Briefly explain how bright and dark fringes are formed on the screen in Young's double slit experiment. Hence, derive the expression for the fringe width.

We get bright fringes, when waves from two coherent sources meet at a point on the screen with a phase difference $$\Delta \phi=2 n \pi \quad(\mathrm{n}=1,2,3 \ldots \ldots) We get dark fringes, when waves from two coherent sources meet at a point on the screen with a phase difference \Delta \phi=(2 n+1) \pi \quad(\mathrm{n}=1,2,3 \ldots \ldots) For maxima, $\displaystyle S_{2} P-S_{1} P=n \lambda \quad n=0,1,2,3 \ldots \ldots$ From figure, $\displaystyle \left(S_{2} P\right)^{2}-\left(S_{1} P\right)^{2}=\left[D^{2}+\left(x+\frac{d}{2}\right)^{2}\right]-\left[D^{2}+\left(x-\frac{d}{2}\right)^{2}\right]=2 x d$ \begin{align*} & S_{2} P-S_{1} P=\frac{2 x d}{S_{2} P-S_{1} P} \\ & S_{2} P \approx S_{1} P \approx D \\ & S_{2} P-S_{1} P=\frac{2 x d}{2 D}=\frac{x d}{D} \tag{2} \end{align*} From equation ($\displaystyle 1$) \& \begin{aligned} n \lambda & =\frac{x d}{D} \\ x_{n} & =\frac{n \lambda D}{d} \quad \text { for } n^{\text {th } \text { max ima }} \end{aligned} Similarly for $\displaystyle (\mathrm{n}+1)^{\mathrm{th}}$ maxima \begin{aligned} & x_{n+1}=\frac{(n+1) \lambda D}{d} \\ & +1-x_{n}=\frac{\lambda D}{d} \end{aligned} $$