Physics - Wave Optics Question with Solution | TestHub

Position of $n$ th bright fringe from central $$\begin{array}{l}\text{ maxima }{x}_{r_1}=\frac{n_1\lambda D}{d}\text{ here }{n}_1=5\\ \therefore x_{n_4}=\frac{5\lambda D}{d}\end{array}$$ Position of $n$ th dark fringe from central maxima $$\begin{array}{rl}{x}_n& =\frac{(2n-1)\lambda D}{2d},\text{ Here }n=3\\ x_n& =\frac{5}{2}\frac{\lambda D}{d}\\ x_{r_4}-x_n& =\frac{2.5\lambda D}{d}=2.5\beta \end{array}$$ Given $\beta =0.4\mathrm{mm}$ $$\Rightarrow x_{n_1}-x_n=1\mathrm{mm}$$