Physics - Wave Optics Question with Solution | TestHub

Here  $\mathrm{d}=0.20 \mathrm{mm}=20\times {10}^{-5} \mathrm{m}$  $\mathrm{\lambda }=600 \mathrm{nm}=600\times {10}^{-9} \mathrm{m}$$\mathrm{D}=1 \mathrm{m}$ For the distance of the second bright fringe ${\mathrm{x}}_2=\frac{2\mathrm{\lambda D}}{\mathrm{d}}$                                                                $[\because {\mathrm{x}}_{\mathrm{n}}=\frac{\mathrm{n\lambda D}}{\mathrm{d}}]$  $=\frac{2\times 600\times {10}^{-9}\times 1.0}{20\times {10}^{-5}}$ $=60\times {10}^{-4}=6\times {10}^{-3} \text{m}=6\text{ mm}$