Physics - Wave Optics Question with Solution | TestHub

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

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

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

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

For path difference $\Delta x=\frac{\lambda }{4}$:

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

$I_1=I_{max}{\cos }^2\left(\frac{\pi /2}{2}\right)=I_{max}{\cos }^2\left(\frac{\pi }{4}\right)=I_{max}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_{max}}{2}$

Given that $I_1=\frac{K}{4}$, we have:

$\frac{I_{max}}{2}=\frac{K}{4}\Rightarrow I_{max}=\frac{K}{2}$

For path difference $\Delta x=\lambda$:

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

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

Therefore, the intensity at this point is $I_2=\frac{K}{2}$.

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