Physics - Wave Optics Question with Solution | TestHub

The intensity at the point due to interference is given as $I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$ For path difference $\frac{\lambda }{4}$, the phase difference is ${\varphi }_1=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$ For path difference $\frac{\lambda }{6}$, the phase difference is ${\varphi }_2=\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}$ Assuming equal intensity of the interfering waves i.e., ${\mathrm{I}}_1={\mathrm{I}}_2={\mathrm{I}}_0$ Equation (i) becomes, $$\begin{array}{rl}& I=I_0+I_0+2I_0\cos \varphi \\ & I=2I_0(1+\cos \varphi )\end{array}$$ For the given path difference, $I_1=2I_0\left(1+\cos \frac{\pi }{2}\right)$. $$\begin{array}{rl}& \text{ and }{I}_2=2I_0\left(1+\cos \frac{\pi }{3}\right)\\ & \therefore \frac{I_1}{I_2}=\frac{1+\cos \frac{\pi }{2}}{1+\cos \frac{\pi }{3}}\end{array}$$ $\begin{array}{rl}\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}& =\frac{1+0}{1+0.5}\\ \therefore \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}& =\frac{1}{1.5}=\frac{2}{3}\end{array}$