In Young's double slit experiment, two coherent sources are $0.8 \mathrm{~mm}$ apart and the interfe

Question

In Young's double slit experiment, two coherent sources are $0.8 \mathrm{~mm}$ apart and the interference pattern is observed on a screen $1.2 \mathrm{~m}$ away. If the third dark fringe is located at a physical distance of $1.5 \mathrm{~mm}$ from the central maximum, the wavelength of the monochromatic light used is

Professor PC Thomas Classes · Answer

(a) : The position of the $n$-th dark fringe from the center is $y_n = (2n-1)\frac{\lambda D}{2d}$. For the 3rd dark fringe ($n=3$), $y_3 = \frac{5\lambda D}{2d}$. Substituting the given values: $1.5 	imes 10^{-3} = \frac{5 \lambda 	imes 1.2}{2 	imes 0.8 	imes 10^{-3}} = \frac{6 \lambda}{1.6 	imes 10^{-3}}$. Rearranging for $\lambda$ yields $\lambda = \frac{1.5 	imes 10^{-3} 	imes 1.6 	imes 10^{-3}}{6} = \frac{2.4 	imes 10^{-6}}{6} = 0.4 	imes 10^{-6} \mathrm{~m} = 400 \mathrm{~nm}$.