In Young's double slit experiment, the distance between the two coherent sources is ' d ' and the distance between the source and screen is ' D '. When the wavelength ( \(\lambda\) ) of light source used is \(\frac{d^2}{3 D}\), then \(n^{\text {th }}\) dark fringe is observed on the screen, exactly in front of one of the slits. The value of ' \(n\) ' is

\(\begin{aligned} \mathrm{S}_2 \mathrm{P} & =\left(\mathrm{D}^2+\mathrm{d}^2\right)^{1 / 2} \\ & =\mathrm{D}\left[1+\frac{\mathrm{d}^2}{\mathrm{D}^2}\right]^{1 / 2}\end{aligned}\)

Using binomial equation,
\(\begin{aligned}
& \mathrm{S}_2 \mathrm{P}=\mathrm{D}\left[1+\frac{1}{2} \frac{\mathrm{~d}^2}{\mathrm{D}^2}\right]^{1 / 2}=\mathrm{D}+\frac{\mathrm{d}^2}{2 \mathrm{D}} \\
& \Rightarrow \text { Path difference }=\frac{\mathrm{d}^2}{2 \mathrm{D}}
\end{aligned}\)
For dark fringe, \(\frac{\mathrm{d}^2}{2 \mathrm{D}}=(2 \mathrm{n}-1) \frac{\lambda}{2}\)
\(\begin{aligned} & \therefore \quad \frac{\mathrm{d}^2}{2 \mathrm{D}}=(2 \mathrm{n}-1) \frac{\mathrm{d}^2}{6 \mathrm{D}} \\ & \therefore \quad 2 \mathrm{n}-1=3 \\ & \therefore \quad n=2\end{aligned}\)