Physics - Wave Optics Question with Solution | TestHub

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 $y_n^{\prime }=(2n-1)\frac{\lambda D}{2d}$ ... (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 $\begin{array}{rl}& {\mathrm{y}}_{13}-{\mathrm{y}}_4^{\prime }\\ & =\frac{13\lambda \mathrm{D}}{\mathrm{d}}-\frac{7}{2}\frac{\lambda \mathrm{D}}{\mathrm{d}}\\ & =\left(13-\frac{7}{2}\right)\frac{\lambda \mathrm{D}}{\mathrm{d}}\\ & =\frac{19}{2}\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{19}{2}\mathrm{W}\\ & \\ & \text{ but }\mathrm{W}=2\mathrm{mm}\\ & \therefore {\mathrm{y}}_{13}-{\mathrm{y}}_4^{\prime }=\frac{19}{2}\times 2=19\mathrm{mm}\end{array}$