Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.

For the limit of angular resolution produced by diffraction at a circular aperture we begin with the well-known Rayleigh criterion. The formula is stated first:

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

Here $$\theta_{\text{min}}$$ denotes the smallest angle (in radians) that the telescope can resolve, $$\lambda$$ is the wavelength of the light, and $$D$$ is the diameter of the objective lens or mirror.

We are given a wavelength $$\lambda = 500 \, \text{nm}$$ and an objective diameter $$D = 200 \, \text{cm}$$. Before substituting, both quantities must be converted to the same system of units (metres in SI).

Recall that $$1 \, \text{nm} = 10^{-9} \, \text{m}$$, so

$$\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}.$$

Also $$1 \, \text{cm} = 10^{-2} \, \text{m}$$, therefore

$$D = 200 \, \text{cm} = 200 \times 10^{-2} \, \text{m} = 2 \, \text{m}.$$

Now we substitute these values into the Rayleigh formula:

$$\theta_{\text{min}} = 1.22 \, \frac{500 \times 10^{-9} \, \text{m}}{2 \, \text{m}}.$$

First carry out the division in the numerator and denominator:

$$\frac{500 \times 10^{-9}}{2} = 250 \times 10^{-9}.$$

Next multiply by the factor $$1.22$$:

$$\theta_{\text{min}} = 1.22 \times 250 \times 10^{-9}.$$

Performing the multiplication of the numerical factors,

$$1.22 \times 250 = 305.$$

So we obtain

$$\theta_{\text{min}} = 305 \times 10^{-9} \, \text{radian}.$$

This value can also be expressed as $$3.05 \times 10^{-7} \, \text{radian}$$, but the form above matches one of the given options exactly.

Hence, the correct answer is Option A.