In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 $$\mathring{\mathrm{A}}$$ is:

In Young's double slit experiment, the fringe width is given by $$\beta = \frac{\lambda D}{d}$$, where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation.

Here, $$d = 0.5$$ mm $$= 0.5 \times 10^{-3}$$ m, $$D = 0.5$$ m, and $$\lambda = 5890$$ angstrom $$= 5890 \times 10^{-10}$$ m. The distance between the first and third bright fringes is $$\Delta y = (3 - 1)\beta = 2\beta = \frac{2\lambda D}{d}$$.

Substituting: $$\Delta y = \frac{2 \times 5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}} = \frac{5890 \times 10^{-10}}{5 \times 10^{-4}} = 1178 \times 10^{-6}$$ m.