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 at any point on the screen is given by $I=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

For a path difference $\Delta x=\lambda$, the phase difference is $\varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$

The intensity 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 $\Delta x=\frac{\lambda }{6}$, the phase difference is $\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}$

The intensity at this point is $I_2=I_{max}{\cos }^2\left(\frac{\pi /3}{2}\right)=I_{max}{\cos }^2\left(\frac{\pi }{6}\right)$

Substituting $I_{max}=X$ and $\cos \left(\frac{\pi }{6}\right)=\cos 30^{\circ }=\frac{\sqrt{3}}{2}$:

$I_2=X{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3X}{4}$

Answer: $\frac{3X}{4}$