A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is______$$\times10^{-5}$$

Wavelengths: $$\lambda_1 = 650$$ nm and $$\lambda_2 = 550$$ nm, slit separation: $$d = 2$$ mm $$= 2 \times 10^{-3}$$ m, and screen distance: $$D = 1.2$$ m.

The bright fringe positions for wavelength $$\lambda$$ in Young's double slit experiment are given by $$y_n = \frac{n \lambda D}{d}$$.

For bright fringes of both wavelengths to coincide at the same point, we require $$n_1 \lambda_1 = n_2 \lambda_2$$.

Since $$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$$, the smallest positive integers satisfying this are $$n_1 = 11$$ and $$n_2 = 13$$.

The distance of their least coincidence from the central maximum is $$y = \frac{n_1 \lambda_1 D}{d} = \frac{11 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$$.

$$y = \frac{11 \times 650 \times 1.2}{2} \times 10^{-9+3}$$

$$y = \frac{8580}{2} \times 10^{-6} = 4290 \times 10^{-6} = 429 \times 10^{-5} \text{ m}$$

The answer is 429.