A telescope of aperture $$3 \times 10^{-2}$$ m diameter is focused on a window at 80 m distance fitted with a wire mesh of spacing $$2 \times 10^{-3}$$ m. Given: $$\lambda = 5.5 \times 10^{-7}$$ m, which of the following is true for observing the mesh through the telescope?

The smallest angular separation that a circular aperture can resolve is given by the Rayleigh criterion: $$\theta_{\min}=1.22\frac{\lambda}{D}\,,-(1)$$ where $$\lambda$$ is the wavelength of light and $$D$$ is the diameter of the aperture.

Data given:
Diameter of the telescope objective: $$D = 3\times 10^{-2}\,\text{m}$$
Wavelength of light: $$\lambda = 5.5\times 10^{-7}\,\text{m}$$

Substituting in $$(1)$$,
$$\theta_{\min}=1.22\frac{5.5\times 10^{-7}}{3\times 10^{-2}}=1.22\times\frac{5.5}{3}\times10^{-5}\,\text{rad}$$
$$\theta_{\min}\approx2.24\times10^{-5}\,\text{rad}$$

The telescope is focused on a mesh that is $$L = 80\,\text{m}$$ away. The minimum linear separation on the mesh that can just be resolved is
$$s_{\min}=L\,\theta_{\min}=80\times2.24\times10^{-5}\,\text{m}$$
$$s_{\min}\approx1.8\times10^{-3}\,\text{m}=1.8\,\text{mm}$$

The actual spacing between adjacent wires of the mesh is $$a = 2\times10^{-3}\,\text{m}=2.0\,\text{mm}$$, which is slightly larger than $$s_{\min}$$.

Since the mesh spacing $$a$$ is greater than the telescope’s resolution limit $$s_{\min}$$, the individual wires can indeed be distinguished. Hence the mesh can be observed (resolved) with the given aperture.

Check for half the aperture (optional):
If $$D$$ is halved, $$\theta_{\min}$$ doubles and $$s_{\min}$$ becomes approximately $$3.6\,\text{mm} \gt 2.0\,\text{mm}$$, so resolution would be lost. Therefore only the full diameter works.

Therefore, the correct statement is:
Option A which is: Yes, it is possible with the same aperture size.