The angular width of the central maximum in a single slit diffraction pattern is 60$$^\circ$$. The width of the slit is 1 $$\mu$$m. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., the distance between the centres of each slit.)

We are told that the angular width of the central maximum in the single-slit diffraction pattern is $$60^\circ$$. By definition, the angular width of the central maximum equals the angle between the first minima on either side of the central bright band. Hence if we denote the angle of the first minimum (on one side) by $$\theta_1$$, we have

$$2\theta_1 = 60^\circ \Longrightarrow \theta_1 = 30^\circ.$$

For a single slit of width $$a$$ illuminated by monochromatic light of wavelength $$\lambda$$, the condition for the first diffraction minimum is stated by the formula

$$a \sin \theta_1 = m\lambda, \quad m = 1.$$

Substituting $$m = 1$$, $$a = 1\;\mu\text{m} = 1 \times 10^{-6}\,\text{m}$$ and $$\theta_1 = 30^\circ$$, we get

$$\lambda = a \sin \theta_1 = 1 \times 10^{-6}\,\text{m}\; \sin 30^\circ = 1 \times 10^{-6}\,\text{m}\; \times \frac12 = 0.5 \times 10^{-6}\,\text{m} = 5 \times 10^{-7}\,\text{m}.$$

Thus the wavelength of the monochromatic light is $$\lambda = 5 \times 10^{-7}\,\text{m} \, (500\;\text{nm}).$$

Now a second identical slit is made, producing Young’s double-slit interference. The distance between the slits is denoted by $$d$$ and the screen is placed at a distance $$D = 50\;\text{cm} = 0.5\;\text{m}$$. In Young’s experiment the fringe width (distance between successive bright or dark fringes) is given by the formula

$$\beta = \frac{\lambda D}{d}.$$

The observed fringe width is $$\beta = 1\;\text{cm} = 0.01\;\text{m}$$. Substituting the known values into the formula, we have

$$0.01 = \frac{(5 \times 10^{-7}) \times 0.5}{d}.$$

First multiply the numerator:

$$(5 \times 10^{-7}) \times 0.5 = 2.5 \times 10^{-7}.$$

Now solve for $$d$$:

$$d = \frac{2.5 \times 10^{-7}}{0.01} = 2.5 \times 10^{-7} \times \frac{1}{10^{-2}} = 2.5 \times 10^{-7} \times 10^{2} = 2.5 \times 10^{-5}\,\text{m}.$$

Convert metres to micrometres (since $$1\;\mu\text{m} = 10^{-6}\,\text{m}$$):

$$d = 2.5 \times 10^{-5}\,\text{m} = \frac{2.5 \times 10^{-5}}{10^{-6}}\;\mu\text{m} = 25\;\mu\text{m}.$$

Hence, the correct answer is Option B.