Physics - Wave Optics Question with Solution | TestHub

If $a$ is the amplitude of the wave then $\frac{I_{\max }}{4}=a^2=a^2+a^2+2aa\cos \varphi$ or $\cos \varphi =-\frac{1}{2}$ or $\varphi =\frac{2\pi }{3}$. Corresponding path difference, $\Delta x=\frac{\varphi \times \lambda }{2\pi }=\frac{(2\pi /3)\times \lambda }{2\pi }=\frac{\lambda }{3}$ So $d\sin \theta =\frac{\lambda }{3}$ or $\theta ={\sin }^{-1}\left(\frac{\lambda }{3d}\right)$