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

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

For a path difference $\Delta x=\frac{\lambda }{4}$, the phase difference is:

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

The intensity at this point is:

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

Given that this intensity is $\frac{K}{2}$, we have:

$\frac{I_0}{2}=\frac{K}{2}\Rightarrow I_0=K$

For a path difference $\Delta x=\lambda$, the phase difference is:

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

The intensity at this point is:

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

Substituting $I_0=K$, the intensity is $K$.

Answer: K