Physics - Wave Optics Question with Solution | TestHub

The phase difference $\varphi$ corresponding to a path difference $\Delta x$ is given by $\varphi =\frac{2\pi }{\lambda }\Delta x$.

Substituting $\Delta x=\frac{\lambda }{6}$, we get $\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}$.

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

Given $I_{max}=K$, we have $I=K{\cos }^2\left(\frac{\pi }{6}\right)$.

Since $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$, we get $I=K{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3K}{4}$.

Answer: $\frac{3\mathrm{K}}{4}$