Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm coming from a distant object, the limit of resolution of the telescope is close to:

We are dealing with the diffraction-limited resolving power of a telescope whose objective is a circular aperture. For such an aperture, the smallest angular separation $$\theta_{\min}$$ that can be resolved is given by the Rayleigh criterion

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

where $$\lambda$$ is the wavelength of the light and $$D$$ is the diameter of the objective.

First we convert all quantities to SI units. The diameter is given as 250 cm:

$$D = 250\;\text{cm} = 250 \times 10^{-2}\;\text{m} = 2.5\;\text{m}$$

The wavelength is 600 nm:

$$\lambda = 600\;\text{nm} = 600 \times 10^{-9}\;\text{m}$$

We simplify the wavelength value for easier calculation:

$$600 \times 10^{-9}\;\text{m} = 6.0 \times 10^{2} \times 10^{-9}\;\text{m} = 6.0 \times 10^{-7}\;\text{m}$$

Now we substitute $$\lambda = 6.0 \times 10^{-7}\;\text{m}$$ and $$D = 2.5\;\text{m}$$ into the Rayleigh formula:

$$\theta_{\min}=1.22 \times \frac{6.0 \times 10^{-7}\;\text{m}}{2.5\;\text{m}}$$

We calculate the numerator first:

$$1.22 \times 6.0 \times 10^{-7} = 7.32 \times 10^{-7}$$

Now we divide by 2.5:

$$\theta_{\min} = \frac{7.32 \times 10^{-7}}{2.5}$$

To divide, we handle the coefficients and powers of ten separately:

$$\frac{7.32}{2.5} = 2.928$$

The power of ten remains $$10^{-7}$$, so

$$\theta_{\min} \approx 2.928 \times 10^{-7}\;\text{rad}$$

Rounding to two significant figures to match the precision of the options, we write

$$\theta_{\min} \approx 3.0 \times 10^{-7}\;\text{rad}$$

This value matches Option B.

Hence, the correct answer is Option B.