Physics - Wave Optics Question with Solution | TestHub

We know that a phase change of $\pi$ occurs when the reflection takes place at the boundary of denser medium. This is equivalent to a path difference of $\lambda /2$. $\therefore$ Total phase difference $=\pi -\pi =0$ Thus, the two waves superimpose in phase. $\begin{array}{rl}& \text{ Resultant amplitude }=\sqrt{I}+\sqrt{\left(\frac{I}{4}\right)}=\frac{3}{2}\sqrt{I}\\ & \text{ Resultant intensity }={\left(\frac{3}{2}\sqrt{I}\right)}^2=\frac{9}{4}I=\frac{9}{4}\text{ fold. }\end{array}$