Physics - Wave Optics Question with Solution | TestHub

For ${\mathrm{n}}^{\text{th }}$ secondary minimum, path difference $=a\sin {\theta }_{\mathrm{n}}=\mathrm{n}\lambda$ For ${\mathrm{n}}^{\text{th }}$ secondary maximum, path difference $=a\sin {\theta }_n=(2n+1)\frac{\lambda }{2}$ $\therefore$ For $1^{\text{st }}$ minimum, a $\sin 30^{\circ }=\lambda$ For $2^{\text{nd }}$ maximum, a $\sin {\theta }_{\mathrm{n}}=(2+1)\frac{\lambda }{2}=\frac{3\lambda }{2}$ $\therefore$ Dividing equation (i) by equation (ii), $\frac{\left(\frac{1}{2}\right)}{\sin {\theta }_n}=\frac{2}{3}\Rightarrow {\theta }_n={\sin }^{-1}\left(\frac{3}{4}\right)$