Physics - Wave Optics Question with Solution | TestHub

Here  $d=0.12$ $\mathrm{mm}$  $=0.12\times {10}^{-3}$ $\mathrm{m}$  $=12\times {10}^{-5}$ $\mathrm{m}$ $\lambda =420$ $\mathrm{nm}$ $=420\times {10}^{-9}$ $\mathrm{m}$, $D=1.0$ $\mathrm{m}$The distance of the second bright fringe $x_2=\frac{2\lambda D}{d}\left[\because x_n=\frac{n\lambda D}{d}\right]$                                                                $\begin{array}{l}=\frac{2\times 420\times {10}^{-9}\times 1.0}{12\times {10}^{-5}}\\ =70\times {10}^{-4}=7\times {10}^{-3}\text{m}=7\text{ mm}\end{array}$