Physics - Wave Optics Question with Solution | TestHub

We know that, intensity, $I=4I_0{\cos }^2\frac{\varphi }{2}$ $\begin{array}{rl}& \text{ Case I }2I_0=4I_0{\cos }^2\frac{\varphi }{2}\\ & \therefore \varphi =\frac{2\pi }{\lambda }\cdot \Delta x\\ & \frac{\pi }{2}=\frac{2\pi }{\lambda }\cdot \frac{Y_{1}d}{D}\left[\because \Delta x=\frac{Y_{1}d}{D}\text{ and }\varphi =\frac{\pi }{2}\right]\\ & \Rightarrow Y_1=\frac{\lambda D}{4d}......(i)\\ & \end{array}$ Case II $I_0=4I_0\cdot {\cos }^2\frac{\varphi }{2}$ $\therefore \varphi =\frac{2\pi }{\lambda }\cdot \Delta x$ $\frac{2\pi }{3}=\frac{2\pi }{\lambda }\cdot \frac{Y_{2}d}{D}[\because \varphi =\frac{2\pi }{3}$ and $\Delta x\frac{y_{2}d}{D}]$ $\therefore Y_2=\frac{\lambda D}{3d}$.....(ii) $\therefore \Delta Y=Y_2-Y_1=\frac{\lambda D}{3d}-\frac{\lambda D}{4d}=\frac{\lambda D}{12d}$ $\because \frac{\lambda D}{d}=b$ [given] $\Delta Y=\frac{b}{12}$