In Young's experiment,obtain the distance between two consecutive bright fringes and two consecutive dark fringes.

(N/A) For fringes of constructive interference (bright fringes):
The position of the $n^{\text{th}}$ bright fringe is given by $x_{n} = \frac{n \lambda D}{d}$.
The position of the $(n+1)^{\text{th}}$ bright fringe is $x_{n+1} = \frac{(n+1) \lambda D}{d}$.
The distance between two consecutive bright fringes is $\beta = x_{n+1} - x_{n} = \frac{(n+1) \lambda D}{d} - \frac{n \lambda D}{d} = \frac{\lambda D}{d} (n+1-n) = \frac{\lambda D}{d}$.
For fringes of destructive interference (dark fringes):
The position of the $n^{\text{th}}$ dark fringe is $x'_{n} = (2n-1) \frac{\lambda D}{2d}$ (for $n=1, 2, 3...$).
The position of the $(n+1)^{\text{th}}$ dark fringe is $x'_{n+1} = (2(n+1)-1) \frac{\lambda D}{2d} = (2n+1) \frac{\lambda D}{2d}$.
The distance between two consecutive dark fringes is $\beta' = x'_{n+1} - x'_{n} = (2n+1) \frac{\lambda D}{2d} - (2n-1) \frac{\lambda D}{2d} = \frac{\lambda D}{2d} (2n+1-2n+1) = \frac{\lambda D}{2d} (2) = \frac{\lambda D}{d}$.
Thus,the distance between two consecutive bright fringes and two consecutive dark fringes is the same,given by $\beta = \frac{\lambda D}{d}$,which is known as the fringe width.