A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is $$x$$ nm. The value of $$x$$ to the nearest integer is ________.

The fringe width in Young's double slit experiment is given by $$\beta = \frac{\lambda D}{d}$$, where $$\lambda$$ is the wavelength of light, $$D$$ is the distance from the slits to the screen, and $$d$$ is the separation between the slits.

We are given $$\beta = 6$$ mm $$= 6 \times 10^{-3}$$ m, $$d = 1$$ mm $$= 1 \times 10^{-3}$$ m, and $$D = 10$$ m. Rearranging for $$\lambda$$, we get $$\lambda = \frac{\beta \cdot d}{D} = \frac{6 \times 10^{-3} \times 1 \times 10^{-3}}{10} = \frac{6 \times 10^{-6}}{10} = 6 \times 10^{-7}$$ m.

Converting to nanometres, $$\lambda = 6 \times 10^{-7} \text{ m} = 600 \text{ nm}$$. Therefore, $$x = 600$$.