Physics - Wave Optics Question with Solution | TestHub

When path difference is $\lambda$, phase difference $\varphi$ is $2\pi$ and when path difference is $\frac{\lambda }{4}$, the phase difference is $\frac{\pi }{2}$. If ${\mathrm{I}}_0$ is the intensity of the two waves then the resultant intensity is given by $$\begin{array}{rl}I& =4I_0{\cos }^2\frac{\varphi }{2}\\ \therefore K& =4I_0{\cos }^2\frac{2\pi }{2}=4I{\cos }^2\pi =4I_0\because \cos \pi =-1\end{array}$$ $$\begin{array}{l}\text{ When }\varphi =\frac{\pi }{2}\\ \begin{array}{rl}\mathrm{I}=4{\mathrm{I}}_0{\cos }^2\cdot \frac{\pi }{4}& \\ & =4{\mathrm{I}}_0\frac{1}{2}=\frac{\mathrm{K}}{2}\because \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\end{array}\end{array}$$