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.)
Answer: (B) $25\ \mu$m
The first minima are at $\pm30^\circ$: $a\sin30^\circ = \lambda$, so $\lambda = 0.5\ \mu$m.
Fringe width $\beta = \dfrac{\lambda D}{d}$:
$$d = \frac{\lambda D}{\beta} = \frac{0.5\times10^{-6}\times0.5}{0.01} = 25\times10^{-6}\ \text{m} = 25\ \mu\text{m}$$
Solution by Sreeraj P, M.Sc Physics