Physics - Wave Optics Question with Solution | TestHub

The fringe width is given as $X$.

The position of the $n$-th bright fringe from the central maximum is $y_n=nX$.

The position of the $n$-th dark fringe from the central maximum is $y_n^{\prime }=\left(n-\frac{1}{2}\right)X$.

The distance of the sixth dark fringe from the central maximum is $y_6^{\prime }=\left(6-\frac{1}{2}\right)X=5.5X$.

The distance of the fourth bright fringe from the central maximum is $y_4=4X$.

Since the two fringes are on opposite sides of the central bright band, the total distance between them is the sum of their distances from the central maximum.

Total distance $=y_6^{\prime }+y_4=5.5X+4X=9.5X$.

Answer: 9.5 X