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 ______ $\times10^{-5}$ m.
Numerical value type. Enter your answer.
Answer: 429
Bright fringes coincide when $n_1\lambda_1 = n_2\lambda_2$:
$$650\,n_1 = 550\,n_2 \Rightarrow 13n_1 = 11n_2$$
Smallest integers: $n_1 = 11$, $n_2 = 13$.
$$y = \frac{n_1\lambda_1D}{d} = \frac{11\times650\times10^{-9}\times1.2}{2\times10^{-3}} = 4.29\times10^{-3}\ \text{m} = 429\times10^{-5}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics