Physics - Wave Optics Question with Solution | TestHub

The phase difference $\varphi$ is related to the path difference $\Delta x$ by the formula:

$\varphi =\frac{2\pi }{\lambda }\Delta x$

For point $P_1$, the path difference is $\Delta x_1=\frac{\lambda }{6}$

${\varphi }_1=\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}=60^{\circ }$

The intensity at a point is given by $I=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

$I_1=I_{max}{\cos }^2(30^{\circ })=I_{max}{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}{I}_{max}$

For point $P_2$, the path difference is $\Delta x_2=\frac{\lambda }{3}$

${\varphi }_2=\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}=120^{\circ }$

$I_2=I_{max}{\cos }^2(60^{\circ })=I_{max}{\left(\frac{1}{2}\right)}^2=\frac{1}{4}{I}_{max}$

The ratio of intensities is:

$\frac{I_1}{I_2}=\frac{\frac{3}{4}{I}_{max}}{\frac{1}{4}{I}_{max}}=\frac{3}{1}$

Answer: $3:1$