Physics - Wave Optics Question with Solution | TestHub

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

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

Given $\Delta x=\frac{\lambda }{3}$, the phase difference is:

$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}$

The intensity $I$ at a point on the screen is given by:

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

Substituting the value of $\varphi$:

$I=I_0{\cos }^2\left(\frac{2\pi /3}{2}\right)=I_0{\cos }^2\left(\frac{\pi }{3}\right)$

Since $\cos \left(\frac{\pi }{3}\right)=\cos 60^{\circ }=\frac{1}{2}$, we get:

$I=I_0{\left(\frac{1}{2}\right)}^2=\frac{I_0}{4}$