In Young's double slit experiment, the fringe width is \(2 \mathrm{~mm}\). The separation between the \(13^{\text {th }}\) bright fringe and the \(4^{\text {th }}\) dark fringe from the centre of the screen on same side will be

Given: Fringe width \(\mathrm{W}=2 \mathrm{~mm}\) The distance of the \(\mathrm{n}^{\text {th }}\) bright fringe from centre of the screen \(\mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) ... (i)
The distance of the \(\mathrm{n}^{\text {th }}\) dark fringe from centre of the screen \(y_n^{\prime}=(2 n-1) \frac{\lambda D}{2 d}\) ... (ii)
Substituting, \(\mathrm{n}=13\) in (i) and \(\mathrm{n}=4\) in (ii) we get \(\mathrm{y}_{13}=\frac{13 \lambda \mathrm{D}}{\mathrm{d}}\) and \(\mathrm{y}_4^{\prime}=\frac{7}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}\)
The separation between the \(13^{\text {th }}\) bright fringe and the \(7^{\text {th }}\) dark fringe is
\(\begin{aligned} & \quad \mathrm{y}_{13}-\mathrm{y}_4^{\prime} \\ & =\frac{13 \lambda \mathrm{D}}{\mathrm{d}}-\frac{7}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & =\left(13-\frac{7}{2}\right) \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & =\frac{19}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{19}{2} \mathrm{~W} \\ & \\ & \text { but } \mathrm{W}=2 \mathrm{~mm} \\ & \therefore \quad \mathrm{y}_{13}-\mathrm{y}_4^{\prime}=\frac{19}{2} \times 2=19 \mathrm{~mm}\end{aligned}\)