The separation between the consecutive dark fringes in a Young's double slit experiment is 1⋅0 mm. The screen is placed at a distance of 2⋅5 m from the slits and the separation between the slits is 1⋅0 mm. Calculate the wavelength of light used for the experiment.

Given:
Separation between consecutive dark fringes = fringe width (β) = 1 mm = 10⁻³ m
Distance between screen and slit (D) = 2.5 m
The separation between slits (d) = 1 mm = 10⁻³ m
Let the wavelength of the light used in experiment be λ.
We know that
$\beta =\frac{\lambda D}{d}$
$$\begin{gathered} {10}^{-3}\mathrm{m}=\frac{2.5\times \lambda }{{10}^{-3}} \\ \Rightarrow \lambda =\frac{1}{2.5}{10}^{-6}\mathrm{m} \\ =4\times {10}^{-7}\mathrm{m}=400\mathrm{nm} \end{gathered}$$

Hence, the wavelength of light used for the experiment is 400 nm.