Physics - Wave Optics Question with Solution | TestHub

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

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

Given $\Delta x=\frac{\lambda }{4}$, we have:

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

The intensity $I$ at a point is given by:

$I=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

Substituting $\varphi =\frac{\pi }{2}$:

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

Since $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$:

$I=I_0{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_0}{2}$

The ratio of $I_0$ to $I$ is:

$\frac{I_0}{I}=\frac{2}{1}$

Answer: $2:1$