text{m}$$, which is closest to $$60\ \text{m}$$ in the given options.

Hence, the correct answer is Option A.