Physics - Wave Optics Question with Solution | TestHub

As, $\varphi =\frac{2\pi }{\lambda }\Delta x\Rightarrow \varphi =\frac{2\pi }{\lambda }\lambda$ $\varphi =2\pi$ $\therefore I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$ $=I^{\prime }+I^{\prime }+2I^{\prime }\cos 2\pi$ $I=2I^{\prime }+2I^{\prime }=4I^{\prime }$ When $\Delta x=\frac{\lambda }{6}$, then $\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{2\pi }{6}$ $\therefore I_1=I^{\prime }+I^{\prime }+2\sqrt{I^{\prime }{I}^{\prime }}\cos \frac{2\pi }{6}$ $I_1=2I^{\prime }+2I^{\prime }(1/2)$ $I_1=3I^{\prime }$ $\therefore \frac{I_1}{I}=\frac{3I^{\prime }}{4I^{\prime }}=\frac{3}{4}$ or $I_1=\frac{3}{4}I$