Physics - Wave Optics Question with Solution | TestHub

The phase difference $\varphi$ is related to the path difference $\Delta x$ by the relation $\varphi =\frac{2\pi }{\lambda }\Delta x$.

The intensity at any point on the screen is given by $I=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$, where $I_0$ is the maximum intensity.

For the first point, the path difference is $\Delta x_1=0$.

${\varphi }_1=0\Rightarrow I_1=I_0{\cos }^2(0)=I_0$

For the second point, the path difference is $\Delta x_2=\frac{\lambda }{4}$.

${\varphi }_2=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$

$I_2=I_0{\cos }^2\left(\frac{\pi }{4}\right)=I_0{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_0}{2}$

The ratio of intensities is $\frac{I_1}{I_2}=\frac{I_0}{I_0/2}=\frac{2}{1}$.

Answer: $2:1$