In Young's double slit experiment, the two slits are \(d\) distance apart. Interference pattern is observed on the screen at a distance \(D\) from the slits. Fist dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of the light is.

Consider the following diagram:

The dark fringe is produced at a point just opposite to the slit \(S_1\), i.e.,
\(y_1=\frac{d}{2}\)
For dark fringe, path difference is odd multiple of half of the wavelength:
\(\Delta x=(2 n-1) \frac{\lambda}{2}\)
From the figure, relation between fringe location and path difference can be obtained as:
\(\tan \theta=\frac{y}{D} \approx \sin \theta=\frac{\Delta x}{d} \)
\( \therefore y=\frac{\Delta x D}{a}=(2 n-1)\left(\frac{\lambda}{2}\right) \cdot\left(\frac{D}{a}\right)\)\(=(2 n-1) \frac{\lambda D}{2 d}\)
where \(n=1,2,3, \ldots \ldots \ldots\)
\(\begin{aligned} & \Rightarrow \frac{d}{2}=(2 n-1) \frac{\lambda D}{2 d} \\ & \Rightarrow \lambda=\frac{d^2}{(2 n-1) D}\end{aligned}\)
For \(n=1\),
\(\Rightarrow \lambda=\frac{d^2}{D}\)