Physics - Wave Optics Question with Solution | TestHub

Width of central bright fringe${\beta }_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}}=\frac{2\lambda D}{a}$

Therefore$\alpha =\frac{{\beta }_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}}}{D}=\frac{2\lambda D}{aD}$

$\alpha =\frac{{\beta }_{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}}}{D}=\frac{2\lambda }{a}$