A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits wit

Question

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 $\_\_\_\_$ $	imes 10^{-5} \mathrm{~m}$.

Professor PC Thomas Classes · Answer

Let the $n_1$-th bright fringe of wavelength $\lambda_1 = 650 	ext{ nm}$ coincide with the $n_2$-th bright fringe of wavelength $\lambda_2 = 550 	ext{ nm}$. The position of a bright fringe is $y = \frac{n\lambda D}{d}$. For coincidence, $y_1 = y_2 \implies \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \implies n_1 \lambda_1 = n_2 \lambda_2$. Thus, $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$. The least distance occurs for the smallest integers, $n_1 = 11$ and $n_2 = 13$. Substituting $n_1$ back into the position formula: $y = \frac{11 	imes (650 	imes 10^{-9}) 	imes 1.2}{2 	imes 10^{-3}} = \frac{7150 	imes 1.2 	imes 10^{-6}}{2} = 3575 	imes 1.2 	imes 10^{-6} = 4290 	imes 10^{-6} 	ext{ m} = 429 	imes 10^{-5} 	ext{ m}$.