The intensity of each of the two slits in Young's double slit...

We know that, intensity, $I=4 I_0 \cos ^2 \frac{\phi}{2}$$\begin{aligned} & \text{ Case I } 2 I_0=4 I_0 \cos ^2 \frac{\phi}{2} \\ & \therefore \quad \phi=\frac{2 \pi}{\lambda} \cdot \Delta x \\ & \frac{\pi}{2}=\frac{2 \pi}{\lambda} \cdot \frac{Y_1 d}{D}\left[\because \Delta x=\frac{Y_1 d}{D} \text{\text{ and } } \phi=\frac{\pi}{2}\right] \\ & \Rightarrow \quad Y_1=\frac{\lambda D}{4 d}......(i) \\ & \end{aligned}$$$Case II$I_0=4 I_0 \cdot \cos ^2 \frac{\phi}{2} \therefore \quad \phi=\frac{2 \pi}{\lambda} \cdot \Delta x \frac{2 \pi}{3}=\frac{2 \pi}{\lambda} \cdot \frac{Y_2 d}{D} \quad\left[\because \phi=\frac{2 \pi}{3}\right. and \left.\Delta x \frac{y_2 d}{D}\right] \therefore \quad Y_2=\frac{\lambda D}{3 d}$.....(ii)$\therefore \quad \Delta Y=Y_2-Y_1=\frac{\lambda D}{3 d}-\frac{\lambda D}{4 d}=\frac{\lambda D}{12 d} \because \quad \frac{\lambda D}{d}=b$ [given] \Delta Y=\frac{b}{12}