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

When the path difference is $\lambda$, the phase difference is $\varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$.

The intensity at this point is $I=I_{max}{\cos }^2\left(\frac{2\pi }{2}\right)=I_{max}{\cos }^2(\pi )=I_{max}$.

Given that this intensity is $K$, we have $I_{max}=K$.

When the path difference is $\frac{\lambda }{6}$, the phase difference is $\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}$.

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

Substituting the value of $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$, we get:

$I^{\prime }=K{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3K}{4}$.

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