In a Young's double slit experiment, a light source emitting two wavelengths $\lambda_1 = 400 \mathr

Question

In a Young's double slit experiment, a light source emitting two wavelengths $\lambda_1 = 400 \mathrm{~nm}$ and $\lambda_2 = 600 \mathrm{~nm}$ is used. The minimum non-zero distance from the central maximum where the bright fringes of both wavelengths perfectly coincide is $x$. This distance $x$ corresponds to which bright fringe of the $600 \mathrm{~nm}$ light?

Professor PC Thomas Classes · Answer

(a) : The condition for overlapping of bright fringes of two different wavelengths is $n_1\lambda_1 = n_2\lambda_2$. Substituting the given wavelengths, $n_1(400) = n_2(600) \implies \frac{n_1}{n_2} = \frac{600}{400} = \frac{3}{2}$. Thus, the 3rd bright fringe of the $400 \mathrm{~nm}$ light coincides exactly with the 2nd bright fringe of the $600 \mathrm{~nm}$ light.