The value of numerical aperture of the objective lens of a microscope is 1.25. If light of wavelength 5000 Å is used, the minimum separation between two points, to be seen as distinct, will be:

We are told that the numerical aperture of the objective lens is $$\text{NA}=1.25$$ and that light of wavelength $$\lambda=5000\ \text{\AA}$$ is used.

First, the wavelength must be expressed in metres. We recall that $$1\ \text{\AA}=10^{-10}\ \text{m}.$$ Hence

$$ \lambda = 5000\ \text{\AA}=5000 \times 10^{-10}\ \text{m}=5 \times 10^{-7}\ \text{m}. $$

For a microscope objective, the Rayleigh criterion gives the minimum resolvable distance (limit of resolution) as

$$ d = \frac{0.61\,\lambda}{\text{NA}}. $$

Substituting the known values, we have

$$ d = \frac{0.61 \times 5 \times 10^{-7}\ \text{m}}{1.25}. $$

First multiply the numerator:

$$ 0.61 \times 5 = 3.05, $$

so

$$ d = \frac{3.05 \times 10^{-7}\ \text{m}}{1.25}. $$

Now divide by 1.25:

$$ \frac{3.05}{1.25}=2.44, $$

hence

$$ d = 2.44 \times 10^{-7}\ \text{m}. $$

To convert this result into micrometres, we remember that $$1\ \mu\text{m}=10^{-6}\ \text{m}.$$ Therefore

$$ d = 2.44 \times 10^{-7}\ \text{m} = 2.44 \times 10^{-7}\ \text{m} \times \frac{1\ \mu\text{m}}{10^{-6}\ \text{m}} = 0.244\ \mu\text{m}. $$

Rounded to two significant figures, this is $$0.24\ \mu\text{m}.$$

Hence, the correct answer is Option B.