Physics - Wave Optics Question with Solution | TestHub

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

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

The intensity of light at a point is given by:

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

For a path difference of $\lambda$, the phase difference is:

${\varphi }_1=\frac{2\pi }{\lambda }\times \lambda =2\pi$

The intensity at this point is:

$I_1=I_{max}{\cos }^2\left(\frac{2\pi }{2}\right)=I_{max}{\cos }^2(\pi )=I_{max}$

Given that $I_1=x$, we have $I_{max}=x$.

For a path difference of $\frac{\lambda }{4}$, the phase difference is:

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

The intensity at this point is:

$I_2=I_{max}{\cos }^2\left(\frac{\pi /2}{2}\right)=x{\cos }^2\left(\frac{\pi }{4}\right)$

$I_2=x{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{x}{2}$

Answer: $\frac{x}{2}$